Discussion
Consider the LP problem :
Max \(Z = x_{1} - x_{2} + 3x_{3}\)
Subject to the constraints :
\(x_{1} + x_{2} + x_{3} \le 10\)
\(2x_{1} - x_{2} - x_{3} \le 2\)
\(2x_{1} - 2x_{2} - 3x_{3} \le 6\) and \(x_{1}, x_{2}, x_{3} \ge 0\).
Then the dual of the above LP problem is :
Min \(W : py_{1} + qy_{2} + ry_{3}\)
Subject to the constraints :
\(ay_{1} + by_{2} + cy_{3} \ge 1\)
\(y_{1} - y_{2} - 2y_{3} \ge -1\)
\(y_{1} - y_{2} - 3y_{3} \ge 3\) and \(y_{1}, y_{2}, y_{3} \ge 0\)
where the values of \(p, q, r, a, b, c\) are :
Answer: B
The values are \(p = 10, q = 2, r = 6, a = 1, b = 2, c = 2\).
For a maximisation problem with \(\le\) constraints and non-negative variables, the dual minimises the right-hand sides times the dual variables, and each primal column gives a \(\ge\) constraint:
- Objective: \(10y_{1} + 2y_{2} + 6y_{3}\), so \(p = 10, q = 2, r = 6\).
- Column of \(x_{1}\): coefficients 1, 2, 2, so \(y_{1} + 2y_{2} + 2y_{3} \ge 1\), giving \(a = 1, b = 2, c = 2\).
- Columns of \(x_{2}\) and \(x_{3}\): \(y_{1} - y_{2} - 2y_{3} \ge -1\) and \(y_{1} - y_{2} - 3y_{3} \ge 3\), as printed.
The other options change signs of the right-hand sides or of the first column, which the primal does not have.
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