Question

A farmer has 1000 acre of land on which he can grow corn, wheat and soyabeans. Each acre of corn cost Rs. 100 for preparation, requires 7 man days of work and yields a profit of Rs. 30. An acre of wheat cost Rs. 120 for preparation, requires 10 man days of work and yields a profit of Rs. 40. An acre of soyabeans cost Rs. 70 to prepare, requires 8 man days of work and yields a profit of Rs, 200. If the farmer has Rs. 1, 00,000 for preparation and count on 8000man days of work, how many acres should be allocated to each crop to maximise profit. (Use Simplex Method) Discuss the profitability of different alternatives.

Answer 1

Solution:

Let x acre of land be allocated for corn

y acre of land be allocated for wheat

z acre of land be allocated for soyabean

Since each acre of land for corn yields a profit of Rs. 30, for wheat yields Rs. 40 and for soyabean Rs. 20. The mathematical formulation of the L.L.P is

Max Z = 30x + 40y + 20z + 0S₁, + OS₂, + 0S₃

Subject to

100 x +120y + 70z ≤ 100000

7x + 10y + 8z ≤ 8000

x + y + z ≤ 1000

x, y, z ≥ 0

Let us convert the inequalities into equations by introducing slack variables S₁, S₂, and S₃. The objective function and the constraint can be written as

In basic variable column the vectors are for variable S₁, (1, 0, 0), S₂,(1, 0, 1) and S₃(0, 0, 1) the initial feasible solution is given by the variables S₁, S₂ and S₃ both total profit =0

Now Z_j and C_j – Z_j, are calculated by Rule 1, 2 and 3. The key column is determined with start marked column and Simplex Table II is prepared as follows.

Table II does not provide optimum solution we proceed further to prepare simple table III and improve the solution as follows:

Hence for Corn = 250 Acre , for wheat = 625 acre , for soyabean = 125 acre land will give maximum profit.