An Inventory Model With Stock Dependent Demand, Weibull Distribution Deterioration

An Inventory Model With Stock Dependent Demand, Weibull Distribution Deterioration

R. Babu Krishnaraj Research Scholar,

Kongunadu Arts & Science College, Coimbatore 641 029.

Tamilnadu, INDIA. &

K. Ramasamy Department of Mathematics

Kongunadu Arts & Science College, Coimbatore 641 029.

Tamilnadu, INDIA.

An inventory problem can be solved by using several methods starting from trial and error methods to mathematical and simulation methods. Mathematical methods help in deriving certain rule and which may suggest how to minimize the total inventory cost in case of deterministic demand. Here an attempt has been made for obtaining a deterministic inventory model for stock dependent demand pattern incorporating two-parameter Weibull distribution deterioration and with reserve inventory.

Key words: EOQ, Deterioration, Stock dependent demand pattern

A number of researchers have worked on inventory with constant demand rate, time varying demand patterns. A few of the researchers have considered the demand of the items as stock dependent demand pattern.

Datta and Pal [3] have developed an order level inventory system with power demand pattern, assuming the deterioration of items governed by a special form of Weibull density function

(t) 0t; 00 1, t0.

They used special form of Weibull density function to sidetrack the mathematical complications in deriving a compact EOQ model. Gupta and Jauhari [4] has developed an EOQ model for deteriorating items with power demand pattern with an additional feature of permissible delay in payments. They also used special form of Weibull density function for deterioration of items.

A step forward to special form of Weibull density function, here we shall develop an EOQ model with stock dependent demand pattern and

using actual form of Weibull density function Z(t) =

t 1,

where

(0 1), 0 for deterioration of items. Despite of all mathematical

intricacies, expressions for various inventory parameters are obtained.

Here we shall develop the same problem with stock dependent demand pattern with reserve inventory.

Inventory model is developed under the following assumptions and notations.

Replenishment rate is infinite. The lead-time is zero.

Shortages in inventory are not allowed.

The demand is given by the stock dependent demand pattern for which, demand upto time t is assumed to be , D(t) = a+bI(t),

Where D is the demand size during the fixed cycle time T, and is the demand rate at time t

The rate of deterioration at any time t 0 follows the two-parameter

Weibull distribution Z(t) = t 1 , where (0

1) , is the scale parameter

and

(0)

is the shape parameter.

T The fixed length of each ordering/production cycle.

C1 The holding cost, per unit time.

C3 The cost of each deteriorated unit.

C2 – The reserve inventory cost per unit time.

Let Q be the number of items produced or purchased at the beginning of the cycle and (Q-S) items be delivered into the reserve inventory stock. Balance of S items as the initial inventory of the cycle. It will be the initial

inventory at time t = 0 and d be the demand during period

t1 . Now, the

inventory level S gradually falls during time period (0, t1 ), due to demand

and deterioration. At time t =

t1 inventory level becomes zero. Shortages are

fulfilled from the reserve stock (Q-S), after the period

t1 .

Let I(t) be the on-hand inventory, then the various states of the system are governed by the following differential equations:

dI (t) t 1I (t) (a bI (t))

dt

Where Z(t) = t 1

0 t t1

(1)

(2)

dI (t) a dt

t1 t T

(3)

Using (2) in (1), the Solution of I(t) is,

e(t 1 b)dt e(t bt )

I(t) e(t bt )

= ae(t bt )dt

I(t) =

Se(t bt ) ae(t bt ) e(t bt )dt

(4)

Solving further, on expanding

e(t bt) 1 (t bt)

, as (

1) gives

I(t) =

Se( t bt ) ae( t bt )[t t bt ]

1 2

1 2

1 2

(5)

Solution of (3) using

I (t1 ) 0

gives,

I(t) =

a(t1 t)

(6)

Using I( t1 ) = 0, in (5) gives,

S = a[t1

t 1 bt 2

1 1 ]

1 2

(7)

The total amount of deteriorated units,

= S –

t1

(a bI (t))dt

0

Using (7) in the above, the total amount of deteriorated units in [0, t1 ]

= a[t1

t 1 bt 2

1 1 ]

1 1 ]

1 2

– [at

• bt t1

2

2

]

]

2 0

t 1

bt 2

bt2

= a[t1 1

1 ] – [at1 1 ]

1

1

= t 1

1

1 2 2

(8)

Average total cost per unit is given by

t 1 C t1 C T

C(S, T) =

C3 1 +

1 I (t)dt – 2 I (t)dt

(9)

1 T 0 T t

1

Substituting the values of I(t) from (5) and (6), eliminating S using (7) and integrating yields,

t 1

3

3

C(S,T) = C 1 +

1

t1 1 2

C1 ( Se( t bt ) ae( t bt )[t t bt ]) dt+

T 0 1 2

C T

2 ( a(t1 t) )dt

T t1

t 1

3

3

= C 1

1

t 1 2

1

1

1 2

T

T

+ C1

( a[t

t1

bt1 ]e( t bt ) ae( t bt )[t t bt

] )dt

0

C T

1 1 2

1 2

– 2

a(t1 t) dt

T t1

t 1

3

3

= C 1 +

1

t 1

bt 2

C t1

a[t1 1 1 ][1 ( t1 bt1 )]

1 2

1

T 0

t 1

bt 2

a[1 ( t1 bt1 )][t 1 2 ]

C T

dt

– 2

a(t1 t) dt

T t1

t 1

C a t1

t 1

bt 1

bt 2

bt 2

3 1

T 1 1 1

1

1 2 2

= C 1 +

1 [t

0

t 1 bt2 1

1 1

1 ] –

t 1 bt2

2 2 2 3

bt bt b t

bt bt b t

1 +

[t 1 2 t

2 1 2

]dt

2

2

+ C2a t T

T [t1 t

]

2 t1

t

t

1

= C 1

+ C1a

bt2t

t t 1

( 1) b)

1 bt 2t

t1

3 1

T 1 2 1 1 2

t t 1

t t 1

0

t2

t 2

bt3

t 2

bt 3

bt 3

b2t 4

t1

–

2 ( 1)( 2) 6 2 2( 3) ( 1)( 3) 8

0

2

2

– C2a t T

T [t1t

]

2 t1

= t 1 +

C a

t2 bt3 b2t4 ( b)

2

C3 1

1 1 1 1

t1

1

T 2 3 8

1 2 ( 1)( 2)

– 1 1

3

C a

• T t

(10)

bt1

2 t1T 1

2 2

2 2

2 2( 3) ( 1)( 3) T 2 2

(neglecting higher powers of

2, 3…. )

Further, for the minimization of the cost, we set

dC(t1 ) 0

dt

C a

2 b t

(( b) 1)( 2)

1

C3 t1

1 t1

bt1

1 t1

2 3

2 3

T 2 1

–

–

( ( 1) 1)( 3)

2( 1)

bt 2

– C2a T t ] 0

[

[

T 1

1

1

1

1

On solving it, we obtain value of value of t

t1 and let this value of t1 be the optimum

1

1

Solving the optimum value of t in (7), optimum value of S is,

S* =

a[t*

t* 1 bt*2

1 1 ]

1 2

1

1

From the above work, An inventory model is studied using Weibull distribution. With Stock dependent demand rate, we have obtained Optimal value of S, with minimum cost .

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