research article eoq model for deteriorating items with ... · eoq model for deteriorating items...

15
Research Article EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit Jie Min, 1 Jian Ou, 1 Yuan-Guang Zhong, 2 and Xin-Bao Liu 3 1 School of Mathematics and Physics, Anhui Jianzhu University, Hefei 230601, China 2 School of Business Administration, South China University of Technology, Guangzhou 510640, China 3 School of Management, Hefei University of Technology, Hefei 230009, China Correspondence should be addressed to Jie Min; [email protected] Received 24 July 2014; Accepted 29 October 2014; Published 16 December 2014 Academic Editor: Zhen-Lai Han Copyright © 2014 Jie Min et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. is paper develops a generalized inventory model for exponentially deteriorating items with current-stock-dependent demand rate and permissible delay in payments. In the model, the payment for the item must be made immediately if the order quantity is less than the predetermined quantity; otherwise, a fixed trade credit period is permitted. e maximization of the average profit per unit of time is taken as the inventory system’s objective. e necessary and sufficient conditions and some properties of the optimal solution to the model are developed. Simple solution procedures are proposed to efficiently determine the optimal ordering policies of the considered problem. Numerical example is also presented to illustrate the solution procedures obtained. 1. Introduction In classical inventory models, the demand rate of items was oſten assumed to be either constant or time dependent. However, many marketing researchers and practitioners have recognized that the demand rate of many retail items is usually influenced by the amount of inventory displayed. For example, Whitin [1] stated, “For retail stores the inventory control problem for style goods is further complicated by the fact that inventory and sales are not independent of one another. An increase in inventories may bring about increased sales of some items.” Levin et al. [2] and Sliver and Peterson [3] also observed that sales at the retail level tend to be proportional to the inventory displayed and that large piles of goods displayed in a supermarket will lead customers to buy more. e reason behind this phenomenon is a typical psychology of customers. ey may have the feeling of obtaining a wide range for selection when a large amount is stored or displayed. Conversely, they may have doubt about the freshness or quality of the product when a small amount is stored. Based on the observed phenomenon, it is clear that in real life the demand rate of items may be influenced by the stock levels. Many researchers have devel- oped lots of inventory models to cover this phenomenon. Gupta and Vrat [4] were the first to develop an inven- tory model with the initial-stock-dependent consumption rate. Padmanabhan and Vrat [5] proposed an EOQ model for initial-stock-dependent demand and exponential decay items. Giri et al. [6] presented a model under inflation for initial-stock-dependent consumption rate and exponential decay. In addition, Baker and Urban [7] investigated another kind of inventory model in which the demand of item is a polynomial function of the instantaneous stock level. Mandal and Phaujdar [8] developed an inventory model for deteriorating items with linearly current-stock-dependent demand. Pal et al. [9] extended the model of Baker and Urban [7] to the case of deteriorating items. Padmanabhan and Vrat [10] further developed three models for deteriorating items under current-stock-level-dependent demand rate: without backorder, with complete backorder, and with partial backo- rder. Chung et al. [11] modified Padmanabhan and Vrat’s [10] models by showing the necessary and sufficient conditions of the existing optimal solution for models with no backorder and complete backorder. Zhou and Yang [12] developed Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2014, Article ID 962128, 14 pages http://dx.doi.org/10.1155/2014/962128

Upload: others

Post on 28-Feb-2021

12 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

Research ArticleEOQ Model for Deteriorating Items with Stock-Level-DependentDemand Rate and Order-Quantity-Dependent Trade Credit

Jie Min1 Jian Ou1 Yuan-Guang Zhong2 and Xin-Bao Liu3

1School of Mathematics and Physics Anhui Jianzhu University Hefei 230601 China2School of Business Administration South China University of Technology Guangzhou 510640 China3School of Management Hefei University of Technology Hefei 230009 China

Correspondence should be addressed to Jie Min minjieahjzueducn

Received 24 July 2014 Accepted 29 October 2014 Published 16 December 2014

Academic Editor Zhen-Lai Han

Copyright copy 2014 Jie Min et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This paper develops a generalized inventory model for exponentially deteriorating items with current-stock-dependent demandrate and permissible delay in payments In the model the payment for the item must be made immediately if the order quantity isless than the predetermined quantity otherwise a fixed trade credit period is permittedThemaximization of the average profit perunit of time is taken as the inventory systemrsquos objectiveThe necessary and sufficient conditions and some properties of the optimalsolution to themodel are developed Simple solution procedures are proposed to efficiently determine the optimal ordering policiesof the considered problem Numerical example is also presented to illustrate the solution procedures obtained

1 Introduction

In classical inventory models the demand rate of items wasoften assumed to be either constant or time dependentHowever manymarketing researchers and practitioners haverecognized that the demand rate of many retail items isusually influenced by the amount of inventory displayed Forexample Whitin [1] stated ldquoFor retail stores the inventorycontrol problem for style goods is further complicated bythe fact that inventory and sales are not independent ofone another An increase in inventories may bring aboutincreased sales of some itemsrdquo Levin et al [2] and Sliverand Peterson [3] also observed that sales at the retail leveltend to be proportional to the inventory displayed and thatlarge piles of goods displayed in a supermarket will leadcustomers to buy moreThe reason behind this phenomenonis a typical psychology of customers They may have thefeeling of obtaining a wide range for selection when a largeamount is stored or displayed Conversely they may havedoubt about the freshness or quality of the product when asmall amount is stored Based on the observed phenomenonit is clear that in real life the demand rate of items may be

influenced by the stock levels Many researchers have devel-oped lots of inventory models to cover this phenomenonGupta and Vrat [4] were the first to develop an inven-tory model with the initial-stock-dependent consumptionrate Padmanabhan and Vrat [5] proposed an EOQ modelfor initial-stock-dependent demand and exponential decayitems Giri et al [6] presented a model under inflation forinitial-stock-dependent consumption rate and exponentialdecay In addition Baker and Urban [7] investigated anotherkind of inventory model in which the demand of itemis a polynomial function of the instantaneous stock levelMandal and Phaujdar [8] developed an inventory modelfor deteriorating itemswith linearly current-stock-dependentdemand Pal et al [9] extended themodel of Baker andUrban[7] to the case of deteriorating items Padmanabhan and Vrat[10] further developed three models for deteriorating itemsunder current-stock-level-dependent demand rate withoutbackorder with complete backorder and with partial backo-rder Chung et al [11] modified Padmanabhan and Vratrsquos [10]models by showing the necessary and sufficient conditions ofthe existing optimal solution for models with no backorderand complete backorder Zhou and Yang [12] developed

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 962128 14 pageshttpdxdoiorg1011552014962128

2 Mathematical Problems in Engineering

an inventory model for a general inventory-level-dependentdemand with limited storage space Moreover many furtherextensive models were developed by researchers such asBalkhi and Benkherouf [13] Dye and Ouyang [14] Zhou andYang [15] Wu et al [16] and Alfares [17]

In all the above inventory models with stock-level-dependent demand rate assumed that the retailer must payfor the items as soon as the items are received Howeverin the real life the supplier sometimes will offer the retailera delay period that is a trade credit period in paying forthe amount of purchasing cost Before the end of the tradecredit period the retailer can sell the goods and accumulatethe revenue to earn interest A higher interest is chargedif the payment is not settled by the end of trade creditperiod In real world the supplier often makes use of thispolicy to promote their commodities During the recent twodecades the effect of a permissible delay in payments onthe optimal inventory systems has received more attentionfrom numerous researchers Goyal [18] explored a single itemEOQ model under permissible delay in payments Aggarwaland Jaggi [19] extended Goyalrsquos [18] model to the case withdeteriorating items Jamal et al [20] further generalized theabove inventory models to allow for shortages Other relatedarticles can be found in Sarker et al [21] Teng [22] Huang[23] Chang et al [24 25] Chung and Liao [26] Teng et al[27] Liao [28] and Teng and Goyal [29] and their referencesMore recently Chen et al [30] studied an economic orderquantity model under conditionally permissible delay inpayments in which the supplier offers the retailer a fullypermissible delay periods if the retailer orders more thanor equal to a predetermined quantity otherwise partialpayments will be provided Chang et al [31] developed anappropriate inventory model for noninstantaneously deteri-orating items where suppliers provide a permissible paymentdelay schedule linked to order quantity However all of theabovementioned papers with trade credit financing did notaddress inventory-level-dependent demand rate

The present paper establishes an EOQ model fordeteriorating items with current-inventory-level-dependentdemand rate and permissible delay in payments which isdependent on the retailerrsquos order quantity That is if theorder quantity is less than the predetermined quantity thepayment for the item must be made immediately otherwisea fixed trade credit period is permitted We then establishthe proper mathematical model and study the necessary andsufficient conditions of the optimal solution to the modelSome properties of the optimal solution to the model areproposed for obtaining the optimal replenishment quantityand replenishment cycle whichmake the average profit of thesystem maximized

The remainder of the paper is organized as below Thenext section introduces some notations and assumptionsused in this paper We then present in Section 3 the for-mulation of the inventory model with current-stock-level-dependent demand rate and order-quantity-dependent tradecredit Section 4 examines the existence and uniqueness ofthe optimal solution to the considered inventory systemand shows some properties of the optimal ordering poli-cies In Section 5 numerical example is given to illustrate

the solution procedure obtained in this paper Some conclu-sions are included in Section 6

2 Notations and Assumptions

The following notations and assumptions are used in thepaper

Notations

119870 ordering cost one order in dollars119888 purchase cost per unit in dollars119904 selling price per unit in dollars (119904 gt 119888)ℎ stock holding cost per year per unit in dollars(excluding interest charges)120579 deteriorating rate of items (0 lt 120579 lt 1)119876119889 the qualified quantity in units at or beyond whichthe delay in payments is permitted119879119889 the time interval in years in which 119876119889 units aredepleted to zero due to both demand and deteriora-tion119868119890 interest which can be earned per $ per year by theretailer119868119901 interest charges per $ in stocks per year by thesupplier119872 the retailerrsquos trade credit period offered by supplierin years119879 inventory cycle length in years (decision variable)119876 the retailerrsquos order quantity in units per cycleΠ(119879) the retailerrsquos average profit function per unittime in dollars

Assumptions

(1) The demand rate 119877(119905) is a known function of retailerrsquosinstantaneous stock level 119868(119905) which is given by

119877 (119905) = 119863 + 120572119868 (119905)

where 119863 and 120572 are positive constants(1)

(2) The time horizon of the inventory system is infinite(3) The lead time is negligible(4) Shortages are not allowed to occur(5) When the order quantity is less than the qualified

quantity 119876119889 the payment for the item must be madeimmediately Otherwise the fixed trade credit period119872 is permitted

(6) If the trade credit period 119872 is offered the retailerwould settle the account at 119905 = 119872 and pay for theinterest charges on items in stock with rate 119868119901 overthe interval [119872 119879] as 119879 ge 119872 whereas the retailersettles the account at 119905 = 119872 and does not need to payany interest charge of items in stock during the wholecycle as 119879 le 119872

Mathematical Problems in Engineering 3

(7) The retailer can accumulate revenue and earn interestfrom the very beginning of inventory cycle until theend of the trade credit period offered by the supplierThat is the retailer can accumulate revenue and earninterest during the period from 119905 = 0 to 119905 = 119872 withrate 119868119890 under the condition of trade credit

(8) 119904 ge 119888 119868119901 ge 119868119890 which are reasonable assumptions inreality

3 Mathematical Formulation of the Model

Based on the above assumptions the considered inventorysystem goes like this at the beginning (say the time 119905 = 0)of each replenishment cycle items of 119876 units are held andthe items are depleted gradually in the interval [0 119879] due tothe combined effects of demand and deterioration At time119905 = 119879 the inventory level reaches zero and the whole processis repeated

The variation of inventory level 119868(119905) with respect to timecan be described by the following differential equation

119889119868 (119905)

119889119905= minus119863 minus 120572119868 (119905) minus 120579119868 (119905) 0 le 119905 le 119879 (2)

with the boundary condition 119868(119879) = 0The solution to the above differential equation is

119868 (119905) =119863

120572 + 120579(119890(120572+120579)(119879minus119905)

minus 1) 0 le 119905 le 119879 (3)

So the retailerrsquos order size per cycle is

119876 = 119868 (0) =119863

120572 + 120579(119890(120572+120579)119879

minus 1) (4)

We observe that if the order quantity 119876 lt 119876119889 then thepayment must be made immediately Otherwise the retailerwill get a certain credit period 119872 Hence from (4) one hasthe fact that the inequality119876 lt 119876119889 holds if and only if119879 lt 119879119889where 119879119889 is given as below

119879119889 =1

120572 + 120579ln [1 + (120572 + 120579)

119876119889

119863] (5)

The elements comprising the profit function per cycle of theretailer are listed below

(a) The ordering cost = 119870(b) The holding cost (excluding interest charges) =

ℎ int119879

0119868(119905)119889119905 = (ℎ119863(120572 + 120579)

2)[119890(120572+120579)119879

minus (120572 + 120579)119879 minus 1]

(c) The purchasing cost = 119888119876 = (119888119863(120572 + 120579))[119890(120572+120579)119879

minus 1]

(d) The sales revenue = 119904 int119879

0119877(119905)119889119905 = 119904[120579119863119879(120572 + 120579) +

(120572119863(120572 + 120579)2)(119890(120572+120579)119879

minus 1)](e) According to the above assumption there are two

cases to occur in interest payable in each order cycle

Case 1 (119879 le 119879119889) In this case the delay in payments isnot permitted That is all products in inventory have to be

financed with annual rate 119868119901 Therefore interest payable ineach order cycle

= 119888119868119901 int

119879

0

119868 (119905) 119889119905 =

119888119868119901119863

(120572 + 120579)2[119890(120572+120579)119879

minus (120572 + 120579) 119879 minus 1] (6)

Case 2 (119879 ge 119879119889) In this case the delay credit period 119872 inpayments is permitted which arouses two subcases Subcase21 where 119879 le 119872 and Subcase 22 where 119879 ge 119872

Subcase 21 (119879 le 119872) Since the cycle time 119879 is shorter thanthe credit period 119872 there is no interest paid for financingthe inventory in stock So the interest payable in each ordercycle = 0

Subcase 22 (119879 ge 119872) When the credit period 119872 is shorterthan or equal to the replenishment cycle time 119879 the retailerstarts paying the interest for the items in stock after time119872with rate 119868119901 Hence the interest payable in each order cycle

= 119888119868119901 int

119879

119872

119868 (119905) 119889119905

=

119888119868119901119863

(120572 + 120579)2[119890(120572+120579)(119879minus119872)

minus (120572 + 120579) (119879 minus119872) minus 1]

(7)

(f) Similarly there are also two cases to occur in interestearned in each order cycle

Case 1 (119879 le 119879119889) In this case the delay in payments is notpermitted so the interest earned in each order cycle = 0

Case 2 (119879 ge 119879119889) In this case the delay in payments ispermitted which causes two subcases as below

Subcase 21 (119879 le 119872) Since the cycle time 119879 is shorter thanthe credit period 119872 from time 0 to 119879 the retailer sells thegoods and continues to accumulate the sales revenue to earninterest 119904119868119890 int

119879

0int119905

0119877(119906)119889119906 119889119905 and from time119879 to119872 the retailer

can use the sales revenue generated in [0 119879] to earn interest119904119868119890 int119879

0119877(119906)119889119906(119872 minus 119879) Therefore the interest earned from

time 0 to119872 per cycle

= 119904119868119890 [int

119879

0

int

119905

0

119877 (119906) 119889119906 119889119905 + int

119879

0

119877 (119906) 119889119906 (119872 minus 119879)]

= 119904119868119890 1205791198631198792

2 (120572 + 120579)minus

120572119863

(120572 + 120579)3(119890(120572+120579)119879

minus 1)

+120572119863119879

(120572 + 120579)2119890(120572+120579)119879

+[120579119863119879

120572 + 120579+

120572119863

(120572 + 120579)2(119890(120572+120579)119879

minus 1)] (119872 minus 119879)

(8)

Subcase 22 (119879 ge 119872) In this case the delay in payments ispermitted and during time 0 through119872 the retailer sells thegoods and continues to accumulate sales revenue and earns

4 Mathematical Problems in Engineering

the interest with rate 119868119890 Therefore the interest earned fromtime 0 to119872 per cycle

= 119904119868119890 int

119872

0

int

119905

0

119877 (119906) 119889119906 119889119905

= 119904119868119890 [1205791198631198722

2 (120572 + 120579)+

120572119863119872

(120572 + 120579)2119890(120572+120579)119879

+120572119863

(120572 + 120579)3119890(120572+120579)119879

(119890minus(120572+120579)119872

minus 1)]

(9)

From the above the profit function for the retailer can beexpressed as

Π (119879) = Sales revenue + interest earned minus ordering cost

minus holding cost minus purchasing cost

minus interest payable times 119879minus1(10)

On simplification and summation we get the profit functionas the following three forms

(A) if 119879 le 119879119889 then

Π (119879) = Π1 (119879)

=1

119879

119863119864

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863119879

120572 + 120579(ℎ + 120579119904 + 119888119868119901) minus 119870

(11)

(B) if 119879 gt 119879119889 but 119879 le 119872 then

Π (119879) = Π2 (119879)

=1

119879

119863119865

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863119879

120572 + 120579(ℎ + 120579119904 + 120579119904119868119890119872+

120572119904119868119890

120572 + 120579)

minus120579119904119868119890119863119879

2

2 (120572 + 120579)minus 119870

(12)

(C) if 119879 gt 119879119889 and 119879 gt 119872 then

Π (119879) = Π3 (119879)

=1

119879

119863119865

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863

(120572 + 120579)2(119890(120572+120579)(119879minus119872)

minus 1) (120572119904119868119890

120572 + 120579minus 119888119868119901)

+119863119879

120572 + 120579(ℎ + 120579119904 + 119888119868119901) +

119863119872

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(13)

where the meaning of notations 119864 and 119865 in the aboveexpressions is

119864 = 120572119904 minus (120572 + 120579) 119888 minus ℎ minus 119888119868119901

119865 = 120572119904 minus (120572 + 120579) 119888 minus ℎ + 120572119904119868119890119872minus120572119904119868119890

120572 + 120579

(14)

For convenience we will consider the problem through thetwo following situations (1)119872 ge 119879119889 and (2)119872 lt 119879119889

(1) Suppose That 119872 ge 119879119889 In the situation of119872 ge 119879119889 Π(119879)has three different expressions as follows

Π (119879) = Π1 (119879) 0 lt 119879 le 119879119889 (15a)

Π (119879) = Π2 (119879) 119879119889 le 119879 le 119872 (15b)

Π (119879) = Π3 (119879) 119879 ge 119872 (15c)

It is easy to verify that in the case of 119872 ge 119879119889 Π(119879)continues except at 119879 = 119879119889 In fact if 119888119868119901 ge 119904119868119890 then we couldprove that Π1(119879119889) lt Π2(119879119889) (Proof of Π1(119879119889) lt Π2(119879119889) isshown in Appendix A)

(2) Suppose That 119872 lt 119879119889 If 119872 lt 119879119889 then Π(119879) has twodifferent expressions as follows

Π (119879) = Π1 (119879) 0 lt 119879 le 119879119889 (16a)

Π (119879) = Π3 (119879) 119879 ge 119879119889 (16b)

Here Π1(119879) and Π3(119879) are given by (11) and (13)respectively Π(119879) continues except at 119879 = 119879119889 In fact wecould prove that Π1(119879119889) lt Π3(119879119889) when 119872 lt 119879119889 (Proof ofΠ1(119879119889) lt Π3(119879119889) is shown in Appendix B)

In the next section we will determine the retailerrsquosoptimal cycle time119879lowast for the above two situations using somealgebraic method

4 The Solution Procedures

In order to decide the optimal solution 119879lowast of function Π(119879)we need first to study the properties of function Π119894(119879) (119894 =1 2 3) on (0 +infin) respectively The first-order necessarycondition forΠ1(119879) in (11) to bemaximum is119889Π1(119879)119889119879 = 0which implies that

minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119890

(120572+120579)119879minus 119890(120572+120579)119879

+ 1] minus 119870 = 0 (17)

For convenience we set the left-hand expression of (17) as1198911(119879) Taking the derivative of 1198911(119879) with respect to 119879 weget

1198891198911 (119879)

119889119879= minus119863119864119879119890

(120572+120579)119879 (18)

From (18) we know that if 119864 lt 0 then 1198891198911(119879)119889119879 gt

0 that is 1198911(119879) is increasing on (0 +infin) It is obvious that1198911(0) = minus119870 lt 0 and lim119879rarr+infin1198911(119879) = +infin therefore theIntermediate Value Theorem implies that 1198911(119879) = 0 that

Mathematical Problems in Engineering 5

is 119889Π1(119879)119889119879 = 0 has a unique root (say 11987901) on (0 +infin)

Moreover taking the second derivative ofΠ1(119879)with respectto 119879 gives

1198892Π1 (119879)

1198891198792= minus

2119870

1198793+

119863119864

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(19)

It is easy to show that the inequality 1199092119890119909minus2119909119890119909+2119890119909minus2 gt0 holds for all 119909 gt 0 so 1198892Π1(119879)119889119879

2 will always be negativeon (0 +infin) if 119864 lt 0 that is Π1(119879) is convex on (0 +infin)Therefore 1198790

1is the unique maximum point of Π1(119879) on

(0 +infin) But if 119864 ge 0 (18) means that 1198911(119879) is decreasingon (0 +infin) since 1198911(0) = minus119870 lt 0 then 1198911(119879) is negative and119889Π1(119879)119889119879 is positive for all 119879 gt 0 hence Π1(119879) is strictlyincreasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π1(119879) on (0 +infin)

Lemma 1 If 119864 lt 0 Π1(119879) is convex on (0 +infin) and 11987901is the

unique maximum point of Π1(119879) Otherwise Π1(119879) is strictlyincreasing on (0 +infin)

Similarly the first-order necessary condition forΠ2(119879) in(12) to be maximized is 119889Π2(119879)119889119879 = 0 which leads to

minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119890

(120572+120579)119879minus 119890(120572+120579)119879

+ 1]

+120579119904119868119890119863119879

2

2 (120572 + 120579)minus 119870 = 0

(20)

We set the left-hand expression of (20) as 1198912(119879) Taking thefirst derivative of 1198912(119879) with respect to 119879 gives

1198891198912 (119879)

119889119879= minus(119865119890

(120572+120579)119879minus

120579119904119868119890

120572 + 120579)119863119879 (21)

Consequently from the sign of 119865 there are two distinct casesfor finding the property of Π2(119879) as follows

(1) When 119865 le 0 we have 1198891198912(119879)119889119879 gt 0 that is 1198912(119879) isstrictly increasing on (0 +infin) Since 1198912(0) = minus119870 lt 0

and lim119879rarr+infin1198912(119879) = +infin 1198912(119879) = 0 has a uniquesolution (say 1198790

2) on (0 +infin) Moreover from (13) we

have

1198892Π2 (119879)

1198891198792= minus

2119870

1198793+

119863119865

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(22)

Since 119865 le 0 and (120572+120579)21198792119890(120572+120579)119879 minus2(120572+120579)119879119890(120572+120579)119879 +2119890(120572+120579)119879

minus 2 gt 0 we get 1198892Π2(119879)1198891198792lt 0 Hence

Π2(119879) is convex on (0 +infin) and 1198790

2is the unique

maximum point of Π2(119879)

(2) When 119865 gt 0 there will exist a unique root to theequation 1198891198912(119879)119889119879 = 0 Denote this root by 119879 then

119879=

1

120572 + 120579[ln 120579119904119868119890 minus ln (120572 + 120579) 119865] (23)

(A) If 119879ge 0 that is 0 lt 119865 le 120579119904119868119890(120572 + 120579) then

1198891198912(119879)119889119879 gt 0 for 119879 isin (0 119879) and 1198891198912(119879)119889119879 lt

0 for 119879 isin (119879 +infin) that is 1198912(119879) is increasing

on (0 119879) and decreasing on (119879

+infin)

(a) When 1198912(119879) gt 0 1198912(119879) = 0 has a unique

root (say11987912) on (0 119879

) due to1198912(0) = minus119870 lt

0 Hence Π2(119879) is increasing on (0 1198791

2)

and decreasing on (1198791

2 119879

) On the other

interval of (119879 +infin) however because of

1198912(119879) gt 0 and lim119879rarr+infin1198912(119879) = minusinfin

for 119865 gt 0 1198912(119879) has a unique zeropoint (say 119879

2

2) on (119879

+infin) Thus Π2(119879)

is decreasing on (119879 1198792

2) and increasing on

(1198792

2 +infin) Hence Π2(119879) is increasing on

(0 1198791

2) decreasing on (1198791

2 1198792

2) and increas-

ing on (11987922 +infin)

(b) When 1198912(119879) le 0 1198912(119879) is negative

on (0 119879) because 1198912(119879) is increasing on

(0 119879) whereas 1198912(119879) is also negative on

(119879 +infin) since 1198912(119879) is decreasing on

(119879 +infin) Therefore 1198912(119879) is always non-

positive on (0 +infin) andΠ2(119879) is increasingon (0 +infin)

(B) If 119879lt 0 that is 119865 gt 120579119904119868119890(120572 + 120579) then for any

