International Journal of Scientific & Engineering Research, Volume 4, Issue 4, April-2013 201

ISSN 2229-5518

Inventory Model for Exponentially Increasing

Demand with Time and Deterioration and

Permissible Delay in Payments

R a j e n d r a S h a r m a 1, Jasvinder Kaur2 , A.P.Singh3

*( Department O f M a t h e m a t i c s G r a p h i c E r a U n i v e r s i t y D e h r a d u n , i n d i a

Email: rs.2103@yahoo.in)

** (Department O f M a t h e m a t i c s , G r a p h i c E r a U n i v e r s i t y D e h r a d u n , I n d i a

Email:jasvinddn@gmail.com)

***( Department O f M a t h e m a t i c s , S.G.R.R.(P.G.).College, Dehradun, India)

ABSTRACT

In this paper, an inventory model is developed in which demand is exponentially increasing with time and deterioration is taken non instantaneous. Realistic situation of permissible delay is also taken into consideration. Three different cases have been discussed for different situations. Expressions are obtained for total optimal cost in different situations. Three different algorithms are given to obtain the optimal solution. Cost minimization technique is used to solve the model.

Keywords - Inventory model, Non instantaneous deterioration shortages, permissible delay in payments Finance,

I. INTRODUCTION

In today’s business transaction it is more and more common to see that the purchases are allowed a fixed time period before they settle the account with the supplier. This provides an advantage to the purchaser, due to the fact that they do not have to pay the supplier immediately after receiving the items but instead can defer their payment until the end of the allowed period. Thus paying later indirectly reduces the purchases cost of the items. On the other hand the permissible delay in payments produces benefit to the supplier such as it should attract new purchasers who consider it to be a type of price reduction. It is important to control and maintain the inventories of deteriorating items for the modern corporation. In
general deterioration is defined as the damage,
spoilage, dryness, vaporization etc. that results in decreases of usefulness of the original one. Inventory problems for deteriorating items have been widely studied by Ghare and Schrader (1963). They presented an EOQ model for an exponentially decaying inventory. Later Covert and Philip (1973) formulated the model with variable deterioration rate with two parameter Weibull distribution. Philip (

1974 ) developed an inventory model with three parameter Weibull distribution rate and no shortages. Shah (1977) extends Philip (1974) model and considered that shortages are allowed. Recently Goyal and Giri (2001) provides a detailed review of deteriorating inventory literature. In the mentioned literature almost all the inventory models for deteriorating items assume that the deterioration occurs as soon as the retailer receives the commodities. However in real life most of goods would have a span of maintaining quality of the original condition, namely, during that period, there was no deterioration occurring. Such deterioration is treated as non instantaneous deterioration.

In the present model, an inventory model is developed in which demand is exponentially increasing with time and deterioration is taken non instantaneous. Realistic situation of permissible delay is also taken into consideration. Three different cases have been discussed for different situations. Expressions are obtained for total optimal cost in different situations. Three different algorithms are given to obtain the optimal solution. Cost minimization technique is used to solve the model.

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II. NOTATIONS

1. C: Ordering cost of inventory per order.
2. C1: Holding cost excluding interest charge per unit per unit time.
3. C2: Shortage cost per unit per unit time.
4. C3: Unit purchase cost.
5. Ir: Interest paid per rupee invested in stock s per year Ir>Ie.
6. Ie: Interest which can be earned per rupee per year.
7. q (t) : Inventory level at time t.
3. The replenishment occurs instantaneously at an infinite rate.
4. When produced or purchased items arrive in stock they are fresh and new. They begin to deteriorate after a fixed time interval  .
The deterioration function θ(t) is taken in the following form
θ(t) = t-1 H (t - ) (0 < <<1 )

  1

t ,  > 0
where H (t - ) is heaviside function defined as
8. M : Permissible Delay period for settling accounts
in time 0<M<1.

