In maximization simplex, the pivot is the smallest element in the column divided by the rightmost corresponding number. I am stumbling with the Example 3 here with solution that choose the pivot with the largest element. Please, explain how to choose the pivot in different cases and why the latter document can choose the largest element, intstead of the smallest?
In the paper, I am puzzled how it can move from this phase: \begin{equation*} \begin{array}{rrrrr|r} x & y & u & v & P & \\ \hline 1 & 2 & 1 & 0 & 0 & 6 \\ 3 & 2 & 0 & 1 & 0 & 12 \\ \hline -2 & 1 & 0 & 0 & 1 & 0 \\ \hline \end{array} \end{equation*}
to this phase: \begin{equation*} \begin{array}{rrrrr|r} x & y & u & v & P & \\ \hline 0 & \frac{4}{3} & 1 & -\frac{1}{3} & 0 & 2 \\ 1 & \frac{2}{3} & 0 & \frac{1}{3} & 0 & 4 \\ \hline 0 & \frac{5}{3} & 0 & \frac{1}{3} & 1 & 8 \\ \hline \end{array} \end{equation*}
I would have choosen 1 as pivot because $\frac{1}{6}<\frac{3}{12}=\frac{1}{4}$. When I plug my solutions back to the original restrictions, I notice that they do not satisfy the results while the paper solution does. It may have something to do with the fact that this problem is a minimization problem, changed to maximization problem with $P=-C$.