119879 ge 0 we have 1198891198912(119879)119889119879 lt 0 that is 1198912(119879)is strictly decreasing on (0 +infin) Since 1198912(0) =minus119870 lt 0 1198912(119879) is negative on (0 +infin) HenceΠ2(119879) is increasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π2(119879) on (0 +infin)

Lemma 2 (1) For 119865 le 0 Π2(119879) is convex on (0 +infin) and 11987902

is the unique maximum point of Π2(119879)(2) For 0 lt 119865 le 120579119904119868119890(120572+120579)Π2(119879) is increasing on (0 11987912 )

decreasing on (11987912 1198792

2) and increasing on (1198792

2 +infin) if1198912(119879

) gt

0 and Π2(119879) is increasing on (0 +infin) otherwise(3) For 119865 gt 120579119904119868119890(120572 + 120579) Π2(119879) is always increasing on

(0 +infin)

Likewise the first-order necessary condition forΠ3(119879) in(13) to be maximum is 119889Π3(119879)119889119879 = 0 which implies that

119863119865(119890(120572+120579)119879

minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879] minus119863119865119879

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879minus119872)

minus 1) (120572119904119868119890

120572 + 120579minus 119888119868119901)

6 Mathematical Problems in Engineering

[1 minus (120572 + 120579) 119879] minus119863 (119879 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870 = 0

(24)

We set the left-hand expression of (24) as 1198913(119879) Taking thefirst derivative of 1198913(119879) with respect to 119879 gives

1198891198913 (119879)

119889119879= minus [119865 + (

120572119904119868119890

120572 + 120579minus 119888119868119901) 119890

minus(120572+120579)119872]119863119879119890

(120572+120579)119879

(25)

We consider the following two cases

(1) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198891198913(119879)119889119879 ge

0 Therefore 1198913(119879) is increasing on (0 +infin) It is easyto get 1198913(0) = (119863(120572 + 120579)

2)(119890minus(120572+120579)119872

minus 1)(120572119904119868119890(120572 +

120579) minus 119888119868119901) + 12057911990411986811989011986311987222(120572 + 120579) + 120572119904119868119890119863119872(120572 + 120579)

2minus

119888119868119901119863119872(120572 + 120579) minus 119870 and lim119879rarr+infin1198913(119879) = +infinThus if 1198913(0) lt 0 then 1198913(119879) = 0 has a uniquesolution (say 1198790

3) on (0 +infin) Hence 1198913(119879) will be

negative on (0 11987903) and positive on (1198790

3 +infin) which

implies Π3(119879) is increasing on (0 11987903) and decreasing

on (11987903 +infin) That is Π3(119879) is unimodal on (0 +infin)

and 1198790

3is the unique maximum point of Π3(119879) if

1198913(0) lt 0 In contrast if 1198913(0) ge 0 then 1198913(119879) willalways be nonnegative on (0 +infin) hence Π3(119879) isalways decreasing on (0 +infin)

(2) If119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 then1198913(119879) is strictly

decreasing on (0 +infin) Noting that lim119879rarr+infin1198913(119879) =minusinfin for 119865 gt minus(120572119904119868119890(120572 + 120579) minus 119888119868119901)119890

minus(120572+120579)119872 one easilyobtains that if 1198913(0) gt 0 there will exist a unique root(say 1198791

3) to the equation 1198913(119879) = 0 hence 1198913(119879) is

positive on (0 11987913) and negative on (1198791

3 +infin) which

implies Π3(119879) is decreasing on (0 11987913) and increasing

on (11987913 +infin) But if1198913(0) le 01198913(119879) is always negative

on (0 +infin) then Π3(119879) is increasing on (0 +infin)

Summarizing the above arguments the following lemmato describe the property ofΠ3(119879) on (0 +infin) will be obtained

Lemma 3 (1) For 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 Π3(119879) is

increasing on (0 11987903) and decreasing on (1198790

3 +infin) if 1198913(0) lt 0

and Π3(119879) is decreasing on (0 +infin) otherwise(2) For119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872Π3(119879) is decreasingon (0 1198791

3) and increasing on (1198791

3 +infin) if 1198913(0) gt 0 andΠ3(119879)

is increasing on (0 +infin) otherwise

Based upon Lemmas 1ndash3 we now will decide the optimalsolution 119879lowast to Π(119879) from the below two situations

41 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 ge

119879119889 When 119872 ge 119879119889 the piecewise profit function Π(119879) has

three different expressions that is Π1(119879) Π2(119879) and Π3(119879)respectively From (17) (20) and (24) we have

1198911 (119879119889) =minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1] minus 119870

(26)

1198912 (119879119889) =minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

+120579119904119868119890119863119879

2

119889

2 (120572 + 120579)minus 119870

(27)

1198912 (119872) = 1198913 (119872)

=minus119863119865

(120572 + 120579)2[(120572 + 120579)119872119890

(120572+120579)119872minus 119890(120572+120579)119872

+ 1]

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(28)

As shown in Appendix C we will find that 1198911(119879119889) gt 1198912(119879119889)For convenience we rewrite the notation 119864 as 119864 = 119865 minus 119888119868119901 +

120572119904119868119890(120572 + 120579) minus 120572119904119868119890119872 and the following two theorems wouldbe obtained to decide the optimal solution 119879lowastwhen119872 ge 119879119889

Theorem 4 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = Π2(119879

3

2) Hence 119879lowast is 1198793

2 where 1198793

2isin

(119879119889119872) and satisfies (20)(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 the following applies

(A) While 1198912(119879) gt 0 one has the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = Π2(1198791

2) Hence 119879lowast is 1198791

2

Mathematical Problems in Engineering 7

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2)

Π3(1198790

3) Hence 119879lowast is 1198790

1 11987912 or 11987903associ-

ated with maximum profit(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889)

Π3(1198790

3) Hence 119879lowast is 1198790

1 119879119889 or 11987903 associ-

ated with maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit(B) While 1198912(119879

) le 0 one has the following(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0

then Π(119879lowast) = Π3(1198790

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 11987903 associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(4) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

Theorem 5 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus

120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the

same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

42 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 lt

119879119889 When 119872 lt 119879119889 the piecewise profit function Π(119879) hasonly two different expressions that is Π1(119879) and Π3(119879)respectively From (24) we have

1198913 (119879119889) =

119863119865 (119890(120572+120579)119879119889 minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879119889] minus119863119865119879119889

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1) (

120572119904119868119890

120572 + 120579minus 119888119868119901)

[1 minus (120572 + 120579) 119879119889] minus119863 (119879119889 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(29)

As shown in Appendix D we will find that 1198911(119879119889) gt

1198913(119879119889) The following theorem would be given to decide thesystemrsquos optimal cycle time 119879lowast when119872 lt 119879119889

Theorem 6 If119872 lt 119879119889 then one has the following results

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the following

applies

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then Π(119879lowast) =

Π3(1198790

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then Π(119879lowast) =

maxΠ1(1198790

1) Π3(119879119889) Hence 119879

lowast is 1198790

1or 119879119889

associated with the maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then Π(119879

lowast) =

maxΠ1(1198790

1) Π3(119879

0

3) Hence 119879

lowast is 1198790

1or 1198790

3

associated with the maximum profit

(2) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix F for detail

5 Numerical Example

In this section we will provide the following numericalexample to illustrate the present model

Example 1 Given 119876119889 = 200 units 119863 = 1500 units 120572 =

04 120579 = 020 119870 = $50 ℎ = $1unityear 119888 = $ 5 perunit 119904 = $9 per unit 119868119890 = 013$year 119868119901 = 019$year and119872 = 03 year it is easy to get 119879119889 = 02123 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 and 119865 le (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872 Consequently by using Bisection method or

Newton-Raphson method we could obtain 1198790

1= 01760

1198790

2= 01892 and 119879

0

3= 01931 Because of 1198911(119879119889) le 0

1198912(119879119889) lt 0 and 1198912(119872) le 0 according to Theorem 4(1) theretailerrsquos optimal cycle time should be 119879lowast = 119879

0

3= 01931

8 Mathematical Problems in Engineering

Table 1 The impact of change of 120572 120579 119876119889 and119872 on the optimal solutions

Parameters 1198790

11198790

21198790

3119879lowast

119876lowast

Πlowast

120572

01 03381 03637 03895 03637 695772 6023591

02 04467 03956 03957 03956 823678 6217691

03 04792 04231 04320 04231 895987 6430335

04 05561 04538 04439 04538 985672 6663862

05 06723 05562 05071 05071 1033781 6931645

06 08552 05781 06047 06032 1367563 7362176

120579

001 05651 04436 05008 04436 1098853 7721683

002 05513 04471 04973 04471 1064786 7439053

005 05167 04441 04812 04441 994372 7268573

010 04783 04368 04791 04368 979335 7082341

050 03365 03568 04659 03568 833563 6887732

100 03561 02896 04543 03561 710856 6625568

119876119889

50 02344 02432 02589 02344 463085 4329953

100 03832 03565 03867 03565 750255 3948304

150 04973 04236 04973 04236 919334 3430035

200 06601 05467 05675 05675 1276297 3215728

250 07756 06873 06981 06981 1633156 2833270

300 08132 07675 07321 07321 1927368 2173673

119872

01 03726 03826 02873 03826 737665 2746542

02 03875 03835 03965 03835 799762 2823871

03 03811 04009 04623 04009 833721 2902323

04 03765 04197 04987 03765 899932 3130369

05 03732 04618 05013 04618 955513 3273771

06 03586 04745 05233 03586 998756 3633689

hence the optimal order quantity and the maximum averageprofit are 119876lowast = 1588763 and Πlowast = 18673562 respectively

Next we further study the effects of changes of parame-ters 120572 120579 119876119889 and 119872 on the optimal solutions The values ofother parameters keep the same as in the above examplewheneach of parameters120572 120579119876119889 and119872 varies Table 1 presents theobserved results with various parameters 120572 120579 119876119889 and119872

The following inferences can be made based on Table 1

(1) The retailerrsquos optimal order quantity 119876lowast and the

optimal average profit Πlowast increase as the value of 120572increases It implies that when the market demand ismore sensitive to the inventory level the retailer willincrease hisher order quantity to make more profitunder permissible delay in payments

(2) The optimal order quantity 119876lowast and the maximum

average profitΠlowast decrease as the value of 120579 increases

(3) When 119876119889 increases the optimal order quantity 119876lowastand the optimal cycle time 119879lowast are increasing but theoptimal average profit Πlowast is decreasing

(4) The retailerrsquos order quantity 119876lowast is increasing as 119872increases This verifies the fact that the retailer couldindeed increase the sales quantity by adopting thetrade credit policy provided by hisher supplier

6 Conclusions

In this paper we develop an EOQ model for deterioratingitems under permissible delay in payments The primarydifference of this paper as compared to previous relatedstudies is the following four aspects (1) the demand rateof items is dependent on the retailerrsquos current stock level(2) many items deteriorate continuously such as fruits andvegetables (3) the retailer who purchases the items enjoys afixed credit period offered by the supplier if the order quantityis greater than or equal to the predetermined quantity and(4) we here use maximizing profit as the objective to find theoptimal replenishment policies which could better describethe essence of the inventory system than a cost-minimizationmodel In addition we have discussed in detail the conditionsof the existence and uniqueness of the optimal solutions tothe model Three easy-to-use theorems are developed to findthe optimal ordering policies for the considered problem andthese theoretical results are illustrated by numerical exampleBy studying the effects of 120572 120579 119876119889 and 119872 on the optimalorder quantity 119876lowast and the optimal average profit Πlowast somemanagerial insights are derived

The presented model can be further extended to somemore practical situations For example we could allow forshortages quantity discounts time value of money andinflation price-sensitive and stock-dependent demand andso forth

Mathematical Problems in Engineering 9

Appendices

A Proof of Π1(119879119889) lt Π2(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (12) we get

Π2 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2(119890(120572+120579)119879119889 minus (120572 + 120579) 119879119889 minus 1)

+ 119904119868119890119872119863119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

(A1)

It is evident that119872 ge 119879119889 and (119890(120572+120579)119879119889 minus (120572+120579)119879119889 minus1) gt 0 for

119879119889 gt 0 then

Π2 (119879119889) minus Π1 (119879119889)

gt1

119879119889

1198631198792

119889

2(120572119904119868119890119872minus

120572119904119868119890

120572 + 120579+ 119888119868119901)

+ 119904119868119890119863119872119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

=119863119879119889

2(120572119904119868119890119872+ 119888119868119901) minus

119904119868119890119863119879119889

2+ 119904119868119890119863119872

ge119863119879119889

2(120572119904119868119890119872+ 119888119868119901) +

119904119868119890119863119872

2gt 0

(A2)

As a result we have Π1(119879119889) lt Π2(119879119889)This completes the proof of Appendix A

B Proof of Π1(119879119889) lt Π3(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (13) we get

Π3 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2[119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872) minus (120572 + 120579)119872]

+120572119904119868119890119872119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1)

+120572119904119868119890119863119872

2

(120572 + 120579)+120579119904119868119890119863119872

2

2 (120572 + 120579)

(B1)

Because of 119879119889 gt 119872 and 119890(120572+120579)119879119889 minus119890(120572+120579)(119879119889minus119872) minus(120572+120579)119872 gt 0we have Π3(119879119889) minus Π1(119879119889) gt 0

This completes the proof of Appendix B

C Proof of 1198911(119879119889) gt 1198912(119879119889)

If 119888119868119901 ge 119904119868119890 from (26) and (27) we have

1198912 (119879119889) minus 1198911 (119879119889)

= minus119863

(120572 + 120579)2(120572119904119868119890119872+ 119888119868119901 minus

120572119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

le minus119863

(120572 + 120579)2(120572119904119868119890119872+

120579119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

= minus119863119904119868119890

(120572 + 120579)2(120572119872 +

120579

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

minus120579 (120572 + 120579) 119879

2

119889

2

(C1)

Let 119866(119879119889) = (120572119872 + 120579(120572 + 120579))[(120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 +

1] minus 120579(120572 + 120579)1198792

1198892 then

1198661015840(119879119889) = (120572119872 +

120579

120572 + 120579) (120572 + 120579)

2119879119889119890(120572+120579)119879119889 minus 120579 (120572 + 120579) 119879119889

= 120579 (120572 + 120579) 119879119889 [(120572119872(120572 + 120579)

120579+ 1) 119890

(120572+120579)119879119889 minus 1] gt 0

(C2)

Hence 119866(119879119889) gt 119866(0) = 0 that is 1198911(119879119889) gt 1198912(119879119889)This completes the proof of Appendix C

D Proof of 1198911(119879119889) gt 1198913(119879119889)

From (26) and (24) we have

1198913 (119879119889) minus 1198911 (119879119889)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

times (120572119904119868119890

120572 + 120579minus 119888119868119901 minus 120572119904119868119890119872) minus

119863

(120572 + 120579)2

[(120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872) minus 119890

(120572+120579)(119879119889minus119872) + 1 minus (120572 + 120579)119872]

times (120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 2: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

2 Mathematical Problems in Engineering

an inventory model for a general inventory-level-dependentdemand with limited storage space Moreover many furtherextensive models were developed by researchers such asBalkhi and Benkherouf [13] Dye and Ouyang [14] Zhou andYang [15] Wu et al [16] and Alfares [17]

In all the above inventory models with stock-level-dependent demand rate assumed that the retailer must payfor the items as soon as the items are received Howeverin the real life the supplier sometimes will offer the retailera delay period that is a trade credit period in paying forthe amount of purchasing cost Before the end of the tradecredit period the retailer can sell the goods and accumulatethe revenue to earn interest A higher interest is chargedif the payment is not settled by the end of trade creditperiod In real world the supplier often makes use of thispolicy to promote their commodities During the recent twodecades the effect of a permissible delay in payments onthe optimal inventory systems has received more attentionfrom numerous researchers Goyal [18] explored a single itemEOQ model under permissible delay in payments Aggarwaland Jaggi [19] extended Goyalrsquos [18] model to the case withdeteriorating items Jamal et al [20] further generalized theabove inventory models to allow for shortages Other relatedarticles can be found in Sarker et al [21] Teng [22] Huang[23] Chang et al [24 25] Chung and Liao [26] Teng et al[27] Liao [28] and Teng and Goyal [29] and their referencesMore recently Chen et al [30] studied an economic orderquantity model under conditionally permissible delay inpayments in which the supplier offers the retailer a fullypermissible delay periods if the retailer orders more thanor equal to a predetermined quantity otherwise partialpayments will be provided Chang et al [31] developed anappropriate inventory model for noninstantaneously deteri-orating items where suppliers provide a permissible paymentdelay schedule linked to order quantity However all of theabovementioned papers with trade credit financing did notaddress inventory-level-dependent demand rate

The present paper establishes an EOQ model fordeteriorating items with current-inventory-level-dependentdemand rate and permissible delay in payments which isdependent on the retailerrsquos order quantity That is if theorder quantity is less than the predetermined quantity thepayment for the item must be made immediately otherwisea fixed trade credit period is permitted We then establishthe proper mathematical model and study the necessary andsufficient conditions of the optimal solution to the modelSome properties of the optimal solution to the model areproposed for obtaining the optimal replenishment quantityand replenishment cycle whichmake the average profit of thesystem maximized

The remainder of the paper is organized as below Thenext section introduces some notations and assumptionsused in this paper We then present in Section 3 the for-mulation of the inventory model with current-stock-level-dependent demand rate and order-quantity-dependent tradecredit Section 4 examines the existence and uniqueness ofthe optimal solution to the considered inventory systemand shows some properties of the optimal ordering poli-cies In Section 5 numerical example is given to illustrate

the solution procedure obtained in this paper Some conclu-sions are included in Section 6

2 Notations and Assumptions

The following notations and assumptions are used in thepaper

Notations

119870 ordering cost one order in dollars119888 purchase cost per unit in dollars119904 selling price per unit in dollars (119904 gt 119888)ℎ stock holding cost per year per unit in dollars(excluding interest charges)120579 deteriorating rate of items (0 lt 120579 lt 1)119876119889 the qualified quantity in units at or beyond whichthe delay in payments is permitted119879119889 the time interval in years in which 119876119889 units aredepleted to zero due to both demand and deteriora-tion119868119890 interest which can be earned per $ per year by theretailer119868119901 interest charges per $ in stocks per year by thesupplier119872 the retailerrsquos trade credit period offered by supplierin years119879 inventory cycle length in years (decision variable)119876 the retailerrsquos order quantity in units per cycleΠ(119879) the retailerrsquos average profit function per unittime in dollars

Assumptions

(1) The demand rate 119877(119905) is a known function of retailerrsquosinstantaneous stock level 119868(119905) which is given by

119877 (119905) = 119863 + 120572119868 (119905)

where 119863 and 120572 are positive constants(1)

(2) The time horizon of the inventory system is infinite(3) The lead time is negligible(4) Shortages are not allowed to occur(5) When the order quantity is less than the qualified

quantity 119876119889 the payment for the item must be madeimmediately Otherwise the fixed trade credit period119872 is permitted

(6) If the trade credit period 119872 is offered the retailerwould settle the account at 119905 = 119872 and pay for theinterest charges on items in stock with rate 119868119901 overthe interval [119872 119879] as 119879 ge 119872 whereas the retailersettles the account at 119905 = 119872 and does not need to payany interest charge of items in stock during the wholecycle as 119879 le 119872

Mathematical Problems in Engineering 3

(7) The retailer can accumulate revenue and earn interestfrom the very beginning of inventory cycle until theend of the trade credit period offered by the supplierThat is the retailer can accumulate revenue and earninterest during the period from 119905 = 0 to 119905 = 119872 withrate 119868119890 under the condition of trade credit

(8) 119904 ge 119888 119868119901 ge 119868119890 which are reasonable assumptions inreality

3 Mathematical Formulation of the Model

Based on the above assumptions the considered inventorysystem goes like this at the beginning (say the time 119905 = 0)of each replenishment cycle items of 119876 units are held andthe items are depleted gradually in the interval [0 119879] due tothe combined effects of demand and deterioration At time119905 = 119879 the inventory level reaches zero and the whole processis repeated

The variation of inventory level 119868(119905) with respect to timecan be described by the following differential equation

119889119868 (119905)

119889119905= minus119863 minus 120572119868 (119905) minus 120579119868 (119905) 0 le 119905 le 119879 (2)

with the boundary condition 119868(119879) = 0The solution to the above differential equation is

119868 (119905) =119863

120572 + 120579(119890(120572+120579)(119879minus119905)

minus 1) 0 le 119905 le 119879 (3)

So the retailerrsquos order size per cycle is

119876 = 119868 (0) =119863

120572 + 120579(119890(120572+120579)119879

minus 1) (4)

We observe that if the order quantity 119876 lt 119876119889 then thepayment must be made immediately Otherwise the retailerwill get a certain credit period 119872 Hence from (4) one hasthe fact that the inequality119876 lt 119876119889 holds if and only if119879 lt 119879119889where 119879119889 is given as below

119879119889 =1

120572 + 120579ln [1 + (120572 + 120579)

119876119889

119863] (5)

The elements comprising the profit function per cycle of theretailer are listed below

(a) The ordering cost = 119870(b) The holding cost (excluding interest charges) =

ℎ int119879

0119868(119905)119889119905 = (ℎ119863(120572 + 120579)

2)[119890(120572+120579)119879

minus (120572 + 120579)119879 minus 1]

(c) The purchasing cost = 119888119876 = (119888119863(120572 + 120579))[119890(120572+120579)119879

minus 1]

(d) The sales revenue = 119904 int119879

0119877(119905)119889119905 = 119904[120579119863119879(120572 + 120579) +

(120572119863(120572 + 120579)2)(119890(120572+120579)119879

minus 1)](e) According to the above assumption there are two

cases to occur in interest payable in each order cycle

Case 1 (119879 le 119879119889) In this case the delay in payments isnot permitted That is all products in inventory have to be

financed with annual rate 119868119901 Therefore interest payable ineach order cycle

= 119888119868119901 int

119879

0

119868 (119905) 119889119905 =

119888119868119901119863

(120572 + 120579)2[119890(120572+120579)119879

minus (120572 + 120579) 119879 minus 1] (6)

Case 2 (119879 ge 119879119889) In this case the delay credit period 119872 inpayments is permitted which arouses two subcases Subcase21 where 119879 le 119872 and Subcase 22 where 119879 ge 119872

Subcase 21 (119879 le 119872) Since the cycle time 119879 is shorter thanthe credit period 119872 there is no interest paid for financingthe inventory in stock So the interest payable in each ordercycle = 0

Subcase 22 (119879 ge 119872) When the credit period 119872 is shorterthan or equal to the replenishment cycle time 119879 the retailerstarts paying the interest for the items in stock after time119872with rate 119868119901 Hence the interest payable in each order cycle

= 119888119868119901 int

119879

119872

119868 (119905) 119889119905

=

119888119868119901119863

(120572 + 120579)2[119890(120572+120579)(119879minus119872)

minus (120572 + 120579) (119879 minus119872) minus 1]

(7)

(f) Similarly there are also two cases to occur in interestearned in each order cycle

Case 1 (119879 le 119879119889) In this case the delay in payments is notpermitted so the interest earned in each order cycle = 0

Case 2 (119879 ge 119879119889) In this case the delay in payments ispermitted which causes two subcases as below

Subcase 21 (119879 le 119872) Since the cycle time 119879 is shorter thanthe credit period 119872 from time 0 to 119879 the retailer sells thegoods and continues to accumulate the sales revenue to earninterest 119904119868119890 int

119879

0int119905

0119877(119906)119889119906 119889119905 and from time119879 to119872 the retailer

can use the sales revenue generated in [0 119879] to earn interest119904119868119890 int119879

0119877(119906)119889119906(119872 minus 119879) Therefore the interest earned from

time 0 to119872 per cycle

= 119904119868119890 [int

119879

0

int

119905

0

119877 (119906) 119889119906 119889119905 + int

119879

0

119877 (119906) 119889119906 (119872 minus 119879)]

= 119904119868119890 1205791198631198792

2 (120572 + 120579)minus

120572119863

(120572 + 120579)3(119890(120572+120579)119879

minus 1)

+120572119863119879

(120572 + 120579)2119890(120572+120579)119879

+[120579119863119879

120572 + 120579+

120572119863

(120572 + 120579)2(119890(120572+120579)119879

minus 1)] (119872 minus 119879)

(8)

Subcase 22 (119879 ge 119872) In this case the delay in payments ispermitted and during time 0 through119872 the retailer sells thegoods and continues to accumulate sales revenue and earns

4 Mathematical Problems in Engineering

the interest with rate 119868119890 Therefore the interest earned fromtime 0 to119872 per cycle

= 119904119868119890 int

119872

0

int

119905

0

119877 (119906) 119889119906 119889119905

= 119904119868119890 [1205791198631198722

2 (120572 + 120579)+

120572119863119872

(120572 + 120579)2119890(120572+120579)119879

+120572119863

(120572 + 120579)3119890(120572+120579)119879

(119890minus(120572+120579)119872

minus 1)]

(9)

From the above the profit function for the retailer can beexpressed as

Π (119879) = Sales revenue + interest earned minus ordering cost

minus holding cost minus purchasing cost

minus interest payable times 119879minus1(10)

On simplification and summation we get the profit functionas the following three forms

(A) if 119879 le 119879119889 then

Π (119879) = Π1 (119879)

=1

119879

119863119864

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863119879

120572 + 120579(ℎ + 120579119904 + 119888119868119901) minus 119870

(11)