  1

H t  μ 

t  μ

  0

t  μ

9. t1: Time at which shortages starts.
10. T: Length of replenishment cycle.
11.  : Life period of item at the end of which deterioration starts.
12. Q: Total amount of inventory produced or purchased at the beginning of each production cycle.
13. S(S<Q): Initial amount of inventory after fulfilling back orders.
14. TC (t1, T): The total average cost of the inventory system per unit time.
15. TC1 (a) (t1, T): The total average cost of the inventory system per unit time for M  t1 and M  .
16. TC1 (b) (t1, T): The total average cost of the inventory system for M  t1 and M  .
17. TC2 (t1, T): The total average cost of the inventory

5. Demand rate (D(t) ) is known and increases exponentially at time t, t  0

D (t) = ae bt a  0

where a is the initial demand and b is a constant governing the increasing rate of demand.

6. Shortages are allowed and only a fraction  ( 0 <

 <1 ) of the demand during the stock out period is backlogged and the remaining fraction ( 1 -  ) is lost.

7. During the fixed credit period M the unit cost of generated sales revenue is deposited in an interest bearing account. The difference between sales price and unit cost is retained by the system to meet the day to day expenses of the system. At the end of the credit period account is to be settled. Then interest is again earned during the period (M, t1). If M  t, interest charges are paid on the stock held beyond the permissible period.

IV. MATHEMATICAL EQUATIONS

system per unit time for M>t1.

III. ASSUMPTIONS

1. The inventory system consists of single item only.
2. There is no repair or replacement of the deteriorated unit.

dqt

d qt

a e bt

 θ0 qt   a ebt

,0t μ

,μ  t  t1

(1)
(2)

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dqt

a ebt

,t1 t T

. (3)

qt eθ0t  a  e b

b θ0 

 θ0  μ  e b

 θ0  t   



a 1  bμ eθ0μ

b 

(6)
Boundary conditions are at

t  0,

q t   S

Using boundary condition at t=t1, Q(t1)=0 from
at t  μ,

q t   q μ 

equation (6) the value of S is given by

at t  t1 ,

q t1   0 .

S  e

b θ 0

 b  θ 0  μ

 e b  θ 0  t

e  θ 0μ 

a 1  bμ 

Solution of equation (1) is given by
(7)

Now substituting the value of S from equation (7)

q t    a  e bt

dt  A

in equation (6), one can get

where A is the constant of integration.

q t  e

b  θ0

 θ0 t1

t  ebt 

μ  t  t1

or q t   

e bt

 A .

Solution of equation (3) is given by

(8)

Using boundary condition at t  0,qt  S ,

q t  

 a e bt  E

we get

where E is a constant of integration. Applying

the boundary condition at

t  t 1

, q t  

0 , one can get

S   a  A

b .

qt   a  ebt1  ebt 

t1  t  T

(9)

Therefore

Total amount of holding units q H  during the period 0, t 1  is

qt S a 1 e bt 

,0 t μ

(4)

q H 

μ q t  dt

 t 1 q t  dt

Also at t  μ
equation (4) reduces to

μ  a  1

  S  1 e dt  

aeθ0t

e b  θ0  t 1  e b  θ0  t dt

q μ   S 

a  1 

e b μ 

0  b

 b

θ0 

Solution of equation (2) is given by
(5)

  S μ 



a μ 

a e b μ  1  

b 2 

b  θ 0 

q t  e θ 0 t

  a  e b  θ 0  t

dt  B

 e bt 1

  

 θ 0

e bt 1



e b  θ 0 t 1  μ 



θ 0

e b μ 



b 

where B is constant of integration.

