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Solve using simplex $$\text{ Max } z=4x_1+3x_2$$ $$\text{s.t } 2x_1+3x_2\leq 6 $$ $$-3x_1+2x_2\leq 3$$ $$2x_1 \leq 5$$ $$2x_1+x_2\leq 4$$ $$ x_1,x_2 \geq 0$$

I have solve it and got:

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$So x_1 = 1.5, x_2 = 1 \text{ and } z=9$

One can see that some of the constrains are redundant as we need at most two equation for two unknowns. Can we have two "unused" slack variable in the basis? we just ignore them?

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  • $\begingroup$ I think your calculation will be more easy if you use Dual Simplex Method. $\endgroup$ – nmasanta Jun 2 '19 at 17:01
  • $\begingroup$ @nmasanta I agree, but I was asked to solve using Simplex only $\endgroup$ – newhere Jun 2 '19 at 17:05
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    $\begingroup$ The basic difference between the regular Simplex Method and the Dual Simplex Method is that whereas the regular Simplex Method starts with basic feasible solution, which is not optimal and it works towards optimality, the dual Simplex Method starts with an infeasible solution which is optimal and works towards feasibility. The following steps are involved in this method. $\endgroup$ – nmasanta Jun 2 '19 at 17:12
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Yes, if the solution in the original variables $x_1$ and $x_2$ is what you're interested in, you can ignore the slack variables in the final tableau. They are necessarily part of the basis since the slack of the second and third restriction is non-zero.

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  • $\begingroup$ This would not be the case if they were artificial variables correct? $\endgroup$ – newhere Jun 4 '19 at 8:51

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