Suppose we have primal and its dual in standard form, that is

\begin{align*} (P) \max z = cx \\ st \; \; Ax = b \\ \; \; \; x \geq 0 \\ \end{align*}

\begin{align*} (D) \min z = by \\ st \; \; yA \geq c \\ \; \; \; y \; \; \; free \\ \end{align*}

Where $A$ is an $m $ by $n$ matrix an $x$ is an n vector and $y$ is an m vector.

Suppose we multiply one of the constraints of the primal by some number $\alpha > 0$. Does this affect the solution of the dual?


Since a constraint is of the form $a_{ij} \cdot x $, take one of the $i's$, say we multiply

$$ a_{i1}x_1 + a_{i2} x_2 + ... + a_{in} x_n $$

by $\alpha $

Once we set up our tableau, once we divide this row by $\alpha$, then in the LFH, we would have

$$ \frac{ b_i}{\alpha} $$

the ith component of the vector $b$. Doesnt it change the solution in the primal tableau? Since solutions are the same for primal and dual???


Multiplication by a non-zero scalar is equivalent to multiplication of an elementary matrix, $E$.

\begin{align*} (P') \max z = c^Tx \\ st \; \; (EA)x = (Eb) \\ \; \; \; x \geq 0 \\ \end{align*}

The dual is \begin{align*} (D') \min z = (Eb)^Ty \\ st \; \; y^T(EA) \geq c \\ \; \; \; y \; \; \; free \\ \end{align*}

Suppose $w$ is the original dual solution, then $y=E^{-T}w.$

For the operation of multiplication by a scalar, we have $E^T=E$.

Hence $y=E^{-1}w$. That is if we multiply $\alpha$ to the $i$-th constraint, now for the dual solution, we would divide $w_i$ by $\alpha$ and we can keep the rest to be the same.

  • $\begingroup$ Can you help me with this related problem: math.stackexchange.com/questions/2989440/… $\endgroup$ – Mikey Spivak Nov 8 '18 at 15:11
  • $\begingroup$ at first glance, it seems that doing that operation might cause the primal problem to become infeasible. $\endgroup$ – Siong Thye Goh Nov 8 '18 at 15:24
  • $\begingroup$ But isnt it doing the same operations as the previous case but this time to the dual? $\endgroup$ – Mikey Spivak Nov 8 '18 at 18:06
  • $\begingroup$ the duality is in inequality form, multiplying an inequality and adding it to another one might change the feasible set. Also, if you multiply by a negative number, the inequality get flipped. $\endgroup$ – Siong Thye Goh Nov 8 '18 at 18:09
  • $\begingroup$ You are right :/ $\endgroup$ – Mikey Spivak Nov 8 '18 at 18:23

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