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Suppose we have the problem $\min c^T x$, subject to $Ax=b, x \geq 0$.

Suppose that this program and its dual are feasible. Let $\lambda$ be the optimal solution of the dual. If the $k$th constraint equation of the primal is multiplied by $\mu \neq 0$, how could we express an optimal solution $w$ to the dual of this new problem?

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The answer is $w_k = \lambda_k/\mu$ and $w_i = \lambda_i$ for $i \neq k$.

Multiplying the $k$th constraint of the primal by $\mu$ doesn't change the primal feasible region and so doesn't change the primal optimal solution. Thus the value $z^*$ of the optimal primal solution doesn't change.

The only change to the dual problem is that the $k$th variable is replaced by $\mu$ times that variable (in the constraints and the objective). Since $z^*$ is the value of the original dual optimal solution, letting $w_i = \lambda_i$ for $i \neq k$ and $w_k = \lambda_k/\mu$ yields a solution that is feasible for the new dual and that also has value $z^*$. Thus, by weak duality, $w$ must be optimal.

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Great answer, the solution was easier that I thought – com Dec 9 '11 at 10:12

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