By32474479

Anil R. Maisuriya, PravinH. Bhatawala / International Journal of Engineering Research and
Applications (IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 3, Issue 2, March -April 2013, pp.474-479
An EOQ Model With Stock Dependent Demand Rate And
Variable time
Anil R. Maisuriya (1) andPravinH. Bhatawala (2)
Assistant professor, Agrawal College, Navsari, Gujarat, India.
Retired Head, Department of Mathematics,VeerNarmad South Gujarat University, Surat, Gujarat, India.

Abstract
Inventories are being considered to be (I) Firstly, with instantaneous rate of replenishment
an important aspect for any organization. EOQ and no shortages allowed
model is defined as a controller of quantity and
total cost per unit. This paper throws light on the (II) Secondly, with finite rate of replenishment with
Optimal Economic Order Quantity with stock no shortages allowed and linear
dependent demand rate and consider variable demand rate
time‘t’ which is used for two constants values α In this EOQ model it is derived with
and β because after long period it changes time “Stock Dependent Demand Rate”. Here for
to time. This model tries to overcome the different functional relationship between demand
limitations of the model given by Gupta and Vrat and supply we have to derive EOQ model with
who have used functional relationship in which α instantaneous rate of replenishment [commodities]
and β were constant but in present model we with no shortages allow and finite rate of
used time‘t’ to make them varies. Here, we used replenishment also with no shortages and with time
instantaneous case of replenishment and finite period [t]. The analysis is given below:
case of replenishment for different functional Analysis
relationship to get optimum value of stock Condition: - EOQ model with
quantity and total cost per unit time. It is easy to instantaneous replenishment with no
use EOQ models to get an optimal order shortages.
quantity level. In addition, we make different In prior EOQ model which is assumed that demand
conditions and propose economic interpretation. rate (r) is constant. For this the total cost of the
model per unit time is given by:
Keywords: EOQ Model, A (Assumption), Calculation of EOQ Model
Instantaneous Demand,Replenishment rate, time (t). C C q
Sr (t ) p c
Introduction E ( q )  r (t )C   ……… (1)
In past inventory model it is assumed that
p q 2
demand rate does not depend upon inventory level.
But it is not true in the cases [for many Where q = the order quantity
commodities] where the consumption pattern is r =
influenced by the inventory maintained. replenishment rate or demand rate
In general situation the demand may increase if the Cp = the unit
level of commodities [inventory or stock] is high purchase cost
and may decrease if the level of commodities S = setup cost
[inventory or stock] is low, because the customer per order
may be attracted with thepresence of large pile of Cc = unit
inventory or stock in the market. This is called carrying charges per unit per unit
“Stock dependent demand rate”. In such cases the time
total cost should be include the direct purchase cost EOQ model has been derived for the different
of the items as certainly the inventory process does functional relationships which are given below:
affect the direct purchase cost of product. So we Case: - (a) r (t )   t   t log q
should include direct purchase cost in total
inventory cost to get actual or accurate result of Substituting the value of r in equation (1) we get,
Inventory model.
S ( t   t logq) C pCc q
Gupta and Vrat presented an EOQ model E(q)  ( t   t logq)C p  
where stock level affects the consumption rate. q 2
They have given different functional relationship
for demand and stock or supply which are as
follows:

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To get the stationary value of q we take
q q
dE(q)  2 t C (log a) a q 2  2 S  t  2 S  t (log a)a q
0 p e e
dq
 tC p  S t  S t S t log q C p Cc 2 S  t a q  C C q 2  0

  
 2  2   0 p c
q  q  q q2 2
  ………....... (8)
………… (2) Now to get optimum solution we can use second
By simplification of this result we get, derivation.

 C pCc q 2  2 tC p q  2S t log q  2st(   )  0 q q
 tC (log a) 2 a q3  2St  S t (log a) 2 a q 2
…… (3)
p e e
The above equation no (2) can be solved by taking q q
second derivation to obtain the value of q 2S t (log a)a q  2S ta
d 2 E (q) St[  (log q  3)  2 ]   tC p q e
2

  0 
dq 2 q3 q3
…………….. (4) ….. (9)
This is positive for all values of q given by, Now to get the optimal value of q we take

 (log q 2  3)  2 C p d 2 E (q)
  ……… (5) 0 because q is always positive and
q S 2
The equation (3) gives an optimal value of q under
dq
the condition (5) non zero.
q q
q   C (log a) 2 a q3  2S [1  (log a) 2 a q 2
Case: - (b) r (t )   t   ta p e e
Putting the value of r in to equation (1), we get, q q
 (log a)a q  a )]  0
q C C q e
q S ( t   ta ) p c ……… (10)
E (q )  ( t   ta )C   Solving equation (10) we get optimal value of q.
p q 2 q
……..… (6) Case: - (c) r (t )   t   t a
By busing first derivation we get stationary values Putting this value of demand rate r into equation (1),
of q which is given below. we get
q
1 S  t q a log a E (q)   t   t a q C p 
S C Cq
   a q   p c
  t C a log a  S  t   
q e q 2
p e  q2  q2
 
q C C S  S  a q C p Cc q
S  t a p c  E (q)   t C p   t C p a 
q
 
  0 q q 2
q 2 2 ... (11)
……………… (7) Now to get stationery value of q, we used first
d E (q)
derivative and take 0
dq