(B) if 119879 gt 119879119889 but 119879 le 119872 then

Π (119879) = Π2 (119879)

=1

119879

119863119865

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863119879

120572 + 120579(ℎ + 120579119904 + 120579119904119868119890119872+

120572119904119868119890

120572 + 120579)

minus120579119904119868119890119863119879

2

2 (120572 + 120579)minus 119870

(12)

(C) if 119879 gt 119879119889 and 119879 gt 119872 then

Π (119879) = Π3 (119879)

=1

119879

119863119865

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863

(120572 + 120579)2(119890(120572+120579)(119879minus119872)

minus 1) (120572119904119868119890

120572 + 120579minus 119888119868119901)

+119863119879

120572 + 120579(ℎ + 120579119904 + 119888119868119901) +

119863119872

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(13)

where the meaning of notations 119864 and 119865 in the aboveexpressions is

119864 = 120572119904 minus (120572 + 120579) 119888 minus ℎ minus 119888119868119901

119865 = 120572119904 minus (120572 + 120579) 119888 minus ℎ + 120572119904119868119890119872minus120572119904119868119890

120572 + 120579

(14)

For convenience we will consider the problem through thetwo following situations (1)119872 ge 119879119889 and (2)119872 lt 119879119889

(1) Suppose That 119872 ge 119879119889 In the situation of119872 ge 119879119889 Π(119879)has three different expressions as follows

Π (119879) = Π1 (119879) 0 lt 119879 le 119879119889 (15a)

Π (119879) = Π2 (119879) 119879119889 le 119879 le 119872 (15b)

Π (119879) = Π3 (119879) 119879 ge 119872 (15c)

It is easy to verify that in the case of 119872 ge 119879119889 Π(119879)continues except at 119879 = 119879119889 In fact if 119888119868119901 ge 119904119868119890 then we couldprove that Π1(119879119889) lt Π2(119879119889) (Proof of Π1(119879119889) lt Π2(119879119889) isshown in Appendix A)

(2) Suppose That 119872 lt 119879119889 If 119872 lt 119879119889 then Π(119879) has twodifferent expressions as follows

Π (119879) = Π1 (119879) 0 lt 119879 le 119879119889 (16a)

Π (119879) = Π3 (119879) 119879 ge 119879119889 (16b)

Here Π1(119879) and Π3(119879) are given by (11) and (13)respectively Π(119879) continues except at 119879 = 119879119889 In fact wecould prove that Π1(119879119889) lt Π3(119879119889) when 119872 lt 119879119889 (Proof ofΠ1(119879119889) lt Π3(119879119889) is shown in Appendix B)

In the next section we will determine the retailerrsquosoptimal cycle time119879lowast for the above two situations using somealgebraic method

4 The Solution Procedures

In order to decide the optimal solution 119879lowast of function Π(119879)we need first to study the properties of function Π119894(119879) (119894 =1 2 3) on (0 +infin) respectively The first-order necessarycondition forΠ1(119879) in (11) to bemaximum is119889Π1(119879)119889119879 = 0which implies that

minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119890

(120572+120579)119879minus 119890(120572+120579)119879

+ 1] minus 119870 = 0 (17)

For convenience we set the left-hand expression of (17) as1198911(119879) Taking the derivative of 1198911(119879) with respect to 119879 weget

1198891198911 (119879)

119889119879= minus119863119864119879119890

(120572+120579)119879 (18)

From (18) we know that if 119864 lt 0 then 1198891198911(119879)119889119879 gt

0 that is 1198911(119879) is increasing on (0 +infin) It is obvious that1198911(0) = minus119870 lt 0 and lim119879rarr+infin1198911(119879) = +infin therefore theIntermediate Value Theorem implies that 1198911(119879) = 0 that

Mathematical Problems in Engineering 5

is 119889Π1(119879)119889119879 = 0 has a unique root (say 11987901) on (0 +infin)

Moreover taking the second derivative ofΠ1(119879)with respectto 119879 gives

1198892Π1 (119879)

1198891198792= minus

2119870

1198793+

119863119864

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(19)

It is easy to show that the inequality 1199092119890119909minus2119909119890119909+2119890119909minus2 gt0 holds for all 119909 gt 0 so 1198892Π1(119879)119889119879

2 will always be negativeon (0 +infin) if 119864 lt 0 that is Π1(119879) is convex on (0 +infin)Therefore 1198790

1is the unique maximum point of Π1(119879) on

(0 +infin) But if 119864 ge 0 (18) means that 1198911(119879) is decreasingon (0 +infin) since 1198911(0) = minus119870 lt 0 then 1198911(119879) is negative and119889Π1(119879)119889119879 is positive for all 119879 gt 0 hence Π1(119879) is strictlyincreasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π1(119879) on (0 +infin)

Lemma 1 If 119864 lt 0 Π1(119879) is convex on (0 +infin) and 11987901is the

unique maximum point of Π1(119879) Otherwise Π1(119879) is strictlyincreasing on (0 +infin)

Similarly the first-order necessary condition forΠ2(119879) in(12) to be maximized is 119889Π2(119879)119889119879 = 0 which leads to

minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119890

(120572+120579)119879minus 119890(120572+120579)119879

+ 1]

+120579119904119868119890119863119879

2

2 (120572 + 120579)minus 119870 = 0

(20)

We set the left-hand expression of (20) as 1198912(119879) Taking thefirst derivative of 1198912(119879) with respect to 119879 gives

1198891198912 (119879)

119889119879= minus(119865119890

(120572+120579)119879minus

120579119904119868119890

120572 + 120579)119863119879 (21)

Consequently from the sign of 119865 there are two distinct casesfor finding the property of Π2(119879) as follows

(1) When 119865 le 0 we have 1198891198912(119879)119889119879 gt 0 that is 1198912(119879) isstrictly increasing on (0 +infin) Since 1198912(0) = minus119870 lt 0

and lim119879rarr+infin1198912(119879) = +infin 1198912(119879) = 0 has a uniquesolution (say 1198790

2) on (0 +infin) Moreover from (13) we

have

1198892Π2 (119879)

1198891198792= minus

2119870

1198793+

119863119865

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(22)

Since 119865 le 0 and (120572+120579)21198792119890(120572+120579)119879 minus2(120572+120579)119879119890(120572+120579)119879 +2119890(120572+120579)119879

minus 2 gt 0 we get 1198892Π2(119879)1198891198792lt 0 Hence

Π2(119879) is convex on (0 +infin) and 1198790

2is the unique

maximum point of Π2(119879)

(2) When 119865 gt 0 there will exist a unique root to theequation 1198891198912(119879)119889119879 = 0 Denote this root by 119879 then

119879=

1

120572 + 120579[ln 120579119904119868119890 minus ln (120572 + 120579) 119865] (23)

(A) If 119879ge 0 that is 0 lt 119865 le 120579119904119868119890(120572 + 120579) then

1198891198912(119879)119889119879 gt 0 for 119879 isin (0 119879) and 1198891198912(119879)119889119879 lt

0 for 119879 isin (119879 +infin) that is 1198912(119879) is increasing

on (0 119879) and decreasing on (119879

+infin)

(a) When 1198912(119879) gt 0 1198912(119879) = 0 has a unique

root (say11987912) on (0 119879

) due to1198912(0) = minus119870 lt

0 Hence Π2(119879) is increasing on (0 1198791

2)

and decreasing on (1198791

2 119879

) On the other

interval of (119879 +infin) however because of

1198912(119879) gt 0 and lim119879rarr+infin1198912(119879) = minusinfin

for 119865 gt 0 1198912(119879) has a unique zeropoint (say 119879

2

2) on (119879

+infin) Thus Π2(119879)

is decreasing on (119879 1198792

2) and increasing on

(1198792

2 +infin) Hence Π2(119879) is increasing on

(0 1198791

2) decreasing on (1198791

2 1198792

2) and increas-

ing on (11987922 +infin)

(b) When 1198912(119879) le 0 1198912(119879) is negative

on (0 119879) because 1198912(119879) is increasing on

(0 119879) whereas 1198912(119879) is also negative on

(119879 +infin) since 1198912(119879) is decreasing on

(119879 +infin) Therefore 1198912(119879) is always non-

positive on (0 +infin) andΠ2(119879) is increasingon (0 +infin)

(B) If 119879lt 0 that is 119865 gt 120579119904119868119890(120572 + 120579) then for any

119879 ge 0 we have 1198891198912(119879)119889119879 lt 0 that is 1198912(119879)is strictly decreasing on (0 +infin) Since 1198912(0) =minus119870 lt 0 1198912(119879) is negative on (0 +infin) HenceΠ2(119879) is increasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π2(119879) on (0 +infin)

Lemma 2 (1) For 119865 le 0 Π2(119879) is convex on (0 +infin) and 11987902

is the unique maximum point of Π2(119879)(2) For 0 lt 119865 le 120579119904119868119890(120572+120579)Π2(119879) is increasing on (0 11987912 )

decreasing on (11987912 1198792

2) and increasing on (1198792

2 +infin) if1198912(119879

) gt

0 and Π2(119879) is increasing on (0 +infin) otherwise(3) For 119865 gt 120579119904119868119890(120572 + 120579) Π2(119879) is always increasing on

(0 +infin)

Likewise the first-order necessary condition forΠ3(119879) in(13) to be maximum is 119889Π3(119879)119889119879 = 0 which implies that

119863119865(119890(120572+120579)119879

minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879] minus119863119865119879

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879minus119872)

minus 1) (120572119904119868119890

120572 + 120579minus 119888119868119901)

6 Mathematical Problems in Engineering

[1 minus (120572 + 120579) 119879] minus119863 (119879 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870 = 0

(24)

We set the left-hand expression of (24) as 1198913(119879) Taking thefirst derivative of 1198913(119879) with respect to 119879 gives

1198891198913 (119879)

119889119879= minus [119865 + (

120572119904119868119890

120572 + 120579minus 119888119868119901) 119890

minus(120572+120579)119872]119863119879119890

(120572+120579)119879

(25)

We consider the following two cases

(1) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198891198913(119879)119889119879 ge

0 Therefore 1198913(119879) is increasing on (0 +infin) It is easyto get 1198913(0) = (119863(120572 + 120579)

2)(119890minus(120572+120579)119872

minus 1)(120572119904119868119890(120572 +

120579) minus 119888119868119901) + 12057911990411986811989011986311987222(120572 + 120579) + 120572119904119868119890119863119872(120572 + 120579)

2minus

119888119868119901119863119872(120572 + 120579) minus 119870 and lim119879rarr+infin1198913(119879) = +infinThus if 1198913(0) lt 0 then 1198913(119879) = 0 has a uniquesolution (say 1198790

3) on (0 +infin) Hence 1198913(119879) will be

negative on (0 11987903) and positive on (1198790

3 +infin) which

implies Π3(119879) is increasing on (0 11987903) and decreasing

on (11987903 +infin) That is Π3(119879) is unimodal on (0 +infin)

and 1198790

3is the unique maximum point of Π3(119879) if

1198913(0) lt 0 In contrast if 1198913(0) ge 0 then 1198913(119879) willalways be nonnegative on (0 +infin) hence Π3(119879) isalways decreasing on (0 +infin)

(2) If119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 then1198913(119879) is strictly

decreasing on (0 +infin) Noting that lim119879rarr+infin1198913(119879) =minusinfin for 119865 gt minus(120572119904119868119890(120572 + 120579) minus 119888119868119901)119890

minus(120572+120579)119872 one easilyobtains that if 1198913(0) gt 0 there will exist a unique root(say 1198791

3) to the equation 1198913(119879) = 0 hence 1198913(119879) is

positive on (0 11987913) and negative on (1198791

3 +infin) which

implies Π3(119879) is decreasing on (0 11987913) and increasing

on (11987913 +infin) But if1198913(0) le 01198913(119879) is always negative

on (0 +infin) then Π3(119879) is increasing on (0 +infin)

Summarizing the above arguments the following lemmato describe the property ofΠ3(119879) on (0 +infin) will be obtained

Lemma 3 (1) For 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 Π3(119879) is

increasing on (0 11987903) and decreasing on (1198790

3 +infin) if 1198913(0) lt 0

and Π3(119879) is decreasing on (0 +infin) otherwise(2) For119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872Π3(119879) is decreasingon (0 1198791

3) and increasing on (1198791

3 +infin) if 1198913(0) gt 0 andΠ3(119879)

is increasing on (0 +infin) otherwise

Based upon Lemmas 1ndash3 we now will decide the optimalsolution 119879lowast to Π(119879) from the below two situations

41 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 ge

119879119889 When 119872 ge 119879119889 the piecewise profit function Π(119879) has

three different expressions that is Π1(119879) Π2(119879) and Π3(119879)respectively From (17) (20) and (24) we have

1198911 (119879119889) =minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1] minus 119870

(26)

1198912 (119879119889) =minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

+120579119904119868119890119863119879

2

119889

2 (120572 + 120579)minus 119870

(27)

1198912 (119872) = 1198913 (119872)

=minus119863119865

(120572 + 120579)2[(120572 + 120579)119872119890

(120572+120579)119872minus 119890(120572+120579)119872

+ 1]

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(28)

As shown in Appendix C we will find that 1198911(119879119889) gt 1198912(119879119889)For convenience we rewrite the notation 119864 as 119864 = 119865 minus 119888119868119901 +

120572119904119868119890(120572 + 120579) minus 120572119904119868119890119872 and the following two theorems wouldbe obtained to decide the optimal solution 119879lowastwhen119872 ge 119879119889

Theorem 4 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = Π2(119879

3

2) Hence 119879lowast is 1198793

2 where 1198793

2isin

(119879119889119872) and satisfies (20)(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 the following applies

(A) While 1198912(119879) gt 0 one has the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = Π2(1198791

2) Hence 119879lowast is 1198791

2

Mathematical Problems in Engineering 7

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2)

Π3(1198790

3) Hence 119879lowast is 1198790

1 11987912 or 11987903associ-

ated with maximum profit(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889)

Π3(1198790

3) Hence 119879lowast is 1198790

1 119879119889 or 11987903 associ-

ated with maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit(B) While 1198912(119879

) le 0 one has the following(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0

then Π(119879lowast) = Π3(1198790

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 11987903 associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(4) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

Theorem 5 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus

120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the

same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

42 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 lt

119879119889 When 119872 lt 119879119889 the piecewise profit function Π(119879) hasonly two different expressions that is Π1(119879) and Π3(119879)respectively From (24) we have

1198913 (119879119889) =

119863119865 (119890(120572+120579)119879119889 minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879119889] minus119863119865119879119889

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1) (

120572119904119868119890

120572 + 120579minus 119888119868119901)

[1 minus (120572 + 120579) 119879119889] minus119863 (119879119889 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(29)

As shown in Appendix D we will find that 1198911(119879119889) gt

1198913(119879119889) The following theorem would be given to decide thesystemrsquos optimal cycle time 119879lowast when119872 lt 119879119889

Theorem 6 If119872 lt 119879119889 then one has the following results

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the following

applies

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then Π(119879lowast) =

Π3(1198790

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then Π(119879lowast) =

maxΠ1(1198790

1) Π3(119879119889) Hence 119879

lowast is 1198790

1or 119879119889

associated with the maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then Π(119879

lowast) =

maxΠ1(1198790

1) Π3(119879

0

3) Hence 119879

lowast is 1198790

1or 1198790

3

associated with the maximum profit

(2) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix F for detail

5 Numerical Example

In this section we will provide the following numericalexample to illustrate the present model

Example 1 Given 119876119889 = 200 units 119863 = 1500 units 120572 =

04 120579 = 020 119870 = $50 ℎ = $1unityear 119888 = $ 5 perunit 119904 = $9 per unit 119868119890 = 013$year 119868119901 = 019$year and119872 = 03 year it is easy to get 119879119889 = 02123 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 and 119865 le (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872 Consequently by using Bisection method or

Newton-Raphson method we could obtain 1198790

1= 01760

1198790

2= 01892 and 119879

0

3= 01931 Because of 1198911(119879119889) le 0

1198912(119879119889) lt 0 and 1198912(119872) le 0 according to Theorem 4(1) theretailerrsquos optimal cycle time should be 119879lowast = 119879

0

3= 01931

8 Mathematical Problems in Engineering

Table 1 The impact of change of 120572 120579 119876119889 and119872 on the optimal solutions

Parameters 1198790

11198790

21198790

3119879lowast

119876lowast

Πlowast

120572

01 03381 03637 03895 03637 695772 6023591

02 04467 03956 03957 03956 823678 6217691

03 04792 04231 04320 04231 895987 6430335

04 05561 04538 04439 04538 985672 6663862

05 06723 05562 05071 05071 1033781 6931645

06 08552 05781 06047 06032 1367563 7362176

120579

001 05651 04436 05008 04436 1098853 7721683

002 05513 04471 04973 04471 1064786 7439053

005 05167 04441 04812 04441 994372 7268573

010 04783 04368 04791 04368 979335 7082341

050 03365 03568 04659 03568 833563 6887732

100 03561 02896 04543 03561 710856 6625568

119876119889

50 02344 02432 02589 02344 463085 4329953

100 03832 03565 03867 03565 750255 3948304

150 04973 04236 04973 04236 919334 3430035

200 06601 05467 05675 05675 1276297 3215728

250 07756 06873 06981 06981 1633156 2833270

300 08132 07675 07321 07321 1927368 2173673

119872

01 03726 03826 02873 03826 737665 2746542

02 03875 03835 03965 03835 799762 2823871

03 03811 04009 04623 04009 833721 2902323

04 03765 04197 04987 03765 899932 3130369

05 03732 04618 05013 04618 955513 3273771

06 03586 04745 05233 03586 998756 3633689

hence the optimal order quantity and the maximum averageprofit are 119876lowast = 1588763 and Πlowast = 18673562 respectively

Next we further study the effects of changes of parame-ters 120572 120579 119876119889 and 119872 on the optimal solutions The values ofother parameters keep the same as in the above examplewheneach of parameters120572 120579119876119889 and119872 varies Table 1 presents theobserved results with various parameters 120572 120579 119876119889 and119872

The following inferences can be made based on Table 1

(1) The retailerrsquos optimal order quantity 119876lowast and the

optimal average profit Πlowast increase as the value of 120572increases It implies that when the market demand ismore sensitive to the inventory level the retailer willincrease hisher order quantity to make more profitunder permissible delay in payments

(2) The optimal order quantity 119876lowast and the maximum

average profitΠlowast decrease as the value of 120579 increases

(3) When 119876119889 increases the optimal order quantity 119876lowastand the optimal cycle time 119879lowast are increasing but theoptimal average profit Πlowast is decreasing

(4) The retailerrsquos order quantity 119876lowast is increasing as 119872increases This verifies the fact that the retailer couldindeed increase the sales quantity by adopting thetrade credit policy provided by hisher supplier

6 Conclusions

In this paper we develop an EOQ model for deterioratingitems under permissible delay in payments The primarydifference of this paper as compared to previous relatedstudies is the following four aspects (1) the demand rateof items is dependent on the retailerrsquos current stock level(2) many items deteriorate continuously such as fruits andvegetables (3) the retailer who purchases the items enjoys afixed credit period offered by the supplier if the order quantityis greater than or equal to the predetermined quantity and(4) we here use maximizing profit as the objective to find theoptimal replenishment policies which could better describethe essence of the inventory system than a cost-minimizationmodel In addition we have discussed in detail the conditionsof the existence and uniqueness of the optimal solutions tothe model Three easy-to-use theorems are developed to findthe optimal ordering policies for the considered problem andthese theoretical results are illustrated by numerical exampleBy studying the effects of 120572 120579 119876119889 and 119872 on the optimalorder quantity 119876lowast and the optimal average profit Πlowast somemanagerial insights are derived

The presented model can be further extended to somemore practical situations For example we could allow forshortages quantity discounts time value of money andinflation price-sensitive and stock-dependent demand andso forth

Mathematical Problems in Engineering 9

Appendices

A Proof of Π1(119879119889) lt Π2(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (12) we get

Π2 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2(119890(120572+120579)119879119889 minus (120572 + 120579) 119879119889 minus 1)

+ 119904119868119890119872119863119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

(A1)

It is evident that119872 ge 119879119889 and (119890(120572+120579)119879119889 minus (120572+120579)119879119889 minus1) gt 0 for

119879119889 gt 0 then

Π2 (119879119889) minus Π1 (119879119889)

gt1

119879119889

1198631198792

119889

2(120572119904119868119890119872minus

120572119904119868119890

120572 + 120579+ 119888119868119901)

+ 119904119868119890119863119872119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

=119863119879119889

2(120572119904119868119890119872+ 119888119868119901) minus

119904119868119890119863119879119889

2+ 119904119868119890119863119872

ge119863119879119889

2(120572119904119868119890119872+ 119888119868119901) +

119904119868119890119863119872

2gt 0

(A2)

As a result we have Π1(119879119889) lt Π2(119879119889)This completes the proof of Appendix A

B Proof of Π1(119879119889) lt Π3(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (13) we get

Π3 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2[119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872) minus (120572 + 120579)119872]

+120572119904119868119890119872119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1)

+120572119904119868119890119863119872

2

(120572 + 120579)+120579119904119868119890119863119872

2

2 (120572 + 120579)

(B1)

Because of 119879119889 gt 119872 and 119890(120572+120579)119879119889 minus119890(120572+120579)(119879119889minus119872) minus(120572+120579)119872 gt 0we have Π3(119879119889) minus Π1(119879119889) gt 0

This completes the proof of Appendix B

C Proof of 1198911(119879119889) gt 1198912(119879119889)

If 119888119868119901 ge 119904119868119890 from (26) and (27) we have

1198912 (119879119889) minus 1198911 (119879119889)

= minus119863

(120572 + 120579)2(120572119904119868119890119872+ 119888119868119901 minus

120572119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

le minus119863

(120572 + 120579)2(120572119904119868119890119872+

120579119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

= minus119863119904119868119890

(120572 + 120579)2(120572119872 +

120579

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

minus120579 (120572 + 120579) 119879

2

119889

2

(C1)

Let 119866(119879119889) = (120572119872 + 120579(120572 + 120579))[(120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 +

1] minus 120579(120572 + 120579)1198792

1198892 then

1198661015840(119879119889) = (120572119872 +

120579

120572 + 120579) (120572 + 120579)

2119879119889119890(120572+120579)119879119889 minus 120579 (120572 + 120579) 119879119889

= 120579 (120572 + 120579) 119879119889 [(120572119872(120572 + 120579)

120579+ 1) 119890

(120572+120579)119879119889 minus 1] gt 0

(C2)

Hence 119866(119879119889) gt 119866(0) = 0 that is 1198911(119879119889) gt 1198912(119879119889)This completes the proof of Appendix C

D Proof of 1198911(119879119889) gt 1198913(119879119889)

From (26) and (24) we have

1198913 (119879119889) minus 1198911 (119879119889)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

times (120572119904119868119890

120572 + 120579minus 119888119868119901 minus 120572119904119868119890119872) minus

119863

(120572 + 120579)2

[(120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872) minus 119890

(120572+120579)(119879119889minus119872) + 1 minus (120572 + 120579)119872]

times (120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

Mathematical Problems in Engineering 3

(7) The retailer can accumulate revenue and earn interestfrom the very beginning of inventory cycle until theend of the trade credit period offered by the supplierThat is the retailer can accumulate revenue and earninterest during the period from 119905 = 0 to 119905 = 119872 withrate 119868119890 under the condition of trade credit

(8) 119904 ge 119888 119868119901 ge 119868119890 which are reasonable assumptions inreality

3 Mathematical Formulation of the Model

Based on the above assumptions the considered inventorysystem goes like this at the beginning (say the time 119905 = 0)of each replenishment cycle items of 119876 units are held andthe items are depleted gradually in the interval [0 119879] due tothe combined effects of demand and deterioration At time119905 = 119879 the inventory level reaches zero and the whole processis repeated

The variation of inventory level 119868(119905) with respect to timecan be described by the following differential equation

119889119868 (119905)

119889119905= minus119863 minus 120572119868 (119905) minus 120579119868 (119905) 0 le 119905 le 119879 (2)

with the boundary condition 119868(119879) = 0The solution to the above differential equation is

119868 (119905) =119863

120572 + 120579(119890(120572+120579)(119879minus119905)

minus 1) 0 le 119905 le 119879 (3)

So the retailerrsquos order size per cycle is

119876 = 119868 (0) =119863

120572 + 120579(119890(120572+120579)119879

minus 1) (4)

We observe that if the order quantity 119876 lt 119876119889 then thepayment must be made immediately Otherwise the retailerwill get a certain credit period 119872 Hence from (4) one hasthe fact that the inequality119876 lt 119876119889 holds if and only if119879 lt 119879119889where 119879119889 is given as below

119879119889 =1

120572 + 120579ln [1 + (120572 + 120579)

119876119889

119863] (5)

The elements comprising the profit function per cycle of theretailer are listed below

(a) The ordering cost = 119870(b) The holding cost (excluding interest charges) =

ℎ int119879

0119868(119905)119889119905 = (ℎ119863(120572 + 120579)

2)[119890(120572+120579)119879

minus (120572 + 120579)119879 minus 1]

(c) The purchasing cost = 119888119876 = (119888119863(120572 + 120579))[119890(120572+120579)119879

minus 1]

