Volume 03 Issue 03: (2014) May-June 2014

(1)

ISSN: 2278-3369

International Journal of Advances in Management and Economics

Available online at

www.managementjournal.info

RESEARCH ARTICLE

Analysis of Inventory System with Quadratic Deterioration and Two

Levels of Trade Credits

Singh S

1

_{, Chauhan HVS}

2*

_{, Vaish B}

1

1_{Department of Mathematics D.N. (P.G) College, Meerut (U.P), India,}

2_{Department of Mathematics, University of Delhi, Delhi, India.}

*Corresponding Author: Email: [email protected]

Abstract

The main purpose of this study is to investigate the effect of quadratic deterioration rate for inventory system with two levels of trade credits. Also the convexity of the retailer's inventory system is developed. Finally, a theorem is developed to determine the retailer's optimal replenishment cycle time efficiently.

Keywords: EOQ, Inventory, Quadratic deterioration, Two levels of trade credit.

Introduction

In the traditional Economic Order Quantity (EOQ) model, it was taoitly assumed that the buyer must pay for the items purchased as soon as the items are received. However, in practice, the supplier frequently offers its retailer the trade credit (or permissible delay in payments) to attract retailer who consider it to be a type of price reduction.

Goyal [1] derived and EOQ model under the conditions of permissible delay in payments. But he implicitly assumed only one level of trade credit. That is the supplier offers its retailer the trade credit by the retailer does not offer its customer the trade credit. Recently, Huang [2] modified this assumption to two levels of trade credit. That is, not only the supplier offers its retailer the trade credit but also the retailer offers its customer the trade credit. But the decay item was ignored in their models. However, many studies related to the inventory considered the decay item under the trade credit could be found. In this paper, we develop a model by considering time dependent decay item under two levels of trade credit. Then we model the retailer's inventory system as a cost minimization problem to determine the retailer's optimal replenishment cycle time.

Assumption

 Demand rate, , is known and constant.

 Replenishment is instantaneous.

 There is no repair or replacement of the deteriorated inventory during a given cycle.

 There is quadratic deterioration rate (𝑎 + 𝑏𝑡 + 𝑐𝑡2 _{∶ 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐}_{constant fraction on hand}

inventory per unit time). 𝐼𝑘≥ 𝐼𝑒, 𝑀 ≥ 𝑁

 When the 𝑇 ≥ 𝑀, the account is settled at 𝑇 = 𝑀 and the retailer starts paying for the

interest charges on the items in stock with rate

𝐼𝑘. When 𝑇 ≤ 𝑀 the account is settled at 𝑇 = 𝑀

and the retailer does not need to pay any interest charge.

 The retailer can accumulate revenue and earn interest after its customer pays of the amount of purchasing cost to the retailer until the end of the trade credit period offered by the supplier. That is, the retailer can accumulate revenue and earn interest during the period N to M with rate 𝐼𝑒 under the condition of trade

credit.

Notations

𝐷 = Demand rate per year 𝐴 = Ordering cost per order

𝑐 = Unit purchasing price per item

ℎ = Unit stock holding cost per item per year

excluding interest charges.

𝑎 + 𝑏𝑡 + 𝑐𝑡2_{= Deterioration rate, a, b, c o, a >}

b, a>c

𝐼 = Interest earned per $ per year

(2)

𝑀 = The retailer's trade credit period offered by

supplier in years.

𝑁 = The customer's trade credit period offered by

retailer in years.

𝑄 = The order quantity 𝑇 = The cycle time in years

TVC (𝑇) = The annual total variable cost, which is

a function of T

𝑇 = The optimal replenishment cycle time which

minimizes TCT (𝑇) when 𝑇 > 0

Mathematical Model

Le 𝑄(𝑡) denote the on-hand inventory level at

time 𝑡, which is depleted by the effect of demand

and quadratic deterioration, then the differential equation which describes the instantaneous states of 𝑄(𝑡) over (𝑜, 𝑇) is given as

  

t²

  

0

dQ t

a bt c Q t D t T

dt        (1)

then with boundary condition 𝑄(𝑇) = 0. The solution of above equation is given by

 

at+bt²/2+ct³/3  at+bt²/2+ct³/3 ) at+bt²/2+ct³/3  aT+bT²/2+cT³/3 )

0

Q t  De



e dtDe



e dT  t T (2)

Noting that 𝑄(0) = 𝑄, the quantity ordered each replenishment cycle is



2_{/ 2+cT³/3}



₂

/ +ct³/3)

aT bT _{at bt}

t o

Q

D

e



dT

e



dt





















(3)

Furthermore, the total variable cost function per cycle consists of the ordering cost, inventory holding cost, cost of deteriorated units and capital

opportunity cost. From now on, the individual cost is evaluated before they are grouped together.

