Question

A chemical company produces two chemicals, denoted by 0 and 1, and only one can be produced at a time. Each month a decision is made as to which chemical to produce that month. Because the demand for each chemical is predictable, it is known that if 1 is produced this month, there is a 70 percent chance that it will also be produced again next month. Similarly, if 0 is produced this month, there is only a 20 percent chance that it will be produced again next month.

To combat the emissions of pollutants, the chemical company has two processes, process A, which is efficient in combating the pollution from the production of 1 but not from 0, and process B, which is efficient in combating the pollution from the production of 0 but not from 1.   Only one process can be used at a time. The amount of pollution from the production of each chemical under each process is

	0	1
A	100	10
B	10	30

Unfortunately, there is a time delay in setting up the pollution control processes, so that a decision as to which process to use must be made in the month prior to the production decision. Management wants to determine a policy for when to use each pollution control process that will minimize the expected total discounted amount of all future pollution with a discount factor of 0.5.

Suppose now that the company will be producing either of these chemicals for only 4 more months, so a decision on which pollution control process to use 1 month hence only needs to be made three more times.

a) Define Stage, states, and alternatives.

b) Formulate the cost matrix.

c) Identify the stationary policies.

d) Formulate the transition matrix.

e) Find an optimal policy for this three-period problem.

Answer 1

Let x and y be number of hours of Process 1 and Process 2 respectively

A. Since the problem is to minimize cost, the objective function will be minimisation

Min Z = 4x + y subject to

For Chemical A, 3x + y >= 10

For Chemical B, x + y >= 5

For Chemical C, x >=3

Non Negative Constraint: x1, x2 >= 0

B. The graphical solution is as below (To make the graphical solution, plot alll the three constraints as straight line, shade the region for the constraint. As all are >= the region will be on the right side of the line. The line is drawn by two point method. Feasible region lies in the area where all the all the lines are feasible. The solution will be the intersection points)

For constraint A, join (3,1) and (2,4)

For constraint B, join (1,4) and (4,1)

Constraint C is a parallel to y axis at x = 3

The feasible region is marked by green dotted lines and the feasible solution are (3,2) and B(5,0)

Testing for the two points,

Z = 4*3 + 1*2 = $14 and Z = 4*5 + 1*0 = 20

The minimum value of Z occurs at x = 3 and y = 2

Process 1 should be run for hours 3 hours

Process 2 should be run for hours 2 hours

Minimum cost = $14

C. How cost per hour for process 1 must change before the optimal solution changes

Since there are only two optimal points, for the solution to change, the optimal points should change. This is given by following equation (Since function is minimization, the greater than sign is put)

3x + 2y > 5y

y > x i.e., Process 1 cost < Process 2 cost. So the optimal solution will change only when Process 1's cost falls below $1. At $1, both the points will be optimal solution

D. From C, we can infer that the optimal solution will change if Process 2's cost becomes greater than Process 1's cost. Hence anything above $4 for Process 2 will change the optimal solution. At $4, both the points will be optimal solution

E. When demand for A falls to 9 and demand for B falls to 3, the process will utilize only one process 1.

It is determined by making the feasible point only is process 1. Only process 2 is not possible as we can't produce C with process 1.

If there is no demand for C, then only process 2 will be used to produce A and B of 10 and 5 units respectively.