(d) The sales revenue = 119904 int119879

0119877(119905)119889119905 = 119904[120579119863119879(120572 + 120579) +

(120572119863(120572 + 120579)2)(119890(120572+120579)119879

minus 1)](e) According to the above assumption there are two

cases to occur in interest payable in each order cycle

Case 1 (119879 le 119879119889) In this case the delay in payments isnot permitted That is all products in inventory have to be

financed with annual rate 119868119901 Therefore interest payable ineach order cycle

= 119888119868119901 int

119879

0

119868 (119905) 119889119905 =

119888119868119901119863

(120572 + 120579)2[119890(120572+120579)119879

minus (120572 + 120579) 119879 minus 1] (6)

Case 2 (119879 ge 119879119889) In this case the delay credit period 119872 inpayments is permitted which arouses two subcases Subcase21 where 119879 le 119872 and Subcase 22 where 119879 ge 119872

Subcase 21 (119879 le 119872) Since the cycle time 119879 is shorter thanthe credit period 119872 there is no interest paid for financingthe inventory in stock So the interest payable in each ordercycle = 0

Subcase 22 (119879 ge 119872) When the credit period 119872 is shorterthan or equal to the replenishment cycle time 119879 the retailerstarts paying the interest for the items in stock after time119872with rate 119868119901 Hence the interest payable in each order cycle

= 119888119868119901 int

119879

119872

119868 (119905) 119889119905

=

119888119868119901119863

(120572 + 120579)2[119890(120572+120579)(119879minus119872)

minus (120572 + 120579) (119879 minus119872) minus 1]

(7)

(f) Similarly there are also two cases to occur in interestearned in each order cycle

Case 1 (119879 le 119879119889) In this case the delay in payments is notpermitted so the interest earned in each order cycle = 0

Case 2 (119879 ge 119879119889) In this case the delay in payments ispermitted which causes two subcases as below

Subcase 21 (119879 le 119872) Since the cycle time 119879 is shorter thanthe credit period 119872 from time 0 to 119879 the retailer sells thegoods and continues to accumulate the sales revenue to earninterest 119904119868119890 int

119879

0int119905

0119877(119906)119889119906 119889119905 and from time119879 to119872 the retailer

can use the sales revenue generated in [0 119879] to earn interest119904119868119890 int119879

0119877(119906)119889119906(119872 minus 119879) Therefore the interest earned from

time 0 to119872 per cycle

= 119904119868119890 [int

119879

0

int

119905

0

119877 (119906) 119889119906 119889119905 + int

119879

0

119877 (119906) 119889119906 (119872 minus 119879)]

= 119904119868119890 1205791198631198792

2 (120572 + 120579)minus

120572119863

(120572 + 120579)3(119890(120572+120579)119879

minus 1)

+120572119863119879

(120572 + 120579)2119890(120572+120579)119879

+[120579119863119879

120572 + 120579+

120572119863

(120572 + 120579)2(119890(120572+120579)119879

minus 1)] (119872 minus 119879)

(8)

Subcase 22 (119879 ge 119872) In this case the delay in payments ispermitted and during time 0 through119872 the retailer sells thegoods and continues to accumulate sales revenue and earns

4 Mathematical Problems in Engineering

the interest with rate 119868119890 Therefore the interest earned fromtime 0 to119872 per cycle

= 119904119868119890 int

119872

0

int

119905

0

119877 (119906) 119889119906 119889119905

= 119904119868119890 [1205791198631198722

2 (120572 + 120579)+

120572119863119872

(120572 + 120579)2119890(120572+120579)119879

+120572119863

(120572 + 120579)3119890(120572+120579)119879

(119890minus(120572+120579)119872

minus 1)]

(9)

From the above the profit function for the retailer can beexpressed as

Π (119879) = Sales revenue + interest earned minus ordering cost

minus holding cost minus purchasing cost

minus interest payable times 119879minus1(10)

On simplification and summation we get the profit functionas the following three forms

(A) if 119879 le 119879119889 then

Π (119879) = Π1 (119879)

=1

119879

119863119864

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863119879

120572 + 120579(ℎ + 120579119904 + 119888119868119901) minus 119870

(11)

(B) if 119879 gt 119879119889 but 119879 le 119872 then

Π (119879) = Π2 (119879)

=1

119879

119863119865

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863119879

120572 + 120579(ℎ + 120579119904 + 120579119904119868119890119872+

120572119904119868119890

120572 + 120579)

minus120579119904119868119890119863119879

2

2 (120572 + 120579)minus 119870

(12)

(C) if 119879 gt 119879119889 and 119879 gt 119872 then

Π (119879) = Π3 (119879)

=1

119879

119863119865

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863

(120572 + 120579)2(119890(120572+120579)(119879minus119872)

minus 1) (120572119904119868119890

120572 + 120579minus 119888119868119901)

+119863119879

120572 + 120579(ℎ + 120579119904 + 119888119868119901) +

119863119872

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(13)

where the meaning of notations 119864 and 119865 in the aboveexpressions is

119864 = 120572119904 minus (120572 + 120579) 119888 minus ℎ minus 119888119868119901

119865 = 120572119904 minus (120572 + 120579) 119888 minus ℎ + 120572119904119868119890119872minus120572119904119868119890

120572 + 120579

(14)

For convenience we will consider the problem through thetwo following situations (1)119872 ge 119879119889 and (2)119872 lt 119879119889

(1) Suppose That 119872 ge 119879119889 In the situation of119872 ge 119879119889 Π(119879)has three different expressions as follows

Π (119879) = Π1 (119879) 0 lt 119879 le 119879119889 (15a)

Π (119879) = Π2 (119879) 119879119889 le 119879 le 119872 (15b)

Π (119879) = Π3 (119879) 119879 ge 119872 (15c)

It is easy to verify that in the case of 119872 ge 119879119889 Π(119879)continues except at 119879 = 119879119889 In fact if 119888119868119901 ge 119904119868119890 then we couldprove that Π1(119879119889) lt Π2(119879119889) (Proof of Π1(119879119889) lt Π2(119879119889) isshown in Appendix A)

(2) Suppose That 119872 lt 119879119889 If 119872 lt 119879119889 then Π(119879) has twodifferent expressions as follows

Π (119879) = Π1 (119879) 0 lt 119879 le 119879119889 (16a)

Π (119879) = Π3 (119879) 119879 ge 119879119889 (16b)

Here Π1(119879) and Π3(119879) are given by (11) and (13)respectively Π(119879) continues except at 119879 = 119879119889 In fact wecould prove that Π1(119879119889) lt Π3(119879119889) when 119872 lt 119879119889 (Proof ofΠ1(119879119889) lt Π3(119879119889) is shown in Appendix B)

In the next section we will determine the retailerrsquosoptimal cycle time119879lowast for the above two situations using somealgebraic method

4 The Solution Procedures

In order to decide the optimal solution 119879lowast of function Π(119879)we need first to study the properties of function Π119894(119879) (119894 =1 2 3) on (0 +infin) respectively The first-order necessarycondition forΠ1(119879) in (11) to bemaximum is119889Π1(119879)119889119879 = 0which implies that

minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119890

(120572+120579)119879minus 119890(120572+120579)119879

+ 1] minus 119870 = 0 (17)

For convenience we set the left-hand expression of (17) as1198911(119879) Taking the derivative of 1198911(119879) with respect to 119879 weget

1198891198911 (119879)

119889119879= minus119863119864119879119890

(120572+120579)119879 (18)

From (18) we know that if 119864 lt 0 then 1198891198911(119879)119889119879 gt

0 that is 1198911(119879) is increasing on (0 +infin) It is obvious that1198911(0) = minus119870 lt 0 and lim119879rarr+infin1198911(119879) = +infin therefore theIntermediate Value Theorem implies that 1198911(119879) = 0 that

Mathematical Problems in Engineering 5

is 119889Π1(119879)119889119879 = 0 has a unique root (say 11987901) on (0 +infin)

Moreover taking the second derivative ofΠ1(119879)with respectto 119879 gives

1198892Π1 (119879)

1198891198792= minus

2119870

1198793+

119863119864

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(19)

It is easy to show that the inequality 1199092119890119909minus2119909119890119909+2119890119909minus2 gt0 holds for all 119909 gt 0 so 1198892Π1(119879)119889119879

2 will always be negativeon (0 +infin) if 119864 lt 0 that is Π1(119879) is convex on (0 +infin)Therefore 1198790

1is the unique maximum point of Π1(119879) on

(0 +infin) But if 119864 ge 0 (18) means that 1198911(119879) is decreasingon (0 +infin) since 1198911(0) = minus119870 lt 0 then 1198911(119879) is negative and119889Π1(119879)119889119879 is positive for all 119879 gt 0 hence Π1(119879) is strictlyincreasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π1(119879) on (0 +infin)

Lemma 1 If 119864 lt 0 Π1(119879) is convex on (0 +infin) and 11987901is the

unique maximum point of Π1(119879) Otherwise Π1(119879) is strictlyincreasing on (0 +infin)

Similarly the first-order necessary condition forΠ2(119879) in(12) to be maximized is 119889Π2(119879)119889119879 = 0 which leads to

minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119890

(120572+120579)119879minus 119890(120572+120579)119879

+ 1]

+120579119904119868119890119863119879

2

2 (120572 + 120579)minus 119870 = 0

(20)

We set the left-hand expression of (20) as 1198912(119879) Taking thefirst derivative of 1198912(119879) with respect to 119879 gives

1198891198912 (119879)

119889119879= minus(119865119890

(120572+120579)119879minus

120579119904119868119890

120572 + 120579)119863119879 (21)

Consequently from the sign of 119865 there are two distinct casesfor finding the property of Π2(119879) as follows

(1) When 119865 le 0 we have 1198891198912(119879)119889119879 gt 0 that is 1198912(119879) isstrictly increasing on (0 +infin) Since 1198912(0) = minus119870 lt 0

and lim119879rarr+infin1198912(119879) = +infin 1198912(119879) = 0 has a uniquesolution (say 1198790

2) on (0 +infin) Moreover from (13) we

have

1198892Π2 (119879)

1198891198792= minus

2119870

1198793+

119863119865

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(22)

Since 119865 le 0 and (120572+120579)21198792119890(120572+120579)119879 minus2(120572+120579)119879119890(120572+120579)119879 +2119890(120572+120579)119879

minus 2 gt 0 we get 1198892Π2(119879)1198891198792lt 0 Hence

Π2(119879) is convex on (0 +infin) and 1198790

2is the unique

maximum point of Π2(119879)

(2) When 119865 gt 0 there will exist a unique root to theequation 1198891198912(119879)119889119879 = 0 Denote this root by 119879 then

119879=

1

120572 + 120579[ln 120579119904119868119890 minus ln (120572 + 120579) 119865] (23)

(A) If 119879ge 0 that is 0 lt 119865 le 120579119904119868119890(120572 + 120579) then

1198891198912(119879)119889119879 gt 0 for 119879 isin (0 119879) and 1198891198912(119879)119889119879 lt

0 for 119879 isin (119879 +infin) that is 1198912(119879) is increasing

on (0 119879) and decreasing on (119879

+infin)

(a) When 1198912(119879) gt 0 1198912(119879) = 0 has a unique

root (say11987912) on (0 119879

) due to1198912(0) = minus119870 lt

0 Hence Π2(119879) is increasing on (0 1198791

2)

and decreasing on (1198791

2 119879

) On the other

interval of (119879 +infin) however because of

1198912(119879) gt 0 and lim119879rarr+infin1198912(119879) = minusinfin

for 119865 gt 0 1198912(119879) has a unique zeropoint (say 119879

2

2) on (119879

+infin) Thus Π2(119879)

is decreasing on (119879 1198792

2) and increasing on

(1198792

2 +infin) Hence Π2(119879) is increasing on

(0 1198791

2) decreasing on (1198791

2 1198792

2) and increas-

ing on (11987922 +infin)

(b) When 1198912(119879) le 0 1198912(119879) is negative

on (0 119879) because 1198912(119879) is increasing on

(0 119879) whereas 1198912(119879) is also negative on

(119879 +infin) since 1198912(119879) is decreasing on

(119879 +infin) Therefore 1198912(119879) is always non-

positive on (0 +infin) andΠ2(119879) is increasingon (0 +infin)

(B) If 119879lt 0 that is 119865 gt 120579119904119868119890(120572 + 120579) then for any

119879 ge 0 we have 1198891198912(119879)119889119879 lt 0 that is 1198912(119879)is strictly decreasing on (0 +infin) Since 1198912(0) =minus119870 lt 0 1198912(119879) is negative on (0 +infin) HenceΠ2(119879) is increasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π2(119879) on (0 +infin)

Lemma 2 (1) For 119865 le 0 Π2(119879) is convex on (0 +infin) and 11987902

is the unique maximum point of Π2(119879)(2) For 0 lt 119865 le 120579119904119868119890(120572+120579)Π2(119879) is increasing on (0 11987912 )

decreasing on (11987912 1198792

2) and increasing on (1198792

2 +infin) if1198912(119879

) gt

0 and Π2(119879) is increasing on (0 +infin) otherwise(3) For 119865 gt 120579119904119868119890(120572 + 120579) Π2(119879) is always increasing on

(0 +infin)

Likewise the first-order necessary condition forΠ3(119879) in(13) to be maximum is 119889Π3(119879)119889119879 = 0 which implies that

119863119865(119890(120572+120579)119879

minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879] minus119863119865119879

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879minus119872)

minus 1) (120572119904119868119890

120572 + 120579minus 119888119868119901)

6 Mathematical Problems in Engineering

[1 minus (120572 + 120579) 119879] minus119863 (119879 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870 = 0

(24)

We set the left-hand expression of (24) as 1198913(119879) Taking thefirst derivative of 1198913(119879) with respect to 119879 gives

1198891198913 (119879)

119889119879= minus [119865 + (

120572119904119868119890

120572 + 120579minus 119888119868119901) 119890

minus(120572+120579)119872]119863119879119890

(120572+120579)119879

(25)

We consider the following two cases

(1) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198891198913(119879)119889119879 ge

0 Therefore 1198913(119879) is increasing on (0 +infin) It is easyto get 1198913(0) = (119863(120572 + 120579)

2)(119890minus(120572+120579)119872

minus 1)(120572119904119868119890(120572 +

120579) minus 119888119868119901) + 12057911990411986811989011986311987222(120572 + 120579) + 120572119904119868119890119863119872(120572 + 120579)

2minus

119888119868119901119863119872(120572 + 120579) minus 119870 and lim119879rarr+infin1198913(119879) = +infinThus if 1198913(0) lt 0 then 1198913(119879) = 0 has a uniquesolution (say 1198790

3) on (0 +infin) Hence 1198913(119879) will be

negative on (0 11987903) and positive on (1198790

3 +infin) which

implies Π3(119879) is increasing on (0 11987903) and decreasing

on (11987903 +infin) That is Π3(119879) is unimodal on (0 +infin)

and 1198790

3is the unique maximum point of Π3(119879) if

1198913(0) lt 0 In contrast if 1198913(0) ge 0 then 1198913(119879) willalways be nonnegative on (0 +infin) hence Π3(119879) isalways decreasing on (0 +infin)

(2) If119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 then1198913(119879) is strictly

decreasing on (0 +infin) Noting that lim119879rarr+infin1198913(119879) =minusinfin for 119865 gt minus(120572119904119868119890(120572 + 120579) minus 119888119868119901)119890

minus(120572+120579)119872 one easilyobtains that if 1198913(0) gt 0 there will exist a unique root(say 1198791

3) to the equation 1198913(119879) = 0 hence 1198913(119879) is

positive on (0 11987913) and negative on (1198791

3 +infin) which

implies Π3(119879) is decreasing on (0 11987913) and increasing

on (11987913 +infin) But if1198913(0) le 01198913(119879) is always negative

on (0 +infin) then Π3(119879) is increasing on (0 +infin)

Summarizing the above arguments the following lemmato describe the property ofΠ3(119879) on (0 +infin) will be obtained

Lemma 3 (1) For 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 Π3(119879) is

increasing on (0 11987903) and decreasing on (1198790

3 +infin) if 1198913(0) lt 0

and Π3(119879) is decreasing on (0 +infin) otherwise(2) For119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872Π3(119879) is decreasingon (0 1198791

3) and increasing on (1198791

3 +infin) if 1198913(0) gt 0 andΠ3(119879)

is increasing on (0 +infin) otherwise

Based upon Lemmas 1ndash3 we now will decide the optimalsolution 119879lowast to Π(119879) from the below two situations

41 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 ge

119879119889 When 119872 ge 119879119889 the piecewise profit function Π(119879) has

three different expressions that is Π1(119879) Π2(119879) and Π3(119879)respectively From (17) (20) and (24) we have

1198911 (119879119889) =minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1] minus 119870

(26)

1198912 (119879119889) =minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

+120579119904119868119890119863119879

2

119889

2 (120572 + 120579)minus 119870

(27)

1198912 (119872) = 1198913 (119872)

=minus119863119865

(120572 + 120579)2[(120572 + 120579)119872119890

(120572+120579)119872minus 119890(120572+120579)119872

+ 1]

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(28)

As shown in Appendix C we will find that 1198911(119879119889) gt 1198912(119879119889)For convenience we rewrite the notation 119864 as 119864 = 119865 minus 119888119868119901 +

120572119904119868119890(120572 + 120579) minus 120572119904119868119890119872 and the following two theorems wouldbe obtained to decide the optimal solution 119879lowastwhen119872 ge 119879119889

Theorem 4 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = Π2(119879

3

2) Hence 119879lowast is 1198793

2 where 1198793

2isin

(119879119889119872) and satisfies (20)(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 the following applies

(A) While 1198912(119879) gt 0 one has the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = Π2(1198791

2) Hence 119879lowast is 1198791

2

Mathematical Problems in Engineering 7

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2)

Π3(1198790

3) Hence 119879lowast is 1198790

1 11987912 or 11987903associ-

ated with maximum profit(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889)

Π3(1198790

3) Hence 119879lowast is 1198790

1 119879119889 or 11987903 associ-

ated with maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit(B) While 1198912(119879

) le 0 one has the following(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0

then Π(119879lowast) = Π3(1198790

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 11987903 associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(4) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

Theorem 5 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus

120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the

same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

42 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 lt

119879119889 When 119872 lt 119879119889 the piecewise profit function Π(119879) hasonly two different expressions that is Π1(119879) and Π3(119879)respectively From (24) we have

1198913 (119879119889) =

119863119865 (119890(120572+120579)119879119889 minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879119889] minus119863119865119879119889

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1) (

120572119904119868119890

120572 + 120579minus 119888119868119901)

[1 minus (120572 + 120579) 119879119889] minus119863 (119879119889 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(29)

As shown in Appendix D we will find that 1198911(119879119889) gt

1198913(119879119889) The following theorem would be given to decide thesystemrsquos optimal cycle time 119879lowast when119872 lt 119879119889

Theorem 6 If119872 lt 119879119889 then one has the following results

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the following

applies

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then Π(119879lowast) =

Π3(1198790

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then Π(119879lowast) =

maxΠ1(1198790

1) Π3(119879119889) Hence 119879

lowast is 1198790

1or 119879119889

associated with the maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then Π(119879

lowast) =

maxΠ1(1198790

1) Π3(119879

0

3) Hence 119879

lowast is 1198790

1or 1198790

3

associated with the maximum profit

(2) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix F for detail

5 Numerical Example

In this section we will provide the following numericalexample to illustrate the present model

Example 1 Given 119876119889 = 200 units 119863 = 1500 units 120572 =

04 120579 = 020 119870 = $50 ℎ = $1unityear 119888 = $ 5 perunit 119904 = $9 per unit 119868119890 = 013$year 119868119901 = 019$year and119872 = 03 year it is easy to get 119879119889 = 02123 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 and 119865 le (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872 Consequently by using Bisection method or

Newton-Raphson method we could obtain 1198790

1= 01760

1198790

2= 01892 and 119879

0

3= 01931 Because of 1198911(119879119889) le 0

1198912(119879119889) lt 0 and 1198912(119872) le 0 according to Theorem 4(1) theretailerrsquos optimal cycle time should be 119879lowast = 119879

0

3= 01931

8 Mathematical Problems in Engineering

Table 1 The impact of change of 120572 120579 119876119889 and119872 on the optimal solutions

Parameters 1198790

11198790

21198790

3119879lowast

119876lowast

Πlowast

120572

01 03381 03637 03895 03637 695772 6023591

02 04467 03956 03957 03956 823678 6217691

03 04792 04231 04320 04231 895987 6430335

04 05561 04538 04439 04538 985672 6663862

05 06723 05562 05071 05071 1033781 6931645

06 08552 05781 06047 06032 1367563 7362176

120579

001 05651 04436 05008 04436 1098853 7721683

002 05513 04471 04973 04471 1064786 7439053

005 05167 04441 04812 04441 994372 7268573

010 04783 04368 04791 04368 979335 7082341

050 03365 03568 04659 03568 833563 6887732

100 03561 02896 04543 03561 710856 6625568

119876119889

50 02344 02432 02589 02344 463085 4329953

100 03832 03565 03867 03565 750255 3948304

150 04973 04236 04973 04236 919334 3430035

200 06601 05467 05675 05675 1276297 3215728

250 07756 06873 06981 06981 1633156 2833270

300 08132 07675 07321 07321 1927368 2173673

119872

01 03726 03826 02873 03826 737665 2746542

02 03875 03835 03965 03835 799762 2823871

03 03811 04009 04623 04009 833721 2902323

04 03765 04197 04987 03765 899932 3130369

05 03732 04618 05013 04618 955513 3273771

06 03586 04745 05233 03586 998756 3633689

hence the optimal order quantity and the maximum averageprofit are 119876lowast = 1588763 and Πlowast = 18673562 respectively

Next we further study the effects of changes of parame-ters 120572 120579 119876119889 and 119872 on the optimal solutions The values ofother parameters keep the same as in the above examplewheneach of parameters120572 120579119876119889 and119872 varies Table 1 presents theobserved results with various parameters 120572 120579 119876119889 and119872

The following inferences can be made based on Table 1

(1) The retailerrsquos optimal order quantity 119876lowast and the

optimal average profit Πlowast increase as the value of 120572increases It implies that when the market demand ismore sensitive to the inventory level the retailer willincrease hisher order quantity to make more profitunder permissible delay in payments

(2) The optimal order quantity 119876lowast and the maximum

average profitΠlowast decrease as the value of 120579 increases

(3) When 119876119889 increases the optimal order quantity 119876lowastand the optimal cycle time 119879lowast are increasing but theoptimal average profit Πlowast is decreasing

(4) The retailerrsquos order quantity 119876lowast is increasing as 119872increases This verifies the fact that the retailer couldindeed increase the sales quantity by adopting thetrade credit policy provided by hisher supplier

6 Conclusions

In this paper we develop an EOQ model for deterioratingitems under permissible delay in payments The primarydifference of this paper as compared to previous relatedstudies is the following four aspects (1) the demand rateof items is dependent on the retailerrsquos current stock level(2) many items deteriorate continuously such as fruits andvegetables (3) the retailer who purchases the items enjoys afixed credit period offered by the supplier if the order quantityis greater than or equal to the predetermined quantity and(4) we here use maximizing profit as the objective to find theoptimal replenishment policies which could better describethe essence of the inventory system than a cost-minimizationmodel In addition we have discussed in detail the conditionsof the existence and uniqueness of the optimal solutions tothe model Three easy-to-use theorems are developed to findthe optimal ordering policies for the considered problem andthese theoretical results are illustrated by numerical exampleBy studying the effects of 120572 120579 119876119889 and 119872 on the optimalorder quantity 119876lowast and the optimal average profit Πlowast somemanagerial insights are derived

The presented model can be further extended to somemore practical situations For example we could allow forshortages quantity discounts time value of money andinflation price-sensitive and stock-dependent demand andso forth

Mathematical Problems in Engineering 9

Appendices

A Proof of Π1(119879119889) lt Π2(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (12) we get

Π2 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2(119890(120572+120579)119879119889 minus (120572 + 120579) 119879119889 minus 1)

+ 119904119868119890119872119863119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

(A1)

It is evident that119872 ge 119879119889 and (119890(120572+120579)119879119889 minus (120572+120579)119879119889 minus1) gt 0 for

119879119889 gt 0 then

Π2 (119879119889) minus Π1 (119879119889)

gt1

119879119889

1198631198792

119889

2(120572119904119868119890119872minus

120572119904119868119890

120572 + 120579+ 119888119868119901)

+ 119904119868119890119863119872119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

=119863119879119889

2(120572119904119868119890119872+ 119888119868119901) minus

119904119868119890119863119879119889

2+ 119904119868119890119863119872

ge119863119879119889

2(120572119904119868119890119872+ 119888119868119901) +

119904119868119890119863119872

2gt 0

(A2)

As a result we have Π1(119879119889) lt Π2(119879119889)This completes the proof of Appendix A

B Proof of Π1(119879119889) lt Π3(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (13) we get

Π3 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2[119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872) minus (120572 + 120579)119872]

+120572119904119868119890119872119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1)

+120572119904119868119890119863119872

2

(120572 + 120579)+120579119904119868119890119863119872

2

2 (120572 + 120579)

(B1)

Because of 119879119889 gt 119872 and 119890(120572+120579)119879119889 minus119890(120572+120579)(119879119889minus119872) minus(120572+120579)119872 gt 0we have Π3(119879119889) minus Π1(119879119889) gt 0