Substituting the value of S from (7) , we have

Using boundary condition at

t  μ

, qt  qμ ,

 aμ

q H  

 e b  θ0  t 1  e b  θ0  μ e  θ0 μ 

aμ 1  bμ   a μ

one can get

 b θ0  b b

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a  a

 ebt 1

e bt 1

e b θ0 t 1  μ 

e bμ 

The case (1) is further divided into two sub cases i.e.

 e b μ

b 2

 1   b θ

  

   θ 0 b θ 0

 

b 

case I(a) and case I(b)

 aμ

b θ0 

 eb θ0  t 1 μ   ebμ  aμ 2 

a ebμ  1

CASE I(a) : M

CASE I(b) : μ

 μ  t 1

 M  t 1

a  ebt1

 

ebt1



eb θ0 t 1μ



ebμ 

 

CASE I(a) :

b θ0  θ0

b θ0

b 

(10)
Since here the length of period with positive inventory stock is larger than the credit period M, the buyer can

Total amount of deteriorated units q D 

during the period 0, t1  is

use sale revenue to earn the interest with an annual rate Ie during the period [0,M]. The unit cost of the generated sales revenue is deposited in an interest
bearing account. The difference between sales price

q  q μ  

t 1 ae bt dt

D  μ

and unit cost is retained by the system to meet the day to day expenses of the system. At the end of the

 S 

a 1  e b μ 

a e bt  t 1

credit period, the account is settled. After setting the account at time M again the unit cost of generated

 a e b  θ0  t 1  e b  θ0  μ eθ0 μ  a ebt 1  eb μ 

sales revenue is deposited in an interest bearing

b θ0 

b 



(11)
account to earn interest with an annual rate Ie during
the period [M,t1]. Beyond the fixed credit period product still in stock is assumed to be financed with

Amount of shortage units q S  during the

an annual rate Ir. Now the total interest earned

period t1 , T  is given by

IE 1 a 

during the period 0, t 1  is given by

 M bt t1

bt 

q S 

  t

q t  dt

IE1(a)  C 3 I e 0

M t ae

dt M t 1  t ae

dt

1   ae bt 

1  

ae bt 

1  t 1 

 C 3 I e  

 M  t   

  t 1  t   

  

a e bt 1 

e bt

 dt

  b 



b   0 b 

b  M 



t 1 b

 aebM

 C3Ie 

a 

  M

1  aebt1

 

aebM 

  t1  M

1 



  bt  

 b2 b 

b  b2 b 

b 

   a

 e bt 1

t  e  

 ae bt 1 a 

1  ae bM 

 b 

b  

 C 3 I e  2

 b

  M 

b 

 

b  b

t 1  M 



 a

  



e bt 1 T  t1  

 e bT  e bt 1 

b 

(12)
Total interest payable IP1(a)_ is given by
(13)

Now there are two possibilities regarding

the period M of permissible delay in payments.

  t 1

IP 1(a)

C 3 I r

q(t) dt

 μ  t 1 

CASE I : M  t 1

C 3 I r

  M

q(t) dt

 μ q(t)

dt 

CASE II : M  t 1

 C 3 I r

 μ 

   S

 

 a (1

 e bt

 dt



CASE I : M  t 1

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  1 e bt 1  θ 0  t 1  t   e bt dt 

  e e

 e

e  

t a

μ b  θ 0 

  1 

  θ 0 

bt 1 

bt 1  θ 0  t 1  μ    1

 b

bt 1 

b μ  

 

 

b t  μ

 b t 1  θ0  t 1 t

b t  1 

 C 3 I e

 ae



bt 1

 a  M  1 

ae bM



t 1  M 

C I  St

a  t e 

a e

e  

T  b

b  b  b 

 3 r 

 b   b   b θ  θ

 b 

 



  0    0

 

(15)



 C3I  S μ  M 

 μ  M  a

μ 

ebμ ebM  

To minimize the total average cost per unit time TC1(a) ( t1, T) the optimal values of t1 and T (say

r  b b2

b θ0

t1* and T*) can be obtained by solving the following
two equations simultaneously

 1 e bt 1  e bt 1  θ 0 t1  μ  

1 e bt 1  e bμ  

 TC

1(a)

(t 1

, T)

 0

(16)

  θ 0

b  

 t 1

 C I

 aμ Meb θ0 t1  μ   ebμ  a μ Mμ

and

3 r 

 b θ0

 TC

1(a)



(t 1 , T)  0

(17)

 a e b μ 

e bM

 a

 

provided they satisfy the sufficient conditions

b 2 b  θ 0

 2 TC

1(a)

