eoq inventory models for deteriorating item with …inventory model for two parameters weibull...

ISSN (e): 2250 – 3005 || Volume, 06 || Issue, 01 ||January – 2016 ||

International Journal of Computational Engineering Research (IJCER)

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EOQ Inventory Models for Deteriorating Item with Weibull

Deterioration and Time-Varying Quadratic Holding Cost

1Naresh Kumar Kaliraman,

2Ritu Raj,

3DrShalini Chandra,

4Dr Harish Chaudhary

1Research Scholar, Department of Mathematics & Statistics, Banasthali University, P.O.

Banasthali Vidyapith, Banasthali – 304022, Rajasthan, India, 2Research Scholar, Department of Mathematics & Statistics, Banasthali University, P.O.

Banasthali Vidyapith, Banasthali – 304022, Rajasthan, India, 3Associate Professor, Department of Mathematics & Statistics, Banasthali University,

P.O. Banasthali Vidyapith,Banasthali – 304022, Rajasthan, India 4Assistant Professor, Department of Management Studies, Indian Institute of Technology,

Delhi, New Delhi-110016, India,

I. Introduction

An economic order quantity model determines the quantity a company or a retailer must order to

minimize the total inventory cost by balancing the inventory holding cost and fixed ordering cost. In the usual

inventory system, it was considered that the buyer pays to vendor as soon as he receives the goods. Goods

deteriorate and their valuesdecreases over time. Electronic equipment may become outdated as technology

changes; fashion trends depreciate the value of clothes over time; batteries die out as they are old. The outcome

of time is even mostimportantfor consumable goods such as foodstuff and drugs.An inventory model for fashion

goods deteriorating at the end of the storage period was developed by Whitin [29]. Ghare and Schrader [7]

developed an exponentially deteriorating inventory model. A replenishment inventory model for radioactive

nuclide generators presented by Emmons [6].An order- level inventory model for decaying items with a

deterministic rate of deterioration was developed by Shah and Jaiswal [24]. The retailer receives the delivery of

goods and deterioration starts at that moment in all inventory models for decaying items. In real life situation,

some customers would like to wait for backlogging during the shortage phase but the others would not.

Therefore, the opportunity cost due to lost sales should be considered in the inventory models. Murdeswar[18];

Goyal et al. [11]; Hariga [15];Chakrabarti and Chaudhuri [2]; Hariga and Alyan [16] developed many inventory

models assuming that shortages are completely backlogged. Chang and Dye [3] considered that backlogging

shortages depends on lead time. An article in the field of deteriorating items with shortages has provided the

economic order quantity with a known market demand rate developed by Wee [28]. Sana [22] and Roy et al.

[19, 20, 21] developed many inventory models considering partial backlogging rates. The length of the lead time

for the next replenishment becomes main factor for determining whether the backlogging will be accepted or

not. A detailed review of deteriorating inventory literatures was provided by Goyal and Giri [12]. They

considered variable demand rate for many inventory models. Silver and Meal [25] considered time-varying

demand rate in their inventory models.

Abstract This paper develops an EOQ inventory model for deterioratingitemswith two parameters Weibull

deterioration. Shortages are permissible andpartially backlogged. In this model we consider time

varying quadratic holding cost and ramp-type demand. The model is developed under two different

replenishment policies: (i) Starting with no shortages (ii) Starting with shortages.The aim of this study

is to find the optimal solutiontominimizing the total inventory costs for both the above mentioned

strategies. To elevate the model a numerical example has been carried out and a sensitivity analysis

occurred to study the result of parameters on essential variables and the entire cost of this model.

Keywords: Inventory, Partial Backlogging, Quadratic Holding Cost, Ramp-Type Demand, Single Item,

Shortages, Weibull Deterioration

EOQ Inventory Models for Deteriorating Item with…

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In recent research, Covert and Philip [4] presented an inventory model where the time to deterioration

is considered with two parameter Weibull distribution deterioration. Ghosh and Chaudhuri [9] developed an

inventory model for two parameters Weibull deteriorating items, with quadratic demand rate and shortages were

permissible. Haley and Higgins [13] extended the inventory policy for two part trade credit, where the vendors

consider cash discount for paying within a specified period and due in a large credit period. Goyal [10]

developed the EOQ model under the conditions of permissible delay in payments. Aggarwal and Jaggi [1]

considered deteriorating items to develop ordering policy under the conditions of permissible delay in

payments.Sarkar et al. [23] developed an EOQ model for deteriorating items with shortages and time value of

money.Teng et al. [27] proposed an EOQ model with deteriorating objects, price perceptive demand, and

different selling and purchasing prices under trade credit policy and discussed the result obtained by Goyal [10].

Donaldson [5] provided ananalytical solution procedure of the basic inventory policy for a case of positive

linear trend in demand.Giri et al. [8] developed an economic order quantity model for single deteriorating item

with ramp type demand and Weibull distribution deterioration. Shortage is allowed and backlogged completely

over an infinite planning horizon. Skouri et al. [26] provided an inventory model with time dependent Weibull

deterioration rate, partial backlogging of unsatisfied demand and general ramp type demand rate. The model is

developed under two different replenishment policies: (a) starting with no shortages and (b) starting with

shortages. The backlogging rate is non-increasing function of the waiting time up to the next replenishment. Wu

[30] presented an economic order quantity model with ramp type demand and Weibull distribution deterioration.

Shortages are allowed partial backlogging and the rate of backlogging is dependent on waiting time for the next

replenishment.

The main purpose of this paper is to show that there exist a unique optimal cycle time to minimize the total

inventory cost.A numerical illustrationand sensitivity analysisis presented to show the result of the proposed

model.

