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I'm working through the examples in my professor's class notes, but I can't figure out how to solve this problem:

Example: In the following standard tableau, mark by * the choices for the pivot entry consistent with the simplex method: (Note last row is the objective function)

$$ \begin{array}{|ccccc|c|} \hline x_1&x_2&x_3&x_4&x_5&-1&\\ \hline 0&1&-2&3&4&2&=-x_6\\ 1&1&-2&3&4&0&=-x_7\\ -3&-1& 0&-3&-4&-2&=-x_8\\ \hline 1&1& 2&1&1&-1&\rightarrow max\\ \hline \end{array} $$

Solution.

$$ \begin{array}{|ccccc|c|} \hline x_1&x_2&x_3&x_4&x_5&-1&\\ \hline 0&1&-2&3*&4&2&=-x_6\\ 1*&1*&-2&3&4*&0&=-x_7\\ -3&-1& 0&-3*&-4&-2&=-x_8\\ \hline 1&-1& -2&1&1&-1&\rightarrow max\\ \hline \end{array} $$

I just don't know how these pivots are being identified. I might just be misunderstanding the question, because I thought that each step of the simplex method only had one possible pivot point.

Thanks for the help!

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  • $\begingroup$ You're sure that the solution tableau is right? It's not even feasible nor optimal. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 18 '17 at 18:01
  • $\begingroup$ Related: math.stackexchange.com/q/1605269/290189 \begin{array}{|c|c|c|} \hline x_j & -1 & \\ \hline a_{1j} & b_{1j} & =-x_6 \\ a_{2j} & b_{2j} & =-x_7 \\ a_{3j} & b_{3j} & =-x_8 \\ \hline \star & \star & \to \max \\ \hline \end{array} For each $j$, choose $i \in \operatorname{argmin} \left\{\dfrac{b_{rj}}{a_{rj}}:a_{rj}>0,r \in \{1,2,3\}\right\} $. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 9 '18 at 18:53

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