Question

Consider the LP formulation given below for a typical two product mix problem in which the resources used are Materials, Labor Hours, and storage. Each ton of product-1 requires 2 hours of labor, 4 tons of materials, and 25 cubic meters of storage. A ton of product-2 requires 1 hours of labor, 3 tons of materials, and 30 cubic meters of storage. In every week the firm has 100 Labor Hours, 240 Tons of Materials, and a storage capacity of 2000 cubic meter. The firm makes a profit of $6 thousands per ton on product-1 and $4 thousands per ton on product-2. If the firm wants to maximize the profit, the following is the graph and the LP formulation:

                  Maximize                6X1 + 4X2

                  Subject to

                                                      2X1 + 1X2                  < 100 Labor Hours                                                     Resource-1

                                                      4X1 + 3X2                  < 240 Tons of Materials                                           Resource-2

                                                      25X1 + 25X2             < 2000 Cubic meters of Storage                           Resource-3

                                                      X1, X2                         > 0

You might use any method including the graph below to solve and analyze the above problem and answer the ten questions (a – m) that follow:

L: 4X1 + 3 -20 W昭阳心础叫m弟亞州州加強型四叫四旧H840 : 51 + 30 = 2000 X1 + V: 0 24 、6 8 ' ' 46 1 2 4 5 8 0 2 4 5 蛇 4 5 A1 Payoff: 6X Optimal Decisions(x1,x2): ( 0, 0) : X1+ 位= 10 : X1 + = 240 : X1 + 302 = 200

How many tons of product-1 would you recommend to produce weekly?

Answer 1

One equation given is not right. The correct equations are

Objective function = Max [ 6X1+4X2]

Constraints

      2X1 + 1X2 <= 100 Labor Hours                                                    

4X1 + 3X2 < =240 Tons of Materials                                          

      25X1 + 30X2 <=2000 Cubic meters of Storage

X1, X2 =>0

Points of optimality are (0, 66.66) ( 26.67,44.44), (30,40) and (50,0)

Value of objective function on thse points are

At (0,66) = 264

at (26,44) =332

at ( 30,40) =340

at (50,0) =300

The optimal value os 340 obtaoned at X1=30, X2 =40

Hence X1 should be produced 30 tonnes