research article optimal replenishment policy for...
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Research ArticleOptimal Replenishment Policy for Weibull-DistributedDeteriorating Items with Trapezoidal Demand Rate andPartial Backlogging
Lianxia Zhao
School of Management Shanghai University Shanghai 200444 China
Correspondence should be addressed to Lianxia Zhao zhaolx2004163com
Received 2 December 2015 Accepted 27 March 2016
Academic Editor Alireza Amirteimoori
Copyright copy 2016 Lianxia ZhaoThis 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
An inventory model for Weibull-distributed deteriorating items is considered so as to minimize the total cost per unit time inthis paper The model starts with shortage allowed partial backlogging and trapezoidal demand rate By analyzing the model anefficient solution procedure is proposed to determine the optimal replenishment and the optimal order quantity and the averagetotal costs are also obtained Finally numerical examples are provided to illustrate the theoretical results and a sensitivity analysisof the major parameters with respect to the stability of optimal solution is also carried out
1 Introduction
Deteriorating is a general phenomenon for many productsin that fruits or vegetables are spoiled directly while alco-hol undergoes physical depletion over time and electronicproducts deteriorate through a gradual loss of potentialutility as time passed Based on the above situation theeffect of deteriorating for items can not be disregarded inmany inventory systems An inventorymodel with a constantdecaying rate for items is proposed by Ghare and Schraderin 1963 [1] while Covert and Philip [2] extended Ghare andSchraderrsquos model to an EOQ model with a two-parameterWeibull-distributed deteriorating rate Since then there aremany researchers who focused on such topic Wee et al [3]considered an inventory model for deteriorating items withquantity discount pricing and partial backlogging andmorerelated articles can be referred to like Yang et al [4] Shah etal [5] Yang et al [6] and so forth
However it is impossible for deteriorating items thatthe demand rate increases continuously during their full lifecycle Based on such reasons Hill [7] proposed an inventorymodel with ramp-type demand rate which extended toallow shortage proposed by Mandal and Pal [8] Further
Wu [9] extended Mandal and Palrsquos model to have Weibull-distributed deterioration and time-dependent backloggingSkouri et al [10] considered a model with general ramp-typedemand rate partial backlogging and Weibull deteriorationrate Tan and Weng [11] considered a discrete inventorymodel for deteriorating items with partial backlogging Palet al [12] considered a production inventory model fordeteriorating items with ramp-type demand rate under theeffect of inflation and shortages under fuzziness There aremany other related literatures about such inventory modelsuch as Ahmed et al [13] Agrawal et al [14] Singh andPattnayak [15] Wu et al [16] and Rabbani et al [17]
While for some short life cycle products the demandrate may increase up to a certain level in the early stage ofmarketing then reach a stabilized period and finally decreaseto zero and reach the end of their life cycle Cheng andWang [18] proposed an inventory model for deterioratingitems with trapezoidal type demand rate Further Cheng etal [19] extended the model to be of time-dependent deterio-rating items with trapezoidal type demand rate and partialbacklogging Uthayakumar and Rameswari [20] studied amodel for defective items with trapezoidal type demand rateto determine the optimal product reliability Lin [21] and
Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2016 Article ID 1490712 10 pageshttpdxdoiorg10115520161490712
2 Mathematical Problems in Engineering
Inventorylevel
Time
S
0t1 T
1205831
1205832
Case 1 0 lt t1 le 1205831
Inventorylevel
Time
S
0
t1 T
1205831
1205832
Case 2 1205831le t1 le 120583
2
Time
Inventorylevel
S
0
t1 T
1205831
1205832
Case 3 1205832le t1 lt T
Figure 1 Graphical representation of inventory level over the cycle starting with shortage
Lin et al [22] considered the inventory model which wasproposed by Cheng and Wang [18] Zhao [23] studied aninventory model for Weibull-distributed deterioration itemswith trapezoidal type demand rate and partial backlogging inwhich themodel begins with no-shortageWhile in real situ-ation customers will respond differently in case of shortageseither to wait until replenishment or to look for alternativeproducts in this paper we study an inventory model begin-ning with shortage Weibull-distributed deterioration itemstrapezoidal type demand rate and time-dependent partialbacklogging By analyzing the studied model we propose anoptimal replenishment strategy for it and also carry out asensitivity analysis of the main parameters
The remainder of the paper is organized as followsNotations and assumptions are described in Section 2 Sec-tion 3 constructs and analyzes the inventory model startingwith shortage Some numerical examples to illustrate thesolution procedure are provided then sensitivity analysis ofthe major parameters is also carried out in Section 4 Finallya conclusion is presented
2 Notations and Assumptions
The fundamental notations and assumptions used in thispaper are given as follows
(i) 119868(119905) is the inventory level at time 119905 0 le 119905 le 119879 where119879 is the length of inventory cycle
(ii) 1199051denotes the replenishment time and 119905lowast
1denotes the
optimal 1199051
(iii) 119878 is the maximum inventory level for each orderingcycle
(iv) 119876lowast denotes the optimal ordering quantity(v) 119860
0denotes the fixed cost per order
(vi) c1is the cost of each deteriorated item c
2is the unit
inventory holding cost per unit time c3is the unit
shortage cost per unit time and c4is the unit lost sales
cost(vii) C
119894(1199051) 119894 = 1 2 3 denotes the average total cost per
unit time under different conditions respectively
(viii) TC(1199051) denotes the average total cost per unit time
(ix) The demand rate119863(119905) which is positive and consec-utive is assumed to be a trapezoidal type function oftime that is
119863 (119905) =
119891 (119905) 119905 le 1205831
1198630 120583
1lt 119905 lt 120583
2
119892 (119905) 1205832le 119905 lt 119879
(1)
where 1205831is time point changing from the increasing
demand function 119891(119905) to the constant demand 1198630
and 1205832is time point changing from the constant
demand1198630to the decreasing demand function 119892(119905)
(x) The replenishment rate is infinite that is replenish-ment is instantaneous
(xi) The time to deterioration of the item is distributed asWeibull (120572 120573) that is the deterioration rate is 120579(119905) =120572120573119905120573minus1 where 120572 is the scale parameter 0 lt 120572 lt 1
and120573 is the shape parameter 0 lt 120573
(xii) Shortage is allowed and let 119890minus120575119905 be fraction of short-ages backlogged where 119905 is the waiting time up to thenext replenishment and 120575 is a tiny positive constantthat is 120575 ≪ 1119879
3 Inventory Model Starting with Shortage
The system of model begins with shortage and until 1199051the
shortage achieves its maximum then replenishment occursat time 119905 = 119905
1 All of the shortage demand during (0 119905
1)
is partial backlogged (Figure 1) At 1199051 the total number of
backlogged items will be satisfied by the replenishment andthe inventory level up to 119878 while the deteriorating occursduring the time of [119905
1 119879] According to the notations and
Mathematical Problems in Engineering 3
assumptions mentioned above the model is described asshown in Figure 1
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119863 (119905) 0 lt 119905 lt 119905
1
minus120579 (119905) 119868 (119905) minus 119863 (119905) 1199051lt 119905 lt 119879
(2)
with boundary condition 119868(0) = 0 119868(1199051) = 119878 119868(119879) = 0
Based on the values of 1199051 1205831 and 120583
2 three possible cases
are presented as follows
Case 1 (0 lt 1199051le 1205831) If the replenishment time 119905
1isin (0 120583
1]
from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 119891 (119905) 1199051lt 119905 lt 120583
1
minus120572120573119905120573minus1
119868 (119905) minus 1198630 120583
1lt 119905 lt 120583
2
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1205832lt 119905 lt 119879
(3)
Solving the differential equations (3) with 119868(0) = 0 we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 119905
1
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 120583
1
1198630int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
1lt 119905 lt 120583
2
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
2lt 119905 lt 119879
(4)
The total replenishment quantity can be computed as
119878 = 119868 (1199051)
= int
1205831
1199051
119890120572(119909120573minus119905120573
1)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905120573
1)119889119909
+ int
119879
1205832
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909
(5)
The total shortage quantity 119861119879during the interval [0 119905
1) is
119861119879= int
1199051
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905 (6)
The total number of perished items during the interval [1199051 119879]
say119863119879 is
119863119879= int
1205831
1199051
(119890120572(119905120573minus119905120573
1)minus 1)119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905
(7)
The total number of inventories carried during the interval[1199051 119879] say119867
119879 is
119867119879= int
1205831
1199051
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 119889119905
+ 1198630int
1205831
1199051
int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ 1198630int
1205832
1205831
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
(8)
The total of lost sales during the interval [0 1199051] say 119871
119879 is
119871119879= int
1199051
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905 (9)
Therefore the average total cost per unit time under thecondition 119905
1le 1205831can be given by
C1(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c3[int
1199051
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905]
+ c4[int
1199051
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905]
+ c1[int
1205831
1199051
(119890120572(119905120573minus119905120573
1)minus 1)119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
4 Mathematical Problems in Engineering
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905]
+ c2[int
1205831
1199051
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 119889119905
+ 1198630int
1205831
1199051
int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ 1198630int
1205832
1205831
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905]
(10)
Taking the first and the second order derivative of C1(1199051)with
respect to 1199051 respectively we have
119889C1(1199051)
1198891199051
=1
119879c3int
1199051
0
(1 minus 1205751199051+ 120575119905) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ c4120575int
1199051
0
119890120575(119905minus1199051)119891 (119905) 119889119905 minus (c
2+ c1120572120573119905120573minus1
1)
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119863 (119905) 119889119905
(11)
1198892C1(1199051)
1198891199052
1
=1
119879c3[119891 (1199051)
+ 120575int
1199051
0
(1205751199051minus 120575119905 minus 2) 119890
120575(119905minus1199051)119891 (119905) 119889119905] + c
4120575 [119891 (119905
1)
minus 120575int
1199051
0
119890120575(119905minus1199051)119891 (119905) 119889119905] + (c
1120572120573119905120573minus1
1+ c2) 119891 (1199051)
+ 120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119863 (119905) 119889119905
(12)
Then we have
119889C1(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=0
= minusc2
119879int
119879
0
119890120572119905120573
119863(119905) 119889119905 lt 0 (13)
and let
Δ1= c3int
1205831
0
(1 minus 1205751205831+ 120575119905) 119890
120575(119905minus1205831)119891 (119905) 119889119905
+ c4120575int
1205831
0
119890120575(119905minus120583
1)119891 (119905) 119889119905 minus (c
2+ c1120572120573120583120573minus1
1)
sdot (1198630int
1205832
1205831
119890120572(119905120573minus120583120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus120583120573
1)119892 (119905) 119889119905)
(14)
Therefore (i) if Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (0 120583
1) satisfying 119889C
1(1199051)1198891199051= 0
and C1(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt 0
and 1198892C1(1199051)1198891199052
1gt 0 it means that C
1(1199051) is a nonincreasing
function and obtains its minimum value at point 119905lowast1= 1205831
(iii) if the sign of 1198892C1(1199051)1198891199052
1is indefinite then the curve
of C1(1199051) is W-shape and C
1(1199051)may obtain its minimum at
more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
1(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= int
1205831
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905lowast
1
120573)119889119909
+ int
119879
1205832
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909
(15)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
119905lowast
1
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 (16)
Case 2 (1205831le 1199051le 1205832) If the replenishment time 119905
1isin [1205831 1205832]
then from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 1198630 119905
1lt 119905 lt 120583
2
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1205832lt 119905 lt 119879
(17)
Solving the differential equation (17) with 119868(0) = 119868(119879) = 0we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
minus1198630
120575(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 119905
1
1198630int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 120583
2
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
2lt 119905 lt 119879
(18)
Mathematical Problems in Engineering 5
From (18) the total replenishment quantity can be computed
119878 = 119868 (1199051) = 119863
0int
1205832
1199051
119890120572(119909120573minus119905120573
1)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (19)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905
(20)
The total number of perished items in the interval [1199051 119879] is
119863119879= 1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905
(21)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= 1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
(22)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
(23)
Therefore the average total cost per unit time under thecondition 120583
1lt 1199051le 1205832can be given by
C2(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1[1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905]
+ c2[1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905]
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905]
(24)
Taking the first and the second order derivative of C2(1199051)
with respect to 1199051 respectively we have
119889C2(1199051)
1198891199051
=1
119879c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630int
1199051
1205831
119890120575(119905minus1199051)119889119905] minus (c
1120572120573119905120573minus1
1+ c2)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 minus 1205751199051+ 120575119905) 119890
120575(119905minus1199051)119891 (119905) 119889119905
minus 1198630(1205831minus 1199051) 119890120575(1205831minus1199051)]
(25)
1198892C2(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1(c1(120572120573119905120573
1minus 120573 + 1) + c
21199051)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ (c1120572120573119905120573minus1
1+ c2)1198630+ c4120575 [1198630119890120575(1205831minus1199051)
minus 120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630(1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)]
(26)
Let
Δ2= c3[int
1205831
0
(1 minus 1205751205832+ 120575119905) 119890
120575(119905minus1205832)119891 (119905) 119889119905
+ 1198630(1205832minus 1205831) 119890120575(1205831minus1205832)]
+ c4120575 [int
1205831
0
119890120575(119905minus120583
2)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus120583
2)119889119905]
minus (c2+ c1120572120573120583120573minus1
2)int
119879
1205832
119890120572(119905120573minus120583120573
2)119892 (119905) 119889119905
(27)
6 Mathematical Problems in Engineering
Therefore (i) if Δ1lt 0 Δ
2gt 0 and 1198892C
2(1199051)1198891199052
1gt 0 there
exists a unique solution 119905lowast1isin (1205831 1205832) satisfying 119889C
2(1199051)1198891199051=
0 and C2(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt
0 Δ2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 it means that C
2(1199051) is
a nonincreasing function and obtains its minimum at point119905lowast
1= 1205832 (iii) if Δ
1ge 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 it
means that C2(1199051) is a nondecreasing function and obtains its
minimum at point 119905lowast1= 1205831 (iv) if the sign of 1198892C
2(1199051)1198891199052
1is
indefinite then the curve of C2(1199051) is W-shape and C
2(1199051)
may obtain its minimum value at more than one pointwithout loss of generality set 119905
1= 119905lowast
1 where 119905lowast
1is one of the
solutions of 119889C2(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= 1198630int
1205832
119905lowast
1
119890120572(119905120573minus119905lowast
1
120573)119889119905 + int
119879
1205832
119890120572(119905120573minus119905lowast
1
120573)119892 (119905) 119889119905 (28)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
119905lowast
1
1205831
119890120575(119905minus119905lowast
1)119889119905 (29)
Case 3 (1205832le 1199051lt 119879) If the replenishment time 119905
1isin [1205832 119879)
from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 120583
2
minus119890minus120575(1199051minus119905)119892 (119905) 120583
2lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1199051lt 119905 lt 119879
(30)
Solving (30) with 119868(0) = 119868(119879) = 0 we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(119905minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 120583
2
minus int
119905
1205832
119890120575(119909minus119905
1)119892 (119909) 119889119909 +
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
2lt 119905 lt 119905
1
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 119879
(31)
From the last equation of (31) the total replenishmentquantity can be computed
119878 = 119868 (1199051) = int
119879
1199051
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (32)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575[(1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051)]
(33)
The total number of perished items in the interval [1199051 119879] is
119863119879= int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909 (34)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905 (35)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905
(36)
Therefore the average total cost per unit time under thecondition 120583
2lt 1199051le 119879 can be given by
C3(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909
+ c2int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
Mathematical Problems in Engineering 7
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(37)
Taking the first and the second order derivative of C3(1199051)
with respect to 1199051 respectively we have
119889C3(1199051)
1198891199051
=1
119879minus (c2+ c1120572120573119905120573minus1
1) int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905
+ c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119892 (119905) 119889119905
+ 1198630((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(38)
1198892C3(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + (c
2+ c1120572120573119905120573minus1
1) 119892 (1199051)
+ c4120575 [120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0120575int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ 120575int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c2[120572120573119905120573minus1
1int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)
+ 1198630((1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)
minus (1 + 1205751205832minus 1205751199051) 119890120575(1205832minus1199051))]
(39)
We know that
119889C3(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=119879
=1
119879c4120575 [int
1205831
0
119890120575(119905minus119879)
119891 (119905) 119889119905
+ 1198630int
1205832
1205831
119890120575(119905minus119879)
119889119905 + int
119879
1205832
119890120575(119905minus119879)
119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119891 (119905) 119889119905
+ int
119879
1205832
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119892 (119905) 119889119905
+ 1198630((119879 minus 120583
1) 119890120575(1205831minus119879)
minus (119879 minus 1205832) 119890120575(1205832minus119879)
)]
gt 0
(40)
Therefore (i) if Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (1205832 119879) satisfying 119889C
3(1199051)1198891199051= 0
and C3(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
2ge 0
and 1198892C3(1199051)1198891199052
1gt 0 it means that C
3(1199051) is a nondecreasing
function and obtains its minimum value at point 119905lowast1= 1205832
(iii) if the sign of 1198892C3(1199051)1198891199052
1is indefinite then the curve of
C3(1199051) is W-shape and C
3(1199051)may obtain its minimum value
at more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
3(1199051)1198891199051= 0
Therefore the optimal value of the order level 119878 = 119868(119905lowast
1)
is
119878lowast= int
119879
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909 (41)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus119905lowast
1)119889119905
+ int
119905lowast
1
1205832
119890120575(119905minus119905lowast
1)119892 (119905) 119889119905
(42)
From the above analysis we obtain that the total averagecost of the system over the interval [0 119879] is
TC (1199051) =
C1(1199051) 0 lt 119905
1le 1205831
C2(1199051) 120583
1lt 1199051le 1205832
C3(1199051) 120583
2lt 1199051lt 119879
(43)
