Production Planning Control

Lecture Notes

Dated: 4/10/2002

 

 

Notes Prepared by: Vikrant Gangwal & Saurabh Singh Rohilla

 

Model with Deterministic Time Varying Demand

 

Let (r_1, r_2, r₃ …. r_n) be the known requirement for a single product over the next n periods. Where  r_n represent the demand of the n^th  period. Period are defined as times of replenishment. Let (Y₁, Y₂… Y_n) denotes order sizes placed in these periods.

 

Wagner-Whiten Model

1. Shortage are not permitted 

2. Starting inventory is zero

3. Only linear costs of holding & fixed order costs are present

 

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.

Inventory at the end of the period t is given by

      

                   " j = 1 to t                    where t = 1 to n.

                    

 Initial Inventory can be assumed to be zero.

 

Let h_t(x) denotes the holding cost incurred in period t, and C_t(Y_t) be the order cost .

 

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

 

Optimal policy is obtained as a solution to the following problem:

 

Minimize å (C_t * Y_t     +   h_t* X_t )

With the constraints:      X_t = å (Y_j - r_j)

                                      X_t ≥ 0     t =1,2,…..n

                                      X_o=0

 

Common form of the cost function is:

          C_t(Y_t) =Kδ(Y_t) + CY_t                 Where   δ(Y_t) = 1      if   Y_t  > 0

                                                                                         = 0         otherwise                   

h_t(x_t) = hx_t

Results: An optimal ordering policy has the property

           

            Y_t*X_t-1 = 0    t =1,2,…..n

We assume that production is made such that it fulfills complete demand, this implies Y_t = 0, r_t , r_t + r_t+1 ,…….. ,r_t + r_t+1 +….+r_n  are the only permissible values.           

 

NETWORK FLOW PROBLEM

 

            Optimal policy may be found by finding shortest path through an acyclic network with the nodes labeled 1,2,…..n and arcs connecting all pairs of the nodes with    i < j.

 

            Length of the arc connecting the nodes i and j ,C_ij , is the cost of ordering in the period i to satisfy requirements through period j-1,where i < j ≤ n+1, plus the inventory cost .

 

Thus                                                        _j-1     

 

 

Example: 4 period problem

 

 

 

 

 

 

 

 

 

 

 

The objective is to find the shortest path when cost of each path is known. This is analogous to find the minimum cost of the ordering plan. 

 

Solution: The network flow problem comes under Dynamic Programming.

Where we find the shortest path between any two (A, B) points by first finding the shortest path between the ‘B’ point and a point near its vicinity and then reaching gradually to the ‘A’ point.

 

Any path from (1-5) [see above fig.] is a feasible plan of action (order).

   C₂₄  (2 to 4) = C₂ (r₂+r₃) + h₂*r₃

   C₂₅  (2 to 5) = C₂ (r₂+ r₃+ r₄) + h₂(r₃ + r₄) + h₃* r₄

 

 

Let f_i denote the cost of optimal policy at the i ^th  period.

 

Then f_n-1 = 0

f_i = min ( C_ij + f_j ) for i = 1,2, ….n

                 ^{j > i} 

 

 Let’s have: r = (52, 82, 23, 56)

       C_t(Y) = 75   if  y > 0

                = 0  otherwise

       h_t(X)  = x

 

Then f₅ =0

 

   f₄ = C₄₅ + f₅ =75     at j=5

   f₃ = min(C₃₅+f_5,C₃₄+f₄)

      =min(131,150)

      =131                  at j=5

   f₂ = min(C₂₅+f_5,C₂₄+f₄, C₂₃+f₃)

      =min(75+23+2*56,75+23+75,75+151)

      =173                     at j=4

   f₁ = min(75+87+2*23+3*56,75+87+2*23+75,75+87+131,75+173)

      =  248                   at j =2

 

    Y₁ = 52 ,Y₂ =110,Y₃=0,Y₄=56

 

 

 

Random Demand Ordering Plan

 

Probability Distribution Function is assumed for this random demand.

 

 

 

 

 

 

                  f(x)

 

 

 

                                                 xà

 

This can be best understood from the problem of a newspaper boy.

           There is some cost which can be recovered by selling extra newspaper (if any) as junk (raddi).

           There is some cost incurred if he doesn’t meet the demand

           There is some cost incurred if he buys newspapers in excess

 

So objective is how many newspapers should he buy to sell them in order to minimize the total cost incurred, where the demand for the newspaper is for that day only (obviously!).

 

Cost price C = 25p

Selling price S = 75p

Unsold (raddi) return = 10p

 

Suppose demand is D and he orders X. Total cost incurred

                         C (X, D) = 15(X-D)⁺  + 50(D-X)⁺

Where

             Over-ordering cost = (Cost price - unsold return) = (25-10) = 15

                 Under-ordering cost = (Selling price – Cost price) = (75-25) = 50

So assigning some expected value to the cost incurred  E(C (X, D)).

We would like to the value of X that minimizes the expected cost C(X).

            = C_oF(X) –C_u(1- F(Q))

                                    for all Q  0

Because the second derivative is non negative, the function C(X) is said to be a convex.

Lets X^* is the optimal solution, where first derivative of C(X) is equal to zero. That   is,

       C’(X^*) = (C_o+ C_u) F(X^*) - C_u

From above equation we get        

F(x) = C_u / (C_u + C_o)

This F (X) represents the probability that the demand is less than ‘X’.