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Possible Duplicate:
Linear programming boundedness

Consider the following LP:

$\max$ $\sum_{i=1}^N b_i \pi_i$

s.t. $\;\;$ $\pi_i-\pi_j\leq 1 $ $\quad$ for each $(i,j) \in \tilde{A}$

$\sum_{i=1}^N a_s^i \pi_i \leq 1$ for each $s \in S$

This LP is always feasible, since zero is a feasible solution. Now, in order to find the optimum value, we relax the LP by multiplying second set of constraints by nonnegative Lagrangian multipliers $\lambda_s, s\in S$ and add them to the objective function. Thereafter, we get the following relaxed dual LP:

$L(\lambda)$: $\max$ $\sum_{i=1}^N (b_i+\sum_{s\in S}\lambda_s a_s^i)\pi_i - \sum_{s\in S}\lambda_s$

s.t. $\pi_i-\pi_j\leq 1$ $\quad$ for each $(i,j) \in \tilde{A}$

Let us consider the Lagrangian dual LP. This LP can be formulated as follows:

LD: $\min$ $L(\lambda)$

s.t. $\lambda \geq 0$

Now, the question is as follows. Both LP and LD problems are feasible (e.g. zero vecor). Moreover, if the LD is bounded so is the LP. What cab be said if the LD is unbounded. Consider that in my case LP is always feasible.

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  • $\begingroup$ With the changes I can make a bit better argument. You can ask "What if the LD is unbounded", but I don't see how the LD can be unbounded. Are you asking theoretically if a similar problem exists with those characteristics? or about this specific problem? $\endgroup$ Commented Oct 1, 2012 at 15:48
  • $\begingroup$ Just this specific problem. Now, what if the LD is unbounded? $\endgroup$
    – Star
    Commented Oct 1, 2012 at 16:09
  • $\begingroup$ I have looked a little more into it and found that the Lagrangian dual you refer to is infact different from the standard LP dual I am used to. I can't make the arguments I wanted to. $\endgroup$ Commented Oct 1, 2012 at 16:27
  • $\begingroup$ I don't see how this question is different than an earlier one you posted; there, you ask "Is it true that if the primal problem is bounded, then the dual problem is bounded?" and the answer is no. Here, you're almost asking the contrapositive: "If the dual problem is unbounded, then the primal problem is unbounded." Well, that's also false, because its contrapositive is false. Nothing can be said about the boundedness of the primal problem. $\endgroup$ Commented Oct 1, 2012 at 20:56

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