where C1(1199051) C2(1199051) and C
3(1199051) are obtained from (10) (24)
and (37) respectively
8 Mathematical Problems in Engineering
Basing on the analyzed results above we provide theresults which ensure the existence of a unique 119905
1 say 119905lowast
1 to
minimize the total average cost for the model system startingwith shortages The procedure is as follows
Step 0 Input the parameters 1205831 1205832 119879 1198861 1198862 1198871 120572 120573 120575
1198630 and c
119894 119894 = 1 2 3 4
Step 1 Calculate Δ1and 1198892C
1(1199051)1198891199052
1
Step 11 If Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 solve a unique 119905lowast
1
from119889C1(1199051)1198891199051= 0 and the total average cost and the order
quantity can be obtained by (10) and (16) respectively
Step 12 If Δ1le 0 and 119889
2C1(1199051)1198891199052
1gt 0 then the optimal
average cost and the optimal order quantity can be obtainedby (10) and (16) at 119905lowast
1= 1205831 respectively
Step 13 If the sign of 1198892C1(1199051)1198891199052
1is indefinite then solve
equation 119889C1(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C1(1199051) to obtain the optimal
replenishment point compute the order quantity by (16)
Step 2 Calculate Δ2and 1198892C
2(1199051)1198891199052
1
Step 21 If Δ1lt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
seek a unique 119905lowast1from equation 119889C
2(1199051)1198891199051= 0 and the
optimal total average cost and the optimal order quantity canbe obtained by (24) and (29) at 119905
1= 119905lowast
1 respectively
Step 22 If Δ1gt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205831 and compute the order
quantity by (29)
Step 23 If Δ1lt 0 Δ
2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205832 and compute the order
quantity by (29)
Step 24 If the sign of 1198892C2(1199051)1198891199052
1is indefinite then solve
equation 119889C2(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C2(1199051) to obtain the optimal
replenishment point compute the optimal order quantity by(29)
Step 3 Calculate 1198892C3(1199051)1198891199052
1
Step 31 If Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 seek the unique
119905lowast
1from equation 119889C
3(1199051)1198891199051= 0 and the total average
cost and the order quantity can be obtained by (37) and (42)respectively
Step 32 If Δ2ge 0 and 1198892C
3(1199051)1198891199052
1gt 0 then C
3(1199051) obtains
its minimum at 119905lowast1= 1205832 and compute the order quantity by
(42)
Step 33 If the sign of 1198892C3(1199051)1198891199052
1is indefinite then
solve equation 119889C3(1199051)1198891199051
= 0 If the solution is notunique by comparing their average costs C
3(1199051) to obtain the
optimal replenishment point compute the order quantity by(42)
Step 4 Find TC(119905lowast1) = minC
1(119905lowast
1)C2(119905lowast
1)C3(119905lowast
1) and
accordingly select the optimum 119905lowast
1 and then obtain corre-
sponding optimal order quantity 119876lowast
4 Numerical Examples andSensitivity Analysis
In order to demonstrate the procedure to obtain the optimalsolution of the model we present four examples for themodel Examples are based on piece-wise demand rate suchas 119891(119905) = 119886
1+ 1198871119905 and 119892(119905) = 119886
2119890minus1198872119905
Example 1 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 004 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $2 c
2= $3 c
3= $12 c
4= $8
Since Δ1
= 18809 gt 0 Δ2
= 222869 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1) 1198892C
2(1199051)1198891199052
1gt 0
for all 1199051
isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
2(1205831) = 763891 in
[1205831 1205832) and C
3(1199062) = 126358 in [120583
2 119879) By solving equation
119889C1(1199051)1198891199051= 0 we have 119905lowast
1= 3664 From (10) we obtain
C1(119905lowast
1) = 762759 Therefore the optimal ordering quantity is
119876lowast= 637839 and the total average cost is TC(119905lowast
1) = 762759
Example 2 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus257889 lt 0 Δ2
= 139269 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin (1205832 119879)
then we get local minimum value C1(1205831) = 1805883 in
(0 1205831] and C
3(1199062) = 158232 in [120583
2 119879) By solving equation
119889C2(1199051)1198891199051= 0 we have 119905lowast
1= 6559 From (24) we obtain
C2(119905lowast
1) = 1480324 Therefore the optimal ordering quantity
is 119876lowast = 571827 and the total average cost is TC(119905lowast1) =
1480324
Example 3 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 2 120575 = 006 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus249329 lt 0 Δ2
= minus121869 lt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin [1205832 119879)
then we get local minimum value C1(1205831) = 1417134 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 9120 From (37) we obtain
C3(119905lowast
1) = 889767 Therefore the optimal ordering quantity is
119876lowast= 278960 and the total average cost is TC(119905lowast
1) = 889767
Example 4 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 16 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus197263 lt 0 Δ2
= minus86675 lt 01198892C4(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0
Mathematical Problems in Engineering 9
Table 1 The sensitivity of 120575 for the model in Example 1
120575 0 001 002 003 004 005 006 007 008119905lowast
134366 34905 35463 36042 36642 37263 37907 38574 39265
119878lowast 5150829 5150829 5114210 5075965 5036066 4994433 4950985 4905641 4858321119876lowast 6481034 6481034 6448369 6414184 6378392 6340900 6301614 6260436 6217267
TC2(119905lowast
1) 7752147 7718437 7684699 7649983 7614252 7577473 7539611 750063 7460498
Table 2 The sensitivity of 120572 for the models in Example 1
120572 0 0001 0002 0003 0004 0005 0006 0007 0008119905lowast
130141 31325 32569 33870 35229 36642 38105 39614 41171
119878lowast 4158882 4320629 4489624 4665543 4847907 5036066 5229206 5426345 5755923119876lowast 5228982 5439279 5659745 5890106 6129883 6378392 6634729 6897782 7230000
TC2(119905lowast
1) 6385314 6614087 6851585 7097639 7351984 7614252 7883972 8160570 8450147
Table 3 The sensitivity of 120573 for the models in Example 1
120573 14 15 16 17 18 19 20 21 22119905lowast
131114 31995 32513 33176 340332 35148 36613 38551 41128
119878lowast 4275053 4401612 4471691 4562029 4679084 4831709 5036066 5295334 5641669119876lowast 5384988 5547959 5639494 5757494 5910495 6110248 6378392 6720262 7179964
TC2(119905lowast
1) 6444596 6703559 6801679 6929359 7095918 7314743 7614252 7976508 8450147
for all 1199051
isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
1(1205831) = 125259 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 8627 From (37) we obtain
TC(119905lowast1) = 842935 Therefore the optimal ordering quantity
is 119876lowast = 241524 obtained from (42) and the minimum costTC(119905lowast1) = 842935
In order to clearly indicate the effects of parameterssuch as 120575 120572 and 120573 on 119878
lowast 119876lowast and TC119894(119905lowast
1) (119894 = 1 2)
respectively we study the sensitivity of different parametersunder the studied inventory modelThe sensitivity analysis isperformed on the base of Example 1 and the results are shownin Tables 1ndash3 respectively
From Table 1 we observe that inventory level 119878lowast orderquantity 119876
lowast and the total average cost TC119894(119905lowast
1) (119894 = 1 2)
gradually decrease as the shortage parameter 120575 increasesfor the model and the replenishment time 119905lowast
1increases as
120575 increases for the model starting with shortage Also weobserve that the values of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) are slightly
sensitive to the changes of 120575From Table 2 we find that when the deterioration
parameter 120572 increases then 119878lowast 119876lowast and TC(119905lowast1) increase for
both models and the replenishment time 119905lowast1increases for the
model starting with shortage We see that the values of 119905lowast1 119878lowast
119876lowast and TC(119905lowast
1) all are temperately sensitive to the changes of
120572 for the modelFrom Table 3 we find that 119878lowast 119876lowast and TC(119905lowast
1) increase
as the deterioration parameter 120573 increases for both modelsAlso we find that the replenishment time 119905lowast
1increases as 120573
increases for themodel startingwith shortageWe see that thevalues of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) all are moderately sensitive
to the changes of 120573 for the model
5 Conclusion
This paper studies an inventory model for Weibull-distributed deterioration with trapezoidal type demand rateand partial backlogging First an optimal replenishmentstrategy for deteriorating items with the model starting withshortage is proposedThen numerical examples are providedto illustrate the theoretical results From the analysis of Tables1ndash3 it can be found that the replenishment time point 119905lowast
1
order quantity 119876lowast and the total average cost TC(119905lowast
1) are
moderately sensitive to the changes of 120572 and 120573 and lowlysensitive to the changes of 120575 respectively However if theorder cycle (119879) is a decision variable such inventory modelwill be of more practical significance which provides us aninteresting topic in future
Competing Interests
The author declares that there are no competing interests
Acknowledgments
The research is supported partly by National Natural ScienceFoundation of China (Grant no 71502100)
References
[1] P M Ghare and G F Schrader ldquoA model for exponentiallydecaying inventoriesrdquo Journal of Industrial Engineering vol 14no 3 pp 238ndash243 1963
[2] R P Covert and G C Philip ldquoAn EOQ model for items withweibull distribution deteriorationrdquo AIIE Transaction vol 5 no4 pp 323ndash326 1973
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 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
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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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
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Mathematical Problems in Engineering
Inventorylevel
Time
S
0t1 T
1205831
1205832
Case 1 0 lt t1 le 1205831
Inventorylevel
Time
S
0
t1 T
1205831
1205832
Case 2 1205831le t1 le 120583
2
Time
Inventorylevel
S
0
t1 T
1205831
1205832
Case 3 1205832le t1 lt T
Figure 1 Graphical representation of inventory level over the cycle starting with shortage
Lin et al [22] considered the inventory model which wasproposed by Cheng and Wang [18] Zhao [23] studied aninventory model for Weibull-distributed deterioration itemswith trapezoidal type demand rate and partial backlogging inwhich themodel begins with no-shortageWhile in real situ-ation customers will respond differently in case of shortageseither to wait until replenishment or to look for alternativeproducts in this paper we study an inventory model begin-ning with shortage Weibull-distributed deterioration itemstrapezoidal type demand rate and time-dependent partialbacklogging By analyzing the studied model we propose anoptimal replenishment strategy for it and also carry out asensitivity analysis of the main parameters
The remainder of the paper is organized as followsNotations and assumptions are described in Section 2 Sec-tion 3 constructs and analyzes the inventory model startingwith shortage Some numerical examples to illustrate thesolution procedure are provided then sensitivity analysis ofthe major parameters is also carried out in Section 4 Finallya conclusion is presented
2 Notations and Assumptions
The fundamental notations and assumptions used in thispaper are given as follows
(i) 119868(119905) is the inventory level at time 119905 0 le 119905 le 119879 where119879 is the length of inventory cycle
(ii) 1199051denotes the replenishment time and 119905lowast
1denotes the
optimal 1199051
(iii) 119878 is the maximum inventory level for each orderingcycle
(iv) 119876lowast denotes the optimal ordering quantity(v) 119860
0denotes the fixed cost per order
(vi) c1is the cost of each deteriorated item c
2is the unit
inventory holding cost per unit time c3is the unit
shortage cost per unit time and c4is the unit lost sales
cost(vii) C
119894(1199051) 119894 = 1 2 3 denotes the average total cost per
unit time under different conditions respectively
(viii) TC(1199051) denotes the average total cost per unit time
(ix) The demand rate119863(119905) which is positive and consec-utive is assumed to be a trapezoidal type function oftime that is
119863 (119905) =
119891 (119905) 119905 le 1205831
1198630 120583
1lt 119905 lt 120583
2
119892 (119905) 1205832le 119905 lt 119879
(1)
where 1205831is time point changing from the increasing
demand function 119891(119905) to the constant demand 1198630
and 1205832is time point changing from the constant
demand1198630to the decreasing demand function 119892(119905)
(x) The replenishment rate is infinite that is replenish-ment is instantaneous
(xi) The time to deterioration of the item is distributed asWeibull (120572 120573) that is the deterioration rate is 120579(119905) =120572120573119905120573minus1 where 120572 is the scale parameter 0 lt 120572 lt 1
and120573 is the shape parameter 0 lt 120573
(xii) Shortage is allowed and let 119890minus120575119905 be fraction of short-ages backlogged where 119905 is the waiting time up to thenext replenishment and 120575 is a tiny positive constantthat is 120575 ≪ 1119879
3 Inventory Model Starting with Shortage
The system of model begins with shortage and until 1199051the
shortage achieves its maximum then replenishment occursat time 119905 = 119905
1 All of the shortage demand during (0 119905
1)
is partial backlogged (Figure 1) At 1199051 the total number of
backlogged items will be satisfied by the replenishment andthe inventory level up to 119878 while the deteriorating occursduring the time of [119905
1 119879] According to the notations and
Mathematical Problems in Engineering 3
assumptions mentioned above the model is described asshown in Figure 1
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119863 (119905) 0 lt 119905 lt 119905
1
minus120579 (119905) 119868 (119905) minus 119863 (119905) 1199051lt 119905 lt 119879
(2)
with boundary condition 119868(0) = 0 119868(1199051) = 119878 119868(119879) = 0
Based on the values of 1199051 1205831 and 120583
2 three possible cases
are presented as follows
Case 1 (0 lt 1199051le 1205831) If the replenishment time 119905
1isin (0 120583
1]
from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 119891 (119905) 1199051lt 119905 lt 120583
1
minus120572120573119905120573minus1
119868 (119905) minus 1198630 120583
1lt 119905 lt 120583
2
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1205832lt 119905 lt 119879
(3)
Solving the differential equations (3) with 119868(0) = 0 we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 119905
1
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 120583
1
1198630int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
1lt 119905 lt 120583
2
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
2lt 119905 lt 119879
(4)
The total replenishment quantity can be computed as
119878 = 119868 (1199051)
= int
1205831
1199051
119890120572(119909120573minus119905120573
1)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905120573
1)119889119909
+ int
119879
1205832
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909
(5)
The total shortage quantity 119861119879during the interval [0 119905
1) is
119861119879= int
1199051
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905 (6)
The total number of perished items during the interval [1199051 119879]
say119863119879 is
119863119879= int
1205831
1199051
(119890120572(119905120573minus119905120573
1)minus 1)119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905
(7)
The total number of inventories carried during the interval[1199051 119879] say119867
119879 is
119867119879= int
1205831
1199051
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 119889119905
+ 1198630int
1205831
1199051
int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ 1198630int
1205832
1205831
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
(8)
The total of lost sales during the interval [0 1199051] say 119871
119879 is
119871119879= int
1199051
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905 (9)
Therefore the average total cost per unit time under thecondition 119905
1le 1205831can be given by
C1(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c3[int
1199051
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905]
+ c4[int
1199051
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905]
+ c1[int
1205831
1199051
(119890120572(119905120573minus119905120573
1)minus 1)119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
4 Mathematical Problems in Engineering
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905]
+ c2[int
1205831
1199051
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 119889119905
+ 1198630int
1205831
1199051
int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ 1198630int
1205832
1205831
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905]
(10)
Taking the first and the second order derivative of C1(1199051)with
respect to 1199051 respectively we have
119889C1(1199051)
1198891199051
=1
119879c3int
1199051
0
(1 minus 1205751199051+ 120575119905) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ c4120575int
1199051
0
119890120575(119905minus1199051)119891 (119905) 119889119905 minus (c
2+ c1120572120573119905120573minus1
1)
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119863 (119905) 119889119905
(11)
1198892C1(1199051)
1198891199052
1
=1
119879c3[119891 (1199051)
+ 120575int
1199051
0
(1205751199051minus 120575119905 minus 2) 119890
120575(119905minus1199051)119891 (119905) 119889119905] + c
4120575 [119891 (119905
1)
minus 120575int
1199051
0
119890120575(119905minus1199051)119891 (119905) 119889119905] + (c
1120572120573119905120573minus1
1+ c2) 119891 (1199051)
+ 120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119863 (119905) 119889119905
(12)
Then we have
119889C1(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=0
= minusc2
119879int
119879
0
119890120572119905120573
119863(119905) 119889119905 lt 0 (13)
and let
Δ1= c3int
1205831
0
(1 minus 1205751205831+ 120575119905) 119890
120575(119905minus1205831)119891 (119905) 119889119905
+ c4120575int
1205831
0
119890120575(119905minus120583
1)119891 (119905) 119889119905 minus (c
2+ c1120572120573120583120573minus1
1)
sdot (1198630int
1205832
1205831
119890120572(119905120573minus120583120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus120583120573
1)119892 (119905) 119889119905)
(14)
Therefore (i) if Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (0 120583
1) satisfying 119889C
1(1199051)1198891199051= 0
and C1(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt 0
and 1198892C1(1199051)1198891199052
1gt 0 it means that C
1(1199051) is a nonincreasing
function and obtains its minimum value at point 119905lowast1= 1205831
(iii) if the sign of 1198892C1(1199051)1198891199052
1is indefinite then the curve
of C1(1199051) is W-shape and C
1(1199051)may obtain its minimum at
more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
1(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= int
1205831
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905lowast
1
120573)119889119909
+ int
119879
1205832
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909
(15)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
119905lowast
1
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 (16)
Case 2 (1205831le 1199051le 1205832) If the replenishment time 119905
1isin [1205831 1205832]
then from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 1198630 119905
1lt 119905 lt 120583
2
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1205832lt 119905 lt 119879
(17)
Solving the differential equation (17) with 119868(0) = 119868(119879) = 0we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
minus1198630
120575(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 119905
1
1198630int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 120583
2
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
2lt 119905 lt 119879
(18)
Mathematical Problems in Engineering 5
From (18) the total replenishment quantity can be computed
119878 = 119868 (1199051) = 119863
0int
1205832
1199051
119890120572(119909120573minus119905120573
1)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (19)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905
(20)
The total number of perished items in the interval [1199051 119879] is
119863119879= 1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905
(21)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= 1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
(22)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
(23)
Therefore the average total cost per unit time under thecondition 120583
1lt 1199051le 1205832can be given by
C2(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1[1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905]
+ c2[1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905]
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905]
(24)
Taking the first and the second order derivative of C2(1199051)
with respect to 1199051 respectively we have
119889C2(1199051)
1198891199051
=1
119879c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630int
1199051
1205831
119890120575(119905minus1199051)119889119905] minus (c
1120572120573119905120573minus1
1+ c2)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 minus 1205751199051+ 120575119905) 119890
120575(119905minus1199051)119891 (119905) 119889119905
minus 1198630(1205831minus 1199051) 119890120575(1205831minus1199051)]
(25)
1198892C2(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1(c1(120572120573119905120573
1minus 120573 + 1) + c
21199051)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ (c1120572120573119905120573minus1