(t 1 , T)  0

 1 e bt 1  e bt 1  θ 0  t1  μ   

1 e bt 1  e b μ   1

  θ 0

b  

 2 TC

1(a)

(t 1 , T)  0

Therefore total average cost in this case is

(14)
and

 T 2

TC (t , T) 

C C1qH

 C3q D

 C2qS

 IP1(a)

 IE1(a)

 2 TC1(a)(t1,T)  2

TC1(a)(t1,T)

 2 TC1(a)(t1,T)

1(a) 1 T

  t2

  T2

  

 t  T

  0

 C 

C 1 



a μ e b  θ 0  t 1  μ  



e b μ 

1  

  

T T  b 

θ 0 

Equations (16) and (17) are equivalent to

 a μ 2  a

e b μ  1

 a  e

bt 1

e bt 1



b 2   b  θ

   θ

b C1  aμ

b  θ  t μ 

a  b e b t1 b t

0  0

T   b θ

  θ 0 e

0 1

b θ 0

 

 e 1 

 θ 0

eb θ0 

t 1  μ

ebμ  C  a

   

e b  θ  t  μ e

bμ 

 a e

C a

 e  

t 1 

T  

θ0 b





3 0 1

T  b θ0 

b θ0  t1

μ  b 

bt 1

b μ  



2 

T  b

bt 1

 e 

 

C3  a

bθ  t μ 

bt 

1  e bT  e bt 1   C 3 I r  a μ  M e b  θ 0  t 1  μ  e bμ 

θ0 2

 



T  b θ

 θ 0 e

0 1  ae 1 



b  

T  b 

θ 0 

 a μ

μ  M

 a e  e

 a T b θ  0

 ae

 t1

T

C  a

θ ebθ t μ

aebt  C2 bt

   b μ

b 2

b  θ   0

0 1  1 



1  

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C 3 I r

 a  μ  M  b  θ  t  μ  a

 e 0 1 θ 0 

(IV) Repeat (II) and (III) until no change occurs in the
value of t1 and T.

T   b  θ 0 

b  θ 0 

STEP 2: To compare t1 and M

 b e b t1   b  θ 0  e b t1  θ 0  t1  μ   e b t1  

(I) If M  t1 , t1 is feasible than go to step (3).

 θ 0

 C

and

3 I e

θ 0

 a e bt 1



 b

a e bM 

 

b 

 

(18)
(II) If M  t1 , t1 is not feasible set t1 = M and evaluate the corresponding values of T from equation (19) and then go to the step (3).

STEP 3: Calculate the corresponding total cost.

TC1(a) ( t1* , T* )

1   aμ  

 C  C1 

e b θ 0  t1

μ   e bμ  aμ 2

T 2 

 b  θ 0 

CASE I(b): M   and M  t1
This case is similar to case I(a). But as M > 

 bt1

bt1

b  θ 0  t 1  μ 

bμ 

 2   b  θ

   θ  b θ

 b 

the interest earned IE1(b) during [ 0, t1 ] is given by

a e bμ 1

a  e e e

0  0 0

e 



IE  C I

M

M t

ae bt   1 t

 t ae bt dt 

 C 3

 a e

 b  θ 0 

b  θ 0 

t 1  μ

 e bμ 

a e

bt 1  e

b μ 





1b 

3 e  0  

 ae bt 1 a  1 



ae bM

 



a 1  a μ M

 C 3 I e  2 

 M 

  t 1  M 

b b

 C 

ebt1 t  T 

 ebT  ebt 1   C I    

 b   

2  1



 3 r



 b

θ0 

(20)

 b θ  t  μ ebμ  aμ μ M  a ebμ  ebM a

Interest payable IP1(b) for the period [ M, t1 ] is given

e 0 1

 1

1 

b θ0  by

 ae bt1 t 1

  e bt1  e bt1  θ0  t 1  μ   

e bt1  ebμ   C I 

IP1b  

C 3 I r 

q t  dt

 θ 

 b 

3 e  2 M

0 b

 a  M  1   ae



t  M

C 2   a e bt 1

a b M  t a

   1

b  b  b

 



