Complicated situation in the simplex method?

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I am having a problem with the simplex method and here is my tableau section

$$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline x1 & x2 & x3 & x4 & x5 & x6 & x7 & s1 & s2 & s3 & s4 & s5 & s6 & s7 & s8 & s9 & s10 & RHS \\ \hline 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 1 & 9 \\ \hline {Other\ rows} \\ \hline 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 9 \\ \hline {Other\ rows} \\ \hline 20 & 0 & -10 & -15 ( most\ negative) & {Unimportant \ values} \\ \hline \end{array}$$ In this case I took 9 and divided by 1 in the pivot column (col x4 since -15 is the most negative value). And now It is quite difficult to decide which row is the pivot row because 9/1 in row 1 and row 3 is now equal.

Any suggestion ?