* Annual ordering Cost = 𝐴/𝑇 (4)

* Annual inventory holding cost (including the capital opportunity cost)

 



2_{/ 2+ct³/3}

 

2_{/ 2+ct³/3}





2_{/ 2+ct³/3}





2_{/ 2+cT³/3}



T

o

T at bt at bt at bt aT bT

o

h

Q t dt T

hD

e e dt e e dT

T

     



 

 _  _

 



(5)

* Annual cost of Deteriorated units







2_{/ 2+cT³/3}





2_{/ 2+ct³/3}



0 T

o

aT bT at bt

t

C Q Ddt

T DC

e dT e dt T

T

 



 

  

 __  _  _

 

 



(6)

* From assumption (8) and (9), there are three

cases to discuss annual capital opportunity cost. Case 𝑇 ≥ 𝑀: The annual capital opportunity cost

 





















2_{/ 2+ct³/3} 2_{/ 2+ct³/3} 2_{/ 2+ct³/3} 2_{/ 2+cT³/3}

2 2

2

T M

K _m _N

T at bt at bt at bt aT bT

K

M

CI Q t dt CIe Dt dt

T CI D

e e dt e e dT dt

T

CIeD M N

T

     

 

 

 _  _

 

 



(7)

(3)





2 2

2

T

M

CIe

Dt dt

DT

M

T

CIeD

MT

N

T



_

_











 





 

_



_



 

8

Case < 𝑁 : The annual capital opportunity cost



M N



D

CIe T

dt DT CIe M

N   







(9)

According to the above arguments, we

have 𝑇𝑉𝐶(𝑇) =

 

1

2

3

(10 )

0 (10 )

TVC

T

if

M

T

a

TVC

T

if

N

T

M

b

TVC

T

if

T

N

c















_



Where

 

























2 2

2 2 2 2

/ 2+ct³/3 / 2+ct³/3

1 ₀

/ 2+ct³/3 / 2+cT³/3 / 2+cT³/3 / 2+ct³/3

\ T at bt at bt

at bt aT bT aT bT at bt

t o

hD

TVC T A T e e dt

T

DC

e e dT dt e dT e dt T

T

  

    





  _



 

 

   _    _

   



















T _  



 _  

M

bt at bt

at bt

at

k _e _e _dt _e

T

DCI 2/2+ct³/3 2/2+ct³/3 2/2+ct³/3





_

_



  /2+cT³/3  2  2

2

N M

T CIeD dt

dT

eaT bT (11)

 



 











_e





_dT

_e





_dt



_T



T

DC

dt

dT

e

dt

e

T

hD

T

A

T

TVC

t bt

at bT

aT bT

aT bt

at

T

bt at bt

at











 

 

 

 



0 ct³/3 + 2 / cT³/3

+ 2 / cT³/3

+ 2 / ct³/3

+ 2 /

0

ct³/3 + 2 / ct³/3

+ 2 / 2

2 2

/



2 2



T N

MT 2 T 2

CIeD _ _



(12)

 



 







_









2 2

2 ₂ ₂ 2

/ 2+ct³/3 / 2+ct³/3 3

/ 2+ct³/3 _{/ 2+cT³/3} _{/ 2+cT³/3} / 2+ct³/3

0

/

T at bt at bt

o

at bt _{aT bT} _{aT bT} at bt

t

hD

TVC T

A T

e

dt

T

DC

e

dT dt

e

dT

e

dt

T

CIeD M

N

  

  _ _ 









_















_



_



_



_









 

13

Since 𝑇𝑉𝐶1(𝑀) = 𝑇𝑉𝐶2(𝑀) and 𝑇𝑉𝐶2(𝑁) = 𝑇𝑉𝐶3(𝑁), 𝑇𝑉𝐶 (𝑇) is continuous and well defined.