T  b

 e

b 

(19)

 C 3 I r M

b  θ 0

 e

bt 1  θ 0 t1  t   e bt

 dt

  bt 1  θ 0 t1  t 

bt  1

To get the optimal value of t1 and T which minimizes

 e

 C 3 I r 

b θ θ

 e  

total cost TC1(a)( t1, T) one need to develop the
following algorithm to find the optimal ( t1, T )

  



0    0  M 

ALGORITHM 1(p):

 a

 C3Ir 

 ebt1



ebt1 θ0 t 1 M



ebt1



ebM 

 

 b θ0    θ0

 θ0 b

b 

STEP 1: Perform (I) to (IV) (I) Start with t1 = M
(II) Substitute t1(1) in equation (18) to obtain T(1)

  

(21)
Now the total average cost TC1(b) ( t1, T ) in this case is given by
(III) Using T(1) determines t1(2) from equation (19)

TC   t , T  

C C1q H

 C3q D

 C 2 q S

 IP1b 

 IE

1b 

1 b 1 T

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C C  a μ

Equation (23) and (24) are equivalent to

  1 

e b  θ 0  t1  μ   e b μ  a μ 2

T T  b  θ 0 

C1  aμ θ

e b  θ  t  μ  

a  b e

bt1

 ebt1 

a a  e bt1

ebt1 e b θ0  t 1  μ 

T   b θ  0

0 1

b θ    θ

 e bμ  1    

b 2 b θ0

  θ0 b θ0

e b  θ 0

 t 1  μ    C  a

  θ

e b  θ 0  t1  μ   ae bt 1 

e b μ  

C 3  a 

b  θ  t

 μ 

b μ 

θ 0 2



 

T  b  θ  0 

   

b  

T  b  θ 0 

0 1  e

 C 2

a e

bt 1

T  t 1 

C 3 I r  a

  

 be bt 1

 

T T  b

θ 0  θ 0

 a

ebt1  e

bμ  

C2  a



ebt

1 t1  T 

1  ebT

 ebt

1 



  et1  b θ0  θ0M b  θ

  ebt1   C3Ie a a ebt1  ebM  0

 C 3 I r 

 T

a  e bt 1



 b

e bt 1  θ 0 t1  M 



b 

e bt 1

 

 

e bM  

 

 T b

(25)

T  b  θ 0    θ 0

θ 0 b

b  

and

 

C 3 I e  ae bt 1 a 

1  ae bM

  1

C  C aμ

e b  θ 0  t 1  μ   e bμ  a μ 2

  

T  b 2

 M 

b 

 

b  b

t 1  M 



T 2 

 b  θ 0 

 

 bt 1

bt 1

b  θ 0

t 1  μ

bμ  

  

b 2

 b  θ

   θ  b θ

 b  

(22)

a e bμ 1

a  e e e

0  0 0

e  

 

To minimize the total average cost per unit

C  a e b  θ0  t 1  μ  ebμ  a  bt1 

bμ 

time TC1( b) ( t1, T) the optimal values of t1 and T (say t1* and T*) can be obtained by solving the following

 3  b θ  b e e 

two equations simultaneously

 a bt

1  bT

bt 1  

 a μ  M 

 C 2  b e

1 t1  T    e  e

   C 3 I r  b  θ 

 TC

1(a)

(t 1 , T)

 0



(23)

    0

 t 1

 TC

1(a)

(t 1 , T)

eb  θ0  t 1  μ  ebμ  aμ μ  M 

a ebμ  ebM 

b θ 0 

and  0
(24)

1  bt

   1 bt

bμ  

 ae

bt1

 T 

 e 1  ebt1  θ0

t 1  μ  

e 1  e



C I 

provided they satisfy the sufficient conditions

 θ0 



 b

 bM 



3 e

 b

 a  M  1   ae

t  M   C 2   a e bt 1  e bT   0

 2 TC

1(a)

(t 1 , T)  0

b  b  b 1

 T  

1 (26)

 2 TC

1(a)

(t 1 , T)  0

Now we develop the algorithm to find the optimal
values of t1 and T.
and

 T 2

ALGORITHM 1(q):

2 STEP1: PERFORM (I) TO (IV)

  TC1(a) (t1, T)    TC1(a) (t1, T) 

  TC1(a) (t1, T)  .