II. Assumptions The following assumptions are mandatory todevelop mathematical model:

1. The Inventory system consider single item.

2. The inventory level defined by I t at time t. i.e.

1

1

0 , 0

0 ,

a

b

I t t t

I t

I t t t T

, where I t is retailer’s stock level.

3. The demand rate D t is a ramp type function of time and is defined as:

,

,

D t tD t

D t

, D t is positive and continuous for time 0,t T

4. The lead time is zero.

5. Shortagesare allowed and partially backlogged.

6. Cycle horizon is considered as T units of time and replenishment rate is infinite.

7. The rate of deterioration is time dependent, which is two parameters Weibulldistribution deterioration,

denotedby1

t

,where 0 1 , 1 and 0t .A value of 1t defines that the failure rate

decreases with time. This happens if imperfect items are deterioratingfirst and the failure rate decreases with

time. A value of 1t defines that the failure rate is deterministic over time. A value of 1t definesthat

the failure rate increases with time.

8. The ordering quantity level is considered as Q .

9. Holding cost H C t is a time dependent quadratic function and is defined as

2

H C t a b t ct ,

where 0a ,

0b and 0c .

III. Notations

The following notations are mandatory to develop mathematical model:

H C : The stock holding cost per year.

cS : The shortage cost per year.

http://en.wikipedia.org/wiki/Failure_rate


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T : Length of time is considered as annually.

1t : Time at which inventory becomes zero.

T C : Total inventory cost.

L S : Lost sale cost.

I t :Inventory level at time t .

: Ramp-type demand function of time.

cD :Deterioration cost.

IV. Mathematical Modelwithout Shortages: The inventory model is starting without shortages. At the start of the cycle, the production starts at

0t and continuesup to1

t t . At this time theinventory level reaches its maximum level and then production

is stopped. The inventory lessens to zero due to demand and deteriorationduring 10, t and falls to zero at

1t t .

Thus, shortages occur during 1,t T which is partially backlogged.Therefore, the inventoryis described by the

followingdifferential equations:

1

, 0 ... 1a

a

d I tI t D t t t

d t

with boundary condition 10

aI t

1

, . . . 2b

d I tD t T t t t T

d t

with boundary condition 10

bI t

Where

1

,,

,

D t tD t a n d t

D t

,

There are two cases arise: i t and ii t

4.1. Case (i) t

From Eq. (1), we have

2 2 2 2

1 1

1... 3

2 2

t

aI t D e t t t t

Eq. (2), solved by the following two ways:

1,

, , I I

b

b

d I tD t T t t t i

d t

d I tD T t t T ii

d t

We have,

2 2 3 3

1 1

1... 4

2 3b

TI t D t t t t

and



2 3 2

1 11

1... 5

2

1 1

2 3b

T t t tI t D B T t

Where

2

1

31

2 6

1B TT

Total amount of deterioration during 10, t

1 1

2

1

0 0

... 62

t t

t

c

DD D te d t D td t t

Total cost of holding during 10, t is

1

4 3 3 2 3 4 2 4

1 1 1 1 1 1 1

0

... 78 3

t

a

b aH C H t I t d t D t t B t C t E t F t

Where

2 2 2

, , ,3 1 3 1 2 3 2 4 2 4 2 2 4

a a a b b bB C E F

The shortage cost during 1,t T is

1 1

2 2 2

2 1 1 1. . .

16 88 3

1 2

T T

c b b b

t t

S I t d t I t d t I B t T T t tt d t D

Where

3 4 3 3 2 2

2 1

1 1

6 1 2 6 2

T TB B T T T

Lost sales cost during 1

,t T is

1

3 2

3 1 1

12 3 1 ... 91 1

6

T

t

L S T t D td t D T t d t B t TD t

Where 2 2 2 3

36 3 3 3B T T T

Ordering quantity during 0 , T is

1

10

2

3 2 2 2 3

1 1 1

1 1 1... 1 0

3 2 2 2 2 2 6

t T

t

t

O Q e D td t D t T t d t D T t d t

TO Q D t t t T T

Total cost during 0 , T is the sum of deterioration cost, holding cost, shortage cost and lost sales cost is given

by

1 1 c cT C t H C S L S D

2

2 4 3 3 2 3

1 1 1 1 1

1 1

4 2 4

1 21

2 3

1

12

2 8 4 3...

11

6

1 1

2

bt t a t B tT

BT T

C t

T C t D

E t F t t B



t

TC1

4.1.1. Solution:

1 3 2 2

1 1 1 11

2 2 3 2 31

11 1 1

2

2 3 2 32 ... 1 2

4 2 4 1

bt t a t B t C

T CD

tt E t

T

Tt T tF

2

1 1 1

2

1 2 1 21

1 1 12

12 2 2

1

1 3 2 2 22

3 2 2 2 3 3 4 ... 1 3

2 13 2 4

bt t a t B

T CD t C t E t

T

t

T

TF t

Main objective is to minimize the total relevant cost for the inventory model starting without shortages. The

essential condition to minimize the total relevant cost is1

1

0T C

t

, we have

1 3 2 2

1 1 1 1

2 2 3 2 3

1 1 1

2

1

2 3 2 32

4 1 0 . 12 .4 . 4

bt t a t B t C

t E t F

T

tt T T

Using the software Mathematica, we can calculate the optimal value of1

t by Eq. (14) and the optimal

value 1 1T C t of the total relevant cost is determined by Eq. (11). The optimal value of

1t satisfy the sufficient

condition for minimizing total relevant cost 1 1T C t is

2

1

2

1

0 ... 1 5T C

t

The sufficient condition is satisfied.

Fig.1. Graphical representation of 1

T C and 1

t .