1+ c2)1198630+ c4120575 [1198630119890120575(1205831minus1199051)
minus 120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630(1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)]
(26)
Let
Δ2= c3[int
1205831
0
(1 minus 1205751205832+ 120575119905) 119890
120575(119905minus1205832)119891 (119905) 119889119905
+ 1198630(1205832minus 1205831) 119890120575(1205831minus1205832)]
+ c4120575 [int
1205831
0
119890120575(119905minus120583
2)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus120583
2)119889119905]
minus (c2+ c1120572120573120583120573minus1
2)int
119879
1205832
119890120572(119905120573minus120583120573
2)119892 (119905) 119889119905
(27)
6 Mathematical Problems in Engineering
Therefore (i) if Δ1lt 0 Δ
2gt 0 and 1198892C
2(1199051)1198891199052
1gt 0 there
exists a unique solution 119905lowast1isin (1205831 1205832) satisfying 119889C
2(1199051)1198891199051=
0 and C2(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt
0 Δ2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 it means that C
2(1199051) is
a nonincreasing function and obtains its minimum at point119905lowast
1= 1205832 (iii) if Δ
1ge 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 it
means that C2(1199051) is a nondecreasing function and obtains its
minimum at point 119905lowast1= 1205831 (iv) if the sign of 1198892C
2(1199051)1198891199052
1is
indefinite then the curve of C2(1199051) is W-shape and C
2(1199051)
may obtain its minimum value at more than one pointwithout loss of generality set 119905
1= 119905lowast
1 where 119905lowast
1is one of the
solutions of 119889C2(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= 1198630int
1205832
119905lowast
1
119890120572(119905120573minus119905lowast
1
120573)119889119905 + int
119879
1205832
119890120572(119905120573minus119905lowast
1
120573)119892 (119905) 119889119905 (28)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
119905lowast
1
1205831
119890120575(119905minus119905lowast
1)119889119905 (29)
Case 3 (1205832le 1199051lt 119879) If the replenishment time 119905
1isin [1205832 119879)
from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 120583
2
minus119890minus120575(1199051minus119905)119892 (119905) 120583
2lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1199051lt 119905 lt 119879
(30)
Solving (30) with 119868(0) = 119868(119879) = 0 we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(119905minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 120583
2
minus int
119905
1205832
119890120575(119909minus119905
1)119892 (119909) 119889119909 +
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
2lt 119905 lt 119905
1
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 119879
(31)
From the last equation of (31) the total replenishmentquantity can be computed
119878 = 119868 (1199051) = int
119879
1199051
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (32)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575[(1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051)]
(33)
The total number of perished items in the interval [1199051 119879] is
119863119879= int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909 (34)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905 (35)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905
(36)
Therefore the average total cost per unit time under thecondition 120583
2lt 1199051le 119879 can be given by
C3(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909
+ c2int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
Mathematical Problems in Engineering 7
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(37)
Taking the first and the second order derivative of C3(1199051)
with respect to 1199051 respectively we have
119889C3(1199051)
1198891199051
=1
119879minus (c2+ c1120572120573119905120573minus1
1) int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905
+ c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119892 (119905) 119889119905
+ 1198630((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(38)
1198892C3(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + (c
2+ c1120572120573119905120573minus1
1) 119892 (1199051)
+ c4120575 [120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0120575int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ 120575int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c2[120572120573119905120573minus1
1int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)
+ 1198630((1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)
minus (1 + 1205751205832minus 1205751199051) 119890120575(1205832minus1199051))]
(39)
We know that
119889C3(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=119879
=1
119879c4120575 [int
1205831
0
119890120575(119905minus119879)
119891 (119905) 119889119905
+ 1198630int
1205832
1205831
119890120575(119905minus119879)
119889119905 + int
119879
1205832
119890120575(119905minus119879)
119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119891 (119905) 119889119905
+ int
119879
1205832
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119892 (119905) 119889119905
+ 1198630((119879 minus 120583
1) 119890120575(1205831minus119879)
minus (119879 minus 1205832) 119890120575(1205832minus119879)
)]
gt 0
(40)
Therefore (i) if Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (1205832 119879) satisfying 119889C
3(1199051)1198891199051= 0
and C3(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
2ge 0
and 1198892C3(1199051)1198891199052
1gt 0 it means that C
3(1199051) is a nondecreasing
function and obtains its minimum value at point 119905lowast1= 1205832
(iii) if the sign of 1198892C3(1199051)1198891199052
1is indefinite then the curve of
C3(1199051) is W-shape and C
3(1199051)may obtain its minimum value
at more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
3(1199051)1198891199051= 0
Therefore the optimal value of the order level 119878 = 119868(119905lowast
1)
is
119878lowast= int
119879
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909 (41)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus119905lowast
1)119889119905
+ int
119905lowast
1
1205832
119890120575(119905minus119905lowast
1)119892 (119905) 119889119905
(42)
From the above analysis we obtain that the total averagecost of the system over the interval [0 119879] is
TC (1199051) =
C1(1199051) 0 lt 119905
1le 1205831
C2(1199051) 120583
1lt 1199051le 1205832
C3(1199051) 120583
2lt 1199051lt 119879
(43)
where C1(1199051) C2(1199051) and C
3(1199051) are obtained from (10) (24)
and (37) respectively
8 Mathematical Problems in Engineering
Basing on the analyzed results above we provide theresults which ensure the existence of a unique 119905
1 say 119905lowast
1 to
minimize the total average cost for the model system startingwith shortages The procedure is as follows
Step 0 Input the parameters 1205831 1205832 119879 1198861 1198862 1198871 120572 120573 120575
1198630 and c
119894 119894 = 1 2 3 4
Step 1 Calculate Δ1and 1198892C
1(1199051)1198891199052
1
Step 11 If Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 solve a unique 119905lowast
1
from119889C1(1199051)1198891199051= 0 and the total average cost and the order
quantity can be obtained by (10) and (16) respectively
Step 12 If Δ1le 0 and 119889
2C1(1199051)1198891199052
1gt 0 then the optimal
average cost and the optimal order quantity can be obtainedby (10) and (16) at 119905lowast
1= 1205831 respectively
Step 13 If the sign of 1198892C1(1199051)1198891199052
1is indefinite then solve
equation 119889C1(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C1(1199051) to obtain the optimal
replenishment point compute the order quantity by (16)
Step 2 Calculate Δ2and 1198892C
2(1199051)1198891199052
1
Step 21 If Δ1lt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
seek a unique 119905lowast1from equation 119889C
2(1199051)1198891199051= 0 and the
optimal total average cost and the optimal order quantity canbe obtained by (24) and (29) at 119905
1= 119905lowast
1 respectively
Step 22 If Δ1gt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205831 and compute the order
quantity by (29)
Step 23 If Δ1lt 0 Δ
2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205832 and compute the order
quantity by (29)
Step 24 If the sign of 1198892C2(1199051)1198891199052
1is indefinite then solve
equation 119889C2(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C2(1199051) to obtain the optimal
replenishment point compute the optimal order quantity by(29)
Step 3 Calculate 1198892C3(1199051)1198891199052
1
Step 31 If Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 seek the unique
119905lowast
1from equation 119889C
3(1199051)1198891199051= 0 and the total average
cost and the order quantity can be obtained by (37) and (42)respectively
Step 32 If Δ2ge 0 and 1198892C
3(1199051)1198891199052
1gt 0 then C
3(1199051) obtains
its minimum at 119905lowast1= 1205832 and compute the order quantity by
(42)
Step 33 If the sign of 1198892C3(1199051)1198891199052
1is indefinite then
solve equation 119889C3(1199051)1198891199051
= 0 If the solution is notunique by comparing their average costs C
3(1199051) to obtain the
optimal replenishment point compute the order quantity by(42)
Step 4 Find TC(119905lowast1) = minC
1(119905lowast
1)C2(119905lowast
1)C3(119905lowast
1) and
accordingly select the optimum 119905lowast
1 and then obtain corre-
sponding optimal order quantity 119876lowast
4 Numerical Examples andSensitivity Analysis
In order to demonstrate the procedure to obtain the optimalsolution of the model we present four examples for themodel Examples are based on piece-wise demand rate suchas 119891(119905) = 119886
1+ 1198871119905 and 119892(119905) = 119886
2119890minus1198872119905
Example 1 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 004 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $2 c
2= $3 c
3= $12 c
4= $8
Since Δ1
= 18809 gt 0 Δ2
= 222869 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1) 1198892C
2(1199051)1198891199052
1gt 0
for all 1199051
isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
2(1205831) = 763891 in
[1205831 1205832) and C
3(1199062) = 126358 in [120583
2 119879) By solving equation
119889C1(1199051)1198891199051= 0 we have 119905lowast
1= 3664 From (10) we obtain
C1(119905lowast
1) = 762759 Therefore the optimal ordering quantity is
119876lowast= 637839 and the total average cost is TC(119905lowast
1) = 762759
Example 2 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus257889 lt 0 Δ2
= 139269 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin (1205832 119879)
then we get local minimum value C1(1205831) = 1805883 in
(0 1205831] and C
3(1199062) = 158232 in [120583
2 119879) By solving equation
119889C2(1199051)1198891199051= 0 we have 119905lowast
1= 6559 From (24) we obtain
C2(119905lowast
1) = 1480324 Therefore the optimal ordering quantity
is 119876lowast = 571827 and the total average cost is TC(119905lowast1) =
1480324
Example 3 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 2 120575 = 006 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus249329 lt 0 Δ2
= minus121869 lt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin [1205832 119879)
then we get local minimum value C1(1205831) = 1417134 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 9120 From (37) we obtain
C3(119905lowast
1) = 889767 Therefore the optimal ordering quantity is
119876lowast= 278960 and the total average cost is TC(119905lowast
1) = 889767
Example 4 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 16 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus197263 lt 0 Δ2
= minus86675 lt 01198892C4(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0
Mathematical Problems in Engineering 9
Table 1 The sensitivity of 120575 for the model in Example 1
120575 0 001 002 003 004 005 006 007 008119905lowast
134366 34905 35463 36042 36642 37263 37907 38574 39265
119878lowast 5150829 5150829 5114210 5075965 5036066 4994433 4950985 4905641 4858321119876lowast 6481034 6481034 6448369 6414184 6378392 6340900 6301614 6260436 6217267
TC2(119905lowast
1) 7752147 7718437 7684699 7649983 7614252 7577473 7539611 750063 7460498
Table 2 The sensitivity of 120572 for the models in Example 1
120572 0 0001 0002 0003 0004 0005 0006 0007 0008119905lowast
130141 31325 32569 33870 35229 36642 38105 39614 41171
119878lowast 4158882 4320629 4489624 4665543 4847907 5036066 5229206 5426345 5755923119876lowast 5228982 5439279 5659745 5890106 6129883 6378392 6634729 6897782 7230000
TC2(119905lowast
1) 6385314 6614087 6851585 7097639 7351984 7614252 7883972 8160570 8450147
Table 3 The sensitivity of 120573 for the models in Example 1
120573 14 15 16 17 18 19 20 21 22119905lowast
131114 31995 32513 33176 340332 35148 36613 38551 41128
119878lowast 4275053 4401612 4471691 4562029 4679084 4831709 5036066 5295334 5641669119876lowast 5384988 5547959 5639494 5757494 5910495 6110248 6378392 6720262 7179964
TC2(119905lowast
1) 6444596 6703559 6801679 6929359 7095918 7314743 7614252 7976508 8450147
for all 1199051
isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
1(1205831) = 125259 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 8627 From (37) we obtain
TC(119905lowast1) = 842935 Therefore the optimal ordering quantity
is 119876lowast = 241524 obtained from (42) and the minimum costTC(119905lowast1) = 842935
In order to clearly indicate the effects of parameterssuch as 120575 120572 and 120573 on 119878
lowast 119876lowast and TC119894(119905lowast
1) (119894 = 1 2)
respectively we study the sensitivity of different parametersunder the studied inventory modelThe sensitivity analysis isperformed on the base of Example 1 and the results are shownin Tables 1ndash3 respectively
From Table 1 we observe that inventory level 119878lowast orderquantity 119876
lowast and the total average cost TC119894(119905lowast
1) (119894 = 1 2)
gradually decrease as the shortage parameter 120575 increasesfor the model and the replenishment time 119905lowast
1increases as
120575 increases for the model starting with shortage Also weobserve that the values of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) are slightly
sensitive to the changes of 120575From Table 2 we find that when the deterioration
parameter 120572 increases then 119878lowast 119876lowast and TC(119905lowast1) increase for
both models and the replenishment time 119905lowast1increases for the
model starting with shortage We see that the values of 119905lowast1 119878lowast
119876lowast and TC(119905lowast
1) all are temperately sensitive to the changes of
120572 for the modelFrom Table 3 we find that 119878lowast 119876lowast and TC(119905lowast
1) increase
as the deterioration parameter 120573 increases for both modelsAlso we find that the replenishment time 119905lowast
1increases as 120573
increases for themodel startingwith shortageWe see that thevalues of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) all are moderately sensitive
to the changes of 120573 for the model
5 Conclusion
This paper studies an inventory model for Weibull-distributed deterioration with trapezoidal type demand rateand partial backlogging First an optimal replenishmentstrategy for deteriorating items with the model starting withshortage is proposedThen numerical examples are providedto illustrate the theoretical results From the analysis of Tables1ndash3 it can be found that the replenishment time point 119905lowast
1
order quantity 119876lowast and the total average cost TC(119905lowast
1) are
moderately sensitive to the changes of 120572 and 120573 and lowlysensitive to the changes of 120575 respectively However if theorder cycle (119879) is a decision variable such inventory modelwill be of more practical significance which provides us aninteresting topic in future
Competing Interests
The author declares that there are no competing interests
Acknowledgments
The research is supported partly by National Natural ScienceFoundation of China (Grant no 71502100)
References
[1] P M Ghare and G F Schrader ldquoA model for exponentiallydecaying inventoriesrdquo Journal of Industrial Engineering vol 14no 3 pp 238ndash243 1963
[2] R P Covert and G C Philip ldquoAn EOQ model for items withweibull distribution deteriorationrdquo AIIE Transaction vol 5 no4 pp 323ndash326 1973
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 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
Mathematical Problems in Engineering 3
assumptions mentioned above the model is described asshown in Figure 1
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119863 (119905) 0 lt 119905 lt 119905
1
minus120579 (119905) 119868 (119905) minus 119863 (119905) 1199051lt 119905 lt 119879
(2)
with boundary condition 119868(0) = 0 119868(1199051) = 119878 119868(119879) = 0
Based on the values of 1199051 1205831 and 120583
2 three possible cases
are presented as follows
Case 1 (0 lt 1199051le 1205831) If the replenishment time 119905
1isin (0 120583
1]
from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 119891 (119905) 1199051lt 119905 lt 120583
1
minus120572120573119905120573minus1
119868 (119905) minus 1198630 120583
1lt 119905 lt 120583
2
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1205832lt 119905 lt 119879
(3)
Solving the differential equations (3) with 119868(0) = 0 we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 119905
1
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 120583
1
1198630int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
1lt 119905 lt 120583
2
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
2lt 119905 lt 119879
(4)
The total replenishment quantity can be computed as
119878 = 119868 (1199051)
= int
1205831
1199051
119890120572(119909120573minus119905120573
1)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905120573
1)119889119909
+ int
119879
1205832
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909
(5)
The total shortage quantity 119861119879during the interval [0 119905
1) is
119861119879= int
1199051
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905 (6)
The total number of perished items during the interval [1199051 119879]
say119863119879 is
119863119879= int
1205831
1199051
(119890120572(119905120573minus119905120573
1)minus 1)119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905
(7)
The total number of inventories carried during the interval[1199051 119879] say119867
119879 is
119867119879= int
1205831
1199051
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 119889119905
+ 1198630int
1205831
1199051
int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ 1198630int
1205832
1205831
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
(8)
The total of lost sales during the interval [0 1199051] say 119871
119879 is
119871119879= int
1199051
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905 (9)
Therefore the average total cost per unit time under thecondition 119905
1le 1205831can be given by
C1(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c3[int
1199051
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905]
+ c4[int
1199051
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905]
+ c1[int
1205831
1199051
(119890120572(119905120573minus119905120573
1)minus 1)119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
4 Mathematical Problems in Engineering
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905]
+ c2[int
1205831
1199051
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 119889119905
+ 1198630int
1205831
1199051
int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ 1198630int
1205832
1205831
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905]
(10)
Taking the first and the second order derivative of C1(1199051)with
respect to 1199051 respectively we have
119889C1(1199051)
1198891199051
=1
119879c3int
1199051
0
(1 minus 1205751199051+ 120575119905) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ c4120575int
1199051
0
119890120575(119905minus1199051)119891 (119905) 119889119905 minus (c
2+ c1120572120573119905120573minus1
1)
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119863 (119905) 119889119905
(11)
1198892C1(1199051)
1198891199052
1
=1
119879c3[119891 (1199051)
+ 120575int
1199051
0
(1205751199051minus 120575119905 minus 2) 119890
120575(119905minus1199051)119891 (119905) 119889119905] + c
4120575 [119891 (119905
1)
minus 120575int
1199051
0
119890120575(119905minus1199051)119891 (119905) 119889119905] + (c
1120572120573119905120573minus1
1+ c2) 119891 (1199051)
+ 120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119863 (119905) 119889119905
(12)
Then we have
119889C1(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=0
= minusc2
119879int
119879
0
119890120572119905120573
119863(119905) 119889119905 lt 0 (13)
and let
Δ1= c3int
1205831
0
(1 minus 1205751205831+ 120575119905) 119890
120575(119905minus1205831)119891 (119905) 119889119905
+ c4120575int
1205831
0
119890120575(119905minus120583
1)119891 (119905) 119889119905 minus (c
2+ c1120572120573120583120573minus1
1)
sdot (1198630int
1205832
1205831
119890120572(119905120573minus120583120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus120583120573
1)119892 (119905) 119889119905)
(14)
Therefore (i) if Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (0 120583
1) satisfying 119889C
1(1199051)1198891199051= 0
and C1(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt 0
and 1198892C1(1199051)1198891199052
1gt 0 it means that C
1(1199051) is a nonincreasing
function and obtains its minimum value at point 119905lowast1= 1205831
(iii) if the sign of 1198892C1(1199051)1198891199052
1is indefinite then the curve
of C1(1199051) is W-shape and C
1(1199051)may obtain its minimum at
more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
1(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= int
1205831
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905lowast
1
120573)119889119909
+ int
119879
1205832
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909
(15)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
119905lowast
1
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 (16)