475 | P a g e

S  S  a loge a A  a q
q
 C C q
   t C p a loge a  2 
q
 2  S ( t   t q ) p c
q q E (q)  ( t   t q )C 
q 
.. (12) p q 2
C p Cc … (18)
 Now to get stationery value of q, we take
2 d E (q)
  2 t C p a  q loge a. q 2  2S   2 S  a  q loge a. q   0 , from equation (18), we get,
dq
.. (13)
2S  a q  C p C c q 2
C C q2
Now to get optimum value of q we can use second derivation. From  p c
equation (12), we  gett q [ C  q  S   S ] 
  S t  0
p 2
 2 t C a (loge a) q  2 A{   a }  A a …q(19)
q 2 3 q q
..
loge a{2  q. loge a }  0 Now to obtain the optimal value of q, we take
…… (14)
d 2 E (q)
Case: - (d) r (t )   t   t q   t q
2 0
dq 2
Putting r (t) in to equation (1), we get, , from equation (19), we get
2
2 )C  S ( t   t q   t q )   1  1
E (q)  ( t   t q   t q   t C  2 q  S t  2 q  S t  q
p q p

C Cq   tC  q C C q
p c p p c

2 To get minimum cost for this model, we must insert
…........ (15) two conditions which are given below:
Now to get stationery value of q we take
d E (q) d 2 E (q)
 0 , from equation (15), we get, (a) If   1then  0 is positive q , and
dq dq 2
 q 2[C C  2 t C  2S t  4 t C q ]  2S t  0
p c P p
d 2 E (q) S t (1   )
.… (16) (b) If 0    1then  0 is positive q 
Equation (16) can be solved by using second order dq2 C t (1   )
p
derivation to get an optimal value of q. If α, β and γare restricted to be positive constant
2 S t then a marginally different form of functional
 q3 .……..…………… (17) r (t )  t   t q
 tC
relationship similar to
can be expressed as
p
Here q is positive, so it gives an optimum quantity. r (t )  t   t q  ……………... [A-1]
Case: - (e) r (t )   t   t q
 Substituting [A-1] in equation (1), we get,

 C C q
Putting r (t) into equation (1), we get,  S ( t   t q ) p c
E (q)  ( t   t q )C  
p q 2

476 | P a g e

Now to get stationery value of q, we take
d E (q) C C q2
0 q p c
dq   t e [C q 2  S q  S ]   S  t  0 . (24)
p 2
  1    Now to get optimum value of q we take
 [2  C t q  2S   t q  2 S  t q ]
p d 2(20)(q)
.... E
C Cq 2 0
dq 2
p c
  S t  0 From equation (24), we get,
2
 1 ……………...…....... (21)  C C  S t e q  2C t e q 
 
Now to get optimal value of q, we take
q    p c p 
2 E (q)  q 
d
  te  …. (25)
dq 2  
Here the value of q become optimum
   1
   t  2C t q  S   2t q
because it is greater than 0.
For positive values of anda marginally
p different forms for functional relationship can be
expressed as
  1  q
 S  t q    C t q  C C q . (22) r (t )  t   t e
p p c ………………. [A-2]
Substituting [A-2] into equation (1), we get,
Insert equation after confirming, q C C q
 q S t S t  e p c
d 2 E (q) E (q)   t C   t C e   
is positive when p p q q 2
dq 2
C t q (1   )  S t (1   )  0
Now to get stationery value of q, we take

p d E (q)
=0
S t (1   ) dq
q 
C t (1   ) C C q2
p q p c
  t e [C q 2  S q  S ]   S  t  0 .…… (26)
0    1 p 2
It implies that here value of q is non negative which Now to get optimum value of q, we take
is define optimum solution.
q d 2 E (q)
Case: - (f) r (t )   t   t e 0
Substituting r (t) into equation (1), we get, dq 2
q C Cq From equation (26), we get,
q S ( t   t e ) p c q q
E (q)  [ t   t e ]C   ... (23) C C  S t e  2 C t e
p q 2 p c p
q …. (27)
Now to get stationery value of q we take
q
d E (q)  te
= 0 from equation (23), we get
Thus equation (29) gives an optimum solution of q.
dq
Analysis:
Finite Replenishment case