The Convexity

Here we shall show that three inventory functions described as above section are convex on their

 𝑇𝑉𝐶1(𝑇) is convex on [𝑀,∞)  𝑇𝑉𝐶2(𝑇) is convex on [0,∞)  𝑇𝑉𝐶3(𝑇) is convex on [0,∞)  𝑇𝑉𝐶(𝑇) is convex on [0,∞)

(4)







_















_









_





at bt2/2+ct³/3 at bt2/2+ct³/3 at bt2/2+ct³/3 aT bT2/2+cT³/3

e

dt

e

P

Suppose

dT

Leema:- Pdt



M N



if T M

T T dt P T Pdt T M T M T

M  

       



0,

2 2 2 2 2 2 2 Proof:

 





2 2 2 2 2

2  

      



_

Pdt M N

T T dt P T T dt P T g T M T M T M

Then we have

 

_

 

 T

M P dt

T T T g ₃ 3 2 ' 2

So 𝑔(𝑇) is increasing on (𝑀,∞)and g

 

T  g

 

M  N  o

2 2





. . 0 2 2 2 2 2 2 2 proof the completes is This M T if N M Pdt T T Pdt T T Pdt ly Consequent M T If T M T M T M           



The Proof of Theorem 1

(1) From equation (11)

 



 



 







2 2



2 2 0 0 2 2 2 ' 1

2T M N

CIeD dt P T I T DC dt P I T DC T DC P T TT DC T P T DC dt P T T hD dt P T hD T A T TVC T M K T M K t T o T o t                  



 

 



 



 









 

14

2 . / 2 2 2 2 0

0 2 0

0 2 2 ' 1 N M T CIeD Pdt T I T DC Pdt I T DC P T T DC P T DC dt P T T hD Pdt T hD T A T TVC T M K T M K t T t T                



 

 







 





 





 

















































    T M K K T M K T M K t t t t T T T T o

T

I

T

DC

Pdt

T

I

T

DC

Pdt

T

I

T

DC

Pdt

I

T

DC

P

T

DC

P

T

DC

P

T

DC

P

T

DC

Pdt

T

hD

Pdt

T

hD

Pdt

T

hD

Pdt

T

Dh

T

A

T

TVC

2 2 2 2 3 0 2 2 0 2 0 2 0 3 0 0 2 2 2 0 2 3 3 " 1

2

2 .

2

2 

2 2



3 2 2 N M T CIeD Pdt T

M  



 





 





 



 





2 2₂

3 0 2 2 0 2 2 0 0 0 3 3 " 1

2

2 /

2 T

T

Pdt

T

Pdt

T

CDI

P

T

C

Pdt

T

h

T

P

T

C

Pdt

T

h

T

P

C

Pdt

h

T

D

T

A

T

TVC

T M T M k t T t T t T o



































































  



2 2



3 2 N M T CIeD Pdt T

M   



(5)

 







 



 





0

2

0 2 2 0 2 2 2 0 0 0 0 3















































_

  



t T t T t T

P

T

C

Pdt

T

h

T

P

T

C

Pdt

T

h

T

P

C

Pdt

h

T

D

Now

 





 

16

2 .

2 2

2 2 2

2 2 2 3 3 "               





Pdt M N

T T Pdt T T Pdt T CDI T A T TVC T M T M T M k

Using lemma ₂1

 

0 , ₁

 

[ , )

2

 

 if T M ThereforeTVC T is convexon M dT

T TVC d

From equation (12)

 



 





 







 

17

2 / 2 2 2 0 0 2 0 0 2 2 ' 2 T N T CIeD P T T DC P T DC Pdt T T hD Pdt T hD T A T

TVC t t

T T            



_ _

 



 





 



 





 

18

2

3 2 0 2 2 0 2 2 2 0 0 0 3 3 " 2

T

CIeDN

P

T

C

Pdt

T

h

T

P

T

C

Pdt

T

h

T

P

C

Pdt

h

T

D

T

A

T

TVC

t T t T t T o















































  



 

2 2 ₃ 0

 

19

2

3 "

2   

T IeDN T A T TVC

From equation (13)

 



 



 





 

20

/ 0 0 2 0 0 2 2 ' 3            



t t T T P T T DC P T DC Pdt T T hD Pdt T hD T A T TVC

 



 





 



 



0 

 

21

2

2 2 0 2 2 2 0 0 0 0 3 3 " 3





























































 

t

P

T

C

Pdt

T

h

T

P

T

C

Pdt

T

h

T

P

C

Pdt

h

T

D

T

A

T

TVC

T t T t T

 

2 /

3

0  

22

"

3

T



A

T



TVC

Therefore, 𝑇𝑉𝐶2(𝑇) and 𝑇𝑉𝐶3(𝑇) is convex on (0,∞) respectively.

(4) Case (1) implies that 𝑇𝑉𝐶1' is increasing on [𝑀,∞). Case (2) and (3) implies that 𝑇𝑉𝐶2(𝑇) and 𝑇𝑉𝐶3(𝑇) is increasing on (0, M]. Since

 

₂'

 

₂'

 

₃'

 

,

'

1

M

TVC

M

and

TVC

N

TVC

N

TVC



then

𝑇𝑉𝐶′(𝑇) in increasing on 𝑇 > 0. Consequently 𝑇𝑉𝐶(𝑇) is convex on 𝑇 > 0. Combing the above

arguments, we have completed the proof.