4.1.2. Numerical example:

Let us consider

1 32

$ 1 / , $ 0 .5 / , $ 0 .5 / , $ 5 / , 0 .1,

0 .3 5 , 0 .1 4 , 1 .4 , 0 .0 2 , 0 .0 0 0 5 , 0 .0 0 2

2 , 1 , 1 0 0 ,

1, 0 .5

, 0 .0 0 0 1 5

c ca u n it b u n it S u n it D u n it T y e a r D

y e a r

B B B C E FB



Thus, the optimal value of1

t is1

0 .4 4t . The optimal ordering quantity is 1 7 .5O Q

. The minimum

relevant cost is1

2 8 .6T C .

4.1.3. SensitivityAnalysis:

To know, how the optimal solution is affected by the values of constraints, we derive the sensitivity analysis for

some constraints. The specific values of some constraints are increased or decreased

by 25% , 25% and 50% , 50% .Thus,we computethe values of1

t

and1

T C

with the help of increased or

decreased values of c

S andc

D .The result of the minimum relevant cost is existing in the following table 1.

Table: 1

From the result of above table, we observe that total relevant cost and ordering quantity is much affected by

deterioration cost and shortage cost, and other parameters are less sensitive.

4.1.4. When Holding Cost is a Quadratic Function:

We have

1 1

2 2 2 2 2

1 1 1

0 0

1

2 2

t t

t

aH C H t I t d t a b t c t D e t t t t d t

5 4 3 3 2 3 4 2 4 2 5

1 1 1 1 1 1 1 1 2 1

5

1. . . 1 6

1 5 8 3

c b aH C D t t t B t C t E t F t F Ft t

Where

2

2 2

1 2

1 1 11 , , ,

3 1 1 2 3 4 2 2

1, ,

2 2 4 3 3 2 5

1

5 3

a a bB C E

cb cF F F

Total cost is the sum of quadratic holding cost, shortage cost, deterioration cost and lost sales cost during

0 , T


2 5 4 3 3 2 3

1 1 1 1 1 1

1 1

4 2 4 2 5

1

5 2 2

1 1 11 2 31 2

22 1 5 8 4 3

1

1 11

...

6

1 7

2

c bt t t a t B t C t

T C t D

T

t T T tE t F t F F t B B

4.1.5. Solution:

1 4

4

3 2 2 2 2

1 1 1 1 1 11

3 2 3 2 41 2

1 1 11 2 1 1

2 3 2 33 2 ... 1 8

4 2 4 5 12 5

c bt t t a tT

t T T t

B t C tT C

Dt

E t F t F F t

Parameters Actual

Values 50% Increased 50% Decreased 25%

Increased

25% Decreased

cS

0.5 0.75 0.25 0.625 0.375

cD

5 7.5 2.5 6.25 3.75

1t

0 .4 4 0 .5 0 .23 0 .5 0 .3 4

1T C

28.6 4 0 .7 5 2 1 .9 8

O Q

1 7 .5 18.9 1 8 .9 1 5 .9 3



3 2 1

1 1 1 1

3 2

1

2

2 1 2 2 21

1 1 12

12 3

1 1 2 1

41 3 2 2 2 3

3 2

2 2 2 3 3 4 2 3 2 4 ... 1 9

2 4 2 54 5 1

c bt t t a t B t

T

T

t

CD C t E t F t

t

F F t T T

Main objective is to minimize the total relevant cost of the inventory model starting without shortages. The

essential condition to minimize the total relevant cost is 1

1

0T C

t

, we have

1 4 3 2 2 2 2

1 1 1 1 1 1

3 2 3 2 4

1 1 1 2 1

4 2

1 1

2 3 2 3 43 2

2 4 25 1 0 ... 25 0

c bt t t a t B t C t E

t F

T

t T TF tt F t






2

1

2

1

0 ... 2 1T C

t



T C and 1

t .

4.1.6. Numerical Example:

Let us consider

$ 1 / , $ 0 .5 / , 1, $ 0 .5 / , $ 5 / , 0 .1, 2 , 1 , 1 0 0 ,

1, 0 .5

c ca u n it b u n it c S u n it D u n it T yea r D

yea r

31 2 1 20 .3 5 , 0 .1 4 , 1 .4 , 0 .0 1, 0 .0 0 0 5 , 0 .0 0 2 , 0 .0 0 0 1 5 , 0 .0 0 2 , 0 .0 0 0 2B B B C E F FB F


t is1

0 .4 2 5t . The optimum ordering quantity is 1 7 .5O Q

. The

minimum relevant cost is1

2 8 .6 9T C .

4.1.7. SensitivityAnalysis: To know, how the optimal solution is affected by the values of constraints, we derive the sensitivity analysis for

some constraints. The specific values of some constraints are increased or decreased


t

and1

T C



S andc

D .The result of minimum relevant cost is existing in the following table 2.

