How to proceed with the simplex tableau? I'm relearning the simplex method and doing this minimization example with variables $y_1,y_2$, but got confused at the following simplex tableau step:
$$\begin{array}{|c|c|c|c|c|c|c|} 
\hline
BV & z & y_1 & y_2 & s_1 & s_2 & rhs \\ \hline
 & 1 & -18 & 0 & 0 & 6 & -6 \\ \hline
y_2 & 0 & 2 & 1 & 0 & -1 & 1 \\ \hline
s_1 & 0 & 3 & 0 & 1 & -2 & 1 \\ \hline
\end{array}$$
I assume that the next variable entering the basis should be $s_2$, since its coefficient in the first row is positive. But multiplying the second or third row by $-1$ or $-1/2$, respectively, would result in negative values in the $rhs$ column. What should be my next step and why?
Also I've been reading up on the online notes, but got quite confused because of the different formats of the tableaux and the mixing up of maximization and minimization problems. For practical purposes, should I always transform a minimization problem to a maximization one? What are some other tips or mental checks for doing this?
 A: There are two options in the case of minimizing problem for choosing a pivot column:

*

*You multiply the objective function $z$ by $(-1) $ and maximize  $-z$, which means that the coefiffients of the objective function have the opposite sign in the table, $+z$.   Then you choose the pivot column by the following rule: Choose  the colomn as pivot colomn with the most negative value at the row of the objective function. You stop pivoting if the row of the objective function constains  non-negative values only.


*You minimize  $z$, which means that the coefiffients of the objective function have the opposite sign in the table, $-z$.   Then you choose the pivot column by the following rule: Choose  the colomn as pivot colomn with the most positive value at the row of the objective function. You stop pivoting if the row of the objective function constains only non-positive values. Then your optimal value for the objective function is the negative value of the RHS.
You have the second case here. So your next pivot column is $s_2$. But you cannot choose it since the ratios are negative. So there exists no solution for the LP if you have mode no flaws in the the steps before.