Determination

of

the

Optimal

Replacement Cycle Time 𝑻

∗

Consider the following equation

If the root of Eq. 23, 24 or 25 exist, then it is unique. For convenience, let 𝑇𝑖∗(𝑖 = 1, 2, 3) denote

(6)























)

(

26

0 )

(

26

0 )

(

26

0 )

(

* 1 * 1 1 1

c

T

if

b

T

if

a

T

M

if

T

TVC























)

(

27

0 )

(

27

0 )

(

27

0 )

(

* 2 * 2 * 2 2

c

T

if

b

T

if

a

T

N

if

T

TVC

and























)

(

28

0 )

(

28

0 )

(

28

0 )

(

* 3 * 3 * 3 3

c

T

if

b

T

if

a

T

N

if

T

TVC

Althoughlim𝑇−0𝑇𝑉𝐶1=∞, we can not make sure

that whether lim𝑇−0𝑇𝑉𝐶1(𝑇) is less than 0,

therefore, one of the following results will be occurred. One is that if 𝑇𝑉𝐶1(𝑀) < 0,. then 𝑇1∗

exists and 𝑇1∗≥ 𝑀, the other is that if 𝑇𝑉𝐶1(𝑀) > 0, then the convexity of 𝑇𝑉𝐶1(𝑇) on [𝑀,∞) implies

that 𝑇𝑉𝐶1(𝑇) is increasing on [𝑀,∞) On the other

hand, it is needless to say that Eq. 15 a-c and

16a-c implies that 𝑇𝑉𝐶1(𝑇) is decreasing on (0, 𝑇1∗] and

increasing on [𝑇1∗,∞) for 𝑖 = 2,3. in addition lim𝑇−0𝑇𝑉𝐶1 𝑇 = −∞ and lim𝑇−0𝑇𝑉𝐶1(𝑇) =∞ the

Intermediate Value Theorem implies that 𝑇2∗ and 𝑇3∗ are exist.

From equation (14), (17) and (20),

 



 







 

29

2 /

2 2 2 2 0 0 2 0 0 2 2 ' 2 ' 1

N

M

D

CI

Pdt

T

I

M

DC

Pdt

I

M

DC

P

T

M

DC

P

M

DC

Pdt

T

M

hD

Pdt

M

hD

M

A

M

TVC

T

TVC

e M T T M K M T T M K M T t M T t M T T M T T



















































































       



and

 

N T T N T T Pdt T N hD Pdt N hD N A N TVC N TVC                    



0 0 2 ' 3 ' 2 /

 





 



N T t N T t P T N DC P N DC             

 2 0 2 0

For convenience, we define

 







 







(

30 )

2 /

2 2 2 0 0 2 0 0 2 2 1

N

M

CIeD

P

T

M

DC

P

M

DC

Pdt

T

M

hD

Pdt

M

hD

M

A

M T t M T t M T T M T T

























































     



and

 







 



(

31 )

/

0 0 2 0 0 2 2 2





















































     



N T t N T t M T T N T T

P

T

N

DC

P

N

DC

Pdt

T

N

hD

Pdt

N

hD

N

A

Since 𝑇𝑉𝐶2(𝑇) is convex on 𝑇 > 0 which implies that 𝑇𝑉𝐶21(𝑇) is increasing on 𝑇 > 0, we have

 

2

' 2 '

2

1  

 TVC M TVC N

Theorem 2

1.

 

*

1 *

1 1

1 0,thenTVC T TVC T .HenceT*isT

If   

2.

 

*

3 * *

3 3

2

0 ,

then

TVC

T

TVC

T

.

Hence

T

is

T

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3.

 

* 2 * *

2 2 *

2

1 0,then 0,thenTVC T TVC T .HenceT isT

If     

Proof

1. If ∆1≤ 0, then ∆2≤ 0 which implies that 𝑇𝑉𝐶1′(𝑀) = 𝑇𝑉𝐶2′(𝑀) ≤ 0 and 𝑇𝑉𝐶2′(𝑁) = 𝑇𝑉𝐶3′(𝑁) < 0

Equation 26a-c28a-c imply that

(i) 𝑇𝑉𝐶1(𝑇) is decreasing on [𝑀, 𝑇1∗) and increasing on [𝑇1∗,∞).

(ii) 𝑇𝑉𝐶2(𝑇) is decreasing on [𝑁, 𝑀).

(iii) 𝑇𝑉𝐶3(𝑇) is decreasing on (0, 𝑁).

