Discussion
Consider the following Linear Programming Problem :
A company manufactures three products P1, P2, P3. The sale of one unit of P1 results in profit of Rs. 5. For one unit of P2 and P3 the profit is Rs. 6 and Rs. 4 respectively. P1 requires 3 hours of machine work and 2 hours of labour. P2 requires 1 hour of machine work and 3 hours of labour and P3 needs 4 hours of machine work and 3 hours of labour. In a week 60 machine hours and 75 hours of labour is available. How many units of P1, P2, P3 should be produced per week to maximize the profit ?
The primal and the dual of this problem are :
Answer: C
The correct pair is the primal Maximize Z with the constraints written as less than or equal and the dual Minimize Z' with the constraints written as greater than or equal.
With \(x_1, x_2, x_3\) units of P1, P2, P3, profit is \(Z = 5x_1 + 6x_2 + 4x_3\), to be maximised subject to machine hours \(3x_1 + x_2 + 4x_3 \le 60\) and labour hours \(2x_1 + 3x_2 + 3x_3 \le 75\), with all \(x \ge 0\).
The dual of a maximisation problem with \(\le\) constraints is a minimisation problem with \(\ge\) constraints: the right-hand sides 60 and 75 become the objective coefficients, and each column of the primal becomes a dual constraint: \(3y_1 + 2y_2 \ge 5\), \(y_1 + 3y_2 \ge 6\), \(4y_1 + 3y_2 \ge 4\).
The first option wrongly maximises the dual, the second gives the dual constraints as \(\le\), and the last minimises the primal with \(\ge\) constraints, which turns the problem upside down.
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