t

TC



Table: 2

Parameters Actual

Values 50%

Increased

50%

Decreased

25%

Increased

25% Decreased

cS

0.5 0.75 0.25 0.625 0.375

cD

5 7.5 2.5 6.25 3.75

1t

0 .425 0 .58 0 .23 0 .51 0 .33

1T C

28.7 40 .61 34 .9 22 .01

O Q

1 7 .5 18.9 1 8 .9 1 5 .9 3



4.2. Case (ii): t

The differential Eq.(1) becomes

1 1

, 0 , I I

, , I 0

a

a

a

a

d I tI t D t t

d t

d I tI t D t t t

d t

We have

2 2 1

1 1 1

1, 0 ... 2 2

2 2 1

t

aI t D e C t t t t t

Where

2 2

1

1 1 1

2 1 2C

and 1 1

1 1 1, ... 2 3

1 1

t

aI t D e t t t t t t

The differential Eq. (2) becomes

We have

1

2 2

1 1 1

1, ... 2 4

2

t

b

t

I t D T t d t D T t t t t t t T

Total amount of deterioration during 10, t

1 1

1

1

0

. .. 2 51

t t

t t

cD D te d t D e d t D d t D S t

Where, 2

21 1

2 2 1S

Total holding cost during 10, t is

1 1

1

0 0

t t

a a aH C H t I t d t a b t I t d t a b t I t d t

1 1

, , 0b

b

d I tD T t t t T I t

d t



3 2 1 2 3 2 2 2 3

1 1 3 5 1 4 8 1 7 1 6 1 9 1 1 0 1 2. . . 2 6

6 2

b aH C D t t C C t C C t C t C t C t C t C

Where,

2 2 4 2 2 3 2 1 4 31 1

2 2 2

4 3 21

1

1 1 1 1 5 1

1 2 3 2 2 3 2 1 2 4 32 2 2 1

1 1 1 1 1 1 1,

2 4 2 4 1 3 3 1 2 2 2 3 2

b C a CC b a b a

b Cb a a C

3 3 2 2 2 3 2 3

2 2 1

3 4 2

1 2 2

5 6

,1 2 2 1 2 1 1 21

, ,1 2 2 2 1 1 3 2 3

a b b a b bC a C

a b b b b b bC a C

2

7 9

2 1 2 2 2 2

8 1 02

1 1 11 , ,

1 2 2 2 1 1

1 1,

1 2 1 2 1 2 3 2 11

a aC C

a a b b bC C

The shortage cost during 1,t T is

1 1

33

2 2 1

1 1

2 2

1 1

1... 2 7

32 3

T T

c b

t t

StT

T t TI t d t D T t t t t d t D t

Lost sales cost during 1,t T is

1

1 1

12 ... 2 8

21

T

t

L S D T t d t D T t T t

Ordering quantity is during 0 , T is

1

10

2 2

2 2

1 11 ... 2 9

2 2 2 2

t T

t

t

O Q D te d t D d t D T t d t

TD t T t

Total cost is the sum of deterioration cost, holding cost, shortage cost and lost sales cost during 0 , T


2 3 2 2 3 2

1 0 1 9 1 6 1 7 1 4 8

3 51 2

2

3

1 1 11

2

3

2 1

2

1

...3 2

3 01

3 2

6 2

C t C t C t C t C C

C Cb aT C t D t

C

Tt T t t

T

TT T S



t

TC

4.2.1.Solution:

1

2 2 2 1 2

1 0 1 9 1 6 1

1 22

7 1 4

2

1

8 1

1

3 5

2 3 2 2 3

2 1 3 ... 3

1

1 31

6

222

C t C t C t

T C bD C t C C t t

T t T

t

aC TC

2 1 2

1 0 1 9 1 6

2

12

1 7 1 4 82

1

1

1

1

2 2 2 3 2 1 2 2

2 3 1 2

2

... 3 21

1 2

C t C t C

T CD t C t C C

t

tt b a T

Main objective is to minimize the total relevant cost of the inventory model starting without shortages. The


1

0T C

t

, we have

2 2 2 1 2 1

1 0 1 9 1 6 1 7 1

2

1 1 3

2

4 8 1

5

3 2

1 0 ... 3

2 3 2 2 3 2

1 3 21 6 2

3

C t C t C t C

t T t

t

b aC

T

C C

T

t

C






2

2

2

1

0 ... 3 4T C

t



T C and 1

t .




Let us consider

1 2 3 4 5 6

7 8 9 1 0

$ 1 / , $ 0 .5 / , $ 0 .5 / , $ 5 / , 0 .1, 2 , 1 ,

1 0 0 , 1

0 .1 2 , 0 .1 2 , 0 .0 2 , 0 .2 8 , 0 .0 0 2 , 0 .5 6 , 0 .0 0 2 5 ,

0 .0 1 7 , 0 .0 1 8

, 0 .5

, 0 .0 0 0 5 , 0 .0 0 0 2

c ca u n it b u n it S u n it D u n it T y e a r

D y e a r

C S C C C C C

C C C C


t is1

0 .5 4t . The optimum ordering quantity is 1 9 .9 5O Q

. The

minimum relevant cost is2

7 7 .0 3T C .


Total Quadratic holding cost during 10, t is

1 1

2 2

1

0 0

t t

a a aH C H t I t d t a b t c t I t d t a b t c t I t d t

4 3 2 1 23

1 1 1 3 4 1 7 1 6 1

3 4 2 2 2 3 2 4

5 1 1 0 1 8 1 9 1 1 1 1 2

1 2 6 2 1 ... 3 5

Cc b at t t C C t C t C t

H C D

C t C t C t C t C t C

Where,

2 2 4 4

3 3 1 3 2 3

2 1 1 1

2 2 2 4 3 2 2 5

4 5 3

1 1 1

2 3

2 5 3

6 2 3 1 2 3 2 2 3 2 8 2 4

2 2 4 2 2 4 3 1 0 3 1 2 4

2 2 5 2 1

a a a a a b b bC a C C C

b b b c c cC c C C

c a a

3 2 2 3 4 4 2 2 4

4

5 1 2 2 3 3

5 5 5

3

2 2 1 2 2 3 1 3 3 1 2 3

11 , ,

4 1 4 2 5 2 5 1 2 2 3 3

a a b b b b

c c c a b b c cC a

1 2 3 2 3

4 5 6

1, , 1 ,

1 2 3 2 3 2 1 1 3 2 3 1 2

a b c b c b b b b aC a C C

2 2 2 2 3 2

7 8 2

2

9 1 0

2

1 1

6 6 3 2, ,

6 1 2 1 1 3 2 1

1 1, ,

2 3 2 1 1 4 4 3 3 1

1 1

2 4 3 1

a a b c b c aC C

b c c c cC C

cC


0 , T


4 3 2 13

1 1 3 4 1 7 1

2 1

2 3 4 2 2 2

2

1

3 2 4

6 1 5 1 1 0 1 8 1 9 1 1

2

1 1 2

3

1 2 6 2 1 1...