  

  

  t 2

   T 2

  

 t  T

  0

(I) Start with t1(1) = M

 1  

  

(II) Substitute t1(1) into equation (25) to evaluate T(1)
(III) Using T(1) to determine t1(2) from equation (26)

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(IV) Repeat (II) and (III) until no change occurs in the
value of t1 and T.

 C 3 I e

 Ma



e bt 1  1 a t

1e bt 1 

1 e bt 1  1

STEP2: To compare t1 and M

T  b

b b 

(28)
(I) If M  t1 , then t1 is feasible than go to step (3).
(II) If M >t1, then t1 is not feasible. Set t1 = M and evaluate the corresponding values of T from equation
(26) and then go to the step (3).
To minimize the total average cost per unit time TC1(a) ( t1, T) the optimal values of t1 and T (say t1* and T*) can be obtained by solving the following two equations simultaneously
STEP 3: Compute the corresponding TC1(b) ( t1* , T* ).
CASE (2): t1 < M
In this case since t1 < M the buyer pays no interest and

 TC

and

2 (t

 t 1

1 , T)  0

(29)
earns the interest during the period [ 0, M ], The interest earned in this case is denoted by IE(2) and is

 TC

2 (t 1 , T)  0

 T

(30)

IE 2  

C 3 I e

t 1 M 

t ae

bt dt

provided they satisfy the sufficient conditions

 Mae bt

tae bt

e bt

 t 1

 2 TC



2 (t t 2

1 , T)  0

 C 3 I e   

 b b

2 

b  0

 2 TC

(t , T)

 Ma

a 1 

2 1  0

 C I

 e bt  1

t e bt1 

e bt1  1

 T 2

3 e  b

b 1 b 2



(27)
and

  2

   2

   2 

The total average cost per unit time TC2 ( t1, T ) in this case is

 TC 2 (t 1 , T)  

 2  

 

TC 2 (t 1 , T)   

 T 2  

TC 2 (t1 , T)   0

 t  T 

TC 2

t1

, T   C  C1q H  C 3 q D  C 2 q S  IE 2

Equations (29) and (30) are equivalent to

bt1

C  C1  aμ

θ eb  θ  t μ 

a b e

 ebt1 

  1 

e b  θ 0  t1  μ   e bμ  aμ 2

T T  b θ  0

0 1

b θ    θ

C C  aμ

T T  b θ 0 

b  θ 0  t1  μ    

  

e  

C 3 a θ

e b  θ 0

t 1  μ

ae bt 1

a a  e bt1

e bt1

e b  θ0  t 1  μ 

θ 0 2



 

T  b  θ  0  

 e bμ  1    

b 2 b θ

  θ b θ

C   

C I 

a   

0 0 0

2 aebt1 T t

 Ma e 1 

ebt1  t bebt1  1 ebt1   0

e b μ  

   

C 3 



a e

b  θ 0

 t 1

 μ 

 e b μ 

T  b 1

b 

(31)

b  

T  b  θ 0 

and

 a e

bt1  e

bμ  

C 2  a



e bt

1 t1  T  

1  e bT

 e bt

1 

b  T  b

b  

1 

 

T 2 

C  C1

 aμ

 b  θ 0 

e b  θ 0  t1

 μ   e

bμ  aμ 2

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 bt1

bt1

b θ0  t 1 μ

bμ 

Interest earned Ie=0.06 Interest charged Ir=0.09

 b2    b θ   θ  b θ  b 

Ordering cost C=50

a ebμ 1

a  e e e



0  0 0

e 





 = 10,000  =2.5  >0,  >1

T = 10, t1 <=t<=T a =100

 a

 C3 

e b  θ0 

t 1  μ

 ebμ 

a e

bt1  e

bμ 

b = 0.5

  0.5

 b θ0 

b  For Case I: t1  m ( credit period)

 a bt

 C 2 

1 t 1 

T  

1  e bT 

e bt 1

  

C 3 I

 ae

e 

bt 1

Table 1

 b b 

   b 2

1 1 1

  bM 

 a  M  1   ae t

 M   C 2   a e bt 1  e bT   0

b  b  b 1

 T  

(32)
Now we develop the following algorithm to find the optimal values of t1 and T.

