The manager of an oil refinery must decide on the optimum mix of two possible blending processes of which the inputs and outputs per production run are as follows: -

The maximum amount available of crudes A and B is 200 units and 150 units respectively. Market requirements show that atleast 100 units of gasoline X and 80 units of gasoline Y must be produced. The profits per production run from process 1 and process 2 are Rs. 300 and Rs. 400 respectively. Solve the LP problem by graphical method. -

We have to find out the number of production runs each of process 1 and process 2, such that the overall profit due to running of these processes is maximized.

Let there be $R_1$ runs of process 1, and $R_2$ runs of process 2.

Each run of process 1 yields a profit of Rs. 300, while each run of process 2 yields a profit of Rs. 400.

To maximize the profit, the maximization function is:

$Z = 300 R_1 + 400 R_2$

Constraints:

• Maximum amounts of crude available:

  Of A, max. 200 units $→ 5 R_1 + 4 R_2 ≤ 200$

  Of B, max. 150 units $→ 3 R_1 + 5 R_2 ≤ 150$

• Minimum amount of gasoline to be produced:

  Of X, min. 100 units $→ 5 R_1 + 4 R_2 ≥ 100$

  Of Y, min. 80 units $→ 8 R_1 + 4 R_2 ≥ 80$

Plotting these constraints on a graph:

We check for optimality at three points, namely A (0,30), B (30.77, 11.54), & C (40,0). Point B represents the intersection of the two lines $5 R_1 + 4 R_2 ≤ 200 \& \ 3 R_1 + 5 R_2 ≤ 150$.

At A (0, 30): Z = 300 (0) + 400 (30) = Rs. 12000

At C (40, 0): Z = 300 (40) + 400 (0) = Rs. 12000

At B (30.77, 11.54): Z = 300 (30.77) + 400 (11.54) = Rs. 13847

Quite clearly, point B represents the optimal solution.

However we cannot have a fraction of a run, i.e. $R_1\ \& \ R_2$ cannot be a fraction, since it has to run completely. So we consider 2 nearby points, B1 (30, 12), which lies on $3 R_1 + 5 R_2 ≤ 150, \& B_2 (32, 10)$ which lies on $5 R_1 + 4 R_2 ≤ 200.$

At B1 (30, 12): Z = 300 (30) + 400 (12) = Rs. 13800

At B2 (32, 10): Z = 300 (32) + 400 (10) = Rs. 13600

So B1 (30, 12) represents the optimal solution.

The manager should undertake 30 runs from process 1, and 12 runs from process 2. The max. profit is Rs. 13800.