13

2

62

1

3

3

Cc b at t C C t C t

T C t D

C

T t T T

TT Tt C t C t C t C t C t C S



4.2.4. Solution:

3 2

1 1 3 4 3 7 1

1 2 3 2 1 2 22

6 1 5

2

1

1 1 0 1 8 1 9 1

12 3

1 1 1

2 13 2

2 3 4 2 2 2 3 ..

1

. 3 7

2 4

c bt t a C C C C t

T CD C t C

T t T

t C t C t C

T

tt

C t

2 1

1 1 3 7 12

1 22

6 1 5 1 1 0 12

12 2 1 2 2

8 1 9 1 1 1 1

2 1

1 2 2 3 3 4 ... 3 8

2 1 2 2 2 2 2 3 2 3

2

2 4

c t b t a C C t

T CD C t C t C t

t

C t C t C t

T



1

0T C

t

, we have

3 2

1 1 3 4 3 7 1

1 2 3 2 1 2 2

6 1 5 1 1 0 1 8 1 9 1

2 3

1 1

2

1

1

2 13 2

2 3 4 2 2 2 3

2 4 0 ... 3 9

1c b

t t a C C C C t

C t C t C t C t C t

T

C

T t T

t






2

2

2

1

0 ... 4 0T C

t



Let us consider

1 2 3 4 5 6 7 8

9 1 0 1 1

0 .1 2 , 0 .0 2 , 0 .6 , 0 .3 9 , 0 .0 0 2 5 , 0 .0 1 7 , 0 .0 1 3, 0 . 0 0 0 5 ,

0 .0 0 0 2 , 0 .0

$ 1 / , $ 0 .5 / , 1, $ 0 .5 / , $ 5 / , 0 .1, 2 , 1 , 1 0 0 , 1,

5 3, 0 .0 0 0 2 5

0 .5

c ca u n it b u n it c S u n it D u n it T y e a r D

y e a r

C C C C C C C C

C C C


t is1

0 .5t . The optimum ordering quantity is 1 9 .9 5O Q

. The minimum

relevant cost is2

8 2 .8T C .

V. Mathematical Model with Shortages:

The cycle starts with shortages that occur during the period 10, t and shortages are partially backlogged.

Replenishment grasps the inventory level up to Q after time1

t . The inventory level depletes and falls to zero

at t T because of demand and deterioration during 1,t T

. Two cases occur

(i) 1

t

(ii)1

t



5.1. Case (i)1

t

Therefore, the inventory I t is described by the system of differential equations

during 0 , T

1

1 1

2

1

3

, 0 , 0 0 ... 4 1

, .. . 4 2

, 0 ... 4 3

d q tD t t t t t q

d t

d q tq t D t t t q q

d t

d q tq t D t T q T

d t


3

2

1 1

1... 4 4

2 3

tq t D t t


2 2

2 1

1... 4 5

2 2

tq t e D D t t

Where,

2 2

1 1

1

2 2 1D T T


1

3 2... 4 6

1

tq t e D D t t

Where,

1

2

1D T T

Total amount of deterioration during 1,t T

1

1 1

1 4

2

2 1

1 3

1

1 ... 4 72

T T

t t t

c

t t

c

D e D te d t D e d t D td t T

tt

D d t

DD D t D

Where,

2

3

22

4

1 1

,1 2

T

T DD T

Total holding cost during 1,t T is

1

T

t

H C a b t q t d t

4 3 2 4 3 2 1 2 3 2 4

5 6 1 1 1 1 1 1 7 1 8 1 9 1 1 0 1 1 1 1 1 2 1

1... 4 8

8 6 2

b aH C D D D t t b D t a D t D t D t D t D t D t D t

Where,



2

2 2 3 3 2 3 2 3

5 2 2

2

2 2 2 2 3 3

2

2

2 2 1 1 2 3 4 22

6 1 1

2 3 1 2 3

11

1 2 2 3 1 2 1

1,

1 2 6 8 2 2 4

b bD a D T b D a T T T

a b aT T a b D

a D a b bT T D a D b D

4

2

4 2 3 3 2 11 1

4

1 1 1 1,

2 2 2 2 3 3 2 2 2 1

b

b D a Da a

1 1

7 8 9 1 0

2 2

1 1 1 2

1 1 1 1, , , ,

4 2 2 3 2 2 2 1

,2 2 3 2 2 4

b D a Db aD D D D

a bD D

The shortage cost during 10, t is

1 1

2 4

1 1 1 1

0 0

... 4 91 2

t t

c

DS t t D t t t d t D t t t d t t

The lost sales cost during 10, t is

1

2 3

1 1 1

0

1 11 ... 5 0

2 6

t

L S t t D td t D t t


1

1

1

1 1

0

'

t T

t t t

t

O Q D t t t d t e D te d t D e d t

3 2 2 2 2 1 1

1 1 1 1 1

1 1' 1 ... 5 1

6 2 2 1O Q D t t t t T T

Total cost during 0 , T is the sum of deterioration cost, holding cost, shortage cost and lost sales cost is given

by

'


2

2 2 2

1

4 3 2 4 2 31

1 1 1 1 1 2 1 1 1 1

'

1

4 3 2 1

7 1 8 1 9 1 1 0 1 1 4 5 6

1

3 3

3

8 1 2 6 2D ... 5 2

2

2

b Db at t a D t D t D t

T C

D t D t D t D t

t t

D Dt D D D



5.1.1. Solution:

3 2 2 3 2 2

1 1 1 1 2 1 1 1 1

'