This completes the proof of Appendix B

C Proof of 1198911(119879119889) gt 1198912(119879119889)

If 119888119868119901 ge 119904119868119890 from (26) and (27) we have

1198912 (119879119889) minus 1198911 (119879119889)

= minus119863

(120572 + 120579)2(120572119904119868119890119872+ 119888119868119901 minus

120572119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

le minus119863

(120572 + 120579)2(120572119904119868119890119872+

120579119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

= minus119863119904119868119890

(120572 + 120579)2(120572119872 +

120579

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

minus120579 (120572 + 120579) 119879

2

119889

2

(C1)

Let 119866(119879119889) = (120572119872 + 120579(120572 + 120579))[(120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 +

1] minus 120579(120572 + 120579)1198792

1198892 then

1198661015840(119879119889) = (120572119872 +

120579

120572 + 120579) (120572 + 120579)

2119879119889119890(120572+120579)119879119889 minus 120579 (120572 + 120579) 119879119889

= 120579 (120572 + 120579) 119879119889 [(120572119872(120572 + 120579)

120579+ 1) 119890

(120572+120579)119879119889 minus 1] gt 0

(C2)

Hence 119866(119879119889) gt 119866(0) = 0 that is 1198911(119879119889) gt 1198912(119879119889)This completes the proof of Appendix C

D Proof of 1198911(119879119889) gt 1198913(119879119889)

From (26) and (24) we have

1198913 (119879119889) minus 1198911 (119879119889)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

times (120572119904119868119890

120572 + 120579minus 119888119868119901 minus 120572119904119868119890119872) minus

119863

(120572 + 120579)2

[(120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872) minus 119890

(120572+120579)(119879119889minus119872) + 1 minus (120572 + 120579)119872]

times (120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

4 Mathematical Problems in Engineering

the interest with rate 119868119890 Therefore the interest earned fromtime 0 to119872 per cycle

= 119904119868119890 int

119872

0

int

119905

0

119877 (119906) 119889119906 119889119905

= 119904119868119890 [1205791198631198722

2 (120572 + 120579)+

120572119863119872

(120572 + 120579)2119890(120572+120579)119879

+120572119863

(120572 + 120579)3119890(120572+120579)119879

(119890minus(120572+120579)119872

minus 1)]

(9)

From the above the profit function for the retailer can beexpressed as

Π (119879) = Sales revenue + interest earned minus ordering cost

minus holding cost minus purchasing cost

minus interest payable times 119879minus1(10)

On simplification and summation we get the profit functionas the following three forms

(A) if 119879 le 119879119889 then

Π (119879) = Π1 (119879)

=1

119879

119863119864

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863119879

120572 + 120579(ℎ + 120579119904 + 119888119868119901) minus 119870

(11)

(B) if 119879 gt 119879119889 but 119879 le 119872 then

Π (119879) = Π2 (119879)

=1

119879

119863119865

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863119879

120572 + 120579(ℎ + 120579119904 + 120579119904119868119890119872+

120572119904119868119890

120572 + 120579)

minus120579119904119868119890119863119879

2

2 (120572 + 120579)minus 119870

(12)

(C) if 119879 gt 119879119889 and 119879 gt 119872 then

Π (119879) = Π3 (119879)

=1

119879

119863119865

(120572 + 120579)2(119890(120572+120579)119879

minus 1)

+119863

(120572 + 120579)2(119890(120572+120579)(119879minus119872)

minus 1) (120572119904119868119890

120572 + 120579minus 119888119868119901)

+119863119879

120572 + 120579(ℎ + 120579119904 + 119888119868119901) +

119863119872

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(13)

where the meaning of notations 119864 and 119865 in the aboveexpressions is

119864 = 120572119904 minus (120572 + 120579) 119888 minus ℎ minus 119888119868119901

119865 = 120572119904 minus (120572 + 120579) 119888 minus ℎ + 120572119904119868119890119872minus120572119904119868119890

120572 + 120579

(14)

For convenience we will consider the problem through thetwo following situations (1)119872 ge 119879119889 and (2)119872 lt 119879119889

(1) Suppose That 119872 ge 119879119889 In the situation of119872 ge 119879119889 Π(119879)has three different expressions as follows

Π (119879) = Π1 (119879) 0 lt 119879 le 119879119889 (15a)

Π (119879) = Π2 (119879) 119879119889 le 119879 le 119872 (15b)

Π (119879) = Π3 (119879) 119879 ge 119872 (15c)

It is easy to verify that in the case of 119872 ge 119879119889 Π(119879)continues except at 119879 = 119879119889 In fact if 119888119868119901 ge 119904119868119890 then we couldprove that Π1(119879119889) lt Π2(119879119889) (Proof of Π1(119879119889) lt Π2(119879119889) isshown in Appendix A)

(2) Suppose That 119872 lt 119879119889 If 119872 lt 119879119889 then Π(119879) has twodifferent expressions as follows

Π (119879) = Π1 (119879) 0 lt 119879 le 119879119889 (16a)

Π (119879) = Π3 (119879) 119879 ge 119879119889 (16b)

Here Π1(119879) and Π3(119879) are given by (11) and (13)respectively Π(119879) continues except at 119879 = 119879119889 In fact wecould prove that Π1(119879119889) lt Π3(119879119889) when 119872 lt 119879119889 (Proof ofΠ1(119879119889) lt Π3(119879119889) is shown in Appendix B)

In the next section we will determine the retailerrsquosoptimal cycle time119879lowast for the above two situations using somealgebraic method

4 The Solution Procedures

In order to decide the optimal solution 119879lowast of function Π(119879)we need first to study the properties of function Π119894(119879) (119894 =1 2 3) on (0 +infin) respectively The first-order necessarycondition forΠ1(119879) in (11) to bemaximum is119889Π1(119879)119889119879 = 0which implies that

minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119890

(120572+120579)119879minus 119890(120572+120579)119879

+ 1] minus 119870 = 0 (17)

For convenience we set the left-hand expression of (17) as1198911(119879) Taking the derivative of 1198911(119879) with respect to 119879 weget

1198891198911 (119879)

119889119879= minus119863119864119879119890

(120572+120579)119879 (18)

From (18) we know that if 119864 lt 0 then 1198891198911(119879)119889119879 gt

0 that is 1198911(119879) is increasing on (0 +infin) It is obvious that1198911(0) = minus119870 lt 0 and lim119879rarr+infin1198911(119879) = +infin therefore theIntermediate Value Theorem implies that 1198911(119879) = 0 that

Mathematical Problems in Engineering 5

is 119889Π1(119879)119889119879 = 0 has a unique root (say 11987901) on (0 +infin)

Moreover taking the second derivative ofΠ1(119879)with respectto 119879 gives

1198892Π1 (119879)

1198891198792= minus

2119870

1198793+

119863119864

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(19)

It is easy to show that the inequality 1199092119890119909minus2119909119890119909+2119890119909minus2 gt0 holds for all 119909 gt 0 so 1198892Π1(119879)119889119879

2 will always be negativeon (0 +infin) if 119864 lt 0 that is Π1(119879) is convex on (0 +infin)Therefore 1198790

1is the unique maximum point of Π1(119879) on

(0 +infin) But if 119864 ge 0 (18) means that 1198911(119879) is decreasingon (0 +infin) since 1198911(0) = minus119870 lt 0 then 1198911(119879) is negative and119889Π1(119879)119889119879 is positive for all 119879 gt 0 hence Π1(119879) is strictlyincreasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π1(119879) on (0 +infin)

Lemma 1 If 119864 lt 0 Π1(119879) is convex on (0 +infin) and 11987901is the

unique maximum point of Π1(119879) Otherwise Π1(119879) is strictlyincreasing on (0 +infin)

Similarly the first-order necessary condition forΠ2(119879) in(12) to be maximized is 119889Π2(119879)119889119879 = 0 which leads to

minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119890

(120572+120579)119879minus 119890(120572+120579)119879

+ 1]

+120579119904119868119890119863119879

2

2 (120572 + 120579)minus 119870 = 0

(20)

We set the left-hand expression of (20) as 1198912(119879) Taking thefirst derivative of 1198912(119879) with respect to 119879 gives

1198891198912 (119879)

119889119879= minus(119865119890

(120572+120579)119879minus

120579119904119868119890

120572 + 120579)119863119879 (21)

Consequently from the sign of 119865 there are two distinct casesfor finding the property of Π2(119879) as follows

(1) When 119865 le 0 we have 1198891198912(119879)119889119879 gt 0 that is 1198912(119879) isstrictly increasing on (0 +infin) Since 1198912(0) = minus119870 lt 0

and lim119879rarr+infin1198912(119879) = +infin 1198912(119879) = 0 has a uniquesolution (say 1198790

2) on (0 +infin) Moreover from (13) we

have

1198892Π2 (119879)

1198891198792= minus

2119870

1198793+

119863119865

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(22)

Since 119865 le 0 and (120572+120579)21198792119890(120572+120579)119879 minus2(120572+120579)119879119890(120572+120579)119879 +2119890(120572+120579)119879

minus 2 gt 0 we get 1198892Π2(119879)1198891198792lt 0 Hence

Π2(119879) is convex on (0 +infin) and 1198790

2is the unique

maximum point of Π2(119879)

(2) When 119865 gt 0 there will exist a unique root to theequation 1198891198912(119879)119889119879 = 0 Denote this root by 119879 then

119879=

1

120572 + 120579[ln 120579119904119868119890 minus ln (120572 + 120579) 119865] (23)

(A) If 119879ge 0 that is 0 lt 119865 le 120579119904119868119890(120572 + 120579) then

1198891198912(119879)119889119879 gt 0 for 119879 isin (0 119879) and 1198891198912(119879)119889119879 lt

0 for 119879 isin (119879 +infin) that is 1198912(119879) is increasing

on (0 119879) and decreasing on (119879

+infin)

(a) When 1198912(119879) gt 0 1198912(119879) = 0 has a unique

root (say11987912) on (0 119879

) due to1198912(0) = minus119870 lt

0 Hence Π2(119879) is increasing on (0 1198791

2)

and decreasing on (1198791

2 119879

) On the other

interval of (119879 +infin) however because of

1198912(119879) gt 0 and lim119879rarr+infin1198912(119879) = minusinfin

for 119865 gt 0 1198912(119879) has a unique zeropoint (say 119879

2

2) on (119879

+infin) Thus Π2(119879)

is decreasing on (119879 1198792

2) and increasing on

(1198792

2 +infin) Hence Π2(119879) is increasing on

(0 1198791

2) decreasing on (1198791

2 1198792

2) and increas-

ing on (11987922 +infin)

(b) When 1198912(119879) le 0 1198912(119879) is negative

on (0 119879) because 1198912(119879) is increasing on

(0 119879) whereas 1198912(119879) is also negative on

(119879 +infin) since 1198912(119879) is decreasing on

(119879 +infin) Therefore 1198912(119879) is always non-

positive on (0 +infin) andΠ2(119879) is increasingon (0 +infin)

(B) If 119879lt 0 that is 119865 gt 120579119904119868119890(120572 + 120579) then for any

119879 ge 0 we have 1198891198912(119879)119889119879 lt 0 that is 1198912(119879)is strictly decreasing on (0 +infin) Since 1198912(0) =minus119870 lt 0 1198912(119879) is negative on (0 +infin) HenceΠ2(119879) is increasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π2(119879) on (0 +infin)

Lemma 2 (1) For 119865 le 0 Π2(119879) is convex on (0 +infin) and 11987902

is the unique maximum point of Π2(119879)(2) For 0 lt 119865 le 120579119904119868119890(120572+120579)Π2(119879) is increasing on (0 11987912 )

decreasing on (11987912 1198792

2) and increasing on (1198792

2 +infin) if1198912(119879

) gt

0 and Π2(119879) is increasing on (0 +infin) otherwise(3) For 119865 gt 120579119904119868119890(120572 + 120579) Π2(119879) is always increasing on

(0 +infin)

Likewise the first-order necessary condition forΠ3(119879) in(13) to be maximum is 119889Π3(119879)119889119879 = 0 which implies that

119863119865(119890(120572+120579)119879

minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879] minus119863119865119879

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879minus119872)

minus 1) (120572119904119868119890

120572 + 120579minus 119888119868119901)

6 Mathematical Problems in Engineering

[1 minus (120572 + 120579) 119879] minus119863 (119879 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870 = 0

(24)

We set the left-hand expression of (24) as 1198913(119879) Taking thefirst derivative of 1198913(119879) with respect to 119879 gives

1198891198913 (119879)

119889119879= minus [119865 + (

120572119904119868119890

120572 + 120579minus 119888119868119901) 119890

minus(120572+120579)119872]119863119879119890

(120572+120579)119879

(25)

We consider the following two cases

(1) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198891198913(119879)119889119879 ge

0 Therefore 1198913(119879) is increasing on (0 +infin) It is easyto get 1198913(0) = (119863(120572 + 120579)

2)(119890minus(120572+120579)119872

minus 1)(120572119904119868119890(120572 +

120579) minus 119888119868119901) + 12057911990411986811989011986311987222(120572 + 120579) + 120572119904119868119890119863119872(120572 + 120579)

2minus

119888119868119901119863119872(120572 + 120579) minus 119870 and lim119879rarr+infin1198913(119879) = +infinThus if 1198913(0) lt 0 then 1198913(119879) = 0 has a uniquesolution (say 1198790

3) on (0 +infin) Hence 1198913(119879) will be

negative on (0 11987903) and positive on (1198790

3 +infin) which

implies Π3(119879) is increasing on (0 11987903) and decreasing

on (11987903 +infin) That is Π3(119879) is unimodal on (0 +infin)

and 1198790

3is the unique maximum point of Π3(119879) if

1198913(0) lt 0 In contrast if 1198913(0) ge 0 then 1198913(119879) willalways be nonnegative on (0 +infin) hence Π3(119879) isalways decreasing on (0 +infin)

(2) If119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 then1198913(119879) is strictly

decreasing on (0 +infin) Noting that lim119879rarr+infin1198913(119879) =minusinfin for 119865 gt minus(120572119904119868119890(120572 + 120579) minus 119888119868119901)119890

minus(120572+120579)119872 one easilyobtains that if 1198913(0) gt 0 there will exist a unique root(say 1198791

3) to the equation 1198913(119879) = 0 hence 1198913(119879) is

positive on (0 11987913) and negative on (1198791

3 +infin) which

implies Π3(119879) is decreasing on (0 11987913) and increasing

on (11987913 +infin) But if1198913(0) le 01198913(119879) is always negative

on (0 +infin) then Π3(119879) is increasing on (0 +infin)

Summarizing the above arguments the following lemmato describe the property ofΠ3(119879) on (0 +infin) will be obtained

Lemma 3 (1) For 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 Π3(119879) is

increasing on (0 11987903) and decreasing on (1198790

3 +infin) if 1198913(0) lt 0

and Π3(119879) is decreasing on (0 +infin) otherwise(2) For119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872Π3(119879) is decreasingon (0 1198791

3) and increasing on (1198791

3 +infin) if 1198913(0) gt 0 andΠ3(119879)

is increasing on (0 +infin) otherwise

Based upon Lemmas 1ndash3 we now will decide the optimalsolution 119879lowast to Π(119879) from the below two situations

41 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 ge

119879119889 When 119872 ge 119879119889 the piecewise profit function Π(119879) has

three different expressions that is Π1(119879) Π2(119879) and Π3(119879)respectively From (17) (20) and (24) we have

1198911 (119879119889) =minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1] minus 119870

(26)

1198912 (119879119889) =minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

+120579119904119868119890119863119879

2

119889

2 (120572 + 120579)minus 119870

(27)

1198912 (119872) = 1198913 (119872)

=minus119863119865

(120572 + 120579)2[(120572 + 120579)119872119890

(120572+120579)119872minus 119890(120572+120579)119872

+ 1]

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(28)

As shown in Appendix C we will find that 1198911(119879119889) gt 1198912(119879119889)For convenience we rewrite the notation 119864 as 119864 = 119865 minus 119888119868119901 +

120572119904119868119890(120572 + 120579) minus 120572119904119868119890119872 and the following two theorems wouldbe obtained to decide the optimal solution 119879lowastwhen119872 ge 119879119889

Theorem 4 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = Π2(119879

3

2) Hence 119879lowast is 1198793

2 where 1198793

2isin

(119879119889119872) and satisfies (20)(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 the following applies

(A) While 1198912(119879) gt 0 one has the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = Π2(1198791

2) Hence 119879lowast is 1198791

2

Mathematical Problems in Engineering 7

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2)

Π3(1198790

3) Hence 119879lowast is 1198790

1 11987912 or 11987903associ-

ated with maximum profit(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889)

Π3(1198790

3) Hence 119879lowast is 1198790

1 119879119889 or 11987903 associ-

ated with maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit(B) While 1198912(119879

) le 0 one has the following(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0

then Π(119879lowast) = Π3(1198790

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 11987903 associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(4) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

Theorem 5 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus

120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the

same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

42 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 lt

119879119889 When 119872 lt 119879119889 the piecewise profit function Π(119879) hasonly two different expressions that is Π1(119879) and Π3(119879)respectively From (24) we have

1198913 (119879119889) =

119863119865 (119890(120572+120579)119879119889 minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879119889] minus119863119865119879119889

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1) (

120572119904119868119890

120572 + 120579minus 119888119868119901)

[1 minus (120572 + 120579) 119879119889] minus119863 (119879119889 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(29)

As shown in Appendix D we will find that 1198911(119879119889) gt

1198913(119879119889) The following theorem would be given to decide thesystemrsquos optimal cycle time 119879lowast when119872 lt 119879119889

Theorem 6 If119872 lt 119879119889 then one has the following results

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the following

applies

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then Π(119879lowast) =

Π3(1198790

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then Π(119879lowast) =

maxΠ1(1198790

1) Π3(119879119889) Hence 119879

lowast is 1198790

1or 119879119889

associated with the maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then Π(119879

lowast) =

maxΠ1(1198790

1) Π3(119879

0

3) Hence 119879

lowast is 1198790

1or 1198790

3

associated with the maximum profit

(2) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix F for detail

5 Numerical Example

In this section we will provide the following numericalexample to illustrate the present model

Example 1 Given 119876119889 = 200 units 119863 = 1500 units 120572 =

04 120579 = 020 119870 = $50 ℎ = $1unityear 119888 = $ 5 perunit 119904 = $9 per unit 119868119890 = 013$year 119868119901 = 019$year and119872 = 03 year it is easy to get 119879119889 = 02123 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 and 119865 le (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872 Consequently by using Bisection method or

Newton-Raphson method we could obtain 1198790

1= 01760

1198790

2= 01892 and 119879

0

3= 01931 Because of 1198911(119879119889) le 0

1198912(119879119889) lt 0 and 1198912(119872) le 0 according to Theorem 4(1) theretailerrsquos optimal cycle time should be 119879lowast = 119879

0

3= 01931

8 Mathematical Problems in Engineering

Table 1 The impact of change of 120572 120579 119876119889 and119872 on the optimal solutions

Parameters 1198790

11198790

21198790

3119879lowast

119876lowast

Πlowast

120572

01 03381 03637 03895 03637 695772 6023591

02 04467 03956 03957 03956 823678 6217691

03 04792 04231 04320 04231 895987 6430335

04 05561 04538 04439 04538 985672 6663862

05 06723 05562 05071 05071 1033781 6931645

06 08552 05781 06047 06032 1367563 7362176

120579

001 05651 04436 05008 04436 1098853 7721683

002 05513 04471 04973 04471 1064786 7439053

005 05167 04441 04812 04441 994372 7268573

010 04783 04368 04791 04368 979335 7082341

050 03365 03568 04659 03568 833563 6887732

100 03561 02896 04543 03561 710856 6625568

119876119889

50 02344 02432 02589 02344 463085 4329953

100 03832 03565 03867 03565 750255 3948304

150 04973 04236 04973 04236 919334 3430035

200 06601 05467 05675 05675 1276297 3215728

250 07756 06873 06981 06981 1633156 2833270

300 08132 07675 07321 07321 1927368 2173673

119872

01 03726 03826 02873 03826 737665 2746542

02 03875 03835 03965 03835 799762 2823871

03 03811 04009 04623 04009 833721 2902323

04 03765 04197 04987 03765 899932 3130369

05 03732 04618 05013 04618 955513 3273771

06 03586 04745 05233 03586 998756 3633689

hence the optimal order quantity and the maximum averageprofit are 119876lowast = 1588763 and Πlowast = 18673562 respectively

Next we further study the effects of changes of parame-ters 120572 120579 119876119889 and 119872 on the optimal solutions The values ofother parameters keep the same as in the above examplewheneach of parameters120572 120579119876119889 and119872 varies Table 1 presents theobserved results with various parameters 120572 120579 119876119889 and119872

The following inferences can be made based on Table 1

(1) The retailerrsquos optimal order quantity 119876lowast and the

optimal average profit Πlowast increase as the value of 120572increases It implies that when the market demand ismore sensitive to the inventory level the retailer willincrease hisher order quantity to make more profitunder permissible delay in payments

(2) The optimal order quantity 119876lowast and the maximum

average profitΠlowast decrease as the value of 120579 increases

(3) When 119876119889 increases the optimal order quantity 119876lowastand the optimal cycle time 119879lowast are increasing but theoptimal average profit Πlowast is decreasing

(4) The retailerrsquos order quantity 119876lowast is increasing as 119872increases This verifies the fact that the retailer couldindeed increase the sales quantity by adopting thetrade credit policy provided by hisher supplier

6 Conclusions

In this paper we develop an EOQ model for deterioratingitems under permissible delay in payments The primarydifference of this paper as compared to previous relatedstudies is the following four aspects (1) the demand rateof items is dependent on the retailerrsquos current stock level(2) many items deteriorate continuously such as fruits andvegetables (3) the retailer who purchases the items enjoys afixed credit period offered by the supplier if the order quantityis greater than or equal to the predetermined quantity and(4) we here use maximizing profit as the objective to find theoptimal replenishment policies which could better describethe essence of the inventory system than a cost-minimizationmodel In addition we have discussed in detail the conditionsof the existence and uniqueness of the optimal solutions tothe model Three easy-to-use theorems are developed to findthe optimal ordering policies for the considered problem andthese theoretical results are illustrated by numerical exampleBy studying the effects of 120572 120579 119876119889 and 119872 on the optimalorder quantity 119876lowast and the optimal average profit Πlowast somemanagerial insights are derived

The presented model can be further extended to somemore practical situations For example we could allow forshortages quantity discounts time value of money andinflation price-sensitive and stock-dependent demand andso forth

Mathematical Problems in Engineering 9

Appendices

A Proof of Π1(119879119889) lt Π2(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (12) we get

Π2 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2(119890(120572+120579)119879119889 minus (120572 + 120579) 119879119889 minus 1)

+ 119904119868119890119872119863119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

(A1)

It is evident that119872 ge 119879119889 and (119890(120572+120579)119879119889 minus (120572+120579)119879119889 minus1) gt 0 for

119879119889 gt 0 then

Π2 (119879119889) minus Π1 (119879119889)

gt1

119879119889

1198631198792

119889

2(120572119904119868119890119872minus

120572119904119868119890

120572 + 120579+ 119888119868119901)

+ 119904119868119890119863119872119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

=119863119879119889

2(120572119904119868119890119872+ 119888119868119901) minus

119904119868119890119863119879119889

2+ 119904119868119890119863119872

ge119863119879119889

2(120572119904119868119890119872+ 119888119868119901) +

119904119868119890119863119872

2gt 0

(A2)

As a result we have Π1(119879119889) lt Π2(119879119889)This completes the proof of Appendix A

B Proof of Π1(119879119889) lt Π3(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (13) we get

Π3 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2[119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872) minus (120572 + 120579)119872]

+120572119904119868119890119872119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1)

+120572119904119868119890119863119872

2

(120572 + 120579)+120579119904119868119890119863119872

2

2 (120572 + 120579)

(B1)

Because of 119879119889 gt 119872 and 119890(120572+120579)119879119889 minus119890(120572+120579)(119879119889minus119872) minus(120572+120579)119872 gt 0we have Π3(119879119889) minus Π1(119879119889) gt 0

This completes the proof of Appendix B

C Proof of 1198911(119879119889) gt 1198912(119879119889)

If 119888119868119901 ge 119904119868119890 from (26) and (27) we have

1198912 (119879119889) minus 1198911 (119879119889)

= minus119863

(120572 + 120579)2(120572119904119868119890119872+ 119888119868119901 minus

120572119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

le minus119863

(120572 + 120579)2(120572119904119868119890119872+

120579119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

= minus119863119904119868119890

(120572 + 120579)2(120572119872 +

120579

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

minus120579 (120572 + 120579) 119879

2

119889

2

(C1)

Let 119866(119879119889) = (120572119872 + 120579(120572 + 120579))[(120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 +

1] minus 120579(120572 + 120579)1198792

1198892 then

1198661015840(119879119889) = (120572119872 +

120579

120572 + 120579) (120572 + 120579)

2119879119889119890(120572+120579)119879119889 minus 120579 (120572 + 120579) 119879119889

= 120579 (120572 + 120579) 119879119889 [(120572119872(120572 + 120579)

120579+ 1) 119890

(120572+120579)119879119889 minus 1] gt 0

(C2)

Hence 119866(119879119889) gt 119866(0) = 0 that is 1198911(119879119889) gt 1198912(119879119889)This completes the proof of Appendix C

