Machine-translated by Google. Questions are authored in English.
Top

Discussion

Consider the LPP :

Minimize \(Z = 4x_{1} + 6x_{2} + 18x_{3}\)

Subject to : \(x_{1} + 3x_{2} \ge 3\),
\(x_{2} + 2x_{3} \ge 5\)
and \(x_{1}, x_{2}, x_{3} \ge 0\).

Then, which of the following is not a constraint in the corresponding dual problem ? (Assume that \(w_{1}\) and \(w_{2}\) are the dual variables.)

  • A.\(w_{1} \le 4\)
  • B.\(w_{2} \le 9\)
  • C.\(3w_{1} + w_{2} \ge 6\)
  • D.\(w_{1} \ge 0,\ w_{2} \ge 0\)

Answer: C

The statement that is not a dual constraint is \(3w_{1} + w_{2} \ge 6\).

For a minimisation primal with \(\ge\) constraints and non-negative variables, the dual is to maximise \(3w_{1} + 5w_{2}\) subject to one \(\le\) constraint per primal variable:

  • \(x_{1}\): \(w_{1} \le 4\)
  • \(x_{2}\): \(3w_{1} + w_{2} \le 6\)
  • \(x_{3}\): \(2w_{2} \le 18\), i.e. \(w_{2} \le 9\)
  • \(w_{1}, w_{2} \ge 0\)

The constraint for \(x_{2}\) is \(3w_{1} + w_{2} \le 6\); option C has the inequality reversed, so it is not a constraint of the dual. The others are.

No comment is present. Be the first to comment.
Loading…

Post your comment

Please log in or register to post a comment.