3 2 11

7 1 8 1 9 1

1

1

1

1

2

2

0 1 1 1

1

1

3

3 2 4 2 32 3 2

D 4 3 2 ..2 2

2 2. 5 3

1

b at t b D D t D t

T C

t

D t D t D tt

D

t

Dt t a D

2 2 2 2 1

1 1 1 2 1 1 1 1

2 '

2 11

7 1 8 12

1

2

2

1

1 2

9 1 1 0 1 1 13

3 2 3 2 4 2 2 2 32 3

D 3 42

2 3 3 ... 5 4

1 2

1 2 2

2

1 12

bt a t D t D t

T CD t D t

t

D t DD t

t

t b D


essential condition to minimize the total relevant cost is

'

1

1

0T C

t

, we have

3 2 2 3 2 2

1 1 1 1 2 1 1 1 1

3 2 1

7 1 8 1

2

2 1

1

9

1

3

1 1 0 1

1

1 1

3 2 4 2 32 3 2

2 2

2

4 3 2 1

. 5 5

2

0 ..

b at t b D D t D t

D t D t D t D

t

t

t

DD t a



value '

1 1T C t of the total relevant cost is determined by Eq. (52). The optimal value of


condition for minimizing total relevant cost '

1 1T C t is

2 '

1

2

1

0 ... 5 6T C

t



Let us consider

1 2 43 5 6 7 8

9 1 0 1 1 1 2

0 .3 6 , 1 .0 3,

$ 1 / , $ 0 .5 / , $ 0 .5 / , $ 5 / , 0 .1, 2 , 1 , 1 0 0 ,

1, 0

0 .1 4 , 0 .2 5 , 0 .0 9 6 , 0 .2 7 , 0 .0 0 6 2 5 , 0 .0 1 5 ,

0 .0 0 4 5 , 0 .0 0 6 , 0 .0 0 0 3 5 , 0 .0 0 0 1 5

.5

c c

D

a u n it b u n it S u n it D u n it T y e a r D

y e a r

D D D D D D D

D D D D


t is1

0 .0 3 5t . The optimum ordering quantity is

'3 9 .0 5O Q

. The

minimum relevant cost is'

13 3 .9T C

.



5.1.3.When Holding Cost is a Quadratic Function:

We have

1

2

T

t

H C a b t c t q t d t

5 4 3 2 3 1

1 1 7 1 1 1 1 1 5 6 8 1 9 1

2 3 4 2 2 4 5 2 51

1 0 1 1 1 1 1 1 2 1 1 3 1 1 4 1

1

1 0 8 2... 5 7

2

c bt t D t b D t a D t D D D t D t

H C Db D

D t D t t D t D t D t

Where,

2

2 2 2 3 3 4 4 2 4 2 42

2

2 2

2 3 2 3 2 2 2 2 4 4

5

3 3 2

2 2

2 3 4 1 2 4

11

1 2 3 1 2 2 4 1

3 1 2 1

b D a c D b c ca D T T T T T

b a cD T T T

b ab c D T a b D T

2 1 12

,

1

a DT

2

1 1

7 8 1 9 1 0

2 2

1 1 1 2 1 3 1 4

, , , ,6 3 3 2 2 1 2 2 3

1 1 1 1, , ,

4 2 2 2 2 4 5 2 2 2 2 5

c D a Da a a aD D c D D D

b b c cD D D D

2 2 5 2

2 3 4 5 2 41

6 1 1

2

2 3 5 4

3 2 11 1

1

1

2 3 6 8 1 0 2 2 5 2 2 4

1 1 1 1

2 2 3 5 2 2 4 2 2 3

,2 2 2 1

c D a b c c bD a D b D

a c b

b D a Da ac D


0 , T

'


5 4 3 2 5 2 4 2 3 2 2

1 1 7 1 1 1 1 1 4 1 1 2 1 1 0 1 1

'

1

5 4 3 2 1

1 3 1 1 1 1 8 1 1 1 9 1

2

2

1

3 31 4 5 6

2

13

1 0 8 1 2 6 2D ... 5 8

12

c bt t D t b D a D t D t D t D t t

T C

D t D t D t b D

t

D Dt D t t D D D

5.1.4. Solution:



4 3 2 2 4 2 3

1 1 7 1 1 1 1 4 1 1 2 1

'

2 2 2 1 4 3 21

1

1

2

0 1 1 1 3 1 1 1 1 8 1

1

1 1

1 1 9 1 13

2

3 3 2 5 2 42 2 3 2

D 2 3 2 2 5 4 3

1 1

c bt t D t b D a D D t D t

T CD t t D t D t D t

t

b D t D t

t

Dt

. . . 5 9

3 2 2 3 2 2

1 1 7 1 1 1 4 1 1 2 1

2 '

2 1 2 3 21

1 0 1 1 1 3 1 1 1 12

1

1 1

8 1 1 1 9 1

2

2 3 2 3 3 2 4 2 5 2 3 2 42 3 2

D 2 2 2 3 1 2 1 4 5 3 4

2 3 1

2

1 1

2

bc t t D t b D D t D t

T CD t t D t D t

t

D t b D t D Dt

2

13

. . . 6 0

1 t



'

1

1

0T C

t

, we have

4 3 2 2 4

1 1 7 1 1 1 1 4 1 1 2

2 3 2 2 2 1 4 3

1 1 0 1 1 1 3 1 1 1 1 8

2 1 1

1 1 1 9 1 1

1

2

3

3 3 2 5 2 42 2 3 2

2 3 2 2 5 4

3 1 1 0 ... 6 1

2

t

D

c bt t D t b D a D D t D

t D t t D t D t D

t b D t D t t



value '