D Proof of 1198911(119879119889) gt 1198913(119879119889)

From (26) and (24) we have

1198913 (119879119889) minus 1198911 (119879119889)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

times (120572119904119868119890

120572 + 120579minus 119888119868119901 minus 120572119904119868119890119872) minus

119863

(120572 + 120579)2

[(120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872) minus 119890

(120572+120579)(119879119889minus119872) + 1 minus (120572 + 120579)119872]

times (120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

Mathematical Problems in Engineering 5

is 119889Π1(119879)119889119879 = 0 has a unique root (say 11987901) on (0 +infin)

Moreover taking the second derivative ofΠ1(119879)with respectto 119879 gives

1198892Π1 (119879)

1198891198792= minus

2119870

1198793+

119863119864

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(19)

It is easy to show that the inequality 1199092119890119909minus2119909119890119909+2119890119909minus2 gt0 holds for all 119909 gt 0 so 1198892Π1(119879)119889119879

2 will always be negativeon (0 +infin) if 119864 lt 0 that is Π1(119879) is convex on (0 +infin)Therefore 1198790

1is the unique maximum point of Π1(119879) on

(0 +infin) But if 119864 ge 0 (18) means that 1198911(119879) is decreasingon (0 +infin) since 1198911(0) = minus119870 lt 0 then 1198911(119879) is negative and119889Π1(119879)119889119879 is positive for all 119879 gt 0 hence Π1(119879) is strictlyincreasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π1(119879) on (0 +infin)

Lemma 1 If 119864 lt 0 Π1(119879) is convex on (0 +infin) and 11987901is the

unique maximum point of Π1(119879) Otherwise Π1(119879) is strictlyincreasing on (0 +infin)

Similarly the first-order necessary condition forΠ2(119879) in(12) to be maximized is 119889Π2(119879)119889119879 = 0 which leads to

minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119890

(120572+120579)119879minus 119890(120572+120579)119879

+ 1]

+120579119904119868119890119863119879

2

2 (120572 + 120579)minus 119870 = 0

(20)

We set the left-hand expression of (20) as 1198912(119879) Taking thefirst derivative of 1198912(119879) with respect to 119879 gives

1198891198912 (119879)

119889119879= minus(119865119890

(120572+120579)119879minus

120579119904119868119890

120572 + 120579)119863119879 (21)

Consequently from the sign of 119865 there are two distinct casesfor finding the property of Π2(119879) as follows

(1) When 119865 le 0 we have 1198891198912(119879)119889119879 gt 0 that is 1198912(119879) isstrictly increasing on (0 +infin) Since 1198912(0) = minus119870 lt 0

and lim119879rarr+infin1198912(119879) = +infin 1198912(119879) = 0 has a uniquesolution (say 1198790

2) on (0 +infin) Moreover from (13) we

have

1198892Π2 (119879)

1198891198792= minus

2119870

1198793+

119863119865

(120572 + 120579)21198793

times [(120572 + 120579)21198792119890(120572+120579)119879

minus 2 (120572 + 120579) 119879119890(120572+120579)119879

+ 2119890(120572+120579)119879

minus 2]

(22)

Since 119865 le 0 and (120572+120579)21198792119890(120572+120579)119879 minus2(120572+120579)119879119890(120572+120579)119879 +2119890(120572+120579)119879

minus 2 gt 0 we get 1198892Π2(119879)1198891198792lt 0 Hence

Π2(119879) is convex on (0 +infin) and 1198790

2is the unique

maximum point of Π2(119879)

(2) When 119865 gt 0 there will exist a unique root to theequation 1198891198912(119879)119889119879 = 0 Denote this root by 119879 then

119879=

1

120572 + 120579[ln 120579119904119868119890 minus ln (120572 + 120579) 119865] (23)

(A) If 119879ge 0 that is 0 lt 119865 le 120579119904119868119890(120572 + 120579) then

1198891198912(119879)119889119879 gt 0 for 119879 isin (0 119879) and 1198891198912(119879)119889119879 lt

0 for 119879 isin (119879 +infin) that is 1198912(119879) is increasing

on (0 119879) and decreasing on (119879

+infin)

(a) When 1198912(119879) gt 0 1198912(119879) = 0 has a unique

root (say11987912) on (0 119879

) due to1198912(0) = minus119870 lt

0 Hence Π2(119879) is increasing on (0 1198791

2)

and decreasing on (1198791

2 119879

) On the other

interval of (119879 +infin) however because of

1198912(119879) gt 0 and lim119879rarr+infin1198912(119879) = minusinfin

for 119865 gt 0 1198912(119879) has a unique zeropoint (say 119879

2

2) on (119879

+infin) Thus Π2(119879)

is decreasing on (119879 1198792

2) and increasing on

(1198792

2 +infin) Hence Π2(119879) is increasing on

(0 1198791

2) decreasing on (1198791

2 1198792

2) and increas-

ing on (11987922 +infin)

(b) When 1198912(119879) le 0 1198912(119879) is negative

on (0 119879) because 1198912(119879) is increasing on

(0 119879) whereas 1198912(119879) is also negative on

(119879 +infin) since 1198912(119879) is decreasing on

(119879 +infin) Therefore 1198912(119879) is always non-

positive on (0 +infin) andΠ2(119879) is increasingon (0 +infin)

(B) If 119879lt 0 that is 119865 gt 120579119904119868119890(120572 + 120579) then for any

119879 ge 0 we have 1198891198912(119879)119889119879 lt 0 that is 1198912(119879)is strictly decreasing on (0 +infin) Since 1198912(0) =minus119870 lt 0 1198912(119879) is negative on (0 +infin) HenceΠ2(119879) is increasing on (0 +infin)

Then we have the following lemma to describe theproperty of Π2(119879) on (0 +infin)

Lemma 2 (1) For 119865 le 0 Π2(119879) is convex on (0 +infin) and 11987902

is the unique maximum point of Π2(119879)(2) For 0 lt 119865 le 120579119904119868119890(120572+120579)Π2(119879) is increasing on (0 11987912 )

decreasing on (11987912 1198792

2) and increasing on (1198792

2 +infin) if1198912(119879

) gt

0 and Π2(119879) is increasing on (0 +infin) otherwise(3) For 119865 gt 120579119904119868119890(120572 + 120579) Π2(119879) is always increasing on

(0 +infin)

Likewise the first-order necessary condition forΠ3(119879) in(13) to be maximum is 119889Π3(119879)119889119879 = 0 which implies that

119863119865(119890(120572+120579)119879

minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879] minus119863119865119879

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879minus119872)

minus 1) (120572119904119868119890

120572 + 120579minus 119888119868119901)

6 Mathematical Problems in Engineering

[1 minus (120572 + 120579) 119879] minus119863 (119879 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870 = 0

(24)

We set the left-hand expression of (24) as 1198913(119879) Taking thefirst derivative of 1198913(119879) with respect to 119879 gives

1198891198913 (119879)

119889119879= minus [119865 + (

120572119904119868119890

120572 + 120579minus 119888119868119901) 119890

minus(120572+120579)119872]119863119879119890

(120572+120579)119879

(25)

We consider the following two cases

(1) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198891198913(119879)119889119879 ge

0 Therefore 1198913(119879) is increasing on (0 +infin) It is easyto get 1198913(0) = (119863(120572 + 120579)

2)(119890minus(120572+120579)119872

minus 1)(120572119904119868119890(120572 +

120579) minus 119888119868119901) + 12057911990411986811989011986311987222(120572 + 120579) + 120572119904119868119890119863119872(120572 + 120579)

2minus

119888119868119901119863119872(120572 + 120579) minus 119870 and lim119879rarr+infin1198913(119879) = +infinThus if 1198913(0) lt 0 then 1198913(119879) = 0 has a uniquesolution (say 1198790

3) on (0 +infin) Hence 1198913(119879) will be

negative on (0 11987903) and positive on (1198790

3 +infin) which

implies Π3(119879) is increasing on (0 11987903) and decreasing

on (11987903 +infin) That is Π3(119879) is unimodal on (0 +infin)

and 1198790

3is the unique maximum point of Π3(119879) if

1198913(0) lt 0 In contrast if 1198913(0) ge 0 then 1198913(119879) willalways be nonnegative on (0 +infin) hence Π3(119879) isalways decreasing on (0 +infin)

(2) If119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 then1198913(119879) is strictly

decreasing on (0 +infin) Noting that lim119879rarr+infin1198913(119879) =minusinfin for 119865 gt minus(120572119904119868119890(120572 + 120579) minus 119888119868119901)119890

minus(120572+120579)119872 one easilyobtains that if 1198913(0) gt 0 there will exist a unique root(say 1198791

3) to the equation 1198913(119879) = 0 hence 1198913(119879) is

positive on (0 11987913) and negative on (1198791

3 +infin) which

implies Π3(119879) is decreasing on (0 11987913) and increasing

on (11987913 +infin) But if1198913(0) le 01198913(119879) is always negative

on (0 +infin) then Π3(119879) is increasing on (0 +infin)

Summarizing the above arguments the following lemmato describe the property ofΠ3(119879) on (0 +infin) will be obtained

Lemma 3 (1) For 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 Π3(119879) is

increasing on (0 11987903) and decreasing on (1198790

3 +infin) if 1198913(0) lt 0

and Π3(119879) is decreasing on (0 +infin) otherwise(2) For119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872Π3(119879) is decreasingon (0 1198791

3) and increasing on (1198791

3 +infin) if 1198913(0) gt 0 andΠ3(119879)

is increasing on (0 +infin) otherwise

Based upon Lemmas 1ndash3 we now will decide the optimalsolution 119879lowast to Π(119879) from the below two situations

41 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 ge

119879119889 When 119872 ge 119879119889 the piecewise profit function Π(119879) has

three different expressions that is Π1(119879) Π2(119879) and Π3(119879)respectively From (17) (20) and (24) we have

1198911 (119879119889) =minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1] minus 119870

(26)

1198912 (119879119889) =minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

+120579119904119868119890119863119879

2

119889

2 (120572 + 120579)minus 119870

(27)

1198912 (119872) = 1198913 (119872)

=minus119863119865

(120572 + 120579)2[(120572 + 120579)119872119890

(120572+120579)119872minus 119890(120572+120579)119872

+ 1]

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(28)

As shown in Appendix C we will find that 1198911(119879119889) gt 1198912(119879119889)For convenience we rewrite the notation 119864 as 119864 = 119865 minus 119888119868119901 +

120572119904119868119890(120572 + 120579) minus 120572119904119868119890119872 and the following two theorems wouldbe obtained to decide the optimal solution 119879lowastwhen119872 ge 119879119889

Theorem 4 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = Π2(119879

3

2) Hence 119879lowast is 1198793

2 where 1198793

2isin

(119879119889119872) and satisfies (20)(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 the following applies

(A) While 1198912(119879) gt 0 one has the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = Π2(1198791

2) Hence 119879lowast is 1198791

2

Mathematical Problems in Engineering 7

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2)

Π3(1198790

3) Hence 119879lowast is 1198790

1 11987912 or 11987903associ-

ated with maximum profit(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889)

Π3(1198790

3) Hence 119879lowast is 1198790

1 119879119889 or 11987903 associ-

ated with maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit(B) While 1198912(119879

) le 0 one has the following(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0

then Π(119879lowast) = Π3(1198790

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 11987903 associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(4) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

Theorem 5 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus

120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the

same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

42 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 lt

119879119889 When 119872 lt 119879119889 the piecewise profit function Π(119879) hasonly two different expressions that is Π1(119879) and Π3(119879)respectively From (24) we have

1198913 (119879119889) =

119863119865 (119890(120572+120579)119879119889 minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879119889] minus119863119865119879119889

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1) (

120572119904119868119890

120572 + 120579minus 119888119868119901)

[1 minus (120572 + 120579) 119879119889] minus119863 (119879119889 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(29)

As shown in Appendix D we will find that 1198911(119879119889) gt

1198913(119879119889) The following theorem would be given to decide thesystemrsquos optimal cycle time 119879lowast when119872 lt 119879119889

Theorem 6 If119872 lt 119879119889 then one has the following results

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the following

applies

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then Π(119879lowast) =

Π3(1198790

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then Π(119879lowast) =

maxΠ1(1198790

1) Π3(119879119889) Hence 119879

lowast is 1198790

1or 119879119889

associated with the maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then Π(119879

lowast) =

maxΠ1(1198790

1) Π3(119879

0

3) Hence 119879

lowast is 1198790

1or 1198790

3

associated with the maximum profit

(2) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix F for detail

5 Numerical Example

In this section we will provide the following numericalexample to illustrate the present model

Example 1 Given 119876119889 = 200 units 119863 = 1500 units 120572 =

04 120579 = 020 119870 = $50 ℎ = $1unityear 119888 = $ 5 perunit 119904 = $9 per unit 119868119890 = 013$year 119868119901 = 019$year and119872 = 03 year it is easy to get 119879119889 = 02123 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 and 119865 le (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872 Consequently by using Bisection method or

Newton-Raphson method we could obtain 1198790

1= 01760

1198790

2= 01892 and 119879

0

3= 01931 Because of 1198911(119879119889) le 0

1198912(119879119889) lt 0 and 1198912(119872) le 0 according to Theorem 4(1) theretailerrsquos optimal cycle time should be 119879lowast = 119879

0

3= 01931

8 Mathematical Problems in Engineering

Table 1 The impact of change of 120572 120579 119876119889 and119872 on the optimal solutions

Parameters 1198790

11198790

21198790

3119879lowast

119876lowast

Πlowast

120572

01 03381 03637 03895 03637 695772 6023591

02 04467 03956 03957 03956 823678 6217691

03 04792 04231 04320 04231 895987 6430335

04 05561 04538 04439 04538 985672 6663862

05 06723 05562 05071 05071 1033781 6931645

06 08552 05781 06047 06032 1367563 7362176

120579

001 05651 04436 05008 04436 1098853 7721683

002 05513 04471 04973 04471 1064786 7439053

005 05167 04441 04812 04441 994372 7268573

010 04783 04368 04791 04368 979335 7082341

050 03365 03568 04659 03568 833563 6887732

100 03561 02896 04543 03561 710856 6625568

119876119889

50 02344 02432 02589 02344 463085 4329953

100 03832 03565 03867 03565 750255 3948304

150 04973 04236 04973 04236 919334 3430035

200 06601 05467 05675 05675 1276297 3215728

250 07756 06873 06981 06981 1633156 2833270

300 08132 07675 07321 07321 1927368 2173673

119872

01 03726 03826 02873 03826 737665 2746542

02 03875 03835 03965 03835 799762 2823871

03 03811 04009 04623 04009 833721 2902323

04 03765 04197 04987 03765 899932 3130369

05 03732 04618 05013 04618 955513 3273771

06 03586 04745 05233 03586 998756 3633689

hence the optimal order quantity and the maximum averageprofit are 119876lowast = 1588763 and Πlowast = 18673562 respectively

Next we further study the effects of changes of parame-ters 120572 120579 119876119889 and 119872 on the optimal solutions The values ofother parameters keep the same as in the above examplewheneach of parameters120572 120579119876119889 and119872 varies Table 1 presents theobserved results with various parameters 120572 120579 119876119889 and119872

The following inferences can be made based on Table 1

(1) The retailerrsquos optimal order quantity 119876lowast and the

optimal average profit Πlowast increase as the value of 120572increases It implies that when the market demand ismore sensitive to the inventory level the retailer willincrease hisher order quantity to make more profitunder permissible delay in payments

(2) The optimal order quantity 119876lowast and the maximum

average profitΠlowast decrease as the value of 120579 increases

(3) When 119876119889 increases the optimal order quantity 119876lowastand the optimal cycle time 119879lowast are increasing but theoptimal average profit Πlowast is decreasing

(4) The retailerrsquos order quantity 119876lowast is increasing as 119872increases This verifies the fact that the retailer couldindeed increase the sales quantity by adopting thetrade credit policy provided by hisher supplier

6 Conclusions

In this paper we develop an EOQ model for deterioratingitems under permissible delay in payments The primarydifference of this paper as compared to previous relatedstudies is the following four aspects (1) the demand rateof items is dependent on the retailerrsquos current stock level(2) many items deteriorate continuously such as fruits andvegetables (3) the retailer who purchases the items enjoys afixed credit period offered by the supplier if the order quantityis greater than or equal to the predetermined quantity and(4) we here use maximizing profit as the objective to find theoptimal replenishment policies which could better describethe essence of the inventory system than a cost-minimizationmodel In addition we have discussed in detail the conditionsof the existence and uniqueness of the optimal solutions tothe model Three easy-to-use theorems are developed to findthe optimal ordering policies for the considered problem andthese theoretical results are illustrated by numerical exampleBy studying the effects of 120572 120579 119876119889 and 119872 on the optimalorder quantity 119876lowast and the optimal average profit Πlowast somemanagerial insights are derived

The presented model can be further extended to somemore practical situations For example we could allow forshortages quantity discounts time value of money andinflation price-sensitive and stock-dependent demand andso forth

Mathematical Problems in Engineering 9

Appendices

A Proof of Π1(119879119889) lt Π2(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (12) we get

Π2 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2(119890(120572+120579)119879119889 minus (120572 + 120579) 119879119889 minus 1)

+ 119904119868119890119872119863119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

(A1)

It is evident that119872 ge 119879119889 and (119890(120572+120579)119879119889 minus (120572+120579)119879119889 minus1) gt 0 for

119879119889 gt 0 then

Π2 (119879119889) minus Π1 (119879119889)

gt1

119879119889

1198631198792

119889

2(120572119904119868119890119872minus

120572119904119868119890

120572 + 120579+ 119888119868119901)

+ 119904119868119890119863119872119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

=119863119879119889

2(120572119904119868119890119872+ 119888119868119901) minus

119904119868119890119863119879119889

2+ 119904119868119890119863119872

ge119863119879119889

2(120572119904119868119890119872+ 119888119868119901) +

119904119868119890119863119872

2gt 0

(A2)

As a result we have Π1(119879119889) lt Π2(119879119889)This completes the proof of Appendix A

B Proof of Π1(119879119889) lt Π3(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (13) we get

Π3 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2[119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872) minus (120572 + 120579)119872]

+120572119904119868119890119872119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1)

+120572119904119868119890119863119872

2

(120572 + 120579)+120579119904119868119890119863119872

2

2 (120572 + 120579)

(B1)

Because of 119879119889 gt 119872 and 119890(120572+120579)119879119889 minus119890(120572+120579)(119879119889minus119872) minus(120572+120579)119872 gt 0we have Π3(119879119889) minus Π1(119879119889) gt 0

This completes the proof of Appendix B

C Proof of 1198911(119879119889) gt 1198912(119879119889)

If 119888119868119901 ge 119904119868119890 from (26) and (27) we have

1198912 (119879119889) minus 1198911 (119879119889)

= minus119863

(120572 + 120579)2(120572119904119868119890119872+ 119888119868119901 minus

120572119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

le minus119863

(120572 + 120579)2(120572119904119868119890119872+

120579119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

= minus119863119904119868119890

(120572 + 120579)2(120572119872 +

120579

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

minus120579 (120572 + 120579) 119879

2

119889

2

(C1)

Let 119866(119879119889) = (120572119872 + 120579(120572 + 120579))[(120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 +

1] minus 120579(120572 + 120579)1198792

1198892 then

1198661015840(119879119889) = (120572119872 +

120579

120572 + 120579) (120572 + 120579)

2119879119889119890(120572+120579)119879119889 minus 120579 (120572 + 120579) 119879119889

= 120579 (120572 + 120579) 119879119889 [(120572119872(120572 + 120579)

120579+ 1) 119890

(120572+120579)119879119889 minus 1] gt 0

(C2)

Hence 119866(119879119889) gt 119866(0) = 0 that is 1198911(119879119889) gt 1198912(119879119889)This completes the proof of Appendix C

D Proof of 1198911(119879119889) gt 1198913(119879119889)

From (26) and (24) we have

1198913 (119879119889) minus 1198911 (119879119889)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

times (120572119904119868119890

120572 + 120579minus 119888119868119901 minus 120572119904119868119890119872) minus

119863

(120572 + 120579)2

[(120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872) minus 119890

(120572+120579)(119879119889minus119872) + 1 minus (120572 + 120579)119872]

times (120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 6: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

6 Mathematical Problems in Engineering

[1 minus (120572 + 120579) 119879] minus119863 (119879 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870 = 0

(24)

We set the left-hand expression of (24) as 1198913(119879) Taking thefirst derivative of 1198913(119879) with respect to 119879 gives

1198891198913 (119879)

119889119879= minus [119865 + (

120572119904119868119890

120572 + 120579minus 119888119868119901) 119890

minus(120572+120579)119872]119863119879119890

(120572+120579)119879

(25)

We consider the following two cases

(1) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198891198913(119879)119889119879 ge

0 Therefore 1198913(119879) is increasing on (0 +infin) It is easyto get 1198913(0) = (119863(120572 + 120579)

2)(119890minus(120572+120579)119872

minus 1)(120572119904119868119890(120572 +

120579) minus 119888119868119901) + 12057911990411986811989011986311987222(120572 + 120579) + 120572119904119868119890119863119872(120572 + 120579)

2minus

119888119868119901119863119872(120572 + 120579) minus 119870 and lim119879rarr+infin1198913(119879) = +infinThus if 1198913(0) lt 0 then 1198913(119879) = 0 has a uniquesolution (say 1198790

3) on (0 +infin) Hence 1198913(119879) will be

negative on (0 11987903) and positive on (1198790

3 +infin) which

implies Π3(119879) is increasing on (0 11987903) and decreasing

on (11987903 +infin) That is Π3(119879) is unimodal on (0 +infin)

and 1198790

3is the unique maximum point of Π3(119879) if

1198913(0) lt 0 In contrast if 1198913(0) ge 0 then 1198913(119879) willalways be nonnegative on (0 +infin) hence Π3(119879) isalways decreasing on (0 +infin)

(2) If119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 then1198913(119879) is strictly

decreasing on (0 +infin) Noting that lim119879rarr+infin1198913(119879) =minusinfin for 119865 gt minus(120572119904119868119890(120572 + 120579) minus 119888119868119901)119890

minus(120572+120579)119872 one easilyobtains that if 1198913(0) gt 0 there will exist a unique root(say 1198791

3) to the equation 1198913(119879) = 0 hence 1198913(119879) is

positive on (0 11987913) and negative on (1198791

3 +infin) which

implies Π3(119879) is decreasing on (0 11987913) and increasing

on (11987913 +infin) But if1198913(0) le 01198913(119879) is always negative

on (0 +infin) then Π3(119879) is increasing on (0 +infin)

Summarizing the above arguments the following lemmato describe the property ofΠ3(119879) on (0 +infin) will be obtained

Lemma 3 (1) For 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 Π3(119879) is

increasing on (0 11987903) and decreasing on (1198790

3 +infin) if 1198913(0) lt 0

and Π3(119879) is decreasing on (0 +infin) otherwise(2) For119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872Π3(119879) is decreasingon (0 1198791

3) and increasing on (1198791

3 +infin) if 1198913(0) gt 0 andΠ3(119879)

is increasing on (0 +infin) otherwise

Based upon Lemmas 1ndash3 we now will decide the optimalsolution 119879lowast to Π(119879) from the below two situations

41 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 ge

119879119889 When 119872 ge 119879119889 the piecewise profit function Π(119879) has

three different expressions that is Π1(119879) Π2(119879) and Π3(119879)respectively From (17) (20) and (24) we have

1198911 (119879119889) =minus119863119864

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1] minus 119870

(26)

1198912 (119879119889) =minus119863119865

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

+120579119904119868119890119863119879

2

119889

2 (120572 + 120579)minus 119870

(27)

1198912 (119872) = 1198913 (119872)

=minus119863119865

(120572 + 120579)2[(120572 + 120579)119872119890

(120572+120579)119872minus 119890(120572+120579)119872

+ 1]

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(28)

As shown in Appendix C we will find that 1198911(119879119889) gt 1198912(119879119889)For convenience we rewrite the notation 119864 as 119864 = 119865 minus 119888119868119901 +

120572119904119868119890(120572 + 120579) minus 120572119904119868119890119872 and the following two theorems wouldbe obtained to decide the optimal solution 119879lowastwhen119872 ge 119879119889

Theorem 4 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = Π2(119879

3

2) Hence 119879lowast is 1198793

2 where 1198793

2isin

(119879119889119872) and satisfies (20)(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 the following applies

(A) While 1198912(119879) gt 0 one has the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = Π2(1198791

2) Hence 119879lowast is 1198791

2

Mathematical Problems in Engineering 7

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2)

Π3(1198790

3) Hence 119879lowast is 1198790

1 11987912 or 11987903associ-

ated with maximum profit(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889)

Π3(1198790

3) Hence 119879lowast is 1198790

1 119879119889 or 11987903 associ-

ated with maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit(B) While 1198912(119879

) le 0 one has the following(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0

then Π(119879lowast) = Π3(1198790

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 11987903 associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(4) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

Theorem 5 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus

120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the

same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

42 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 lt

119879119889 When 119872 lt 119879119889 the piecewise profit function Π(119879) hasonly two different expressions that is Π1(119879) and Π3(119879)respectively From (24) we have

1198913 (119879119889) =

119863119865 (119890(120572+120579)119879119889 minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879119889] minus119863119865119879119889

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1) (

120572119904119868119890

120572 + 120579minus 119888119868119901)