Case 2 (1205831le 1199051le 1205832) If the replenishment time 119905
1isin [1205831 1205832]
then from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 1198630 119905
1lt 119905 lt 120583
2
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1205832lt 119905 lt 119879
(17)
Solving the differential equation (17) with 119868(0) = 119868(119879) = 0we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
minus1198630
120575(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 119905
1
1198630int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 120583
2
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
2lt 119905 lt 119879
(18)
Mathematical Problems in Engineering 5
From (18) the total replenishment quantity can be computed
119878 = 119868 (1199051) = 119863
0int
1205832
1199051
119890120572(119909120573minus119905120573
1)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (19)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905
(20)
The total number of perished items in the interval [1199051 119879] is
119863119879= 1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905
(21)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= 1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
(22)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
(23)
Therefore the average total cost per unit time under thecondition 120583
1lt 1199051le 1205832can be given by
C2(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1[1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905]
+ c2[1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905]
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905]
(24)
Taking the first and the second order derivative of C2(1199051)
with respect to 1199051 respectively we have
119889C2(1199051)
1198891199051
=1
119879c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630int
1199051
1205831
119890120575(119905minus1199051)119889119905] minus (c
1120572120573119905120573minus1
1+ c2)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 minus 1205751199051+ 120575119905) 119890
120575(119905minus1199051)119891 (119905) 119889119905
minus 1198630(1205831minus 1199051) 119890120575(1205831minus1199051)]
(25)
1198892C2(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1(c1(120572120573119905120573
1minus 120573 + 1) + c
21199051)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ (c1120572120573119905120573minus1
1+ c2)1198630+ c4120575 [1198630119890120575(1205831minus1199051)
minus 120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630(1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)]
(26)
Let
Δ2= c3[int
1205831
0
(1 minus 1205751205832+ 120575119905) 119890
120575(119905minus1205832)119891 (119905) 119889119905
+ 1198630(1205832minus 1205831) 119890120575(1205831minus1205832)]
+ c4120575 [int
1205831
0
119890120575(119905minus120583
2)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus120583
2)119889119905]
minus (c2+ c1120572120573120583120573minus1
2)int
119879
1205832
119890120572(119905120573minus120583120573
2)119892 (119905) 119889119905
(27)
6 Mathematical Problems in Engineering
Therefore (i) if Δ1lt 0 Δ
2gt 0 and 1198892C
2(1199051)1198891199052
1gt 0 there
exists a unique solution 119905lowast1isin (1205831 1205832) satisfying 119889C
2(1199051)1198891199051=
0 and C2(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt
0 Δ2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 it means that C
2(1199051) is
a nonincreasing function and obtains its minimum at point119905lowast
1= 1205832 (iii) if Δ
1ge 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 it
means that C2(1199051) is a nondecreasing function and obtains its
minimum at point 119905lowast1= 1205831 (iv) if the sign of 1198892C
2(1199051)1198891199052
1is
indefinite then the curve of C2(1199051) is W-shape and C
2(1199051)
may obtain its minimum value at more than one pointwithout loss of generality set 119905
1= 119905lowast
1 where 119905lowast
1is one of the
solutions of 119889C2(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= 1198630int
1205832
119905lowast
1
119890120572(119905120573minus119905lowast
1
120573)119889119905 + int
119879
1205832
119890120572(119905120573minus119905lowast
1
120573)119892 (119905) 119889119905 (28)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
119905lowast
1
1205831
119890120575(119905minus119905lowast
1)119889119905 (29)
Case 3 (1205832le 1199051lt 119879) If the replenishment time 119905
1isin [1205832 119879)
from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 120583
2
minus119890minus120575(1199051minus119905)119892 (119905) 120583
2lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1199051lt 119905 lt 119879
(30)
Solving (30) with 119868(0) = 119868(119879) = 0 we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(119905minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 120583
2
minus int
119905
1205832
119890120575(119909minus119905
1)119892 (119909) 119889119909 +
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
2lt 119905 lt 119905
1
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 119879
(31)
From the last equation of (31) the total replenishmentquantity can be computed
119878 = 119868 (1199051) = int
119879
1199051
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (32)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575[(1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051)]
(33)
The total number of perished items in the interval [1199051 119879] is
119863119879= int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909 (34)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905 (35)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905
(36)
Therefore the average total cost per unit time under thecondition 120583
2lt 1199051le 119879 can be given by
C3(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909
+ c2int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
Mathematical Problems in Engineering 7
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(37)
Taking the first and the second order derivative of C3(1199051)
with respect to 1199051 respectively we have
119889C3(1199051)
1198891199051
=1
119879minus (c2+ c1120572120573119905120573minus1
1) int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905
+ c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119892 (119905) 119889119905
+ 1198630((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(38)
1198892C3(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + (c
2+ c1120572120573119905120573minus1
1) 119892 (1199051)
+ c4120575 [120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0120575int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ 120575int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c2[120572120573119905120573minus1
1int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)
+ 1198630((1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)
minus (1 + 1205751205832minus 1205751199051) 119890120575(1205832minus1199051))]
(39)
We know that
119889C3(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=119879
=1
119879c4120575 [int
1205831
0
119890120575(119905minus119879)
119891 (119905) 119889119905
+ 1198630int
1205832
1205831
119890120575(119905minus119879)
119889119905 + int
119879
1205832
119890120575(119905minus119879)
119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119891 (119905) 119889119905
+ int
119879
1205832
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119892 (119905) 119889119905
+ 1198630((119879 minus 120583
1) 119890120575(1205831minus119879)
minus (119879 minus 1205832) 119890120575(1205832minus119879)
)]
gt 0
(40)
Therefore (i) if Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (1205832 119879) satisfying 119889C
3(1199051)1198891199051= 0
and C3(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
2ge 0
and 1198892C3(1199051)1198891199052
1gt 0 it means that C
3(1199051) is a nondecreasing
function and obtains its minimum value at point 119905lowast1= 1205832
(iii) if the sign of 1198892C3(1199051)1198891199052
1is indefinite then the curve of
C3(1199051) is W-shape and C
3(1199051)may obtain its minimum value
at more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
3(1199051)1198891199051= 0
Therefore the optimal value of the order level 119878 = 119868(119905lowast
1)
is
119878lowast= int
119879
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909 (41)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus119905lowast
1)119889119905
+ int
119905lowast
1
1205832
119890120575(119905minus119905lowast
1)119892 (119905) 119889119905
(42)
From the above analysis we obtain that the total averagecost of the system over the interval [0 119879] is
TC (1199051) =
C1(1199051) 0 lt 119905
1le 1205831
C2(1199051) 120583
1lt 1199051le 1205832
C3(1199051) 120583
2lt 1199051lt 119879
(43)
where C1(1199051) C2(1199051) and C
3(1199051) are obtained from (10) (24)
and (37) respectively
8 Mathematical Problems in Engineering
Basing on the analyzed results above we provide theresults which ensure the existence of a unique 119905
1 say 119905lowast
1 to
minimize the total average cost for the model system startingwith shortages The procedure is as follows
Step 0 Input the parameters 1205831 1205832 119879 1198861 1198862 1198871 120572 120573 120575
1198630 and c
119894 119894 = 1 2 3 4
Step 1 Calculate Δ1and 1198892C
1(1199051)1198891199052
1
Step 11 If Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 solve a unique 119905lowast
1
from119889C1(1199051)1198891199051= 0 and the total average cost and the order
quantity can be obtained by (10) and (16) respectively
Step 12 If Δ1le 0 and 119889
2C1(1199051)1198891199052
1gt 0 then the optimal
average cost and the optimal order quantity can be obtainedby (10) and (16) at 119905lowast
1= 1205831 respectively
Step 13 If the sign of 1198892C1(1199051)1198891199052
1is indefinite then solve
equation 119889C1(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C1(1199051) to obtain the optimal
replenishment point compute the order quantity by (16)
Step 2 Calculate Δ2and 1198892C
2(1199051)1198891199052
1
Step 21 If Δ1lt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
seek a unique 119905lowast1from equation 119889C
2(1199051)1198891199051= 0 and the
optimal total average cost and the optimal order quantity canbe obtained by (24) and (29) at 119905
1= 119905lowast
1 respectively
Step 22 If Δ1gt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205831 and compute the order
quantity by (29)
Step 23 If Δ1lt 0 Δ
2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205832 and compute the order
quantity by (29)
Step 24 If the sign of 1198892C2(1199051)1198891199052
1is indefinite then solve
equation 119889C2(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C2(1199051) to obtain the optimal
replenishment point compute the optimal order quantity by(29)
Step 3 Calculate 1198892C3(1199051)1198891199052
1
Step 31 If Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 seek the unique
119905lowast
1from equation 119889C
3(1199051)1198891199051= 0 and the total average
cost and the order quantity can be obtained by (37) and (42)respectively
Step 32 If Δ2ge 0 and 1198892C
3(1199051)1198891199052
1gt 0 then C
3(1199051) obtains
its minimum at 119905lowast1= 1205832 and compute the order quantity by
(42)
Step 33 If the sign of 1198892C3(1199051)1198891199052
1is indefinite then
solve equation 119889C3(1199051)1198891199051
= 0 If the solution is notunique by comparing their average costs C
3(1199051) to obtain the
optimal replenishment point compute the order quantity by(42)
Step 4 Find TC(119905lowast1) = minC
1(119905lowast
1)C2(119905lowast
1)C3(119905lowast
1) and
accordingly select the optimum 119905lowast
1 and then obtain corre-
sponding optimal order quantity 119876lowast
4 Numerical Examples andSensitivity Analysis
In order to demonstrate the procedure to obtain the optimalsolution of the model we present four examples for themodel Examples are based on piece-wise demand rate suchas 119891(119905) = 119886
1+ 1198871119905 and 119892(119905) = 119886
2119890minus1198872119905
Example 1 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 004 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $2 c
2= $3 c
3= $12 c
4= $8
Since Δ1
= 18809 gt 0 Δ2
= 222869 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1) 1198892C
2(1199051)1198891199052
1gt 0
for all 1199051
isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
2(1205831) = 763891 in
[1205831 1205832) and C
3(1199062) = 126358 in [120583
2 119879) By solving equation
119889C1(1199051)1198891199051= 0 we have 119905lowast
1= 3664 From (10) we obtain
C1(119905lowast
1) = 762759 Therefore the optimal ordering quantity is
119876lowast= 637839 and the total average cost is TC(119905lowast
1) = 762759
Example 2 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus257889 lt 0 Δ2
= 139269 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin (1205832 119879)
then we get local minimum value C1(1205831) = 1805883 in
(0 1205831] and C
3(1199062) = 158232 in [120583
2 119879) By solving equation
119889C2(1199051)1198891199051= 0 we have 119905lowast
1= 6559 From (24) we obtain
C2(119905lowast
1) = 1480324 Therefore the optimal ordering quantity
is 119876lowast = 571827 and the total average cost is TC(119905lowast1) =
1480324
Example 3 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 2 120575 = 006 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus249329 lt 0 Δ2
= minus121869 lt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin [1205832 119879)
then we get local minimum value C1(1205831) = 1417134 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 9120 From (37) we obtain
C3(119905lowast
1) = 889767 Therefore the optimal ordering quantity is
119876lowast= 278960 and the total average cost is TC(119905lowast
1) = 889767
Example 4 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 16 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus197263 lt 0 Δ2
= minus86675 lt 01198892C4(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0
Mathematical Problems in Engineering 9
Table 1 The sensitivity of 120575 for the model in Example 1
120575 0 001 002 003 004 005 006 007 008119905lowast
134366 34905 35463 36042 36642 37263 37907 38574 39265
119878lowast 5150829 5150829 5114210 5075965 5036066 4994433 4950985 4905641 4858321119876lowast 6481034 6481034 6448369 6414184 6378392 6340900 6301614 6260436 6217267
TC2(119905lowast
1) 7752147 7718437 7684699 7649983 7614252 7577473 7539611 750063 7460498
Table 2 The sensitivity of 120572 for the models in Example 1
120572 0 0001 0002 0003 0004 0005 0006 0007 0008119905lowast
130141 31325 32569 33870 35229 36642 38105 39614 41171
119878lowast 4158882 4320629 4489624 4665543 4847907 5036066 5229206 5426345 5755923119876lowast 5228982 5439279 5659745 5890106 6129883 6378392 6634729 6897782 7230000
TC2(119905lowast
1) 6385314 6614087 6851585 7097639 7351984 7614252 7883972 8160570 8450147
Table 3 The sensitivity of 120573 for the models in Example 1
120573 14 15 16 17 18 19 20 21 22119905lowast
131114 31995 32513 33176 340332 35148 36613 38551 41128
119878lowast 4275053 4401612 4471691 4562029 4679084 4831709 5036066 5295334 5641669119876lowast 5384988 5547959 5639494 5757494 5910495 6110248 6378392 6720262 7179964
TC2(119905lowast
1) 6444596 6703559 6801679 6929359 7095918 7314743 7614252 7976508 8450147
for all 1199051
isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
1(1205831) = 125259 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 8627 From (37) we obtain
TC(119905lowast1) = 842935 Therefore the optimal ordering quantity
is 119876lowast = 241524 obtained from (42) and the minimum costTC(119905lowast1) = 842935
In order to clearly indicate the effects of parameterssuch as 120575 120572 and 120573 on 119878
lowast 119876lowast and TC119894(119905lowast
1) (119894 = 1 2)
respectively we study the sensitivity of different parametersunder the studied inventory modelThe sensitivity analysis isperformed on the base of Example 1 and the results are shownin Tables 1ndash3 respectively
From Table 1 we observe that inventory level 119878lowast orderquantity 119876
lowast and the total average cost TC119894(119905lowast
1) (119894 = 1 2)
gradually decrease as the shortage parameter 120575 increasesfor the model and the replenishment time 119905lowast
1increases as
120575 increases for the model starting with shortage Also weobserve that the values of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) are slightly
sensitive to the changes of 120575From Table 2 we find that when the deterioration
parameter 120572 increases then 119878lowast 119876lowast and TC(119905lowast1) increase for
both models and the replenishment time 119905lowast1increases for the
model starting with shortage We see that the values of 119905lowast1 119878lowast
119876lowast and TC(119905lowast
1) all are temperately sensitive to the changes of
120572 for the modelFrom Table 3 we find that 119878lowast 119876lowast and TC(119905lowast
1) increase
as the deterioration parameter 120573 increases for both modelsAlso we find that the replenishment time 119905lowast
1increases as 120573
increases for themodel startingwith shortageWe see that thevalues of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) all are moderately sensitive
to the changes of 120573 for the model
5 Conclusion
This paper studies an inventory model for Weibull-distributed deterioration with trapezoidal type demand rateand partial backlogging First an optimal replenishmentstrategy for deteriorating items with the model starting withshortage is proposedThen numerical examples are providedto illustrate the theoretical results From the analysis of Tables1ndash3 it can be found that the replenishment time point 119905lowast
1
order quantity 119876lowast and the total average cost TC(119905lowast
1) are
moderately sensitive to the changes of 120572 and 120573 and lowlysensitive to the changes of 120575 respectively However if theorder cycle (119879) is a decision variable such inventory modelwill be of more practical significance which provides us aninteresting topic in future
Competing Interests
The author declares that there are no competing interests
Acknowledgments
The research is supported partly by National Natural ScienceFoundation of China (Grant no 71502100)
References
[1] P M Ghare and G F Schrader ldquoA model for exponentiallydecaying inventoriesrdquo Journal of Industrial Engineering vol 14no 3 pp 238ndash243 1963
[2] R P Covert and G C Philip ldquoAn EOQ model for items withweibull distribution deteriorationrdquo AIIE Transaction vol 5 no4 pp 323ndash326 1973
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 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
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Mathematical Problems in Engineering
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905]
+ c2[int
1205831
1199051
int
1205831
119905
119890120572(119909120573minus119905120573)119891 (119909) 119889119909 119889119905
+ 1198630int
1205831
1199051
int
1205832
1205831
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ 1198630int
1205832
1205831
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905]
(10)
Taking the first and the second order derivative of C1(1199051)with
respect to 1199051 respectively we have
119889C1(1199051)
1198891199051
=1
119879c3int
1199051
0
(1 minus 1205751199051+ 120575119905) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ c4120575int
1199051
0
119890120575(119905minus1199051)119891 (119905) 119889119905 minus (c
2+ c1120572120573119905120573minus1
1)
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119863 (119905) 119889119905
(11)
1198892C1(1199051)
1198891199052
1
=1
119879c3[119891 (1199051)
+ 120575int
1199051
0
(1205751199051minus 120575119905 minus 2) 119890
120575(119905minus1199051)119891 (119905) 119889119905] + c
4120575 [119891 (119905
1)
minus 120575int
1199051
0
119890120575(119905minus1199051)119891 (119905) 119889119905] + (c
1120572120573119905120573minus1
1+ c2) 119891 (1199051)
+ 120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119863 (119905) 119889119905
(12)
Then we have
119889C1(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=0
= minusc2
119879int
119879
0
119890120572119905120573
119863(119905) 119889119905 lt 0 (13)
and let
Δ1= c3int
1205831
0
(1 minus 1205751205831+ 120575119905) 119890
120575(119905minus1205831)119891 (119905) 119889119905
+ c4120575int
1205831
0
119890120575(119905minus120583
1)119891 (119905) 119889119905 minus (c
2+ c1120572120573120583120573minus1
1)
sdot (1198630int
1205832
1205831
119890120572(119905120573minus120583120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus120583120573
1)119892 (119905) 119889119905)
(14)
Therefore (i) if Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (0 120583
1) satisfying 119889C
1(1199051)1198891199051= 0
and C1(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt 0
and 1198892C1(1199051)1198891199052
1gt 0 it means that C
1(1199051) is a nonincreasing
function and obtains its minimum value at point 119905lowast1= 1205831
(iii) if the sign of 1198892C1(1199051)1198891199052
1is indefinite then the curve
of C1(1199051) is W-shape and C
1(1199051)may obtain its minimum at
more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
1(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= int
1205831
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119891 (119909) 119889119909 + 119863
0int
1205832
1205831
119890120572(119909120573minus119905lowast
1
120573)119889119909
+ int
119879
1205832
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909
(15)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
119905lowast
1
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 (16)
Case 2 (1205831le 1199051le 1205832) If the replenishment time 119905
1isin [1205831 1205832]
then from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 1198630 119905
1lt 119905 lt 120583
2
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1205832lt 119905 lt 119879
(17)
Solving the differential equation (17) with 119868(0) = 119868(119879) = 0we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