1 1T C t is

2 '

1

2

1

0 ... 6 2T C

t



Let us consider

1 2 4 5 6 7 8 9

1 0 1 1 1

3

2

0 .3 6 ,

$ 1 / , $ 0

1 .0 3, 0 .1 4 ,

.5 / , $ 1

0 .2 5 , 0 .1 1 2 , 0 .3, 0 .0 4 6 , 0 .0 2 2 , 0 .0

/ ,

1 2 ,

0 .0 0 0 3, 0 .0 0 6 2 5 , 0 .0

$ 0 .5 / , $ 5 / , 0 .1, 2 , 1 , 1 0 0 ,

1, 0 .5

0

c ca u n it b u n it c u n it S u n it D u n it T y e a r D

y e a r

D D D D D D D D

D D

D

D

1 3 1 40 1 5 , 0 .0 1, 0 .0 0 0 3D D


t is1

0 .0 3 5t . The optimum ordering quantity is

'3 9 .0 5O Q

. The


13 3 .9T C

.

5.2. Case (ii)1

t

Therefore, the inventory I t is described by the system of differential equations

during 0 , T



1

1, 0 , 0 0 ... 6 3

d q tD t t t t q

d t

2

1 1, , . . . 6 4

d q tD t t t t q q

d t

3

1, , 0 ... 6 5

d q tq t D t t T q T

d t

From Eq.(63), we have

3

2

1 1

1, 0 ... 6 6

2 3

tq t D t t t

From Eq.(64), we have

2

2

2 1 1 1

1 1, ... 6 7

2 6 2q t D t t t t t t

FromEq.(65), we have

1

3 2 1, ... 6 8

1

tq t e D D t t t t T

Where,

1

2

1D T T

Total amount of deterioration during 1,t T

1

1 1

1

1 2 1 1 11 ... 6 9

1

T T

t t

c

t t

D e D e d t D d t D t D t t T t

Total holding cost during 1,t T is

1

3

2 3 3 2 1 2 2 2 3

1 2 1 2 1 1 3 1 4 1 5 1 6 1 7 1... 7 0

3

T

t

H C a b t q t d t

bD F a D t F t t F t F t F t F t F t

Where,

2

2 3 2 2 32

1 2

2 1 3 22

2 2 2 2 3

2

4 5 6

11

1 2 3 2 3 1 22 1

1 1 11 , ,

1 1 3 2 2 3

1 11 , 1 , ,

1 2 1 1

b Db a b aF T T T

a D b bT T T b D a T a D T F a b D F

a DaF F F

2

2

72,

2 1 2 32 1

b D a bF

The shortage cost during 10, t is



1 2 3 4

3 2

1 2 1 1 1

0

.. . 7 13 2 3 1 2

t

cS q t d t q t d t D t t t

The lost sales cost during 10, t is

1

1 1

0

2

9 1 0 1 11 1

13 ... 7 2

6

t

L S D t t t d t D t t td t F F tD

Where,

9 103 , 3 2F F


1

1

1

1 1

0

1

t T

t t

t

O Q D t t t d t D t t d t e D e d t

2

2 1 1 2 2

1 1 2 1 1 1 2

1 3 5 11 ... 7 3

1 1 2 2 6 2O Q D t t D t t t D

Total cost is the sum of deterioration cost, holding cost, shortage cost and lost sales cost during 0 , T

'


3 2 2

1 2 1 2 1 2 1 5

'

22 3

1 2 3 2 1 2 2 2 3

1 4 1 3 1 1 6 1 7 1 2

1 0

9 1

12 2 6

3 2 1... 7 4

1 1

6

1

61 1 2

b t F t a D t D t F

T C

F

F

D

t F t F t t F t F t D F T

5.2.1. Solution:

2 2 1

1 2 1 2 2 1

' 2

1 2 22

5 1 4 1 3 1 1

1

1 0

2 1 2 2

1 7 16

2 2 6

1 2 3 2 1 ... 7 51

2 2 2

6

1

3

1b t FF t a D D t

T CD F t F t F t t

t

F t F t

2 1

1 2 2 1 5 1

22 '

1 2 12

4 1 3 1 12

1

2 2 1

6 1 7 1

2 2 1 1

2 2 11 2 2 3 ... 7 6

1

2 1 2 2 2 1 2 3

b t F D t F t

T CD F t F t t

t

F t F t



'

2

1

0T C

t

, we have



t

TC

0.76,13.92

2 2 1

1 2 1 2 2 1 5 1

2

1 2 2 2 1 2 2

4 1 3 1 1 1 7 1

1 0

6

2 2 6 11

2 3 2 1 2 2 2 3 0 ... 7 71

1

6b t F t a D D t F t

F t F t t F t F

F

t



value '




1 1T C t is

2 '

2

2

1

0 ... 7 8T C

t


Fig.4. Graphical representation of '

2T C and

1t .


Let us consider

2 1 2 3 4 5 6 7

9 1 0

1 .0 3, 0 .5 9 7 , 0 .2 4 , 0 .0 0 6 7 , 0 .0 1 7 , 0 .0 3 4 , 0 .

$ 1 / , $ 0 .5 / , $ 0 .5 / , $ 5 / , 0 .1, 2 , 1 , 1 0 0 , 1,

0 .5

0 1 2 , 0 .0 0 0 2 3,

1 .7 5 , 7 .5


y e a r

D F F F F F F F

F F


t is1

0 .7 6t . The optimum ordering quantity is

'2 0 .9O Q

. The


21 3 .9 2T C

.