[1 minus (120572 + 120579) 119879119889] minus119863 (119879119889 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(29)

As shown in Appendix D we will find that 1198911(119879119889) gt

1198913(119879119889) The following theorem would be given to decide thesystemrsquos optimal cycle time 119879lowast when119872 lt 119879119889

Theorem 6 If119872 lt 119879119889 then one has the following results

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the following

applies

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then Π(119879lowast) =

Π3(1198790

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then Π(119879lowast) =

maxΠ1(1198790

1) Π3(119879119889) Hence 119879

lowast is 1198790

1or 119879119889

associated with the maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then Π(119879

lowast) =

maxΠ1(1198790

1) Π3(119879

0

3) Hence 119879

lowast is 1198790

1or 1198790

3

associated with the maximum profit

(2) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix F for detail

5 Numerical Example

In this section we will provide the following numericalexample to illustrate the present model

Example 1 Given 119876119889 = 200 units 119863 = 1500 units 120572 =

04 120579 = 020 119870 = $50 ℎ = $1unityear 119888 = $ 5 perunit 119904 = $9 per unit 119868119890 = 013$year 119868119901 = 019$year and119872 = 03 year it is easy to get 119879119889 = 02123 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 and 119865 le (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872 Consequently by using Bisection method or

Newton-Raphson method we could obtain 1198790

1= 01760

1198790

2= 01892 and 119879

0

3= 01931 Because of 1198911(119879119889) le 0

1198912(119879119889) lt 0 and 1198912(119872) le 0 according to Theorem 4(1) theretailerrsquos optimal cycle time should be 119879lowast = 119879

0

3= 01931

8 Mathematical Problems in Engineering

Table 1 The impact of change of 120572 120579 119876119889 and119872 on the optimal solutions

Parameters 1198790

11198790

21198790

3119879lowast

119876lowast

Πlowast

120572

01 03381 03637 03895 03637 695772 6023591

02 04467 03956 03957 03956 823678 6217691

03 04792 04231 04320 04231 895987 6430335

04 05561 04538 04439 04538 985672 6663862

05 06723 05562 05071 05071 1033781 6931645

06 08552 05781 06047 06032 1367563 7362176

120579

001 05651 04436 05008 04436 1098853 7721683

002 05513 04471 04973 04471 1064786 7439053

005 05167 04441 04812 04441 994372 7268573

010 04783 04368 04791 04368 979335 7082341

050 03365 03568 04659 03568 833563 6887732

100 03561 02896 04543 03561 710856 6625568

119876119889

50 02344 02432 02589 02344 463085 4329953

100 03832 03565 03867 03565 750255 3948304

150 04973 04236 04973 04236 919334 3430035

200 06601 05467 05675 05675 1276297 3215728

250 07756 06873 06981 06981 1633156 2833270

300 08132 07675 07321 07321 1927368 2173673

119872

01 03726 03826 02873 03826 737665 2746542

02 03875 03835 03965 03835 799762 2823871

03 03811 04009 04623 04009 833721 2902323

04 03765 04197 04987 03765 899932 3130369

05 03732 04618 05013 04618 955513 3273771

06 03586 04745 05233 03586 998756 3633689

hence the optimal order quantity and the maximum averageprofit are 119876lowast = 1588763 and Πlowast = 18673562 respectively

Next we further study the effects of changes of parame-ters 120572 120579 119876119889 and 119872 on the optimal solutions The values ofother parameters keep the same as in the above examplewheneach of parameters120572 120579119876119889 and119872 varies Table 1 presents theobserved results with various parameters 120572 120579 119876119889 and119872

The following inferences can be made based on Table 1

(1) The retailerrsquos optimal order quantity 119876lowast and the

optimal average profit Πlowast increase as the value of 120572increases It implies that when the market demand ismore sensitive to the inventory level the retailer willincrease hisher order quantity to make more profitunder permissible delay in payments

(2) The optimal order quantity 119876lowast and the maximum

average profitΠlowast decrease as the value of 120579 increases

(3) When 119876119889 increases the optimal order quantity 119876lowastand the optimal cycle time 119879lowast are increasing but theoptimal average profit Πlowast is decreasing

(4) The retailerrsquos order quantity 119876lowast is increasing as 119872increases This verifies the fact that the retailer couldindeed increase the sales quantity by adopting thetrade credit policy provided by hisher supplier

6 Conclusions

In this paper we develop an EOQ model for deterioratingitems under permissible delay in payments The primarydifference of this paper as compared to previous relatedstudies is the following four aspects (1) the demand rateof items is dependent on the retailerrsquos current stock level(2) many items deteriorate continuously such as fruits andvegetables (3) the retailer who purchases the items enjoys afixed credit period offered by the supplier if the order quantityis greater than or equal to the predetermined quantity and(4) we here use maximizing profit as the objective to find theoptimal replenishment policies which could better describethe essence of the inventory system than a cost-minimizationmodel In addition we have discussed in detail the conditionsof the existence and uniqueness of the optimal solutions tothe model Three easy-to-use theorems are developed to findthe optimal ordering policies for the considered problem andthese theoretical results are illustrated by numerical exampleBy studying the effects of 120572 120579 119876119889 and 119872 on the optimalorder quantity 119876lowast and the optimal average profit Πlowast somemanagerial insights are derived

The presented model can be further extended to somemore practical situations For example we could allow forshortages quantity discounts time value of money andinflation price-sensitive and stock-dependent demand andso forth

Mathematical Problems in Engineering 9

Appendices

A Proof of Π1(119879119889) lt Π2(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (12) we get

Π2 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2(119890(120572+120579)119879119889 minus (120572 + 120579) 119879119889 minus 1)

+ 119904119868119890119872119863119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

(A1)

It is evident that119872 ge 119879119889 and (119890(120572+120579)119879119889 minus (120572+120579)119879119889 minus1) gt 0 for

119879119889 gt 0 then

Π2 (119879119889) minus Π1 (119879119889)

gt1

119879119889

1198631198792

119889

2(120572119904119868119890119872minus

120572119904119868119890

120572 + 120579+ 119888119868119901)

+ 119904119868119890119863119872119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

=119863119879119889

2(120572119904119868119890119872+ 119888119868119901) minus

119904119868119890119863119879119889

2+ 119904119868119890119863119872

ge119863119879119889

2(120572119904119868119890119872+ 119888119868119901) +

119904119868119890119863119872

2gt 0

(A2)

As a result we have Π1(119879119889) lt Π2(119879119889)This completes the proof of Appendix A

B Proof of Π1(119879119889) lt Π3(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (13) we get

Π3 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2[119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872) minus (120572 + 120579)119872]

+120572119904119868119890119872119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1)

+120572119904119868119890119863119872

2

(120572 + 120579)+120579119904119868119890119863119872

2

2 (120572 + 120579)

(B1)

Because of 119879119889 gt 119872 and 119890(120572+120579)119879119889 minus119890(120572+120579)(119879119889minus119872) minus(120572+120579)119872 gt 0we have Π3(119879119889) minus Π1(119879119889) gt 0

This completes the proof of Appendix B

C Proof of 1198911(119879119889) gt 1198912(119879119889)

If 119888119868119901 ge 119904119868119890 from (26) and (27) we have

1198912 (119879119889) minus 1198911 (119879119889)

= minus119863

(120572 + 120579)2(120572119904119868119890119872+ 119888119868119901 minus

120572119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

le minus119863

(120572 + 120579)2(120572119904119868119890119872+

120579119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

= minus119863119904119868119890

(120572 + 120579)2(120572119872 +

120579

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

minus120579 (120572 + 120579) 119879

2

119889

2

(C1)

Let 119866(119879119889) = (120572119872 + 120579(120572 + 120579))[(120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 +

1] minus 120579(120572 + 120579)1198792

1198892 then

1198661015840(119879119889) = (120572119872 +

120579

120572 + 120579) (120572 + 120579)

2119879119889119890(120572+120579)119879119889 minus 120579 (120572 + 120579) 119879119889

= 120579 (120572 + 120579) 119879119889 [(120572119872(120572 + 120579)

120579+ 1) 119890

(120572+120579)119879119889 minus 1] gt 0

(C2)

Hence 119866(119879119889) gt 119866(0) = 0 that is 1198911(119879119889) gt 1198912(119879119889)This completes the proof of Appendix C

D Proof of 1198911(119879119889) gt 1198913(119879119889)

From (26) and (24) we have

1198913 (119879119889) minus 1198911 (119879119889)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

times (120572119904119868119890

120572 + 120579minus 119888119868119901 minus 120572119904119868119890119872) minus

119863

(120572 + 120579)2

[(120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872) minus 119890

(120572+120579)(119879119889minus119872) + 1 minus (120572 + 120579)119872]

times (120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 7: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

Mathematical Problems in Engineering 7

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2)

Π3(1198790

3) Hence 119879lowast is 1198790

1 11987912 or 11987903associ-

ated with maximum profit(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le

0 then Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889)

Π3(1198790

3) Hence 119879lowast is 1198790

1 119879119889 or 11987903 associ-

ated with maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0

then Π(119879lowast) = maxΠ1(1198790

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit(B) While 1198912(119879

) le 0 one has the following(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0

then Π(119879lowast) = Π3(1198790

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 the following applies

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0 thenΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 11987903 associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879lowast is 11987901

or 119879119889 associated with maximum profit

(4) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

Theorem 5 If 119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus

120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then one has the following results

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the

same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix E for detail

42 Decision Rule of the Optimal Cycle Time 119879lowast When 119872 lt

119879119889 When 119872 lt 119879119889 the piecewise profit function Π(119879) hasonly two different expressions that is Π1(119879) and Π3(119879)respectively From (24) we have

1198913 (119879119889) =

119863119865 (119890(120572+120579)119879119889 minus 1)

(120572 + 120579)2

[1 minus (120572 + 120579) 119879119889] minus119863119865119879119889

120572 + 120579

+119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1) (

120572119904119868119890

120572 + 120579minus 119888119868119901)

[1 minus (120572 + 120579) 119879119889] minus119863 (119879119889 minus119872)

120572 + 120579(120572119904119868119890

120572 + 120579minus 119888119868119901)

+120579119904119868119890119863119872

2

2 (120572 + 120579)minus 119870

(29)

As shown in Appendix D we will find that 1198911(119879119889) gt

1198913(119879119889) The following theorem would be given to decide thesystemrsquos optimal cycle time 119879lowast when119872 lt 119879119889

Theorem 6 If119872 lt 119879119889 then one has the following results

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the following

applies

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then Π(119879lowast) =

Π3(1198790

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then Π(119879lowast) =

maxΠ1(1198790

1) Π3(119879119889) Hence 119879

lowast is 1198790

1or 119879119889

associated with the maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then Π(119879

lowast) =

maxΠ1(1198790

1) Π3(119879

0

3) Hence 119879

lowast is 1198790

1or 1198790

3

associated with the maximum profit

(2) When 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 119879lowast is +infin

Proof See Appendix F for detail

5 Numerical Example

In this section we will provide the following numericalexample to illustrate the present model

Example 1 Given 119876119889 = 200 units 119863 = 1500 units 120572 =

04 120579 = 020 119870 = $50 ℎ = $1unityear 119888 = $ 5 perunit 119904 = $9 per unit 119868119890 = 013$year 119868119901 = 019$year and119872 = 03 year it is easy to get 119879119889 = 02123 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 and 119865 le (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872 Consequently by using Bisection method or

Newton-Raphson method we could obtain 1198790

1= 01760

1198790

2= 01892 and 119879

0

3= 01931 Because of 1198911(119879119889) le 0

1198912(119879119889) lt 0 and 1198912(119872) le 0 according to Theorem 4(1) theretailerrsquos optimal cycle time should be 119879lowast = 119879

0

3= 01931

8 Mathematical Problems in Engineering

Table 1 The impact of change of 120572 120579 119876119889 and119872 on the optimal solutions

Parameters 1198790

11198790

21198790

3119879lowast

119876lowast

Πlowast

120572

01 03381 03637 03895 03637 695772 6023591

02 04467 03956 03957 03956 823678 6217691

03 04792 04231 04320 04231 895987 6430335

04 05561 04538 04439 04538 985672 6663862

05 06723 05562 05071 05071 1033781 6931645

06 08552 05781 06047 06032 1367563 7362176

120579

001 05651 04436 05008 04436 1098853 7721683

002 05513 04471 04973 04471 1064786 7439053

005 05167 04441 04812 04441 994372 7268573

010 04783 04368 04791 04368 979335 7082341

050 03365 03568 04659 03568 833563 6887732

100 03561 02896 04543 03561 710856 6625568

119876119889

50 02344 02432 02589 02344 463085 4329953

100 03832 03565 03867 03565 750255 3948304

150 04973 04236 04973 04236 919334 3430035

200 06601 05467 05675 05675 1276297 3215728

250 07756 06873 06981 06981 1633156 2833270

300 08132 07675 07321 07321 1927368 2173673

119872

01 03726 03826 02873 03826 737665 2746542

02 03875 03835 03965 03835 799762 2823871

03 03811 04009 04623 04009 833721 2902323

04 03765 04197 04987 03765 899932 3130369

05 03732 04618 05013 04618 955513 3273771

06 03586 04745 05233 03586 998756 3633689

hence the optimal order quantity and the maximum averageprofit are 119876lowast = 1588763 and Πlowast = 18673562 respectively

Next we further study the effects of changes of parame-ters 120572 120579 119876119889 and 119872 on the optimal solutions The values ofother parameters keep the same as in the above examplewheneach of parameters120572 120579119876119889 and119872 varies Table 1 presents theobserved results with various parameters 120572 120579 119876119889 and119872

The following inferences can be made based on Table 1

(1) The retailerrsquos optimal order quantity 119876lowast and the

optimal average profit Πlowast increase as the value of 120572increases It implies that when the market demand ismore sensitive to the inventory level the retailer willincrease hisher order quantity to make more profitunder permissible delay in payments

(2) The optimal order quantity 119876lowast and the maximum

average profitΠlowast decrease as the value of 120579 increases

(3) When 119876119889 increases the optimal order quantity 119876lowastand the optimal cycle time 119879lowast are increasing but theoptimal average profit Πlowast is decreasing

(4) The retailerrsquos order quantity 119876lowast is increasing as 119872increases This verifies the fact that the retailer couldindeed increase the sales quantity by adopting thetrade credit policy provided by hisher supplier

6 Conclusions

In this paper we develop an EOQ model for deterioratingitems under permissible delay in payments The primarydifference of this paper as compared to previous relatedstudies is the following four aspects (1) the demand rateof items is dependent on the retailerrsquos current stock level(2) many items deteriorate continuously such as fruits andvegetables (3) the retailer who purchases the items enjoys afixed credit period offered by the supplier if the order quantityis greater than or equal to the predetermined quantity and(4) we here use maximizing profit as the objective to find theoptimal replenishment policies which could better describethe essence of the inventory system than a cost-minimizationmodel In addition we have discussed in detail the conditionsof the existence and uniqueness of the optimal solutions tothe model Three easy-to-use theorems are developed to findthe optimal ordering policies for the considered problem andthese theoretical results are illustrated by numerical exampleBy studying the effects of 120572 120579 119876119889 and 119872 on the optimalorder quantity 119876lowast and the optimal average profit Πlowast somemanagerial insights are derived

The presented model can be further extended to somemore practical situations For example we could allow forshortages quantity discounts time value of money andinflation price-sensitive and stock-dependent demand andso forth

Mathematical Problems in Engineering 9

Appendices

A Proof of Π1(119879119889) lt Π2(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (12) we get

Π2 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2(119890(120572+120579)119879119889 minus (120572 + 120579) 119879119889 minus 1)

+ 119904119868119890119872119863119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

(A1)

It is evident that119872 ge 119879119889 and (119890(120572+120579)119879119889 minus (120572+120579)119879119889 minus1) gt 0 for

119879119889 gt 0 then

Π2 (119879119889) minus Π1 (119879119889)

gt1

119879119889

1198631198792

119889

2(120572119904119868119890119872minus

120572119904119868119890

120572 + 120579+ 119888119868119901)

+ 119904119868119890119863119872119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

=119863119879119889

2(120572119904119868119890119872+ 119888119868119901) minus

119904119868119890119863119879119889

2+ 119904119868119890119863119872

ge119863119879119889

2(120572119904119868119890119872+ 119888119868119901) +

119904119868119890119863119872

2gt 0

(A2)

As a result we have Π1(119879119889) lt Π2(119879119889)This completes the proof of Appendix A

B Proof of Π1(119879119889) lt Π3(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (13) we get

Π3 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2[119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872) minus (120572 + 120579)119872]

+120572119904119868119890119872119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1)

+120572119904119868119890119863119872

2

(120572 + 120579)+120579119904119868119890119863119872

2

2 (120572 + 120579)

(B1)

Because of 119879119889 gt 119872 and 119890(120572+120579)119879119889 minus119890(120572+120579)(119879119889minus119872) minus(120572+120579)119872 gt 0we have Π3(119879119889) minus Π1(119879119889) gt 0

This completes the proof of Appendix B

C Proof of 1198911(119879119889) gt 1198912(119879119889)

If 119888119868119901 ge 119904119868119890 from (26) and (27) we have

1198912 (119879119889) minus 1198911 (119879119889)

= minus119863

(120572 + 120579)2(120572119904119868119890119872+ 119888119868119901 minus

120572119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

le minus119863

(120572 + 120579)2(120572119904119868119890119872+

120579119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

= minus119863119904119868119890

(120572 + 120579)2(120572119872 +

120579

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

minus120579 (120572 + 120579) 119879

2

119889

2

(C1)

Let 119866(119879119889) = (120572119872 + 120579(120572 + 120579))[(120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 +

1] minus 120579(120572 + 120579)1198792

1198892 then

1198661015840(119879119889) = (120572119872 +

120579

120572 + 120579) (120572 + 120579)

2119879119889119890(120572+120579)119879119889 minus 120579 (120572 + 120579) 119879119889

= 120579 (120572 + 120579) 119879119889 [(120572119872(120572 + 120579)

120579+ 1) 119890

(120572+120579)119879119889 minus 1] gt 0

(C2)

Hence 119866(119879119889) gt 119866(0) = 0 that is 1198911(119879119889) gt 1198912(119879119889)This completes the proof of Appendix C

D Proof of 1198911(119879119889) gt 1198913(119879119889)

From (26) and (24) we have

1198913 (119879119889) minus 1198911 (119879119889)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

times (120572119904119868119890

120572 + 120579minus 119888119868119901 minus 120572119904119868119890119872) minus

119863

(120572 + 120579)2

[(120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872) minus 119890

(120572+120579)(119879119889minus119872) + 1 minus (120572 + 120579)119872]

times (120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

8 Mathematical Problems in Engineering

Table 1 The impact of change of 120572 120579 119876119889 and119872 on the optimal solutions

Parameters 1198790

11198790

21198790

3119879lowast

119876lowast

Πlowast

120572

01 03381 03637 03895 03637 695772 6023591

02 04467 03956 03957 03956 823678 6217691

03 04792 04231 04320 04231 895987 6430335

04 05561 04538 04439 04538 985672 6663862

05 06723 05562 05071 05071 1033781 6931645

06 08552 05781 06047 06032 1367563 7362176

120579

001 05651 04436 05008 04436 1098853 7721683

002 05513 04471 04973 04471 1064786 7439053

005 05167 04441 04812 04441 994372 7268573

010 04783 04368 04791 04368 979335 7082341

050 03365 03568 04659 03568 833563 6887732

100 03561 02896 04543 03561 710856 6625568

119876119889

50 02344 02432 02589 02344 463085 4329953

100 03832 03565 03867 03565 750255 3948304

150 04973 04236 04973 04236 919334 3430035

200 06601 05467 05675 05675 1276297 3215728

250 07756 06873 06981 06981 1633156 2833270

300 08132 07675 07321 07321 1927368 2173673

119872

01 03726 03826 02873 03826 737665 2746542

02 03875 03835 03965 03835 799762 2823871

03 03811 04009 04623 04009 833721 2902323

04 03765 04197 04987 03765 899932 3130369

05 03732 04618 05013 04618 955513 3273771

06 03586 04745 05233 03586 998756 3633689

hence the optimal order quantity and the maximum averageprofit are 119876lowast = 1588763 and Πlowast = 18673562 respectively

Next we further study the effects of changes of parame-ters 120572 120579 119876119889 and 119872 on the optimal solutions The values ofother parameters keep the same as in the above examplewheneach of parameters120572 120579119876119889 and119872 varies Table 1 presents theobserved results with various parameters 120572 120579 119876119889 and119872

The following inferences can be made based on Table 1

(1) The retailerrsquos optimal order quantity 119876lowast and the

optimal average profit Πlowast increase as the value of 120572increases It implies that when the market demand ismore sensitive to the inventory level the retailer willincrease hisher order quantity to make more profitunder permissible delay in payments

(2) The optimal order quantity 119876lowast and the maximum

average profitΠlowast decrease as the value of 120579 increases

(3) When 119876119889 increases the optimal order quantity 119876lowastand the optimal cycle time 119879lowast are increasing but theoptimal average profit Πlowast is decreasing

(4) The retailerrsquos order quantity 119876lowast is increasing as 119872increases This verifies the fact that the retailer couldindeed increase the sales quantity by adopting thetrade credit policy provided by hisher supplier

6 Conclusions

In this paper we develop an EOQ model for deterioratingitems under permissible delay in payments The primarydifference of this paper as compared to previous relatedstudies is the following four aspects (1) the demand rateof items is dependent on the retailerrsquos current stock level(2) many items deteriorate continuously such as fruits andvegetables (3) the retailer who purchases the items enjoys afixed credit period offered by the supplier if the order quantityis greater than or equal to the predetermined quantity and(4) we here use maximizing profit as the objective to find theoptimal replenishment policies which could better describethe essence of the inventory system than a cost-minimizationmodel In addition we have discussed in detail the conditionsof the existence and uniqueness of the optimal solutions tothe model Three easy-to-use theorems are developed to findthe optimal ordering policies for the considered problem andthese theoretical results are illustrated by numerical exampleBy studying the effects of 120572 120579 119876119889 and 119872 on the optimalorder quantity 119876lowast and the optimal average profit Πlowast somemanagerial insights are derived

The presented model can be further extended to somemore practical situations For example we could allow forshortages quantity discounts time value of money andinflation price-sensitive and stock-dependent demand andso forth

Mathematical Problems in Engineering 9

Appendices

A Proof of Π1(119879119889) lt Π2(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (12) we get

Π2 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2(119890(120572+120579)119879119889 minus (120572 + 120579) 119879119889 minus 1)

+ 119904119868119890119872119863119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

(A1)

It is evident that119872 ge 119879119889 and (119890(120572+120579)119879119889 minus (120572+120579)119879119889 minus1) gt 0 for

119879119889 gt 0 then

Π2 (119879119889) minus Π1 (119879119889)

gt1

119879119889

1198631198792

119889

2(120572119904119868119890119872minus

120572119904119868119890

120572 + 120579+ 119888119868119901)

+ 119904119868119890119863119872119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

=119863119879119889

2(120572119904119868119890119872+ 119888119868119901) minus

119904119868119890119863119879119889

2+ 119904119868119890119863119872

ge119863119879119889

2(120572119904119868119890119872+ 119888119868119901) +

119904119868119890119863119872

2gt 0

(A2)

As a result we have Π1(119879119889) lt Π2(119879119889)This completes the proof of Appendix A

B Proof of Π1(119879119889) lt Π3(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (13) we get

Π3 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2[119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872) minus (120572 + 120579)119872]

+120572119904119868119890119872119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1)

+120572119904119868119890119863119872

2

(120572 + 120579)+120579119904119868119890119863119872

2

2 (120572 + 120579)

(B1)

Because of 119879119889 gt 119872 and 119890(120572+120579)119879119889 minus119890(120572+120579)(119879119889minus119872) minus(120572+120579)119872 gt 0we have Π3(119879119889) minus Π1(119879119889) gt 0

This completes the proof of Appendix B

C Proof of 1198911(119879119889) gt 1198912(119879119889)

If 119888119868119901 ge 119904119868119890 from (26) and (27) we have

1198912 (119879119889) minus 1198911 (119879119889)

= minus119863

(120572 + 120579)2(120572119904119868119890119872+ 119888119868119901 minus

120572119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

le minus119863

(120572 + 120579)2(120572119904119868119890119872+

120579119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

= minus119863119904119868119890

(120572 + 120579)2(120572119872 +

120579

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

minus120579 (120572 + 120579) 119879

2

119889

2

(C1)

Let 119866(119879119889) = (120572119872 + 120579(120572 + 120579))[(120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 +

1] minus 120579(120572 + 120579)1198792

1198892 then

1198661015840(119879119889) = (120572119872 +

120579

120572 + 120579) (120572 + 120579)

2119879119889119890(120572+120579)119879119889 minus 120579 (120572 + 120579) 119879119889

= 120579 (120572 + 120579) 119879119889 [(120572119872(120572 + 120579)

120579+ 1) 119890

(120572+120579)119879119889 minus 1] gt 0

(C2)

Hence 119866(119879119889) gt 119866(0) = 0 that is 1198911(119879119889) gt 1198912(119879119889)This completes the proof of Appendix C

D Proof of 1198911(119879119889) gt 1198913(119879119889)