ALGORITHM 2

STEP 1: PERFORM (I) TO (IV)

(I) Start with (t1)1 = M
(II) Substituting t1(1) = M into equation (31) to evaluate
T(1)
(III) Using T(1) to determine t1(2) from equation (32)
(IV) Repeat (II) and (III) until no change occurs in the value of t1 and T.

STEP 2: Compare t1 and M

(I) If t1 < M, t1 is feasible than go to step (3).
(II) If t1  M, t1 is not feasible set t1 = M and evaluate the corresponding values of T from equation (32) and

For Case II: t1  m

Table 2

then go to the step (3).
STEP 3: As stated earlier, the objective of this problem is to determine the optimal values of t1 and T so that TC ( t1, T ) is minimum. As the discussion carried out so far one can get

VI .OBSERVATION:

From Table 1, we can say that if permissible delay period is increase then the time of inventory period is also increase. But selling price is decrease and the profit is increase.
From Table 2, we can say that if permissible

TCt*,T*  MinTC   t*, T* , TC   t*,T* , TC

t*, T* 

delay period is increase then the time of inventory

1 1 a 1

1 b 1

2 1

period is also decrease. And selling price is decrease and the profit is increase.

V. EXAMPLE AND TABLES

NUMERICAL ILLUSTRATION :

Shortage cost C2=3 Unit purchase cost C3=12

Backlogging rate =0.75 Holding cost C1=2.5

VII. CONCLUSION

In this model, an appropriate pricing and lot sizing model for a retailer when the supplier provides a

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permissible delay in payments is developed and discussed. We desire the first and second order conditions for finding the optimal cost and then developed an algorithm to solve the problem. The model is proposed for non instantaneous deteriorating items with exponential demand. Shortages are allowed and they are partially backlogged. During the shortage period only a fraction of the demand is left. The algebraic procedure and cost minimization procedure is applied to find the different optimal values. In the end some particular cases are also given.

REFERENCES

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Abad P.L. and Jaggi C.K. (2003). A joint approach for setting unit price and the length of the credit period for a seller when end demand is price sensitive. International J. of Production Economics, 83,

115-122.

Chang H.C. et al. (2009). The optimal pricing and ordering policy for an integrated inventory model when trade credit linked to order quantity. Applied Mathematical Modeling, 33(7), 2978-2991.

Chen L. H. and Kang F.S. (2007). Integrated vendor–buyer cooperative inventory models with variant permissible delay in payments. European Journal of Operational Research, 183(2), 658-673.

Chu P., Chung K. J. and Lan S. P. (1998). Economic order quantity of deteriorating items under permissible delay in payments. Computers and Operations Research, 25, 817-824.

Goyal S.K. (1985). Economic order quantity under conditions of permissible delay in payments. J.O.R.S., 36, 335-338.

Manna S. K. and Chaudhuri K.S. (2008). An optimal pricing and lot sizing decisions under Weibull distribution deterioration and trade credit policy. Yugoslav J. of Operations Research, 18(2), 221-

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in Payments. International Review of Pure and Applied

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Teng J.T. et al. (2005). Optimal pricing and ordering policy under permissible delay in payments. International Journal of Production Economics, 97(2), 121-129.

Ouyang L.Y. et al. (2008). Retailer’s ordering policy for non-instantaneous deteriorating items with quantity discount, stock dependent demand and stochastic backorder rate. J. of Chinese Institute of Industrial Engineers, 25(1), 62-72.

Singh S. R. (2008). Production Model for Items under the Effect of Inflation and Permissible Delay

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