To know, how the optimal solution is affected by the values ofconstraints, we derive the sensitivity analysis for

some constraints. The specific values of some constraintsare increased or decreased


t

and1

T C



S andc

D .The result of minimum relevant cost is existing in the following table 3.

Table: 3

Parameters Actual

Values

+50%

Increased

-50%

Decreased

+25%

Increased

-25%

Decreased

cS

0.5 0.75 0.25 0.625 0.375

cD

5 7.5 2.5 6.25 3.75

1t

0 .7 6 0 .6 9 0 .8 4 0 .7 2 0 .7 9

1T C

1 3 .9 2 19 .71 7 .5 1 6 .8 8 1 0 .8 2

O Q

2 0 .9 2 4 .6 16 .97 2 2 .9 8 19 .37






We have

1

2

3

2 3 4 4 3 2 12

1 2 1 2 1 3 1 1 4 1 5 1 6 1 1

2 2 2 4 2 3

7 1 8 1 9 1

4 1 ... 7 9

T

t

H C a b t c t q t d t

a DcF a D t F t F t t F t F t F t t

D

F t F t F t

Where,

2 2

2 4 2 3 2 2 42

1 22

11

1 2 4 1 2 3 2 4 12 1

b Dc b a cF T T T T a D T

3 2 1 4 3 22

2 2 2

1 1 1 11 1

3 1 2 1 1 4 3 2

a Db a cD T T T T c D b T b D a T

2 2 3 2 4 5 2

2 2

2

6 7 8 92

1 1 1, , 1 , ,

2 3 4 1 3 1

11 , , ,

2 1 2 1 2 4 1 2 32 1

c bF a b D F b c D F F c D b

b Da a c bF F F F


0 , T

'


4 3 2 2 1

1 3 1 2 1 2 2 1 2 1

2 2 3

2 3 4 2 1 2 2 2 3 2 4

6 1 5 1 4 1 1 7

1 0 1

9

1 9 1 8 1 2 1

1 12 2 6

4 3 2 1' .. . 8 0

1

6

61 2

ct F t F t a D D t a D t

T C D

F t F t F t t F

F t

Ft F t F t D F T

5.2.5. Solution:



t

TC

3 2 2 1

1 3 1 2 1 2 2 1

1 2 32

2 1 6 1 5 1 4 1

12

2 2 1 2 2 2 3

1 7 1

1 0

9 1 8 1

13 2 2 6

'2 3 4 ... 8 1

2 12 2 2 3 2 4

1

6c t F t F t a D D t

T CD a D t F t F t F t

t

t F t F t F t

F

2 2 1

1 3 1 2 2 1 2 1

22

1 22

6 1 5 1 4 12

1

2 1 2 2 1 2 2

1 7 1 9 1 8 1

3 2 3 2 1

2 2 1'1 2 2 3 3 4 ... 8 2

1

2 1 2 2 2 2 2 3 2 3 2 4

c t F t F D t a D t

T CD F t F t F t

t

t F t F t F t



'

2

1

0T C

t

, we have

3 2 2 1

1 3 1 2 1 2 2 1 2 1

2

1 2 3 2 2 1

6 1 5 1 4 1 1 7 1

2 2 2

1 0

3

9 1 8 1

13 2 2 6

2 12 3

6

4 2 21

2 3 2 4 0 ... 8 3

c t F t F t a D D t a D t

F t F t F t t F t

F t F

F

t



value '




1 1T C t is

2 '

2

2

1

0 ... 8 4T C

t


Fig.5. Graphical representation of '

2T C and

1t .




Let us consider

2 1 2 3 4 5 6 7

1 08 9

1 .0 3, 0 .6 8 2 , 0 .2 4 , 0 .1 7 6 , 0 .0 1 1, 0 .0 1 4 , 0 .0 1 6 , 0 .0 1 2 ,

0 .0 0 0 4 , 0 .0 0 0 2 4

$ 1 / , $ 0 .5 / , $ 0 .5 / , $ 5 / , 0 .1, 2 , 1 , 1 0 0 ,

1, 0 .5

, 7 .5


y e a r

D F F F F F F F

F FF


t is1

0 .8t . The optimum ordering quantity is

'2 0 .9O Q

. The minimum

relevant cost is'

22 2 .1 5T C

.


To know, how the optimal solution is affected by the values of constraints, we derive the sensitivity analysis for

some constraints. The specific values of some constraints areincreased or decreased

by 25% , 25% and 50% , 50% .Thus,we compute the values of1

t

and1

T C



S andc

D . The result of the minimum relevant cost is existing in the following table 4.

Table: 4 Parameters Actual

Values

+50%

Increased

-50%

Decreased

+25%

Increased

-25%

Decreased

cS

0.5 0.75 0.25 0.625 0.375

cD

5 7.5 2.5 6.25 3.75

1t

0 .8 0 .7 4 0 .88 0 .7 7 0 .8 4

1T C

2 2 .1 5 31 .99 1 2 .1 3 2 7 .4 8 1 7 .0 0 4

O Q

2 0 .9 2 4 .6 16 .97 2 2 .9 8 19 .37



VI. Conclusion We presented an order level inventory model for deteriorating items with two parameters Weibull

deterioration. The model developed under two replenishment policies: (i) With no shortages and (ii) With

shortages which are partially backlogged. We considered ramp-type demand rate and time varying linear and

quadratic holding costs, and found that total relevant cost with linear holding cost is less than total relevant cost

with quadratic holding cost. Therefore, the linear time-dependent holding cost is more realistic than quadratic

time-dependent holding cost. The proposed model can be extended in numerous ways like permissible delay in

payments, time value of money,quantity discounts etc.

Acknowledgement:We are very thankful to the referees for their convenient suggestions.

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eoq inventory models for deteriorating item with …inventory model for two parameters weibull...

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