From (26) and (24) we have

1198913 (119879119889) minus 1198911 (119879119889)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

times (120572119904119868119890

120572 + 120579minus 119888119868119901 minus 120572119904119868119890119872) minus

119863

(120572 + 120579)2

[(120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872) minus 119890

(120572+120579)(119879119889minus119872) + 1 minus (120572 + 120579)119872]

times (120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 9: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

Mathematical Problems in Engineering 9

Appendices

A Proof of Π1(119879119889) lt Π2(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (12) we get

Π2 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2(119890(120572+120579)119879119889 minus (120572 + 120579) 119879119889 minus 1)

+ 119904119868119890119872119863119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

(A1)

It is evident that119872 ge 119879119889 and (119890(120572+120579)119879119889 minus (120572+120579)119879119889 minus1) gt 0 for

119879119889 gt 0 then

Π2 (119879119889) minus Π1 (119879119889)

gt1

119879119889

1198631198792

119889

2(120572119904119868119890119872minus

120572119904119868119890

120572 + 120579+ 119888119868119901)

+ 119904119868119890119863119872119879119889 minus120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

=119863119879119889

2(120572119904119868119890119872+ 119888119868119901) minus

119904119868119890119863119879119889

2+ 119904119868119890119863119872

ge119863119879119889

2(120572119904119868119890119872+ 119888119868119901) +

119904119868119890119863119872

2gt 0

(A2)

As a result we have Π1(119879119889) lt Π2(119879119889)This completes the proof of Appendix A

B Proof of Π1(119879119889) lt Π3(119879119889)

If 119888119868119901 ge 119904119868119890 we have 119865 minus 119864 = 120572119904119868119890119872+ 119888119868119901 minus 120572119904119868119890(120572 + 120579) gt 0From (11) and (13) we get

Π3 (119879119889) minus Π1 (119879119889)

=1

119879119889

119863 (119865 minus 119864)

(120572 + 120579)2[119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872) minus (120572 + 120579)119872]

+120572119904119868119890119872119863

(120572 + 120579)2(119890(120572+120579)(119879119889minus119872) minus 1)

+120572119904119868119890119863119872

2

(120572 + 120579)+120579119904119868119890119863119872

2

2 (120572 + 120579)

(B1)

Because of 119879119889 gt 119872 and 119890(120572+120579)119879119889 minus119890(120572+120579)(119879119889minus119872) minus(120572+120579)119872 gt 0we have Π3(119879119889) minus Π1(119879119889) gt 0

This completes the proof of Appendix B

C Proof of 1198911(119879119889) gt 1198912(119879119889)

If 119888119868119901 ge 119904119868119890 from (26) and (27) we have

1198912 (119879119889) minus 1198911 (119879119889)

= minus119863

(120572 + 120579)2(120572119904119868119890119872+ 119888119868119901 minus

120572119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

le minus119863

(120572 + 120579)2(120572119904119868119890119872+

120579119904119868119890

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1] +120579119904119868119890119863119879

2

119889

2 (120572 + 120579)

= minus119863119904119868119890

(120572 + 120579)2(120572119872 +

120579

120572 + 120579)

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

minus120579 (120572 + 120579) 119879

2

119889

2

(C1)

Let 119866(119879119889) = (120572119872 + 120579(120572 + 120579))[(120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 +

1] minus 120579(120572 + 120579)1198792

1198892 then

1198661015840(119879119889) = (120572119872 +

120579

120572 + 120579) (120572 + 120579)

2119879119889119890(120572+120579)119879119889 minus 120579 (120572 + 120579) 119879119889

= 120579 (120572 + 120579) 119879119889 [(120572119872(120572 + 120579)

120579+ 1) 119890

(120572+120579)119879119889 minus 1] gt 0

(C2)

Hence 119866(119879119889) gt 119866(0) = 0 that is 1198911(119879119889) gt 1198912(119879119889)This completes the proof of Appendix C

D Proof of 1198911(119879119889) gt 1198913(119879119889)

From (26) and (24) we have

1198913 (119879119889) minus 1198911 (119879119889)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

times (120572119904119868119890

120572 + 120579minus 119888119868119901 minus 120572119904119868119890119872) minus

119863

(120572 + 120579)2

[(120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872) minus 119890

(120572+120579)(119879119889minus119872) + 1 minus (120572 + 120579)119872]

times (120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)

=119863

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 10: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

10 Mathematical Problems in Engineering

minus (120572 + 120579) 119879119889119890(120572+120579)(119879119889minus119872)

+ 119890(120572+120579)(119879119889minus119872) + (120572 + 120579)119872]

(120572119904119868119890

120572 + 120579minus 119888119868119901) +

1205791199041198681198901198631198722

2 (120572 + 120579)minus120572119904119868119890119863119872

(120572 + 120579)2

times [(120572 + 120579) 119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 + 1]

(D1)

For 119888119868119901 ge 119904119868119890 and (120572 + 120579)119879119889119890(120572+120579)119879119889 minus 119890

(120572+120579)119879119889 minus (120572 +

120579)119879119889119890(120572+120579)(119879119889minus119872) + 119890

(120572+120579)(119879119889minus119872) + (120572 + 120579)119872 gt 0 then

1198913 (119879119889) minus 1198911 (119879119889)

le minus120579119904119868119890119863

(120572 + 120579)3(119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872)) [(120572 + 120579) 119879119889 minus 1]

+ (120572 + 120579)119872 minus(120572 + 120579)

21198722

2

minus120572119904119868119890119863119872

(120572 + 120579)2[(120572 + 120579) 119879119889119890

(120572+120579)119879119889 minus 119890(120572+120579)119879119889 + 1]

(D2)

Let 119867(119879119889) = (119890(120572+120579)119879119889 minus 119890

(120572+120579)(119879119889minus119872))[(120572 + 120579)119879119889 minus 1] + (120572 +

120579)119872 minus (120572 + 120579)211987222 then it is easy to obtain1198671015840(119879119889) = (120572 +

120579)21198792

119889119890(120572+120579)119879119889(1minus119890

(120572+120579)119872) gt 0 for119879119889 gt 0 so119867(119879119889) gt 119867(119872) gt

0 hence 1198911(119879119889) gt 1198913(119879119889)This completes the proof of Appendix D

E Proof of Theorems 4 and 5

Before the proof of Theorems 4 and 5 according to theanterior analysis in Section 4 we can get the following results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If119865 le 01198912(119879) is increasing on (119879119889119872) Further if119865 gt

0 and 119879ge 119872 that is 0 lt 119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)1198721198912(119879) is also increasing on (119879119889119872) As a result if119865 le (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 then 1198912(119879) is alwaysincreasing on (119879119889119872) and in this case if 1198912(119879119889) lt 0

and 1198912(119872) gt 0 there must exist a unique root to theequation 1198912(119879) = 0 on (119879119889119872) and we denote thisroot by 1198793

2to differentiate with 119879

0

2isin (0 +infin) and

1198791

2isin (0 119879

) But if 119879119889 le 119879

lt 119872 that is (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 then

1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879) and 1198891198912(119879)119889119879 lt 0

for 119879 isin (119879119872) However when 119879

lt 119879119889 that is

119865 gt (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 we have 1198891198912(119879)119889119879 lt 0 on

(119879119889119872) and in this case if 1198912(119879119889) gt 0 and 1198912(119872) lt 0

theremust exist a unique root (say11987942) to the equation

1198912(119879) = 0 on (119879119889119872)(c) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872 1198913(119879) is increas-ing on (119872+infin) Otherwise 1198913(119879) is decreasing on(119872+infin)

Therefore to solve the optimal order cycle time 119879lowast over(0 +infin) the possible interval which 119865 belongs to needsbe considered hence we should compare the size relationsamong values of 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119872 and (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 with (119888119868119901 minus 120572119904119868119890(120572 +120579))119890minus(120572+120579)119872 Since 119872 ge 119879119889 and 119888119868119901 ge 119904119868119890 one will derive

(120579119904119868119890(120572+120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 and (120579119904119868119890(120572+

120579))119890minus(120572+120579)119872

le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 +

120579) + 120572119904119868119890119872 So there are three cases to occur (i) (120579119904119868119890(120572 +120579))119890minus(120572+120579)119879119889 le (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 (ii) (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 and (iii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt 119888119868119901 minus 120572119904119868119890(120572 + 120579) +

120572119904119868119890119872 Notice that once 119865 gt (119888119868119901 minus120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the

subfunctionΠ3(119879) is always increasing over (119872+infin) so theprofit function Π(119879) has no optimal solution over (0 +infin)Hence we need only to study the two cases as below (i)(120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119879119889 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 and

(ii) (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890

minus(120572+120579)119872According to these two cases and from some analysis theproof of Theorems 4 and 5 would be obtained

Proof of Theorem 4 If119872 ge 119879119889 and (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le

(119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then we could obtain that

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 Hencewe have the following

(1) When 119865 le (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt

1198912(119879119889) and 1198912(119879) is increasing on (119879119889119872) there arethe following five possible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0 then1198790

1ge 119879119889 and 119879

0

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 119879

0

1ge 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt

119872 These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

3

2) and

decreasing on (11987932119872) (iii) Π3(119879) is decreasing

on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) ltΠ2(119879119889) we conclude that Π(119879lowast) = Π2(119879

3

2)

hence 119879lowast is 11987932

(C) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

3

2lt 119872 and 119879

0

3lt 119872

These yield that (i)Π1(119879) is increasing on (0 1198790

1)

and decreasing (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889 1198793

2) and decreasing on (1198793

2119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

3

2) Hence 119879lowast is 1198790

1

or 11987932associated with maximum profit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 11: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

Mathematical Problems in Engineering 11

(D) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3ge 119872 These

yield that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increasing on

(119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we could concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast

is 11987901or 11987903associated with maximum profit

(E) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt 0 then0 lt 119879

0

1lt 119879119889 and 119879

0

3lt 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(2) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt 119865 le (120579119904119868119890(120572 +

120579))119890minus(120572+120579)119879119889 then 1198891198912(119879)119889119879 gt 0 for 119879 isin (119879119889 119879

) and

1198891198912(119879)119889119879 lt 0 for 119879 isin (119879119872)

(A) While 1198912(119879) gt 0 we have the following

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 119879

0

1ge 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt

1198792

2le 119872 and 119879

0

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2) decreasing

on (11987912 1198792

2) and increasing on (1198792

2119872)

(iii) Π3(119879) is increasing on (1198721198790

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ2(119879

1

2) Π3(119879

0

3) Hence

119879lowast is 1198791

2or 11987903associated with maximum

profit(b) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0

then 11987901ge 119879119889 119879119889 lt 119879

1

2lt 119879

and 11987903lt 119872

These yield that (i) Π1(119879) is increasing on(0 119879119889) (ii) Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = Π2(119879

1

2) Hence 119879lowast is 1198791

2

(c) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) le 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

119879lt 1198792

2le 119872 and 1198790

3ge 119872 These yield

that (i) Π1(119879) is increasing on (0 11987901) and

decreasing (11987901 119879119889) (ii) Π2(119879) is increas-

ing on (119879119889 1198791

2) decreasing on (1198791

2 1198792

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Π3(119879

0

3)

Hence 119879lowast is 11987901 11987912 or 1198790

3associated with

maximum profit

(d) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) gt 0then 0 lt 119879

0

1lt 119879119889 119879119889 lt 119879

1

2lt 119879

and 119879

0

3lt 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is increasing on (119879119889 119879

1

2)

and decreasing on (11987912119872) (iii) Π3(119879) is

decreasing on (119872+infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879

1

2) Hence

119879lowast is 1198790

1or 11987912associated with maximum

profit(e) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) le 0

then 0 lt 1198790

1lt 119879119889 119879

lt 1198792

2le 119872

and 1198790

3ge 119872 These yield that (i) Π1(119879)

is increasing on (0 11987901) and decreasing

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889 119879

2

2)

and increasing on (11987922119872) (iii) Π3(119879)

is increasing on (1198721198790

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3)

Hence 119879lowast is 11987901 119879119889 or 119879

0

3associated with

maximum profit(f) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) gt

0 then 0 lt 1198790

1lt 119879119889 and 119879

0

3lt 119872

These yield that (i) Π1(119879) is increasing on(0 11987901) and decreasing (1198790

1 119879119889) (ii) Π2(119879)

is decreasing on (119879119889119872) (iii) Π3(119879) isdecreasing on (119872 +infin) Combining (i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence

119879lowast is 1198790

1or 119879119889 associated with maximum

profit

(B) While 1198912(119879) le 0 then 1198912(119879119889) lt 0 and 1198912(119872) lt

0

(a) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that

(i) Π1(119879) is increasing on (0 119879119889) (ii)Π2(119879) is increasing on (119879119889119872) (iii) Π3(119879)is increasing on (119872119879

0

3) and decreas-

ing on (11987903 +infin) Combining (i)ndash(iii) and

Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(b) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872

These yield that (i) Π1(119879) is increasingon (0 1198790

1) and decreasing on (1198790

1 119879119889)

(ii) Π2(119879) is increasing on (119879119889119872) (iii)Π3(119879) is increasing on (119872119879

0

3) and

decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we concludethat Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence

119879lowast is 1198790

1or 11987903associated with maximum

profit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 12: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

12 Mathematical Problems in Engineering

(3) When (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 lt 119865 le (119888119868119901 minus 120572119904119868119890(120572 +

120579))119890minus(120572+120579)119872 since 1198911(119879119889) gt 1198912(119879119889) and 1198912(119879) is

decreasing on (119879119889119872) there are the following fourpossible cases to occur

(A) If 1198911(119879119889) le 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0 then1198790

1ge 119879119889 and 119879

0

3gt 119872 These yield that (i) Π1(119879)

is increasing on (0 119879119889) (ii) Π2(119879) is increasingon (119879119889119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 1198912(119879119889) lt 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 and 119879

0

3gt 119872 These

yield that (i) Π1(119879) is increasing on (0 1198790

1) and

decreasing on (11987901 119879119889) (ii) Π2(119879) is increasing

on (119879119889119872) (iii) Π3(119879) is increasing on (1198721198790

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with maximum profit

(C) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) lt 0then 0 lt 119879

0

1lt 119879119889 119879119889 le 119879

4

2lt 119872 and

1198790

3gt 119872 These yield that (i) Π1(119879) is increasing

on (0 11987901) and decreasing on (1198790

1 119879119889) (ii)Π2(119879)

is decreasing on (119879119889 1198794

2) and increasing on

(11987942119872) (iii) Π3(119879) is increasing on (119872119879

0

3)

and decreasing on (11987903 +infin) Combining (i)ndash

(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Π3(119879

0

3) Hence

119879lowast is 1198790

1 119879119889 or 119879

0

3associated with maximum

profit(D) If 1198911(119879119889) gt 0 1198912(119879119889) ge 0 and 1198912(119872) ge 0 then

0 lt 1198790

1lt 119879119889 and 119879

0

3le 119872 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii)Π2(119879) is decreasing on (119879119889119872) (iii)

Π3(119879) is decreasing on (119872+infin) Combining(i)ndash(iii) and Π1(119879119889) lt Π2(119879119889) we conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π2(119879119889) Hence 119879

lowast is 11987901

or 119879119889 associated with maximum profit

(4) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 4

Proof ofTheorem 5 If (120579119904119868119890(120572+120579))119890minus(120572+120579)119879119889 gt (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872 then one has (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

lt (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 le 119888119868119901 minus

120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 or (120579119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 le

(120579119904119868119890(120572 + 120579))119890minus(120572+120579)119879119889 Since Π(119879) has no optimal solution

when 119865 gt (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 so there are three

possible situations to occur as below

(1) When119865 le (120579119904119868119890(120572+120579))119890minus(120572+120579)119872 it is obvious that the

results are the same as (1) in Theorem 4(2) When (120579119904119868119890(120572 + 120579))119890

minus(120572+120579)119872lt 119865 le (119888119868119901 minus

120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 the results are the same as (2)

in Theorem 4(3) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890

minus(120572+120579)119872 because on theinterval of (119872+infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 5

F Proof of Theorem 6

According to the anterior analysis in Section 4 we could getthe below results

(a) If 119865 lt 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 1198911(119879) is increasingon (0 119879119889) Otherwise 1198911(119879) is decreasing on (0 119879119889)

(b) If 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 then 1198913(119879) is

increasing on (119879119889 +infin) Otherwise 1198913(119879) is strictlydecreasing on (119879119889 +infin)

Similarly in order to find the systemrsquos optimal cycletime 119879lowast the possible interval which 119865 belongs to needs beconsidered if 119872 lt 119879119889 hence we should compare the sizerelations between 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872 and (119888119868119901 minus

120572119904119868119890(120572+120579))119890minus(120572+120579)119872 If 119888119868119901 ge 119904119868119890 then one has (119888119868119901minus120572119904119868119890(120572+

120579))119890minus(120572+120579)119872

le 119888119868119901 minus 120572119904119868119890(120572 + 120579) + 120572119904119868119890119872

(1) When 119865 le (119888119868119901 minus 120572119904119868119890(120572 + 120579))119890minus(120572+120579)119872 we have the

following

(A) If 1198911(119879119889) le 0 and 1198913(119879119889) lt 0 then 11987901ge 119879119889 and

1198790

3gt 119879119889 These yield that (i) Π1(119879) is increasing

on (0 119879119889) (ii) Π3(119879) is increasing on (119879119889 1198790

3)

and decreasing on (11987903 +infin) Combining (i)-

(ii) and Π1(119879119889) lt Π3(119879119889) we conclude thatΠ(119879lowast) = Π3(119879

0

3) Hence 119879lowast is 1198790

3

(B) If 1198911(119879119889) gt 0 and 1198913(119879119889) ge 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3le 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing on

(11987901 119879119889) (ii) Π3(T) is decreasing on (119879119889 +infin)

Combining (i)-(ii) and Π1(119879119889) lt Π3(119879119889) weconclude that Π(119879lowast) = maxΠ1(119879

0

1) Π3(119879119889)

Hence 119879lowast is 119879

0

1or 119879119889 associated with the

maximum profit(C) If 1198911(119879119889) gt 0 and 1198913(119879119889) lt 0 then 0 lt

1198790

1lt 119879119889 and 119879

0

3gt 119879119889 These yield that (i)

Π1(119879) is increasing on (0 1198790

1) and decreasing

on (11987901 119879119889) (ii) Π3(119879) is increasing on (119879119889 119879

0

3)

and decreasing on (11987903 +infin) Combining (i)-(ii)

and Π1(119879119889) lt Π3(119879119889) we can conclude thatΠ(119879lowast) = maxΠ1(119879

0

1) Π3(119879

0

3) Hence 119879lowast is 1198790

1

or 11987903associated with the maximum profit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 13: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

Mathematical Problems in Engineering 13

(2) When119865 gt (119888119868119901minus120572119904119868119890(120572+120579))119890minus(120572+120579)119872 because on the

interval of (119872 +infin) the average profit function Π(119879)is always increasing the optimal solution 119879

lowast is +infinevidently

This completes the proof of Theorem 6

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported in part by the National NaturalScience Foundation of China Project under project Grantnos 71101002 71072165 71201044 71131003 and 71231004The authors have benefited substantially from the insightfulcomments of the editor and anonymous referees which wereinstrumental for elevating the quality of this work

References

[1] T M Whitin The Theory of Inventory Management PrincetonUniversity Press Princeton NJ USA 1957

[2] R I Levin C P McLaughlin R P Lamone and J F KottasProductionOperations Management Contemporary Policy forManaging Operating Systems McGraw-Hill New York NYUSA 1972

[3] E A Silver and R Peterson Decision Systems for InventoryManagement and Production Planning Wiley New York NYUSA 1985

[4] R Gupta and P Vrat ldquoInventory model for stock-dependentconsumption raterdquo Opsearch vol 23 no 1 pp 19ndash24 1986

[5] G Padmanabhan and P Vrat ldquoAn EOQ model for items withstock dependent consumption rate and exponential decayrdquoEngineering Costs and Production Economics vol 18 no 3 pp241ndash246 1990

[6] B C Giri S Pal A Goswami and K S Chaudhuri ldquoAninventory model for deteriorating items with stock-dependentdemand raterdquo European Journal of Operational Research vol 95no 3 pp 604ndash610 1996

[7] R C Baker and T L Urban ldquoA deterministic inventory systemwith an inventory level-dependent demand raterdquo Journal of theOperational Research Society vol 39 no 9 pp 823ndash831 1988

[8] B N Mandal and S Phaujdar ldquoAn inventory model fordeteriorating items and stock-dependent consumption raterdquoJournal of the Operational Research Society vol 40 no 5 pp483ndash488 1989

[9] S Pal A Goswami and K S Chaudhuri ldquoA deterministicinventory model for deteriorating items with stock-dependentdemand raterdquo International Journal of Production Economicsvol 32 no 3 pp 291ndash299 1993

[10] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[11] K-J Chung P Chu and S-P Lan ldquoA note on EOQ modelsfor deteriorating items under stock dependent selling raterdquoEuropean Journal of Operational Research vol 124 no 3 pp550ndash559 2000

[12] Y-W Zhou and S-L Yang ldquoAn optimal replenishment policyfor items with inventory-level-dependent demand and fixedlifetime under the LIFO policyrdquo Journal of the OperationalResearch Society vol 54 no 6 pp 585ndash593 2003

[13] Z T Balkhi and L Benkherouf ldquoOn an inventory model fordeteriorating items with stock dependent and time-varyingdemand ratesrdquo Computers amp Operations Research vol 31 no2 pp 223ndash240 2004

[14] C-Y Dye and L-Y Ouyang ldquoAn EOQ model for perishableitems under stock-dependent selling rate and time-dependentpartial backloggingrdquo European Journal of Operational Researchvol 163 no 3 pp 776ndash783 2005

[15] Y-W Zhou and S-L Yang ldquoA two-warehouse inventory modelfor items with stock-level-dependent demand raterdquo Interna-tional Journal of Production Economics vol 95 no 2 pp 215ndash228 2005

[16] K-S Wu L-Y Ouyang and C-T Yang ldquoAn optimal replen-ishment policy for non-instantaneous deteriorating items withstock-dependent demand and partial backloggingrdquo Interna-tional Journal of Production Economics vol 101 no 2 pp 369ndash384 2006

[17] H K Alfares ldquoInventory model with stock-level dependentdemand rate and variable holding costrdquo International Journalof Production Economics vol 108 no 1-2 pp 259ndash265 2007

[18] S K Goyal ldquoEconomic order quantity under conditions ofpermissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 36 no 4 pp 335ndash338 1985

[19] S P Aggarwal and C K Jaggi ldquoOrdering policies of deteriorat-ing items under permissible delay in paymentsrdquo Journal of theOperational Research Society vol 46 no 5 pp 658ndash662 1995

[20] A M M Jamal B R Sarker and S Wang ldquoAn ordering policyfor deteriorating items with allowable shortage and permissibledelay in paymentrdquo Journal of the Operational Research Societyvol 48 no 8 pp 826ndash833 1997

[21] B R Sarker AMM Jamal and SWang ldquoSupply chainmodelsfor perishable products under inflation and permissible delay inpaymentrdquo Computers amp Operations Research vol 27 no 1 pp59ndash75 2000

[22] J-T Teng ldquoOn the economic order quantity under conditionsof permissible delay in paymentsrdquo Journal of the OperationalResearch Society vol 53 no 8 pp 915ndash918 2002

[23] Y-F Huang ldquoOptimal retailerrsquos ordering policies in the EOQmodel under trade credit financingrdquo Journal of the OperationalResearch Society vol 54 no 9 pp 1011ndash1015 2003

[24] C-T Chang L-Y Ouyang and J-T Teng ldquoAn EOQ model fordeteriorating items under supplier credits linked to orderingquantityrdquo Applied Mathematical Modelling vol 27 no 12 pp983ndash996 2003

[25] C-T Chang ldquoAn EOQ model with deteriorating items underinflation when supplier credits linked to order quantityrdquo Inter-national Journal of Production Economics vol 88 no 3 pp 307ndash316 2004

[26] K-J Chung and J-J Liao ldquoLot-sizing decisions under tradecredit depending on the ordering quantityrdquo Computers ampOperations Research vol 31 no 6 pp 909ndash928 2004

[27] J-T Teng C-T Chang and S K Goyal ldquoOptimal pricingand ordering policy under permissible delay in paymentsrdquoInternational Journal of Production Economics vol 97 no 2 pp121ndash129 2005

[28] J-J Liao ldquoA note on an EOQ model for deteriorating itemsunder supplier credit linked to ordering quantityrdquo AppliedMathematical Modelling vol 31 no 8 pp 1690ndash1699 2007

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 14: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

14 Mathematical Problems in Engineering

[29] J-T Teng and S K Goyal ldquoOptimal ordering policies for aretailer in a supply chain with up-stream and down-streamtrade creditsrdquo Journal of the Operational Research Society vol58 no 9 pp 1252ndash1255 2007

[30] S-C Chen L E Cardenas-Barron and J-T Teng ldquoRetailerrsquoseconomic order quantity when the supplier offers conditionallypermissible delay in payments link to order quantityrdquo Interna-tional Journal of Production Economics vol 155 pp 284ndash2912013

[31] C-T Chang M-C Cheng and L-Y Ouyang ldquoOptimal pricingand ordering policies for non-instantaneously deterioratingitems under order-size-dependent delay in paymentsrdquo AppliedMathematical Modelling 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 15: Research Article EOQ Model for Deteriorating Items with ... · EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of