minus1198630
120575(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 119905
1
1198630int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 120583
2
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 120583
2lt 119905 lt 119879
(18)
Mathematical Problems in Engineering 5
From (18) the total replenishment quantity can be computed
119878 = 119868 (1199051) = 119863
0int
1205832
1199051
119890120572(119909120573minus119905120573
1)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (19)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905
(20)
The total number of perished items in the interval [1199051 119879] is
119863119879= 1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905
(21)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= 1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
(22)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
(23)
Therefore the average total cost per unit time under thecondition 120583
1lt 1199051le 1205832can be given by
C2(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1[1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905]
+ c2[1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905]
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905]
(24)
Taking the first and the second order derivative of C2(1199051)
with respect to 1199051 respectively we have
119889C2(1199051)
1198891199051
=1
119879c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630int
1199051
1205831
119890120575(119905minus1199051)119889119905] minus (c
1120572120573119905120573minus1
1+ c2)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 minus 1205751199051+ 120575119905) 119890
120575(119905minus1199051)119891 (119905) 119889119905
minus 1198630(1205831minus 1199051) 119890120575(1205831minus1199051)]
(25)
1198892C2(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1(c1(120572120573119905120573
1minus 120573 + 1) + c
21199051)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ (c1120572120573119905120573minus1
1+ c2)1198630+ c4120575 [1198630119890120575(1205831minus1199051)
minus 120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630(1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)]
(26)
Let
Δ2= c3[int
1205831
0
(1 minus 1205751205832+ 120575119905) 119890
120575(119905minus1205832)119891 (119905) 119889119905
+ 1198630(1205832minus 1205831) 119890120575(1205831minus1205832)]
+ c4120575 [int
1205831
0
119890120575(119905minus120583
2)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus120583
2)119889119905]
minus (c2+ c1120572120573120583120573minus1
2)int
119879
1205832
119890120572(119905120573minus120583120573
2)119892 (119905) 119889119905
(27)
6 Mathematical Problems in Engineering
Therefore (i) if Δ1lt 0 Δ
2gt 0 and 1198892C
2(1199051)1198891199052
1gt 0 there
exists a unique solution 119905lowast1isin (1205831 1205832) satisfying 119889C
2(1199051)1198891199051=
0 and C2(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt
0 Δ2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 it means that C
2(1199051) is
a nonincreasing function and obtains its minimum at point119905lowast
1= 1205832 (iii) if Δ
1ge 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 it
means that C2(1199051) is a nondecreasing function and obtains its
minimum at point 119905lowast1= 1205831 (iv) if the sign of 1198892C
2(1199051)1198891199052
1is
indefinite then the curve of C2(1199051) is W-shape and C
2(1199051)
may obtain its minimum value at more than one pointwithout loss of generality set 119905
1= 119905lowast
1 where 119905lowast
1is one of the
solutions of 119889C2(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= 1198630int
1205832
119905lowast
1
119890120572(119905120573minus119905lowast
1
120573)119889119905 + int
119879
1205832
119890120572(119905120573minus119905lowast
1
120573)119892 (119905) 119889119905 (28)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
119905lowast
1
1205831
119890120575(119905minus119905lowast
1)119889119905 (29)
Case 3 (1205832le 1199051lt 119879) If the replenishment time 119905
1isin [1205832 119879)
from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 120583
2
minus119890minus120575(1199051minus119905)119892 (119905) 120583
2lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1199051lt 119905 lt 119879
(30)
Solving (30) with 119868(0) = 119868(119879) = 0 we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(119905minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 120583
2
minus int
119905
1205832
119890120575(119909minus119905
1)119892 (119909) 119889119909 +
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
2lt 119905 lt 119905
1
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 119879
(31)
From the last equation of (31) the total replenishmentquantity can be computed
119878 = 119868 (1199051) = int
119879
1199051
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (32)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575[(1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051)]
(33)
The total number of perished items in the interval [1199051 119879] is
119863119879= int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909 (34)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905 (35)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905
(36)
Therefore the average total cost per unit time under thecondition 120583
2lt 1199051le 119879 can be given by
C3(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909
+ c2int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
Mathematical Problems in Engineering 7
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(37)
Taking the first and the second order derivative of C3(1199051)
with respect to 1199051 respectively we have
119889C3(1199051)
1198891199051
=1
119879minus (c2+ c1120572120573119905120573minus1
1) int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905
+ c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119892 (119905) 119889119905
+ 1198630((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(38)
1198892C3(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + (c
2+ c1120572120573119905120573minus1
1) 119892 (1199051)
+ c4120575 [120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0120575int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ 120575int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c2[120572120573119905120573minus1
1int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)
+ 1198630((1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)
minus (1 + 1205751205832minus 1205751199051) 119890120575(1205832minus1199051))]
(39)
We know that
119889C3(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=119879
=1
119879c4120575 [int
1205831
0
119890120575(119905minus119879)
119891 (119905) 119889119905
+ 1198630int
1205832
1205831
119890120575(119905minus119879)
119889119905 + int
119879
1205832
119890120575(119905minus119879)
119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119891 (119905) 119889119905
+ int
119879
1205832
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119892 (119905) 119889119905
+ 1198630((119879 minus 120583
1) 119890120575(1205831minus119879)
minus (119879 minus 1205832) 119890120575(1205832minus119879)
)]
gt 0
(40)
Therefore (i) if Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (1205832 119879) satisfying 119889C
3(1199051)1198891199051= 0
and C3(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
2ge 0
and 1198892C3(1199051)1198891199052
1gt 0 it means that C
3(1199051) is a nondecreasing
function and obtains its minimum value at point 119905lowast1= 1205832
(iii) if the sign of 1198892C3(1199051)1198891199052
1is indefinite then the curve of
C3(1199051) is W-shape and C
3(1199051)may obtain its minimum value
at more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
3(1199051)1198891199051= 0
Therefore the optimal value of the order level 119878 = 119868(119905lowast
1)
is
119878lowast= int
119879
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909 (41)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus119905lowast
1)119889119905
+ int
119905lowast
1
1205832
119890120575(119905minus119905lowast
1)119892 (119905) 119889119905
(42)
From the above analysis we obtain that the total averagecost of the system over the interval [0 119879] is
TC (1199051) =
C1(1199051) 0 lt 119905
1le 1205831
C2(1199051) 120583
1lt 1199051le 1205832
C3(1199051) 120583
2lt 1199051lt 119879
(43)
where C1(1199051) C2(1199051) and C
3(1199051) are obtained from (10) (24)
and (37) respectively
8 Mathematical Problems in Engineering
Basing on the analyzed results above we provide theresults which ensure the existence of a unique 119905
1 say 119905lowast
1 to
minimize the total average cost for the model system startingwith shortages The procedure is as follows
Step 0 Input the parameters 1205831 1205832 119879 1198861 1198862 1198871 120572 120573 120575
1198630 and c
119894 119894 = 1 2 3 4
Step 1 Calculate Δ1and 1198892C
1(1199051)1198891199052
1
Step 11 If Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 solve a unique 119905lowast
1
from119889C1(1199051)1198891199051= 0 and the total average cost and the order
quantity can be obtained by (10) and (16) respectively
Step 12 If Δ1le 0 and 119889
2C1(1199051)1198891199052
1gt 0 then the optimal
average cost and the optimal order quantity can be obtainedby (10) and (16) at 119905lowast
1= 1205831 respectively
Step 13 If the sign of 1198892C1(1199051)1198891199052
1is indefinite then solve
equation 119889C1(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C1(1199051) to obtain the optimal
replenishment point compute the order quantity by (16)
Step 2 Calculate Δ2and 1198892C
2(1199051)1198891199052
1
Step 21 If Δ1lt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
seek a unique 119905lowast1from equation 119889C
2(1199051)1198891199051= 0 and the
optimal total average cost and the optimal order quantity canbe obtained by (24) and (29) at 119905
1= 119905lowast
1 respectively
Step 22 If Δ1gt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205831 and compute the order
quantity by (29)
Step 23 If Δ1lt 0 Δ
2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205832 and compute the order
quantity by (29)
Step 24 If the sign of 1198892C2(1199051)1198891199052
1is indefinite then solve
equation 119889C2(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C2(1199051) to obtain the optimal
replenishment point compute the optimal order quantity by(29)
Step 3 Calculate 1198892C3(1199051)1198891199052
1
Step 31 If Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 seek the unique
119905lowast
1from equation 119889C
3(1199051)1198891199051= 0 and the total average
cost and the order quantity can be obtained by (37) and (42)respectively
Step 32 If Δ2ge 0 and 1198892C
3(1199051)1198891199052
1gt 0 then C
3(1199051) obtains
its minimum at 119905lowast1= 1205832 and compute the order quantity by
(42)
Step 33 If the sign of 1198892C3(1199051)1198891199052
1is indefinite then
solve equation 119889C3(1199051)1198891199051
= 0 If the solution is notunique by comparing their average costs C
3(1199051) to obtain the
optimal replenishment point compute the order quantity by(42)
Step 4 Find TC(119905lowast1) = minC
1(119905lowast
1)C2(119905lowast
1)C3(119905lowast
1) and
accordingly select the optimum 119905lowast
1 and then obtain corre-
sponding optimal order quantity 119876lowast
4 Numerical Examples andSensitivity Analysis
In order to demonstrate the procedure to obtain the optimalsolution of the model we present four examples for themodel Examples are based on piece-wise demand rate suchas 119891(119905) = 119886
1+ 1198871119905 and 119892(119905) = 119886
2119890minus1198872119905
Example 1 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 004 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $2 c
2= $3 c
3= $12 c
4= $8
Since Δ1
= 18809 gt 0 Δ2
= 222869 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1) 1198892C
2(1199051)1198891199052
1gt 0
for all 1199051
isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
2(1205831) = 763891 in
[1205831 1205832) and C
3(1199062) = 126358 in [120583
2 119879) By solving equation
119889C1(1199051)1198891199051= 0 we have 119905lowast
1= 3664 From (10) we obtain
C1(119905lowast
1) = 762759 Therefore the optimal ordering quantity is
119876lowast= 637839 and the total average cost is TC(119905lowast
1) = 762759
Example 2 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus257889 lt 0 Δ2
= 139269 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin (1205832 119879)
then we get local minimum value C1(1205831) = 1805883 in
(0 1205831] and C
3(1199062) = 158232 in [120583
2 119879) By solving equation
119889C2(1199051)1198891199051= 0 we have 119905lowast
1= 6559 From (24) we obtain
C2(119905lowast
1) = 1480324 Therefore the optimal ordering quantity
is 119876lowast = 571827 and the total average cost is TC(119905lowast1) =
1480324
Example 3 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 2 120575 = 006 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus249329 lt 0 Δ2
= minus121869 lt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin [1205832 119879)
then we get local minimum value C1(1205831) = 1417134 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 9120 From (37) we obtain
C3(119905lowast
1) = 889767 Therefore the optimal ordering quantity is
119876lowast= 278960 and the total average cost is TC(119905lowast
1) = 889767
Example 4 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 16 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus197263 lt 0 Δ2
= minus86675 lt 01198892C4(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0
Mathematical Problems in Engineering 9
Table 1 The sensitivity of 120575 for the model in Example 1
120575 0 001 002 003 004 005 006 007 008119905lowast
134366 34905 35463 36042 36642 37263 37907 38574 39265
119878lowast 5150829 5150829 5114210 5075965 5036066 4994433 4950985 4905641 4858321119876lowast 6481034 6481034 6448369 6414184 6378392 6340900 6301614 6260436 6217267
TC2(119905lowast
1) 7752147 7718437 7684699 7649983 7614252 7577473 7539611 750063 7460498
Table 2 The sensitivity of 120572 for the models in Example 1
120572 0 0001 0002 0003 0004 0005 0006 0007 0008119905lowast
130141 31325 32569 33870 35229 36642 38105 39614 41171
119878lowast 4158882 4320629 4489624 4665543 4847907 5036066 5229206 5426345 5755923119876lowast 5228982 5439279 5659745 5890106 6129883 6378392 6634729 6897782 7230000
TC2(119905lowast
1) 6385314 6614087 6851585 7097639 7351984 7614252 7883972 8160570 8450147
Table 3 The sensitivity of 120573 for the models in Example 1
120573 14 15 16 17 18 19 20 21 22119905lowast
131114 31995 32513 33176 340332 35148 36613 38551 41128
119878lowast 4275053 4401612 4471691 4562029 4679084 4831709 5036066 5295334 5641669119876lowast 5384988 5547959 5639494 5757494 5910495 6110248 6378392 6720262 7179964
TC2(119905lowast
1) 6444596 6703559 6801679 6929359 7095918 7314743 7614252 7976508 8450147
for all 1199051
isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
1(1205831) = 125259 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 8627 From (37) we obtain
TC(119905lowast1) = 842935 Therefore the optimal ordering quantity
is 119876lowast = 241524 obtained from (42) and the minimum costTC(119905lowast1) = 842935
In order to clearly indicate the effects of parameterssuch as 120575 120572 and 120573 on 119878
lowast 119876lowast and TC119894(119905lowast
1) (119894 = 1 2)
respectively we study the sensitivity of different parametersunder the studied inventory modelThe sensitivity analysis isperformed on the base of Example 1 and the results are shownin Tables 1ndash3 respectively
From Table 1 we observe that inventory level 119878lowast orderquantity 119876
lowast and the total average cost TC119894(119905lowast
1) (119894 = 1 2)
gradually decrease as the shortage parameter 120575 increasesfor the model and the replenishment time 119905lowast
1increases as
120575 increases for the model starting with shortage Also weobserve that the values of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) are slightly
sensitive to the changes of 120575From Table 2 we find that when the deterioration
parameter 120572 increases then 119878lowast 119876lowast and TC(119905lowast1) increase for
both models and the replenishment time 119905lowast1increases for the
model starting with shortage We see that the values of 119905lowast1 119878lowast
119876lowast and TC(119905lowast
1) all are temperately sensitive to the changes of
120572 for the modelFrom Table 3 we find that 119878lowast 119876lowast and TC(119905lowast
1) increase
as the deterioration parameter 120573 increases for both modelsAlso we find that the replenishment time 119905lowast
1increases as 120573
increases for themodel startingwith shortageWe see that thevalues of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) all are moderately sensitive
to the changes of 120573 for the model
5 Conclusion
This paper studies an inventory model for Weibull-distributed deterioration with trapezoidal type demand rateand partial backlogging First an optimal replenishmentstrategy for deteriorating items with the model starting withshortage is proposedThen numerical examples are providedto illustrate the theoretical results From the analysis of Tables1ndash3 it can be found that the replenishment time point 119905lowast
1
order quantity 119876lowast and the total average cost TC(119905lowast
1) are
moderately sensitive to the changes of 120572 and 120573 and lowlysensitive to the changes of 120575 respectively However if theorder cycle (119879) is a decision variable such inventory modelwill be of more practical significance which provides us aninteresting topic in future
Competing Interests
The author declares that there are no competing interests
Acknowledgments
The research is supported partly by National Natural ScienceFoundation of China (Grant no 71502100)
References
[1] P M Ghare and G F Schrader ldquoA model for exponentiallydecaying inventoriesrdquo Journal of Industrial Engineering vol 14no 3 pp 238ndash243 1963
[2] R P Covert and G C Philip ldquoAn EOQ model for items withweibull distribution deteriorationrdquo AIIE Transaction vol 5 no4 pp 323ndash326 1973
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 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
Mathematical Problems in Engineering 5
From (18) the total replenishment quantity can be computed
119878 = 119868 (1199051) = 119863
0int
1205832
1199051
119890120572(119909120573minus119905120573
1)119889119909 + int
119879
1205832
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (19)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905
(20)
The total number of perished items in the interval [1199051 119879] is
119863119879= 1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905
(21)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= 1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
(22)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
(23)
Therefore the average total cost per unit time under thecondition 120583
1lt 1199051le 1205832can be given by
C2(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1[1198630int
1205832
1199051
(119890120572(119905120573minus119905120573
1)minus 1) 119889119905
+ int
119879
1205832
(119890120572(119905120573minus119905120573
1)minus 1)119892 (119905) 119889119905]
+ c2[1198630int
1205832
1199051
int
1205832
119905
119890120572(119909120573minus119905120573)119889119909 119889119905
+ int
1205832
1199051
int
119879
1205832
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ int
119879
1205832
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905]
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1199051
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+1198630
120575int
1199051
1205831
(119890120575(119905minus1199051)minus 119890120575(1205831minus1199051)) 119889119905]
(24)
Taking the first and the second order derivative of C2(1199051)
with respect to 1199051 respectively we have
119889C2(1199051)
1198891199051
=1
119879c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630int
1199051
1205831
119890120575(119905minus1199051)119889119905] minus (c
1120572120573119905120573minus1
1+ c2)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 minus 1205751199051+ 120575119905) 119890
120575(119905minus1199051)119891 (119905) 119889119905
minus 1198630(1205831minus 1199051) 119890120575(1205831minus1199051)]
(25)
1198892C2(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1(c1(120572120573119905120573
1minus 120573 + 1) + c
21199051)
sdot [1198630int
1205832
1199051
119890120572(119905120573minus119905120573
1)119889119905 + int
119879
1205832
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905]
+ (c1120572120573119905120573minus1
1+ c2)1198630+ c4120575 [1198630119890120575(1205831minus1199051)
minus 120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ 1198630(1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)]
(26)
Let
Δ2= c3[int
1205831
0
(1 minus 1205751205832+ 120575119905) 119890
120575(119905minus1205832)119891 (119905) 119889119905
+ 1198630(1205832minus 1205831) 119890120575(1205831minus1205832)]
+ c4120575 [int
1205831
0
119890120575(119905minus120583
2)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus120583
2)119889119905]
minus (c2+ c1120572120573120583120573minus1
2)int
119879
1205832
119890120572(119905120573minus120583120573
2)119892 (119905) 119889119905
(27)
6 Mathematical Problems in Engineering
Therefore (i) if Δ1lt 0 Δ
2gt 0 and 1198892C
2(1199051)1198891199052
1gt 0 there
exists a unique solution 119905lowast1isin (1205831 1205832) satisfying 119889C
2(1199051)1198891199051=
0 and C2(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt
0 Δ2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 it means that C
2(1199051) is
a nonincreasing function and obtains its minimum at point119905lowast
1= 1205832 (iii) if Δ
1ge 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 it
