2
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maximize 10𝑥1 + 12𝑥2 +12𝑥3

subject to

𝑥1 + 2𝑥2 + 2𝑥3 + 𝑥4= 20

2𝑥1 + 𝑥2 + 2𝑥3+𝑥5= 20

2𝑥1 + 2𝑥2 + 𝑥3 +𝑥6= 20

𝑥1, … , 𝑥6 ≥ 0

This is my first step for simplex tableau

x1 x2 x3 x4 x5 x6

0 4 1 2 2 1 0 0 20 20

x5 0 5 2 1 2 0 1 0 20 10

0 6 2 2 1 0 0 1 20 10

10 12 12 0 0 0 0

Now which variable should be entering variable?

What confuses me is since we have two 12 on cj-zj column, I suppose we should choose x2 or x3 as entering variable, but simplex calculator proceeds with x1 as entering variable.

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  • 1
    $\begingroup$ I would choose $x_2$ or $x_3$ as entering variable as well. But there exist many other rules for choosing the entering variable. B.t.w the column with the objective function has to be multiplied by (-1). $\endgroup$ – callculus May 1 '16 at 9:38
  • $\begingroup$ @callculus You mean the last row? I think some books use $c_j - z_j$ instead of $z_j - c_j$ $\endgroup$ – BCLC May 6 '16 at 15:47
  • $\begingroup$ It looks like $x_2$ or $x_3$. You can try using this $\endgroup$ – BCLC May 6 '16 at 15:48

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