Lecture No. 16

Held on: Monday, September 18, 2000.

Notes Prepared by: Sarish Jain & Rohit Gulati

 

Models with deterministic time varying demand:

Let the demands in successive periods be (r_1, r₂, …, r_n) where r_n represents the demand in the n^th period and let the ordered amount in successive periods be (Y_1, Y₂, …, Y_n) where Y_n is the order in the n^th period.

WAGNER-WHITE MODEL:

Assumptions:

·         Lead times are zero (This is not really necessary).

• Shortages are not permitted.
• Starting inventory is zero.
• Only linear cost of holding and fixed order cost.

Note: More generally the holding cost is a concave function of the ending inventory in each period and the ordering cost is a concave function of ordered amount.

Concave function: If we take any two points on the function and join the line the line always remains below the function. Its double derivative is non-positive.

 

 

 

 

 

Convex function: It's just vice versa of concave i. e. If we take any two points on the function and join the line the line always remains above the function.

Let the total inventory at time t be X_t.

_t

Then X_t = å (Y_i – r_{i )}

ⁱ⁼¹

Also the holding cost in the period t is a function of X_t: h (X_t) and the ordering cost in period t is a function of Y_t: C_t (Y_t).

So our objective is to minimize the following function for minimum cost

_n

Min. å (C_t (Y_t)+ h (X_t))

^t=1

With the following constraints

X_{t ³} 0 means that no backordering is allowed and

Y_{t ³} 0 means that order cannot be negative.

Typically we have

C_t (Y_t) =K*d (Y_t) + c*Y_t

Where d (Y_t)= 1 if Y_t >0

= 0 otherwise.

and

h (X_t)= h* X_t where h is constant.

 

RESULT:

An ordering policy has the property that if you have inventory at time 't-1', then you don't produce at time 't'. If you don't have inventory at 't-1' then and only then you produce at time 't'. i. e.

Y_t * X_t-1 =0

X₀=0, for t= 1,2…n.

What this means is you want to produce only in terms of consecutive demands i. e. 0, r₁, r₁₊r₂, r₁₊r₂₊r₃ etc. and not in fraction of these demands, the logic being that you can always get a lower cost using the whole demands then using fractional demands.

Now we demonstrate the above logic for a two-period problem.

 

 _____________________________________________________________

Eg. 2 period problem

 

 

\ We have an inventory of d after the first period as we produce r₁+d and demand is r₁ so we are left with an inventory of d after the first period.

Total cost = K₁ + K₂ + C₁ (r₁+d ) + C₂ (r₂-d ) +h*d

= K₁ + K₂ + C₁ * r₁ + C₂ * r₂ + d (C₁ - C₂ + h)

 

Now

·         If (C₁ - C₂ + h) >0 then if you want to minimize cost keep d =0 i. e. produce r₁ in t₁ and r₂ in t₂.

• If (C₁ - C₂ + h) < 0 then you want to maximize d and here you produce r₁+r₂ together and then K₂ vanishes.
• If (C₁ - C₂ + h) =0 then again K₂ vanishes.

_____________________________________________________________

 

Thus according to the above argument Y_t can be,

Y_t = 0, (r_t), (r_t+r_t+1)…(r_t+r_t+1+…+r_t+n)

 

4 period problem:

The problem is that is there are four periods with varying demands, we want to optimize on the total cost that is incurred in meeting the demands. The problem can be seen as somewhat like the shortest path problem. Suppose there are five points in the problem 1,2,3,4,5 and there are several ways of going from point 1 to any other point. Now what is different in all these paths is the distance to be traveled.

For our problem the distance is represented by the costs involved. Here travelling from one point to the other point is like producing the demands for those periods in the starting period and the cost involved is thus the measure of distance. For e.g.C₁₃ means that producing the demands for periods 1 and 2 in the period 1.

Now there are several ways to go from point 1 to 5, like directly from 1 to 5,1 to 2 2 to 4 4 to 5,1 to 3 3 to 5 and so on. What we want to do is to minimize the cost. So let us first represent the costs in terms of terminology we have used so far.

Consider C₁₃ (cost of producing the demands of periods one and two in period 1) it equals

C₁₃= C₁ (r₁+r₂) + h₁*r₂

Here the first term is cost of producing the quantity r₁ + r₂ in the period one and the other term is the cost of holding the inventory of r2 in the period 1.

Similarly

C₂₅ = C₂ (r₂+r₃+r₄) + h₂ (r₃₊r₄) + h₃ (r₄).

In general we can write

_{j-1 j-2 j-1}

C_ij = C_i (å r_k) + å h_k * ( å r_k )

^{k=i k=i l=k+1}

Now that we have defined the cost factors, we need to define the solution procedure for this shortest path problem.

·         We start with f₅ = 0

• Then we try finding out the best way to reach 5 from 4
• Then we try finding out the best way to reach 5 from 3
• Then we try finding out the best way to reach 5 from 2
• Then we try finding out the best way to reach 5 from 1

In general for n periods

f_n+1=0

and we find f_i =min [ C_ij + f_j ]

^j>i

for i= n, n-1, …,1

 

 

 

 

___________________________________________________________

Eg. Demands for cosecutive periods r = (52,87,23,56)

C_t(y) = k d (y) where k =75 for t=1, 2, 3, 4

h_t(x) = h* x where h=1

Solution:

f₅ = 0

f₄ = 75

{cost at period 4 is for just satisfying demand of period 4 which is 'k')

f₃ =min. C₃₄ + f₄

C₃₅

= min. 75+75

75+56

=131 at j=5

f₂ = min C₂₃ + F₃

C₂₄+ f₄

C₂₅

= min 75+131

(75+23)+75

75+23+2*56

=173 at j =4

f_{1 =} min 75+73

75+87+131

75+87+2*23+75

75+87+2*23+3*56

=248 at j=2

Means that it's best to go to 2 from 1; then 2 to 4 and then from 4 to 5

Where values are

Y₁₌r₁=52;

Y₂=r₂₊r₃=110;

Y₃=0;

Y₄=r₄=56;

And final cost = f₁ = 248

 

 

 

 

Extending to allow backordering:

Y_t * X_t-1 =0 in the previous case extends to

Y_t > 0 only if X_{t-1 £} 0

i.e. here you will never produce if you have inventory since backordering is allowed.

 

There exist integers

0= S_0£ S_1£ …£ S_n=n

such that Y_t=0 if S_t-1 = S_t

_St

& Y_t = å r_j 1 £ t £ n

^i=1+st-1

you produce full amounts but you may produce full amounts to satisfy previous demands.

If

S_t = t then zero inventory.

S_t < t then backordered demand at time t.

S_t > t then inventory exists.