means that C2(1199051) is a nondecreasing function and obtains its
minimum at point 119905lowast1= 1205831 (iv) if the sign of 1198892C
2(1199051)1198891199052
1is
indefinite then the curve of C2(1199051) is W-shape and C
2(1199051)
may obtain its minimum value at more than one pointwithout loss of generality set 119905
1= 119905lowast
1 where 119905lowast
1is one of the
solutions of 119889C2(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= 1198630int
1205832
119905lowast
1
119890120572(119905120573minus119905lowast
1
120573)119889119905 + int
119879
1205832
119890120572(119905120573minus119905lowast
1
120573)119892 (119905) 119889119905 (28)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
119905lowast
1
1205831
119890120575(119905minus119905lowast
1)119889119905 (29)
Case 3 (1205832le 1199051lt 119879) If the replenishment time 119905
1isin [1205832 119879)
from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 120583
2
minus119890minus120575(1199051minus119905)119892 (119905) 120583
2lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1199051lt 119905 lt 119879
(30)
Solving (30) with 119868(0) = 119868(119879) = 0 we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(119905minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 120583
2
minus int
119905
1205832
119890120575(119909minus119905
1)119892 (119909) 119889119909 +
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
2lt 119905 lt 119905
1
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 119879
(31)
From the last equation of (31) the total replenishmentquantity can be computed
119878 = 119868 (1199051) = int
119879
1199051
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (32)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575[(1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051)]
(33)
The total number of perished items in the interval [1199051 119879] is
119863119879= int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909 (34)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905 (35)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905
(36)
Therefore the average total cost per unit time under thecondition 120583
2lt 1199051le 119879 can be given by
C3(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909
+ c2int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
Mathematical Problems in Engineering 7
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(37)
Taking the first and the second order derivative of C3(1199051)
with respect to 1199051 respectively we have
119889C3(1199051)
1198891199051
=1
119879minus (c2+ c1120572120573119905120573minus1
1) int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905
+ c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119892 (119905) 119889119905
+ 1198630((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(38)
1198892C3(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + (c
2+ c1120572120573119905120573minus1
1) 119892 (1199051)
+ c4120575 [120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0120575int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ 120575int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c2[120572120573119905120573minus1
1int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)
+ 1198630((1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)
minus (1 + 1205751205832minus 1205751199051) 119890120575(1205832minus1199051))]
(39)
We know that
119889C3(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=119879
=1
119879c4120575 [int
1205831
0
119890120575(119905minus119879)
119891 (119905) 119889119905
+ 1198630int
1205832
1205831
119890120575(119905minus119879)
119889119905 + int
119879
1205832
119890120575(119905minus119879)
119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119891 (119905) 119889119905
+ int
119879
1205832
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119892 (119905) 119889119905
+ 1198630((119879 minus 120583
1) 119890120575(1205831minus119879)
minus (119879 minus 1205832) 119890120575(1205832minus119879)
)]
gt 0
(40)
Therefore (i) if Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (1205832 119879) satisfying 119889C
3(1199051)1198891199051= 0
and C3(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
2ge 0
and 1198892C3(1199051)1198891199052
1gt 0 it means that C
3(1199051) is a nondecreasing
function and obtains its minimum value at point 119905lowast1= 1205832
(iii) if the sign of 1198892C3(1199051)1198891199052
1is indefinite then the curve of
C3(1199051) is W-shape and C
3(1199051)may obtain its minimum value
at more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
3(1199051)1198891199051= 0
Therefore the optimal value of the order level 119878 = 119868(119905lowast
1)
is
119878lowast= int
119879
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909 (41)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus119905lowast
1)119889119905
+ int
119905lowast
1
1205832
119890120575(119905minus119905lowast
1)119892 (119905) 119889119905
(42)
From the above analysis we obtain that the total averagecost of the system over the interval [0 119879] is
TC (1199051) =
C1(1199051) 0 lt 119905
1le 1205831
C2(1199051) 120583
1lt 1199051le 1205832
C3(1199051) 120583
2lt 1199051lt 119879
(43)
where C1(1199051) C2(1199051) and C
3(1199051) are obtained from (10) (24)
and (37) respectively
8 Mathematical Problems in Engineering
Basing on the analyzed results above we provide theresults which ensure the existence of a unique 119905
1 say 119905lowast
1 to
minimize the total average cost for the model system startingwith shortages The procedure is as follows
Step 0 Input the parameters 1205831 1205832 119879 1198861 1198862 1198871 120572 120573 120575
1198630 and c
119894 119894 = 1 2 3 4
Step 1 Calculate Δ1and 1198892C
1(1199051)1198891199052
1
Step 11 If Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 solve a unique 119905lowast
1
from119889C1(1199051)1198891199051= 0 and the total average cost and the order
quantity can be obtained by (10) and (16) respectively
Step 12 If Δ1le 0 and 119889
2C1(1199051)1198891199052
1gt 0 then the optimal
average cost and the optimal order quantity can be obtainedby (10) and (16) at 119905lowast
1= 1205831 respectively
Step 13 If the sign of 1198892C1(1199051)1198891199052
1is indefinite then solve
equation 119889C1(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C1(1199051) to obtain the optimal
replenishment point compute the order quantity by (16)
Step 2 Calculate Δ2and 1198892C
2(1199051)1198891199052
1
Step 21 If Δ1lt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
seek a unique 119905lowast1from equation 119889C
2(1199051)1198891199051= 0 and the
optimal total average cost and the optimal order quantity canbe obtained by (24) and (29) at 119905
1= 119905lowast
1 respectively
Step 22 If Δ1gt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205831 and compute the order
quantity by (29)
Step 23 If Δ1lt 0 Δ
2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205832 and compute the order
quantity by (29)
Step 24 If the sign of 1198892C2(1199051)1198891199052
1is indefinite then solve
equation 119889C2(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C2(1199051) to obtain the optimal
replenishment point compute the optimal order quantity by(29)
Step 3 Calculate 1198892C3(1199051)1198891199052
1
Step 31 If Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 seek the unique
119905lowast
1from equation 119889C
3(1199051)1198891199051= 0 and the total average
cost and the order quantity can be obtained by (37) and (42)respectively
Step 32 If Δ2ge 0 and 1198892C
3(1199051)1198891199052
1gt 0 then C
3(1199051) obtains
its minimum at 119905lowast1= 1205832 and compute the order quantity by
(42)
Step 33 If the sign of 1198892C3(1199051)1198891199052
1is indefinite then
solve equation 119889C3(1199051)1198891199051
= 0 If the solution is notunique by comparing their average costs C
3(1199051) to obtain the
optimal replenishment point compute the order quantity by(42)
Step 4 Find TC(119905lowast1) = minC
1(119905lowast
1)C2(119905lowast
1)C3(119905lowast
1) and
accordingly select the optimum 119905lowast
1 and then obtain corre-
sponding optimal order quantity 119876lowast
4 Numerical Examples andSensitivity Analysis
In order to demonstrate the procedure to obtain the optimalsolution of the model we present four examples for themodel Examples are based on piece-wise demand rate suchas 119891(119905) = 119886
1+ 1198871119905 and 119892(119905) = 119886
2119890minus1198872119905
Example 1 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 004 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $2 c
2= $3 c
3= $12 c
4= $8
Since Δ1
= 18809 gt 0 Δ2
= 222869 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1) 1198892C
2(1199051)1198891199052
1gt 0
for all 1199051
isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
2(1205831) = 763891 in
[1205831 1205832) and C
3(1199062) = 126358 in [120583
2 119879) By solving equation
119889C1(1199051)1198891199051= 0 we have 119905lowast
1= 3664 From (10) we obtain
C1(119905lowast
1) = 762759 Therefore the optimal ordering quantity is
119876lowast= 637839 and the total average cost is TC(119905lowast
1) = 762759
Example 2 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus257889 lt 0 Δ2
= 139269 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin (1205832 119879)
then we get local minimum value C1(1205831) = 1805883 in
(0 1205831] and C
3(1199062) = 158232 in [120583
2 119879) By solving equation
119889C2(1199051)1198891199051= 0 we have 119905lowast
1= 6559 From (24) we obtain
C2(119905lowast
1) = 1480324 Therefore the optimal ordering quantity
is 119876lowast = 571827 and the total average cost is TC(119905lowast1) =
1480324
Example 3 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 2 120575 = 006 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus249329 lt 0 Δ2
= minus121869 lt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin [1205832 119879)
then we get local minimum value C1(1205831) = 1417134 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 9120 From (37) we obtain
C3(119905lowast
1) = 889767 Therefore the optimal ordering quantity is
119876lowast= 278960 and the total average cost is TC(119905lowast
1) = 889767
Example 4 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 16 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus197263 lt 0 Δ2
= minus86675 lt 01198892C4(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0
Mathematical Problems in Engineering 9
Table 1 The sensitivity of 120575 for the model in Example 1
120575 0 001 002 003 004 005 006 007 008119905lowast
134366 34905 35463 36042 36642 37263 37907 38574 39265
119878lowast 5150829 5150829 5114210 5075965 5036066 4994433 4950985 4905641 4858321119876lowast 6481034 6481034 6448369 6414184 6378392 6340900 6301614 6260436 6217267
TC2(119905lowast
1) 7752147 7718437 7684699 7649983 7614252 7577473 7539611 750063 7460498
Table 2 The sensitivity of 120572 for the models in Example 1
120572 0 0001 0002 0003 0004 0005 0006 0007 0008119905lowast
130141 31325 32569 33870 35229 36642 38105 39614 41171
119878lowast 4158882 4320629 4489624 4665543 4847907 5036066 5229206 5426345 5755923119876lowast 5228982 5439279 5659745 5890106 6129883 6378392 6634729 6897782 7230000
TC2(119905lowast
1) 6385314 6614087 6851585 7097639 7351984 7614252 7883972 8160570 8450147
Table 3 The sensitivity of 120573 for the models in Example 1
120573 14 15 16 17 18 19 20 21 22119905lowast
131114 31995 32513 33176 340332 35148 36613 38551 41128
119878lowast 4275053 4401612 4471691 4562029 4679084 4831709 5036066 5295334 5641669119876lowast 5384988 5547959 5639494 5757494 5910495 6110248 6378392 6720262 7179964
TC2(119905lowast
1) 6444596 6703559 6801679 6929359 7095918 7314743 7614252 7976508 8450147
for all 1199051
isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
1(1205831) = 125259 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 8627 From (37) we obtain
TC(119905lowast1) = 842935 Therefore the optimal ordering quantity
is 119876lowast = 241524 obtained from (42) and the minimum costTC(119905lowast1) = 842935
In order to clearly indicate the effects of parameterssuch as 120575 120572 and 120573 on 119878
lowast 119876lowast and TC119894(119905lowast
1) (119894 = 1 2)
respectively we study the sensitivity of different parametersunder the studied inventory modelThe sensitivity analysis isperformed on the base of Example 1 and the results are shownin Tables 1ndash3 respectively
From Table 1 we observe that inventory level 119878lowast orderquantity 119876
lowast and the total average cost TC119894(119905lowast
1) (119894 = 1 2)
gradually decrease as the shortage parameter 120575 increasesfor the model and the replenishment time 119905lowast
1increases as
120575 increases for the model starting with shortage Also weobserve that the values of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) are slightly
sensitive to the changes of 120575From Table 2 we find that when the deterioration
parameter 120572 increases then 119878lowast 119876lowast and TC(119905lowast1) increase for
both models and the replenishment time 119905lowast1increases for the
model starting with shortage We see that the values of 119905lowast1 119878lowast
119876lowast and TC(119905lowast
1) all are temperately sensitive to the changes of
120572 for the modelFrom Table 3 we find that 119878lowast 119876lowast and TC(119905lowast
1) increase
as the deterioration parameter 120573 increases for both modelsAlso we find that the replenishment time 119905lowast
1increases as 120573
increases for themodel startingwith shortageWe see that thevalues of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) all are moderately sensitive
to the changes of 120573 for the model
5 Conclusion
This paper studies an inventory model for Weibull-distributed deterioration with trapezoidal type demand rateand partial backlogging First an optimal replenishmentstrategy for deteriorating items with the model starting withshortage is proposedThen numerical examples are providedto illustrate the theoretical results From the analysis of Tables1ndash3 it can be found that the replenishment time point 119905lowast
1
order quantity 119876lowast and the total average cost TC(119905lowast
1) are
moderately sensitive to the changes of 120572 and 120573 and lowlysensitive to the changes of 120575 respectively However if theorder cycle (119879) is a decision variable such inventory modelwill be of more practical significance which provides us aninteresting topic in future
Competing Interests
The author declares that there are no competing interests
Acknowledgments
The research is supported partly by National Natural ScienceFoundation of China (Grant no 71502100)
References
[1] P M Ghare and G F Schrader ldquoA model for exponentiallydecaying inventoriesrdquo Journal of Industrial Engineering vol 14no 3 pp 238ndash243 1963
[2] R P Covert and G C Philip ldquoAn EOQ model for items withweibull distribution deteriorationrdquo AIIE Transaction vol 5 no4 pp 323ndash326 1973
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 2014
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Mathematical Problems in Engineering
Therefore (i) if Δ1lt 0 Δ
2gt 0 and 1198892C
2(1199051)1198891199052
1gt 0 there
exists a unique solution 119905lowast1isin (1205831 1205832) satisfying 119889C
2(1199051)1198891199051=
0 and C2(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
1lt
0 Δ2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 it means that C
2(1199051) is
a nonincreasing function and obtains its minimum at point119905lowast
1= 1205832 (iii) if Δ
1ge 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 it
means that C2(1199051) is a nondecreasing function and obtains its
minimum at point 119905lowast1= 1205831 (iv) if the sign of 1198892C
2(1199051)1198891199052
1is
indefinite then the curve of C2(1199051) is W-shape and C
2(1199051)
may obtain its minimum value at more than one pointwithout loss of generality set 119905
1= 119905lowast
1 where 119905lowast
1is one of the
solutions of 119889C2(1199051)1198891199051= 0
Thus the optimal value of the order level 119878 = 119868(119905lowast
1) is
119878lowast= 1198630int
1205832
119905lowast
1
119890120572(119905120573minus119905lowast
1
120573)119889119905 + int
119879
1205832
119890120572(119905120573minus119905lowast
1
120573)119892 (119905) 119889119905 (28)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
119905lowast
1
1205831
119890120575(119905minus119905lowast
1)119889119905 (29)
Case 3 (1205832le 1199051lt 119879) If the replenishment time 119905
1isin [1205832 119879)
from (2) we have
119889119868 (119905)
119889119905=
minus119890minus120575(1199051minus119905)119891 (119905) 0 lt 119905 lt 120583
1
minus119890minus120575(1199051minus119905)1198630 120583
1lt 119905 lt 120583
2
minus119890minus120575(1199051minus119905)119892 (119905) 120583
2lt 119905 lt 119905
1
minus120572120573119905120573minus1
119868 (119905) minus 119892 (119905) 1199051lt 119905 lt 119879
(30)
Solving (30) with 119868(0) = 119868(119879) = 0 we have
119868 (119905) =
minusint
119905
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 0 lt 119905 lt 120583
1
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(119905minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
1lt 119905 lt 120583
2
minus int
119905
1205832
119890120575(119909minus119905
1)119892 (119909) 119889119909 +
1198630
120575(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051)) minus int
1205831
0
119890120575(119909minus119905
1)119891 (119909) 119889119909 120583
2lt 119905 lt 119905
1
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119905
1lt 119905 lt 119879
(31)
From the last equation of (31) the total replenishmentquantity can be computed
119878 = 119868 (1199051) = int
119879
1199051
119890120572(119909120573minus119905120573
1)119892 (119909) 119889119909 (32)
The total shortage quantity during the interval (0 1199051] is
119861119879= int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575[(1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051)]
(33)
The total number of perished items in the interval [1199051 119879] is
119863119879= int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909 (34)
The total number of inventories carried during the interval[1199051 119879] is
119867119879= int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905 (35)
The total of lost sales during the interval [0 1199051] is
119871119879= int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905
(36)
Therefore the average total cost per unit time under thecondition 120583
2lt 1199051le 119879 can be given by
C3(1199051) =
1
119879[1198600+ c1119863119879+ c2119867119879+ c3119861119879+ c4119871119879]
=1
1198791198600+ c1int
119879
1199051
(119890120572(119909120573minus119905120573
1)minus 1)119892 (119909) 119889119909
+ c2int
119879
1199051
int
119879
119905
119890120572(119909120573minus119905120573)119892 (119909) 119889119909 119889119905
+ c4[int
1205831
0
(1 minus 119890120575(119905minus1199051)) 119891 (119905) 119889119905
+ 1198630int
1205832
1205831
(1 minus 119890120575(119905minus1199051)) 119889119905
Mathematical Problems in Engineering 7
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(37)
Taking the first and the second order derivative of C3(1199051)
with respect to 1199051 respectively we have
119889C3(1199051)
1198891199051
=1
119879minus (c2+ c1120572120573119905120573minus1
1) int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905
+ c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119892 (119905) 119889119905
+ 1198630((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(38)
1198892C3(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + (c
2+ c1120572120573119905120573minus1
1) 119892 (1199051)
+ c4120575 [120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0120575int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ 120575int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c2[120572120573119905120573minus1
1int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)
+ 1198630((1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)
minus (1 + 1205751205832minus 1205751199051) 119890120575(1205832minus1199051))]
(39)
We know that
119889C3(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=119879
=1
119879c4120575 [int
1205831
0
119890120575(119905minus119879)
119891 (119905) 119889119905
+ 1198630int
1205832
1205831
119890120575(119905minus119879)
119889119905 + int
119879
1205832
119890120575(119905minus119879)
119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119891 (119905) 119889119905
+ int
119879
1205832
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119892 (119905) 119889119905
+ 1198630((119879 minus 120583
1) 119890120575(1205831minus119879)
minus (119879 minus 1205832) 119890120575(1205832minus119879)
)]
gt 0
(40)
Therefore (i) if Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (1205832 119879) satisfying 119889C
3(1199051)1198891199051= 0
and C3(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
2ge 0
and 1198892C3(1199051)1198891199052
1gt 0 it means that C
3(1199051) is a nondecreasing
function and obtains its minimum value at point 119905lowast1= 1205832
(iii) if the sign of 1198892C3(1199051)1198891199052
1is indefinite then the curve of
C3(1199051) is W-shape and C
3(1199051)may obtain its minimum value
at more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
3(1199051)1198891199051= 0
Therefore the optimal value of the order level 119878 = 119868(119905lowast
1)
is
119878lowast= int
119879
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909 (41)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus119905lowast
1)119889119905
+ int
119905lowast
1
1205832
119890120575(119905minus119905lowast
1)119892 (119905) 119889119905
(42)
From the above analysis we obtain that the total averagecost of the system over the interval [0 119879] is
TC (1199051) =
C1(1199051) 0 lt 119905
1le 1205831
C2(1199051) 120583
1lt 1199051le 1205832
C3(1199051) 120583
2lt 1199051lt 119879
(43)
where C1(1199051) C2(1199051) and C
