0
$\begingroup$

Suppose we have a primal-dual pair in standard form, Add a scalar multiple of one primal constraint to another primal constraint. Does this change the dual solution?

Try

Supose we have primal $ \max cx $ subjeect to $Ax = b $, $x \geq 0$ and the ${\bf dual}$ then is given by $\min yb $ subject to $yA \geq c $ and $y \; free$. Consider rows $i$ and $j$ of $A$ :

$$ a^i x_k = b_i \; \; \; \; and \; \; \; \; \; a^j x_k = b_j $$

$k=1,...,n$. Let's perform what we are asked: Let $\alpha$ be scalar so that

$$ (a^i + \alpha a^j) x_k = b_i+\alpha b_j $$

So the ith dual variable coefficient changes. But, is it a multiple of $\alpha$ or not?

$\endgroup$
2
$\begingroup$

Let $E$ be the corresponding elementary matrix corresponding to adding $\alpha$ times the $j$-th row to the $i$-th row.

Let $w$ be the original dual variable, then we have

$$y=E^{-T}w$$

from the working here.

$E^T$ be the corresponding elementary matrix corresponding to adding $\alpha$ times the $i$-th row to the $j$-th row.

$E^{-T}$ be the corresponding elementary matrix corresponding to adding $-\alpha$ times the $i$-th row to the $j$-th row.

Hence the $j$-th entry of the dual would change, $$y_j = w_j-\alpha w_i.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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