Combining (i) and (iii), we conclude that 𝑇𝑉𝐶(𝑇) has the minimum value at 𝑇 = 𝑇1∗ on (0,∞). Hence, we

conclude that 𝑇𝑉𝐶 𝑇∗ = 𝑇𝑉𝐶₁(𝑇1∗). Consequently, 𝑇∗is 𝑇1∗.

2. If ∆2≥ 0, then ∆1> 0, which implies that 𝑇𝑉𝐶1′(𝑀) = 𝑇𝑉𝐶2′(𝑀) > 0 and 𝑇𝑉𝐶2′(𝑁) = 𝑇𝑉𝐶3′(𝑁) ≥ 0.

Equation 26a-c-28a-c imply that. (i) 𝑇𝑉𝐶1(𝑇) is increasing on [𝑀,∞).

(ii) 𝑇𝑉𝐶2(𝑇) is increasing on [𝑁, 𝑀).

(iii) 𝑇𝑉𝐶3(𝑇) is decreasing on (𝑜, 𝑇3∗] and increasing on [𝑇3∗, 𝑁).

Combining (i), (ii) and (iii), we conclude that 𝑇𝑉𝐶 (𝑇) has the minimum value at 𝑇 = 𝑇₃∗_on_(𝑜,_∞_).

Hence, we conclude that 𝑇𝑉𝐶(𝑇∗_{) = 𝑇𝑉𝐶3(𝑇}

3∗). Consequently, 𝑇∗ is 𝑇3∗.

3. If ∆1> 0 and ∆2< 0which implies that 𝑇𝑉𝐶1′(𝑀) = 𝑇𝑉𝐶2′(𝑀) > 0 and 𝑇𝑉𝐶2′(𝑁) = 𝑇𝑉𝐶3′(𝑁) < 0.

Equation 26a-c-28-a-c imply that. (i) 𝑇𝑉𝐶1(𝑇) is increasing on [𝑀,∞).

(ii) 𝑇𝑉𝐶2(𝑇) is decreasing on [𝑁, 𝑇2∗) and increasing on [𝑁2∗, 𝑀).

(iii) 𝑇𝑉𝐶3(𝑇) is decreasing on (𝑜, 𝑁).

Combining (i), (ii) and (iii) we conclude that 𝑇𝑉𝐶(𝑇) has the minimum value at 𝑇 = 𝑇2∗ on (0,∞).

Hence, we conclude that 𝑇𝑉𝐶(𝑇∗) = 𝑇𝑉𝐶2(𝑇2∗). Consequently, 𝑇∗ is 𝑇2∗.

Combining the above arguments, we have completed the proof.

Special Case:- Deterioration = 𝑎 + 𝑏𝑡 + 𝑐𝑡2

if 𝑏 = 0 = 𝑐, then deterioration = 𝑎

Lemma:- eaTM  aTeaTM  a T eaTM aM  a



M  N



 0 if T  M

2 2

1

2 2

2

Proof:-

 

     





2 2

1

2 2

2

N M

a aM e

T a aTe

e T g

Let  a TM   a TM  a TM   

then we have

 

aT M

e T a T

g  

2

2 2 '

So 𝑔(𝑇) is increasing on (𝑀,∞) and 𝑔(𝑇) > 𝑔(𝑀) = 0 2

2 2



N a

     





0 2

2 1

,

2 2

2

 

 

 

  

 a M N

aM e

T a e

aT e

ly consequent

It a T M a T M a T M

If 𝑇 ≥ 𝑀. This completes the proof.

Conclusions

In this paper, we study a model by considering time dependent decay item to find the retailer's optimal replenishment cycle time under two levels of trade credit. In addition, we develop an easy-to-use procedure to find the optimal replenishment

cycle time for the retailer. This procedure is the main contribution of this study.

Acknowledgement

The first author is very much thankful to the Council of Scientific and Industrial Research, New Delhi, India for the financial assistance in the form of Senior Research Fellowship.

Reference

1. Goyal SK (1985) Economic order quantity under conditions of permissible delay in payments. J. Oper. Res. Soc. 36: 35-38.

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when a delay in payment is permissible. Intl. J.Prod. Econ. 63: 270-214.

5. Chang HJ, Dye CY (2001) An inventory model for deteriorating items with partial backlogging and permissible delay in payments. Intl. J. sys. Sci. 32:345-52.

6. Chang CT, Ouyang LY, Teng JT (2003) An EOQ model for deteriorating items under supplier credits linked to ordering quantity. Applied Math. Modeling. 27:983-96.

7. Thomas GB, Finnney RL (1996) Calculus with Analytic Geometry. 9th_{Edn. Addison-Wesley}