Problems Linear Programming Models
 Problems for Linear Programming Models - Chemical Mix

Decision Variables:

Define the variables A, B, and C to be the pounds of raw materials used, M is the pounds of mix, and X and Y are the pounds of product.

Constraints:

Conservation of Raw Materials: A + B+ C = M

Raw Material A restriction: A 0.45M

Raw Material C restriction: C 0.3M

Product X Definition: X = 0.4M

Product Y Definition: Y = 0.2M

Simple Upper Bounds:

X 1000, Y 2000, B 2500, C 1500

Objective: Maximize Profit

Maximize 12X + 18Y - 1.5M - 6A - 3B

Nonegativity:

A 0, B 0, C 0, X 0, Y 0, M 0.

a. The optimum Plan has:

 A B C X Y M Z 1125 625 750 1000 500 2500 8625

b. Product X is at its upper bound and limiting profit. The raw material percentage requirements for A and C are also limiting the profit.

c. From the dual variables, the simple bound with the largest magnitude is the x limitation with value 8.625 where all others are zero. From this we see that we would like to increase the limit on x. We predict the profit will go up by 8.625 for each unit increase. This return is available as if the allowed amount of X goes up to 2000.

d. The sensitivity analysis says that the range on this coefficient is from -6 to 0. Since the coefficients are the negative of cost, this means that the cost can increase to 6 before changing the solution.

e. From the sensitivity analysis we see that the upper range on the x limitation is 2000.

f. Raw Material C restriction:

The mix becomes 0. This is the only solution for which the restrictions are feasible.

Note that it is wrong to say the solution becomes infeasible. There is no constraint that requires any product to be made and if no product is made the content restrictions are not effective.

g. The mix is limited by the available amounts of B and C.

 A B C X Y M Z 21000 2500 1500 10000 5000 25000 39000

Operations Research Models and Methods
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by Paul A. Jensen