MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's say we have a maximization linear program that looks like this: maximize $\vec{c}\vec{x}$, subject to $\matrix{A}\vec{x} \leq 0$, $\vec{x} \geq 0$. If we take the dual, we have "minimize $0\vec{y}$, subject to $\vec{y}\matrix{A}\geq\vec{c}, \vec{y}\geq 0$". I'm particularly bothered by the "minimize $0$" part of the dual program - but does the duality theorem still hold - that is: is it true that if there is a $\vec{y}$ that is feasible for the dual program, then for all $\vec{x}$ that is feasible for the primal program, $\vec{c}\vec{x} \leq 0$?


share|cite|improve this question
up vote 1 down vote accepted

Yes, the duality theorem holds. "Minimize 0" makes your life much easier, because you know exactly what the optimal value of the dual program is...

share|cite|improve this answer

Igor is correct, and more can be said about this primal-dual pair of linear programs: because the primal maximization problem is homogeneous (i.e. zero constant terms in the constraints), the common optimal objecive value for both problems is either zero or $+\infty$:

  1. If there exists a feasible solution $\vec{x}$ with a positive objective value, i.e. with $A \vec{x} \le 0$ and $\vec{c}\vec{x} > 0$, then this solution can be scaled by any positive constant $\lambda > 0$ (because $A (\lambda \vec{x}) \le 0$), which shows that the problem is unbounded ($\vec{c}(\lambda \vec{x})$ tends to $+\infty$ as $\lambda \to +\infty$).
  2. If no feasible solution with negative objective value exists, then $\vec{x}=0$ is an optimal solution, and the optimal objective value is zero.

Using duality theory, one can check that the first case corresponds to an infeasible dual problem (no $\vec{y}\ge 0$ such that $\vec{y} A \ge \vec{c}$), while the second situation happens as soon as the dual problem admits a feasible solution.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.