3(1199051) are obtained from (10) (24)
and (37) respectively
8 Mathematical Problems in Engineering
Basing on the analyzed results above we provide theresults which ensure the existence of a unique 119905
1 say 119905lowast
1 to
minimize the total average cost for the model system startingwith shortages The procedure is as follows
Step 0 Input the parameters 1205831 1205832 119879 1198861 1198862 1198871 120572 120573 120575
1198630 and c
119894 119894 = 1 2 3 4
Step 1 Calculate Δ1and 1198892C
1(1199051)1198891199052
1
Step 11 If Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 solve a unique 119905lowast
1
from119889C1(1199051)1198891199051= 0 and the total average cost and the order
quantity can be obtained by (10) and (16) respectively
Step 12 If Δ1le 0 and 119889
2C1(1199051)1198891199052
1gt 0 then the optimal
average cost and the optimal order quantity can be obtainedby (10) and (16) at 119905lowast
1= 1205831 respectively
Step 13 If the sign of 1198892C1(1199051)1198891199052
1is indefinite then solve
equation 119889C1(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C1(1199051) to obtain the optimal
replenishment point compute the order quantity by (16)
Step 2 Calculate Δ2and 1198892C
2(1199051)1198891199052
1
Step 21 If Δ1lt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
seek a unique 119905lowast1from equation 119889C
2(1199051)1198891199051= 0 and the
optimal total average cost and the optimal order quantity canbe obtained by (24) and (29) at 119905
1= 119905lowast
1 respectively
Step 22 If Δ1gt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205831 and compute the order
quantity by (29)
Step 23 If Δ1lt 0 Δ
2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205832 and compute the order
quantity by (29)
Step 24 If the sign of 1198892C2(1199051)1198891199052
1is indefinite then solve
equation 119889C2(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C2(1199051) to obtain the optimal
replenishment point compute the optimal order quantity by(29)
Step 3 Calculate 1198892C3(1199051)1198891199052
1
Step 31 If Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 seek the unique
119905lowast
1from equation 119889C
3(1199051)1198891199051= 0 and the total average
cost and the order quantity can be obtained by (37) and (42)respectively
Step 32 If Δ2ge 0 and 1198892C
3(1199051)1198891199052
1gt 0 then C
3(1199051) obtains
its minimum at 119905lowast1= 1205832 and compute the order quantity by
(42)
Step 33 If the sign of 1198892C3(1199051)1198891199052
1is indefinite then
solve equation 119889C3(1199051)1198891199051
= 0 If the solution is notunique by comparing their average costs C
3(1199051) to obtain the
optimal replenishment point compute the order quantity by(42)
Step 4 Find TC(119905lowast1) = minC
1(119905lowast
1)C2(119905lowast
1)C3(119905lowast
1) and
accordingly select the optimum 119905lowast
1 and then obtain corre-
sponding optimal order quantity 119876lowast
4 Numerical Examples andSensitivity Analysis
In order to demonstrate the procedure to obtain the optimalsolution of the model we present four examples for themodel Examples are based on piece-wise demand rate suchas 119891(119905) = 119886
1+ 1198871119905 and 119892(119905) = 119886
2119890minus1198872119905
Example 1 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 004 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $2 c
2= $3 c
3= $12 c
4= $8
Since Δ1
= 18809 gt 0 Δ2
= 222869 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1) 1198892C
2(1199051)1198891199052
1gt 0
for all 1199051
isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
2(1205831) = 763891 in
[1205831 1205832) and C
3(1199062) = 126358 in [120583
2 119879) By solving equation
119889C1(1199051)1198891199051= 0 we have 119905lowast
1= 3664 From (10) we obtain
C1(119905lowast
1) = 762759 Therefore the optimal ordering quantity is
119876lowast= 637839 and the total average cost is TC(119905lowast
1) = 762759
Example 2 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus257889 lt 0 Δ2
= 139269 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin (1205832 119879)
then we get local minimum value C1(1205831) = 1805883 in
(0 1205831] and C
3(1199062) = 158232 in [120583
2 119879) By solving equation
119889C2(1199051)1198891199051= 0 we have 119905lowast
1= 6559 From (24) we obtain
C2(119905lowast
1) = 1480324 Therefore the optimal ordering quantity
is 119876lowast = 571827 and the total average cost is TC(119905lowast1) =
1480324
Example 3 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 2 120575 = 006 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus249329 lt 0 Δ2
= minus121869 lt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin [1205832 119879)
then we get local minimum value C1(1205831) = 1417134 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 9120 From (37) we obtain
C3(119905lowast
1) = 889767 Therefore the optimal ordering quantity is
119876lowast= 278960 and the total average cost is TC(119905lowast
1) = 889767
Example 4 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 16 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus197263 lt 0 Δ2
= minus86675 lt 01198892C4(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0
Mathematical Problems in Engineering 9
Table 1 The sensitivity of 120575 for the model in Example 1
120575 0 001 002 003 004 005 006 007 008119905lowast
134366 34905 35463 36042 36642 37263 37907 38574 39265
119878lowast 5150829 5150829 5114210 5075965 5036066 4994433 4950985 4905641 4858321119876lowast 6481034 6481034 6448369 6414184 6378392 6340900 6301614 6260436 6217267
TC2(119905lowast
1) 7752147 7718437 7684699 7649983 7614252 7577473 7539611 750063 7460498
Table 2 The sensitivity of 120572 for the models in Example 1
120572 0 0001 0002 0003 0004 0005 0006 0007 0008119905lowast
130141 31325 32569 33870 35229 36642 38105 39614 41171
119878lowast 4158882 4320629 4489624 4665543 4847907 5036066 5229206 5426345 5755923119876lowast 5228982 5439279 5659745 5890106 6129883 6378392 6634729 6897782 7230000
TC2(119905lowast
1) 6385314 6614087 6851585 7097639 7351984 7614252 7883972 8160570 8450147
Table 3 The sensitivity of 120573 for the models in Example 1
120573 14 15 16 17 18 19 20 21 22119905lowast
131114 31995 32513 33176 340332 35148 36613 38551 41128
119878lowast 4275053 4401612 4471691 4562029 4679084 4831709 5036066 5295334 5641669119876lowast 5384988 5547959 5639494 5757494 5910495 6110248 6378392 6720262 7179964
TC2(119905lowast
1) 6444596 6703559 6801679 6929359 7095918 7314743 7614252 7976508 8450147
for all 1199051
isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
1(1205831) = 125259 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 8627 From (37) we obtain
TC(119905lowast1) = 842935 Therefore the optimal ordering quantity
is 119876lowast = 241524 obtained from (42) and the minimum costTC(119905lowast1) = 842935
In order to clearly indicate the effects of parameterssuch as 120575 120572 and 120573 on 119878
lowast 119876lowast and TC119894(119905lowast
1) (119894 = 1 2)
respectively we study the sensitivity of different parametersunder the studied inventory modelThe sensitivity analysis isperformed on the base of Example 1 and the results are shownin Tables 1ndash3 respectively
From Table 1 we observe that inventory level 119878lowast orderquantity 119876
lowast and the total average cost TC119894(119905lowast
1) (119894 = 1 2)
gradually decrease as the shortage parameter 120575 increasesfor the model and the replenishment time 119905lowast
1increases as
120575 increases for the model starting with shortage Also weobserve that the values of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) are slightly
sensitive to the changes of 120575From Table 2 we find that when the deterioration
parameter 120572 increases then 119878lowast 119876lowast and TC(119905lowast1) increase for
both models and the replenishment time 119905lowast1increases for the
model starting with shortage We see that the values of 119905lowast1 119878lowast
119876lowast and TC(119905lowast
1) all are temperately sensitive to the changes of
120572 for the modelFrom Table 3 we find that 119878lowast 119876lowast and TC(119905lowast
1) increase
as the deterioration parameter 120573 increases for both modelsAlso we find that the replenishment time 119905lowast
1increases as 120573
increases for themodel startingwith shortageWe see that thevalues of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) all are moderately sensitive
to the changes of 120573 for the model
5 Conclusion
This paper studies an inventory model for Weibull-distributed deterioration with trapezoidal type demand rateand partial backlogging First an optimal replenishmentstrategy for deteriorating items with the model starting withshortage is proposedThen numerical examples are providedto illustrate the theoretical results From the analysis of Tables1ndash3 it can be found that the replenishment time point 119905lowast
1
order quantity 119876lowast and the total average cost TC(119905lowast
1) are
moderately sensitive to the changes of 120572 and 120573 and lowlysensitive to the changes of 120575 respectively However if theorder cycle (119879) is a decision variable such inventory modelwill be of more practical significance which provides us aninteresting topic in future
Competing Interests
The author declares that there are no competing interests
Acknowledgments
The research is supported partly by National Natural ScienceFoundation of China (Grant no 71502100)
References
[1] P M Ghare and G F Schrader ldquoA model for exponentiallydecaying inventoriesrdquo Journal of Industrial Engineering vol 14no 3 pp 238ndash243 1963
[2] R P Covert and G C Philip ldquoAn EOQ model for items withweibull distribution deteriorationrdquo AIIE Transaction vol 5 no4 pp 323ndash326 1973
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 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
Mathematical Problems in Engineering 7
+ int
1199051
1205832
(1 minus 119890120575(119905minus1199051)) 119892 (119905) 119889119905]
+ c3[int
1205831
0
(1199051minus 119905) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1199051minus 119905) 119890120575(119905minus1199051)119892 (119905) 119889119905
+1198630
1205752(119890120575(1205831minus1199051)minus 119890120575(1205832minus1199051))
+1198630
120575((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(37)
Taking the first and the second order derivative of C3(1199051)
with respect to 1199051 respectively we have
119889C3(1199051)
1198891199051
=1
119879minus (c2+ c1120572120573119905120573minus1
1) int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905
+ c4120575 [int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(1 + 120575119905 minus 1205751199051) 119890120575(119905minus1199051)119892 (119905) 119889119905
+ 1198630((1199051minus 1205831) 119890120575(1205831minus1199051)minus (1199051minus 1205832) 119890120575(1205832minus1199051))]
(38)
1198892C3(1199051)
1198891199052
1
=1
119879120572120573119905120573minus2
1[c1(120572120573119905120573
1minus 120573 + 1) + c
21199051]
sdot int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + (c
2+ c1120572120573119905120573minus1
1) 119892 (1199051)
+ c4120575 [120575int
1205831
0
119890120575(119905minus1199051)119891 (119905) 119889119905 + 119863
0120575int
1205832
1205831
119890120575(119905minus1199051)119889119905
+ 120575int
1199051
1205832
119890120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c2[120572120573119905120573minus1
1int
119879
1199051
119890120572(119905120573minus119905120573
1)119892 (119905) 119889119905 + 119892 (119905
1)]
+ c3[int
1205831
0
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119891 (119905) 119889119905
+ int
1199051
1205832
(12057521199051minus 1205752119905 minus 2120575) 119890
120575(119905minus1199051)119892 (119905) 119889119905 + 119892 (119905
1)
+ 1198630((1 + 120575120583
1minus 1205751199051) 119890120575(1205831minus1199051)
minus (1 + 1205751205832minus 1205751199051) 119890120575(1205832minus1199051))]
(39)
We know that
119889C3(1199051)
1198891199051
1003816100381610038161003816100381610038161003816100381610038161199051=119879
=1
119879c4120575 [int
1205831
0
119890120575(119905minus119879)
119891 (119905) 119889119905
+ 1198630int
1205832
1205831
119890120575(119905minus119879)
119889119905 + int
119879
1205832
119890120575(119905minus119879)
119892 (119905) 119889119905]
+ c3[int
1205831
0
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119891 (119905) 119889119905
+ int
119879
1205832
(1 + 120575119905 minus 120575119879) 119890120575(119905minus119879)
119892 (119905) 119889119905
+ 1198630((119879 minus 120583
1) 119890120575(1205831minus119879)
minus (119879 minus 1205832) 119890120575(1205832minus119879)
)]
gt 0
(40)
Therefore (i) if Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 there exists
a unique solution 119905lowast
1isin (1205832 119879) satisfying 119889C
3(1199051)1198891199051= 0
and C3(1199051) obtains its minimum at 119905
1= 119905lowast
1 (ii) if Δ
2ge 0
and 1198892C3(1199051)1198891199052
1gt 0 it means that C
3(1199051) is a nondecreasing
function and obtains its minimum value at point 119905lowast1= 1205832
(iii) if the sign of 1198892C3(1199051)1198891199052
1is indefinite then the curve of
C3(1199051) is W-shape and C
3(1199051)may obtain its minimum value
at more than one point without loss of generality set 1199051= 119905lowast
1
where 119905lowast1is one of the solutions of 119889C
3(1199051)1198891199051= 0
Therefore the optimal value of the order level 119878 = 119868(119905lowast
1)
is
119878lowast= int
119879
119905lowast
1
119890120572(119909120573minus119905lowast
1
120573)119892 (119909) 119889119909 (41)
and the optimal order quantity 119876lowast is
119876lowast= 119878lowast+ int
1205831
0
119890120575(119905minus119905lowast
1)119891 (119905) 119889119905 + 119863
0int
1205832
1205831
119890120575(119905minus119905lowast
1)119889119905
+ int
119905lowast
1
1205832
119890120575(119905minus119905lowast
1)119892 (119905) 119889119905
(42)
From the above analysis we obtain that the total averagecost of the system over the interval [0 119879] is
TC (1199051) =
C1(1199051) 0 lt 119905
1le 1205831
C2(1199051) 120583
1lt 1199051le 1205832
C3(1199051) 120583
2lt 1199051lt 119879
(43)
where C1(1199051) C2(1199051) and C
3(1199051) are obtained from (10) (24)
and (37) respectively
8 Mathematical Problems in Engineering
Basing on the analyzed results above we provide theresults which ensure the existence of a unique 119905
1 say 119905lowast
1 to
minimize the total average cost for the model system startingwith shortages The procedure is as follows
Step 0 Input the parameters 1205831 1205832 119879 1198861 1198862 1198871 120572 120573 120575
1198630 and c
119894 119894 = 1 2 3 4
Step 1 Calculate Δ1and 1198892C
1(1199051)1198891199052
1
Step 11 If Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 solve a unique 119905lowast
1
from119889C1(1199051)1198891199051= 0 and the total average cost and the order
quantity can be obtained by (10) and (16) respectively
Step 12 If Δ1le 0 and 119889
2C1(1199051)1198891199052
1gt 0 then the optimal
average cost and the optimal order quantity can be obtainedby (10) and (16) at 119905lowast
1= 1205831 respectively
Step 13 If the sign of 1198892C1(1199051)1198891199052
1is indefinite then solve
equation 119889C1(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C1(1199051) to obtain the optimal
replenishment point compute the order quantity by (16)
Step 2 Calculate Δ2and 1198892C
2(1199051)1198891199052
1
Step 21 If Δ1lt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
seek a unique 119905lowast1from equation 119889C
2(1199051)1198891199051= 0 and the
optimal total average cost and the optimal order quantity canbe obtained by (24) and (29) at 119905
1= 119905lowast
1 respectively
Step 22 If Δ1gt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205831 and compute the order
quantity by (29)
Step 23 If Δ1lt 0 Δ
2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205832 and compute the order
quantity by (29)
Step 24 If the sign of 1198892C2(1199051)1198891199052
1is indefinite then solve
equation 119889C2(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C2(1199051) to obtain the optimal
replenishment point compute the optimal order quantity by(29)
Step 3 Calculate 1198892C3(1199051)1198891199052
1
Step 31 If Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 seek the unique
119905lowast
1from equation 119889C
3(1199051)1198891199051= 0 and the total average
cost and the order quantity can be obtained by (37) and (42)respectively
Step 32 If Δ2ge 0 and 1198892C
3(1199051)1198891199052
1gt 0 then C
3(1199051) obtains
its minimum at 119905lowast1= 1205832 and compute the order quantity by
(42)
Step 33 If the sign of 1198892C3(1199051)1198891199052
1is indefinite then
solve equation 119889C3(1199051)1198891199051
= 0 If the solution is notunique by comparing their average costs C
3(1199051) to obtain the
optimal replenishment point compute the order quantity by(42)
Step 4 Find TC(119905lowast1) = minC
1(119905lowast
1)C2(119905lowast
1)C3(119905lowast
1) and
accordingly select the optimum 119905lowast
1 and then obtain corre-
sponding optimal order quantity 119876lowast
4 Numerical Examples andSensitivity Analysis
In order to demonstrate the procedure to obtain the optimalsolution of the model we present four examples for themodel Examples are based on piece-wise demand rate suchas 119891(119905) = 119886
1+ 1198871119905 and 119892(119905) = 119886
2119890minus1198872119905
Example 1 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 004 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $2 c
2= $3 c
3= $12 c
4= $8
Since Δ1
= 18809 gt 0 Δ2
= 222869 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1) 1198892C
2(1199051)1198891199052
1gt 0
for all 1199051
isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
2(1205831) = 763891 in
[1205831 1205832) and C
3(1199062) = 126358 in [120583
2 119879) By solving equation
119889C1(1199051)1198891199051= 0 we have 119905lowast
1= 3664 From (10) we obtain
C1(119905lowast
1) = 762759 Therefore the optimal ordering quantity is
119876lowast= 637839 and the total average cost is TC(119905lowast
1) = 762759
Example 2 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus257889 lt 0 Δ2
= 139269 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin (1205832 119879)
then we get local minimum value C1(1205831) = 1805883 in
(0 1205831] and C
3(1199062) = 158232 in [120583
2 119879) By solving equation
119889C2(1199051)1198891199051= 0 we have 119905lowast
1= 6559 From (24) we obtain
C2(119905lowast
1) = 1480324 Therefore the optimal ordering quantity
is 119876lowast = 571827 and the total average cost is TC(119905lowast1) =
1480324
Example 3 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 2 120575 = 006 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus249329 lt 0 Δ2
= minus121869 lt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin [1205832 119879)
then we get local minimum value C1(1205831) = 1417134 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 9120 From (37) we obtain
C3(119905lowast
1) = 889767 Therefore the optimal ordering quantity is
119876lowast= 278960 and the total average cost is TC(119905lowast
1) = 889767
Example 4 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 16 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus197263 lt 0 Δ2
= minus86675 lt 01198892C4(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0
Mathematical Problems in Engineering 9
Table 1 The sensitivity of 120575 for the model in Example 1
120575 0 001 002 003 004 005 006 007 008119905lowast
134366 34905 35463 36042 36642 37263 37907 38574 39265
119878lowast 5150829 5150829 5114210 5075965 5036066 4994433 4950985 4905641 4858321119876lowast 6481034 6481034 6448369 6414184 6378392 6340900 6301614 6260436 6217267
TC2(119905lowast
1) 7752147 7718437 7684699 7649983 7614252 7577473 7539611 750063 7460498
Table 2 The sensitivity of 120572 for the models in Example 1
120572 0 0001 0002 0003 0004 0005 0006 0007 0008119905lowast
130141 31325 32569 33870 35229 36642 38105 39614 41171
119878lowast 4158882 4320629 4489624 4665543 4847907 5036066 5229206 5426345 5755923119876lowast 5228982 5439279 5659745 5890106 6129883 6378392 6634729 6897782 7230000
TC2(119905lowast
1) 6385314 6614087 6851585 7097639 7351984 7614252 7883972 8160570 8450147
Table 3 The sensitivity of 120573 for the models in Example 1
120573 14 15 16 17 18 19 20 21 22119905lowast
131114 31995 32513 33176 340332 35148 36613 38551 41128
119878lowast 4275053 4401612 4471691 4562029 4679084 4831709 5036066 5295334 5641669119876lowast 5384988 5547959 5639494 5757494 5910495 6110248 6378392 6720262 7179964
TC2(119905lowast
1) 6444596 6703559 6801679 6929359 7095918 7314743 7614252 7976508 8450147
for all 1199051
isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
1(1205831) = 125259 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 8627 From (37) we obtain
TC(119905lowast1) = 842935 Therefore the optimal ordering quantity
is 119876lowast = 241524 obtained from (42) and the minimum costTC(119905lowast1) = 842935
In order to clearly indicate the effects of parameterssuch as 120575 120572 and 120573 on 119878
lowast 119876lowast and TC119894(119905lowast
1) (119894 = 1 2)
respectively we study the sensitivity of different parametersunder the studied inventory modelThe sensitivity analysis isperformed on the base of Example 1 and the results are shownin Tables 1ndash3 respectively
From Table 1 we observe that inventory level 119878lowast orderquantity 119876
lowast and the total average cost TC119894(119905lowast
1) (119894 = 1 2)
gradually decrease as the shortage parameter 120575 increasesfor the model and the replenishment time 119905lowast
1increases as
120575 increases for the model starting with shortage Also weobserve that the values of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) are slightly
sensitive to the changes of 120575From Table 2 we find that when the deterioration
