Pivot Row in Simplex Method If I am trying to solve a minimization problem without converting it to a maximization problem how do I decide which variable to pivot ? 
I think it involves looking at the ratio of that variable with the RHS but unsure if I should choose the variable. Is the one with the smallest ratio or the largest. 
Here is the problem that I have:
Minimize: 
$x_1+x_2-4x_3$
Subject to: 
$$
\begin{align}
x_1+x_2+2x_3+x_4=9\\
x_1+x_2-x_3+x_5=2\\
-x_1+x_2+x_3+x_6=4\\
x_1,x_2,x_3,x_4,x_5,x_6 \geq 0\\
\end{align}$$
In this case I know the 4 needs to be pivoted to made negative but not sure if I need to use the $x_4,x_5,x_6$ as the first pivot. 

Minimising $x_1+x_2-4x_3$ is equivalent to maximising it's additive inverse, so we can simply copy the coefficients in the original problem to the last row of the inital simplex tableau:
\begin{array}{r|rrrrrr|rr}
& x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & \text{RHS} & \text{ratio}\\ \hline
x_4 & 1 & 1 & 2 & 1 & 0 & 0 & 9 & 9/2 \\
x_5 & 1 & 1 & -1 & 0 & 1 & 0 & 2 & - \\
x_6 & -1 & 1 & 1^* & 0 & 0 & 1 & 4 & 4\\ \hline
& 1 & 1 & -4 & 0 & 0 & 0 & 0
\end{array}
Choose the most negative number at the $z$-row (-4 in this case.), just like what we do for a standard simplex maximisation problem.
Then pick the least nonnegative number at the "ratio" column.  (You may consult my other post on choosing the leaving variable for further explanation.)
\begin{array}{r|rrrrrr|rr}
& x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & \text{RHS} & \text{ratio}\\ \hline
x_4 & 3^* & -1 & 0 & 1 & 0 & -2 & 1 & 1/3\\
x_5 & 0 & 2 & 0 & 0 & 1 & 1 & 6 & - \\
x_3 & -1 & 1 & 1 & 0 & 0 & 1 & 4 & - \\ \hline
& -3 & 5 & 0 & 0 & 0 & 4 & 16
\end{array}
\begin{array}{r|rrrrrr|r}
& x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & \text{RHS} \\ \hline
x_1 & 1 & -1/3 & 0 & 1/3 & 0 & -2/3 &  1/3 \\
x_5 & 0 &    2 & 0 &   0 & 1 &    1 &    6 \\
x_3 & 0 &  2/3 & 1 & 1/3 & 0 &  1/3 & 13/3 \\ \hline
    & 0 &    4 & 0 &   1 & 0 &    2 &   17
\end{array}
Hence our optimal solution is $(x_1,x_2,x_3,x_4,x_5,x_6) = (1/3, 0, 13/3, 0 ,6, 0)$ with optimal value -17.
