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Suppose following tableau came after one iteration in first phase of a two phase method problem, here $s_1$ is a surplus variable and $s_2$ is a slack variable $w$ is a artificial variable.

I tried to minimize $W=w$ $\implies$ maximize $-W=-w$

$$ \begin{array}{c|c} & x_1 & x_2 & x_3 & s_1 & s_2 & w & \text{solution} \\ \hline x_2 & 1/2 & 1 & -1/2 & -1/2 & 0 & 1/2 & 5/2 \\ s_2 & -5/2 & 0 & 1/2 & -1/2 & 1 & 1/2 & 13/2 \\ \hline & 0 & 0 & 0 & 1 & 0 & -1 & 0 \\ \end{array} $$

Now we can clearly see that artificial variable is not in basic variable set.but the solution is not optimal. in fact solution is unbounded.

My problem is can we proceed to second phase after this step?

Even though the $w$ is not in basic solution set $(w=0)$ since the solution is not optimal and unbounded should we stop here?

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  • $\begingroup$ Who is $W$? Presumably $W=c_1x_1+c_2x_2+c_3x_3$ for some $c_1,c_2,c_2\in\mathbb R$? $\endgroup$ – Math1000 Jun 24 '16 at 23:29
  • $\begingroup$ In phase one,we have to minimize the sum of artificial variables.it is $W$ $\endgroup$ – Gune Jun 25 '16 at 0:45
  • $\begingroup$ What is the initial tableau ? $\endgroup$ – callculus Jun 25 '16 at 15:15
  • $\begingroup$ @Gune what's the initial problem here? $\endgroup$ – baxx Feb 22 at 0:25

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