parameter 120572 increases then 119878lowast 119876lowast and TC(119905lowast1) increase for
both models and the replenishment time 119905lowast1increases for the
model starting with shortage We see that the values of 119905lowast1 119878lowast
119876lowast and TC(119905lowast
1) all are temperately sensitive to the changes of
120572 for the modelFrom Table 3 we find that 119878lowast 119876lowast and TC(119905lowast
1) increase
as the deterioration parameter 120573 increases for both modelsAlso we find that the replenishment time 119905lowast
1increases as 120573
increases for themodel startingwith shortageWe see that thevalues of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) all are moderately sensitive
to the changes of 120573 for the model
5 Conclusion
This paper studies an inventory model for Weibull-distributed deterioration with trapezoidal type demand rateand partial backlogging First an optimal replenishmentstrategy for deteriorating items with the model starting withshortage is proposedThen numerical examples are providedto illustrate the theoretical results From the analysis of Tables1ndash3 it can be found that the replenishment time point 119905lowast
1
order quantity 119876lowast and the total average cost TC(119905lowast
1) are
moderately sensitive to the changes of 120572 and 120573 and lowlysensitive to the changes of 120575 respectively However if theorder cycle (119879) is a decision variable such inventory modelwill be of more practical significance which provides us aninteresting topic in future
Competing Interests
The author declares that there are no competing interests
Acknowledgments
The research is supported partly by National Natural ScienceFoundation of China (Grant no 71502100)
References
[1] P M Ghare and G F Schrader ldquoA model for exponentiallydecaying inventoriesrdquo Journal of Industrial Engineering vol 14no 3 pp 238ndash243 1963
[2] R P Covert and G C Philip ldquoAn EOQ model for items withweibull distribution deteriorationrdquo AIIE Transaction vol 5 no4 pp 323ndash326 1973
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 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
8 Mathematical Problems in Engineering
Basing on the analyzed results above we provide theresults which ensure the existence of a unique 119905
1 say 119905lowast
1 to
minimize the total average cost for the model system startingwith shortages The procedure is as follows
Step 0 Input the parameters 1205831 1205832 119879 1198861 1198862 1198871 120572 120573 120575
1198630 and c
119894 119894 = 1 2 3 4
Step 1 Calculate Δ1and 1198892C
1(1199051)1198891199052
1
Step 11 If Δ1gt 0 and 119889
2C1(1199051)1198891199052
1gt 0 solve a unique 119905lowast
1
from119889C1(1199051)1198891199051= 0 and the total average cost and the order
quantity can be obtained by (10) and (16) respectively
Step 12 If Δ1le 0 and 119889
2C1(1199051)1198891199052
1gt 0 then the optimal
average cost and the optimal order quantity can be obtainedby (10) and (16) at 119905lowast
1= 1205831 respectively
Step 13 If the sign of 1198892C1(1199051)1198891199052
1is indefinite then solve
equation 119889C1(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C1(1199051) to obtain the optimal
replenishment point compute the order quantity by (16)
Step 2 Calculate Δ2and 1198892C
2(1199051)1198891199052
1
Step 21 If Δ1lt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
seek a unique 119905lowast1from equation 119889C
2(1199051)1198891199051= 0 and the
optimal total average cost and the optimal order quantity canbe obtained by (24) and (29) at 119905
1= 119905lowast
1 respectively
Step 22 If Δ1gt 0 Δ
2gt 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205831 and compute the order
quantity by (29)
Step 23 If Δ1lt 0 Δ
2le 0 and 119889
2C2(1199051)1198891199052
1gt 0 then
C2(1199051) obtains itsminimumat 119905lowast
1= 1205832 and compute the order
quantity by (29)
Step 24 If the sign of 1198892C2(1199051)1198891199052
1is indefinite then solve
equation 119889C2(1199051)1198891199051= 0 If the solution is not unique by
comparing their average costs C2(1199051) to obtain the optimal
replenishment point compute the optimal order quantity by(29)
Step 3 Calculate 1198892C3(1199051)1198891199052
1
Step 31 If Δ2lt 0 and 119889
2C3(1199051)1198891199052
1gt 0 seek the unique
119905lowast
1from equation 119889C
3(1199051)1198891199051= 0 and the total average
cost and the order quantity can be obtained by (37) and (42)respectively
Step 32 If Δ2ge 0 and 1198892C
3(1199051)1198891199052
1gt 0 then C
3(1199051) obtains
its minimum at 119905lowast1= 1205832 and compute the order quantity by
(42)
Step 33 If the sign of 1198892C3(1199051)1198891199052
1is indefinite then
solve equation 119889C3(1199051)1198891199051
= 0 If the solution is notunique by comparing their average costs C
3(1199051) to obtain the
optimal replenishment point compute the order quantity by(42)
Step 4 Find TC(119905lowast1) = minC
1(119905lowast
1)C2(119905lowast
1)C3(119905lowast
1) and
accordingly select the optimum 119905lowast
1 and then obtain corre-
sponding optimal order quantity 119876lowast
4 Numerical Examples andSensitivity Analysis
In order to demonstrate the procedure to obtain the optimalsolution of the model we present four examples for themodel Examples are based on piece-wise demand rate suchas 119891(119905) = 119886
1+ 1198871119905 and 119892(119905) = 119886
2119890minus1198872119905
Example 1 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 004 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $2 c
2= $3 c
3= $12 c
4= $8
Since Δ1
= 18809 gt 0 Δ2
= 222869 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1) 1198892C
2(1199051)1198891199052
1gt 0
for all 1199051
isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
2(1205831) = 763891 in
[1205831 1205832) and C
3(1199062) = 126358 in [120583
2 119879) By solving equation
119889C1(1199051)1198891199051= 0 we have 119905lowast
1= 3664 From (10) we obtain
C1(119905lowast
1) = 762759 Therefore the optimal ordering quantity is
119876lowast= 637839 and the total average cost is TC(119905lowast
1) = 762759
Example 2 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 8 weeks
120572 = 0005 120573 = 2 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus257889 lt 0 Δ2
= 139269 gt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin [1205831 1205832) and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin (1205832 119879)
then we get local minimum value C1(1205831) = 1805883 in
(0 1205831] and C
3(1199062) = 158232 in [120583
2 119879) By solving equation
119889C2(1199051)1198891199051= 0 we have 119905lowast
1= 6559 From (24) we obtain
C2(119905lowast
1) = 1480324 Therefore the optimal ordering quantity
is 119876lowast = 571827 and the total average cost is TC(119905lowast1) =
1480324
Example 3 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 2 120575 = 006 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus249329 lt 0 Δ2
= minus121869 lt 01198892C1(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0 for
all 1199051isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin [1205832 119879)
then we get local minimum value C1(1205831) = 1417134 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 9120 From (37) we obtain
C3(119905lowast
1) = 889767 Therefore the optimal ordering quantity is
119876lowast= 278960 and the total average cost is TC(119905lowast
1) = 889767
Example 4 The parameter values of an inventory system aregiven as follows 119879 = 12 weeks 120583
1= 4 weeks 120583
2= 6 weeks
120572 = 0005 120573 = 16 120575 = 002 1198861= 30 units 119887
1= 5 units
1198862= 100 units 119860
0= $500 c
1= $5 c
2= $10 c
3= $12 c
4= $8
Since Δ1
= minus197263 lt 0 Δ2
= minus86675 lt 01198892C4(1199051)1198891199052
1gt 0 for all 119905
1isin (0 120583
1] 1198892C
2(1199051)1198891199052
1gt 0
Mathematical Problems in Engineering 9
Table 1 The sensitivity of 120575 for the model in Example 1
120575 0 001 002 003 004 005 006 007 008119905lowast
134366 34905 35463 36042 36642 37263 37907 38574 39265
119878lowast 5150829 5150829 5114210 5075965 5036066 4994433 4950985 4905641 4858321119876lowast 6481034 6481034 6448369 6414184 6378392 6340900 6301614 6260436 6217267
TC2(119905lowast
1) 7752147 7718437 7684699 7649983 7614252 7577473 7539611 750063 7460498
Table 2 The sensitivity of 120572 for the models in Example 1
120572 0 0001 0002 0003 0004 0005 0006 0007 0008119905lowast
130141 31325 32569 33870 35229 36642 38105 39614 41171
119878lowast 4158882 4320629 4489624 4665543 4847907 5036066 5229206 5426345 5755923119876lowast 5228982 5439279 5659745 5890106 6129883 6378392 6634729 6897782 7230000
TC2(119905lowast
1) 6385314 6614087 6851585 7097639 7351984 7614252 7883972 8160570 8450147
Table 3 The sensitivity of 120573 for the models in Example 1
120573 14 15 16 17 18 19 20 21 22119905lowast
131114 31995 32513 33176 340332 35148 36613 38551 41128
119878lowast 4275053 4401612 4471691 4562029 4679084 4831709 5036066 5295334 5641669119876lowast 5384988 5547959 5639494 5757494 5910495 6110248 6378392 6720262 7179964
TC2(119905lowast
1) 6444596 6703559 6801679 6929359 7095918 7314743 7614252 7976508 8450147
for all 1199051
isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
1(1205831) = 125259 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 8627 From (37) we obtain
TC(119905lowast1) = 842935 Therefore the optimal ordering quantity
is 119876lowast = 241524 obtained from (42) and the minimum costTC(119905lowast1) = 842935
In order to clearly indicate the effects of parameterssuch as 120575 120572 and 120573 on 119878
lowast 119876lowast and TC119894(119905lowast
1) (119894 = 1 2)
respectively we study the sensitivity of different parametersunder the studied inventory modelThe sensitivity analysis isperformed on the base of Example 1 and the results are shownin Tables 1ndash3 respectively
From Table 1 we observe that inventory level 119878lowast orderquantity 119876
lowast and the total average cost TC119894(119905lowast
1) (119894 = 1 2)
gradually decrease as the shortage parameter 120575 increasesfor the model and the replenishment time 119905lowast
1increases as
120575 increases for the model starting with shortage Also weobserve that the values of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) are slightly
sensitive to the changes of 120575From Table 2 we find that when the deterioration
parameter 120572 increases then 119878lowast 119876lowast and TC(119905lowast1) increase for
both models and the replenishment time 119905lowast1increases for the
model starting with shortage We see that the values of 119905lowast1 119878lowast
119876lowast and TC(119905lowast
1) all are temperately sensitive to the changes of
120572 for the modelFrom Table 3 we find that 119878lowast 119876lowast and TC(119905lowast
1) increase
as the deterioration parameter 120573 increases for both modelsAlso we find that the replenishment time 119905lowast
1increases as 120573
increases for themodel startingwith shortageWe see that thevalues of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) all are moderately sensitive
to the changes of 120573 for the model
5 Conclusion
This paper studies an inventory model for Weibull-distributed deterioration with trapezoidal type demand rateand partial backlogging First an optimal replenishmentstrategy for deteriorating items with the model starting withshortage is proposedThen numerical examples are providedto illustrate the theoretical results From the analysis of Tables1ndash3 it can be found that the replenishment time point 119905lowast
1
order quantity 119876lowast and the total average cost TC(119905lowast
1) are
moderately sensitive to the changes of 120572 and 120573 and lowlysensitive to the changes of 120575 respectively However if theorder cycle (119879) is a decision variable such inventory modelwill be of more practical significance which provides us aninteresting topic in future
Competing Interests
The author declares that there are no competing interests
Acknowledgments
The research is supported partly by National Natural ScienceFoundation of China (Grant no 71502100)
References
[1] P M Ghare and G F Schrader ldquoA model for exponentiallydecaying inventoriesrdquo Journal of Industrial Engineering vol 14no 3 pp 238ndash243 1963
[2] R P Covert and G C Philip ldquoAn EOQ model for items withweibull distribution deteriorationrdquo AIIE Transaction vol 5 no4 pp 323ndash326 1973
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 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
Mathematical Problems in Engineering 9
Table 1 The sensitivity of 120575 for the model in Example 1
120575 0 001 002 003 004 005 006 007 008119905lowast
134366 34905 35463 36042 36642 37263 37907 38574 39265
119878lowast 5150829 5150829 5114210 5075965 5036066 4994433 4950985 4905641 4858321119876lowast 6481034 6481034 6448369 6414184 6378392 6340900 6301614 6260436 6217267
TC2(119905lowast
1) 7752147 7718437 7684699 7649983 7614252 7577473 7539611 750063 7460498
Table 2 The sensitivity of 120572 for the models in Example 1
120572 0 0001 0002 0003 0004 0005 0006 0007 0008119905lowast
130141 31325 32569 33870 35229 36642 38105 39614 41171
119878lowast 4158882 4320629 4489624 4665543 4847907 5036066 5229206 5426345 5755923119876lowast 5228982 5439279 5659745 5890106 6129883 6378392 6634729 6897782 7230000
TC2(119905lowast
1) 6385314 6614087 6851585 7097639 7351984 7614252 7883972 8160570 8450147
Table 3 The sensitivity of 120573 for the models in Example 1
120573 14 15 16 17 18 19 20 21 22119905lowast
131114 31995 32513 33176 340332 35148 36613 38551 41128
119878lowast 4275053 4401612 4471691 4562029 4679084 4831709 5036066 5295334 5641669119876lowast 5384988 5547959 5639494 5757494 5910495 6110248 6378392 6720262 7179964
TC2(119905lowast
1) 6444596 6703559 6801679 6929359 7095918 7314743 7614252 7976508 8450147
for all 1199051
isin (1205831 1205832] and 119889
2C3(1199051)1198891199052
1gt 0 for all 119905
1isin
[1205832 119879) thenwe get localminimumvalueC
1(1205831) = 125259 in
(0 1205831] and C
2(1199062) = 1052016 in (120583
1 1205832] By solving equation
119889C3(1199051)1198891199051= 0 we have 119905lowast
1= 8627 From (37) we obtain
TC(119905lowast1) = 842935 Therefore the optimal ordering quantity
is 119876lowast = 241524 obtained from (42) and the minimum costTC(119905lowast1) = 842935
In order to clearly indicate the effects of parameterssuch as 120575 120572 and 120573 on 119878
lowast 119876lowast and TC119894(119905lowast
1) (119894 = 1 2)
respectively we study the sensitivity of different parametersunder the studied inventory modelThe sensitivity analysis isperformed on the base of Example 1 and the results are shownin Tables 1ndash3 respectively
From Table 1 we observe that inventory level 119878lowast orderquantity 119876
lowast and the total average cost TC119894(119905lowast
1) (119894 = 1 2)
gradually decrease as the shortage parameter 120575 increasesfor the model and the replenishment time 119905lowast
1increases as
120575 increases for the model starting with shortage Also weobserve that the values of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) are slightly
sensitive to the changes of 120575From Table 2 we find that when the deterioration
parameter 120572 increases then 119878lowast 119876lowast and TC(119905lowast1) increase for
both models and the replenishment time 119905lowast1increases for the
model starting with shortage We see that the values of 119905lowast1 119878lowast
119876lowast and TC(119905lowast
1) all are temperately sensitive to the changes of
120572 for the modelFrom Table 3 we find that 119878lowast 119876lowast and TC(119905lowast
1) increase
as the deterioration parameter 120573 increases for both modelsAlso we find that the replenishment time 119905lowast
1increases as 120573
increases for themodel startingwith shortageWe see that thevalues of 119905lowast
1 119878lowast 119876lowast and TC(119905lowast
1) all are moderately sensitive
to the changes of 120573 for the model
5 Conclusion
This paper studies an inventory model for Weibull-distributed deterioration with trapezoidal type demand rateand partial backlogging First an optimal replenishmentstrategy for deteriorating items with the model starting withshortage is proposedThen numerical examples are providedto illustrate the theoretical results From the analysis of Tables1ndash3 it can be found that the replenishment time point 119905lowast
1
order quantity 119876lowast and the total average cost TC(119905lowast
1) are
moderately sensitive to the changes of 120572 and 120573 and lowlysensitive to the changes of 120575 respectively However if theorder cycle (119879) is a decision variable such inventory modelwill be of more practical significance which provides us aninteresting topic in future
Competing Interests
The author declares that there are no competing interests
Acknowledgments
The research is supported partly by National Natural ScienceFoundation of China (Grant no 71502100)
References
[1] P M Ghare and G F Schrader ldquoA model for exponentiallydecaying inventoriesrdquo Journal of Industrial Engineering vol 14no 3 pp 238ndash243 1963
[2] R P Covert and G C Philip ldquoAn EOQ model for items withweibull distribution deteriorationrdquo AIIE Transaction vol 5 no4 pp 323ndash326 1973
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 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
10 Mathematical Problems in Engineering
[3] H-MWee J C P Yu and S T Law ldquoTwo-warehouse inventorymodel with partial backordering and Weibull distributiondeterioration under inflationrdquo Journal of the Chinese Institute ofIndustrial Engineers vol 22 no 6 pp 451ndash462 2005
[4] G K Yang R Lin J Lin K-C Hung P Chu and WChouhuang ldquoNote on inventory models with Weibull distribu-tion deteriorationrdquo Production Planning amp Control vol 22 no4 pp 437ndash444 2011
[5] N H Shah H N Soni and K A Patel ldquoOptimizing inventoryandmarketing policy for non-instantaneous deteriorating itemswith generalized type deterioration and holding cost ratesrdquoOmega vol 41 no 2 pp 421ndash430 2013
[6] C-T Yang L-YOuyangH-F Yen andK-L Lee ldquoJoint pricingand ordering policies for deteriorating item with retail price-dependent demand in response to announced supply priceincreaserdquo Journal of Industrial and Management Optimizationvol 9 no 2 pp 437ndash454 2013
[7] R M Hill ldquoInventory models for increasing demand followedby level demandrdquo Journal of the Operational Research Societyvol 46 no 10 pp 1250ndash1259 1995
[8] B Mandal and A K Pal ldquoOrder level inventory system withramp type demand rate for deteriorating itemsrdquo Journal ofInterdisciplinary Mathematics vol 1 no 1 pp 49ndash66 1998
[9] K-S Wu ldquoAn EOQ inventory model for items with Weibulldistribution deterioration ramp type demand rate and partialbackloggingrdquo Production Planning amp Control vol 12 no 8 pp787ndash793 2001
[10] K Skouri I Konstantaras S Papachristos and I Ganas ldquoInven-tory models with ramp type demand rate partial backloggingandWeibull deterioration raterdquoEuropean Journal ofOperationalResearch vol 192 no 1 pp 79ndash92 2009
[11] Y Tan and M X Weng ldquoA discrete-in-time deterioratinginventory model with time-varying demand variable deterio-ration rate and waiting-time-dependent partial backloggingrdquoInternational Journal of Systems Science vol 44 no 8 pp 1483ndash1493 2013
[12] S Pal G S Mahapatra and G P Samanta ldquoA productioninventory model for deteriorating item with ramp type demandallowing inflation and shortages under fuzzinessrdquo EconomicModelling vol 46 no 2 pp 334ndash345 2015
[13] M A Ahmed T A Al-Khamis and L Benkherouf ldquoInventorymodels with ramp type demand rate partial backlogging andgeneral deterioration raterdquo Applied Mathematics and Computa-tion vol 219 no 9 pp 4288ndash4307 2013
[14] S Agrawal S Banerjee and S Papachristos ldquoInventory modelwith deteriorating items ramp-type demand and partiallybacklogged shortages for a two warehouse systemrdquo AppliedMathematical Modelling vol 37 no 20-21 pp 8912ndash8929 2013
[15] T Singh and H Pattnayak ldquoAn EOQ inventory model fordeteriorating items with varying trapezoidal type demand rateand Weibull distribution deteriorationrdquo Journal of Informationand Optimization Sciences vol 34 no 6 pp 341ndash360 2013
[16] J Wu L-Y Ouyang L E Cardenas-Barron and S K GoyalldquoOptimal credit period and lot size for deteriorating itemswith expiration dates under two-level trade credit financingrdquoEuropean Journal of Operational Research vol 237 no 3 pp898ndash908 2014
[17] M Rabbani N P Zia and H Rafiei ldquoCoordinated replen-ishment and marketing policies for non-instantaneous stockdeterioration problemrdquo Computers amp Industrial Engineeringvol 88 pp 49ndash62 2015
[18] M B Cheng and G Q Wang ldquoA note on the inventory modelfor deteriorating items with trapezoidal type demand raterdquoComputers amp Industrial Engineering vol 56 no 4 pp 1296ndash1300 2009
[19] M B Cheng B X Zhang and G QWang ldquoOptimal policy fordeteriorating items with trapezoidal type demand and partialbackloggingrdquoAppliedMathematical Modelling vol 35 no 7 pp3552ndash3560 2011
[20] R Uthayakumar and M Rameswari ldquoAn economic produc-tion quantity model for defective items with trapezoidal typedemand raterdquo Journal of Optimization Theory and Applicationsvol 154 no 3 pp 1055ndash1079 2012
[21] K-P Lin ldquoAn extended inventory models with trapezoidal typedemandsrdquo Applied Mathematics and Computation vol 219 no24 pp 11414ndash11419 2013
[22] J Lin K-C Hung and P Julian ldquoTechnical note on inventorymodel with trapezoidal type demandrdquo Applied MathematicalModelling vol 38 no 19-20 pp 4941ndash4948 2014
[23] L Zhao ldquoAn inventory model under trapezoidal type demandWeibull-distributed deterioration and partial backloggingrdquoJournal of Applied Mathematics vol 2014 Article ID 747419 10pages 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
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