A manager of a small fabrication plant must decide on a production schedule of t

Question

A manager of a small fabrication plant must decide on a production schedule of two new products for the automobile industry. The unit profit is $20 for product 1 and $70 for product 2.

The manufacture of these products depends largely on the availability of certain subassemblies the plant receives daily from a local distributor. It takes two of these subassemblies for each unit of product 1 and also two for each unit of product 2. Twenty such subassemblies are delivered daily.

Further, it takes four hours to make a unit of product 1 and six hours to make a unit of product 2. The plant has assigned only six workers, each one with an 8-hour shift, for these new products. Due to limited demand, the manager does not want more than six units of product 2 produced daily.

1) Formulate this problem as a math program.

2) Solve graphically for the optimal solution. Describe the optimal solution.

3) What is the profit generated by using the above optimal solution?

After you finish with this problem, you received a memo from the accounting department. According to the accountant, the revised unit profit is $30 for product 1 and $40 for product 2 based on the new marketing information.

4) What will be your modified optimal solution based on the information from the accountant?

5) What is the modified profit?

Explanation / Answer

Let the decision variables are P1 for product1 and P2 for product2.

Objective is to have Maximum Profit Z = 20P1 + 70P2

Constraints are : 2P1 + 2P2 <= 20 (sub-assembleies), 4P1 + 6P2 <= 48 (labor hours), P2 <= 6 (demand of product2) Lastly P1, P2 >= 0 (being quantities)

Formulation and solution is as follows:

Solution is as follows:

Three units of product 1 and six units of product 2 giving maximum profit 480 (20*3+70*6)

Solution to revised problem is as follows:

Product 1 will have six units whereas product 2 will have four units with maximum profit 340 (30*6+40*4)

P1 P2 RHS Equation form Maximize 20 70 Max 20P1 + 70P2 Subassemblies 2 2 <= 20 2P1 + 2P2 <= 20 labor hours 4 6 <= 48 4P1 + 6P2 <= 48 Demand 0 1 <= 6 P2 <= 6

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