477 | P a g e

Here EOQ model has been derived under finite rate
d 2 E (q)
of replenishment and for this for this following Clearly  0 for all values of q
functional relationship will be used. d q2
t
(a) r ' (t )  r   2Ak  t 2Ak 2  t 2 A k 3 t 
q  2
 q (  t  k q)  q (  t  k q) 2  (  t  k q)3 

 q   . (29)
(b) r ' (t )  r   t q 2Ak r 3

(c) r ' (t )  r   t e
q (  t  k q)3
Here clearly q > 0 for all values of q which is called
Where β and γ are positive constants and r’ (t) is the optimum quantity.
consumption rate during depletion time. The total
Case (b) r ' (t )  r   t q

cycle time T consists of two parts:
Trc = the time during which replenishment Here the total cost per unit time is given by
come and
Td = the time during which stock depletion Ak r A k  t q  1 C pCc q  r 
takes place E (q)    1  
Thus, T = Trc+ Td q (k   t q  ) q (k   t q  ) 2  k
The finite replenishment rate and the consumption
rate are taken k and r respectively such that k > r.
d E (q)
obviously net rate of procuring of items in inventory Condition to obtain value of q is 0
is k– r. dq
t
Case (a): r ' (t )  r   Ak r A k r  t  q  1 A k  t (  1) q  2 
q   2
 q (k   t q  )  q (k   t q  ) 2  (k   t q  )  
Following Gupta and Vrat the inventory carrying  
cost per cycle is given by

A k  2t 2  q 2 2 C p Cc  r 
C p Cc q 2  r    1    0 ………. (30)
1   , where (k   t q  ) 2 2  k
2r  k  Equation (30) can be solved by using second
1 1 1 r derivation for optimum value of q.
  1   d 2 E (q) 2Ak r A k  t (  1)(  2) q  3
r* k r'  k   3  
The total cost per unit time of the system is d q2 q (k   t q  ) (k   t q  )
1 C p Cc q 2  r 
E (q)   A  1  
T
 2r *  k    A k  t  q  3 [r (3   )  3 t (  1)q  ]
(k   t q  ) 2
Ak  t Ak r C p Cc q  r 
   1   (28)
q ( t  k g ) ( t  k g ) 2  k 2 A k (  t  ) 2 q 2  3 ( r   t q  )
For which optimum condition is  …………….. (31)
(k   t q  ) 3
d E (q) Ak  t Ak2  t A k 2
 2   d 2 E (q)
dq q ( t  k g ) q ( t  k g )2 ( t  k g )2 Now  0, if
d q2
C p Cc  r  r (3   )  3 t (  1)q >0 or
 1    0 1
2  k  r (3   )  
q and 2    3 ........................ (32)
3 t (  1) 
This equation can be solved for q by taking second
derivation.
 

478 | P a g e

Hence solution of equation (31) under by Gupta and (36) Vrat under specific conditions.
condition of equation (32) gives an optimal solution Some more functional relations between the
of q demand rate and stock quantity can be constructed
q to obtain the new results.
Case: - (c) r ' (t )  r   t e
In this case the total cost per unit time of
the system is: References:
(1) Aggarwal, S.P., and Jaggi, C.K., “Ordering
policies of deteriorating items under
A k ( r   t e q ) C p Cc q  r  permissible Delay in payments”, Journal
E (q)   1   of the Operational Research Society, 46
q (k   t e q ) 2  k (1995) 658- 662.
(2) Chu, P., Chung, K. J., and S. P.,
d E (q) “Economic order quantity of deteriorating
Condition to obtain value of q is 0 items under permissible delay in
dq payments”, Computers and Operations
 A k  t eq A k (r   t e q A k  t (r e q   t e 2 q )  Research, 25 (1998) 817 – 824.

 q(k   t e q )  q 2 (k   t e q )  q(k   t e q ) 2  
(3) Chung, K. J., “A theorem on the
 
determination of economics order quantity
under permissible delay in payments”,
Computers and Operations Research, 25
C p Cc  r  (1998) 49 – 52.
 1    0 ……………….. (33)
(4) Shah, N.H., “A probabilistic order level
2  k system when delay in payment is
permissible” Journal of Korean OR / MS
To obtain optimum solution, we use second Society (Korea), 18 (1993b) 175 – 182.
derivation: (5) Teng, J. T., “On economic order quantity
d E (q) A k  t e [( q  1)  1
2 q
2Ak r conditions of permissible delay in
  3  payments”, to appear in Journal of the
d q2 q 3 (k   t e q ) q (k   t e q ) Operational Research Society, 2002.

2 A k  t (r e q   t e 2 q )
q 2 (k   t e q ) 2
A k t e q (  r )(k   t e q )

q(k   t e wq )3 ………… (34)
Itis clear that
2
d E (q) k
 0, if (k   t e q )  0 or q  log …..… (35)
dq 2
t
Equation (34) gives an optimum value of order
quantity q under condition of equation (35).

Conclusion:
In EOQ model derived here, not allowing
shortages, demand rate depends upon the stocks i.e.
more the stock, the consumption or demand is also
more i.e. if stock increases, the demand will also be
increased. First model is discussed for instantaneous
case of replenishment and then after for finite
replenishment case for assumed demand rates. A
solution procedure is explained to get optimum
value of stock quantity and substituting this in cost
equation the total cost of the system per unit time
can be forecasted.
An EOQ model formulated in second case
matches with the corresponding EOQ model given

479 | P a g e

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