I saw this simple problem and realized I dont know how the simplex method solves this one in particular -
MIN c1x2 + c2x2 + c3x3
ST x1 + x2 + x3 = 1
xi >= 0, ci >= 0
(Where c3>c1>c2)
Typically with the simplex method we are supposed to make the objective function negative. Since c2 is the smallest, the answer is surely x2=1, x1=0, x3=0 but I dont know how to get there.
In order to solve this model, we would need to add an artificial variable into the first constraint. For simplicity this is done via the Big-M approach as such:
$$\min z - c_1x_1-c_2x_2-c_3x_3-Ma_1=0$$ Subject to: $$x_1+x_2+x_3+a_1=1$$ $$x_1,x_2,x_3\ge0$$ $$c_2<c_1<c_3$$ Where $M$ is an arbitrarily "largest" number in $\Bbb R$. This would look like such in the tableau:
\begin{array} {|c|c|}\hline BV & z & x_1 & x_2 & x_3 & a_1 & RHS & RT \\ \hline z & 1 & -c_1 & -c_2 & -c_3 & -M & 0 & - \\ \hline ? & 0 & 1 & 1 & 1 & 1 & 1 & - \\ \hline \end{array}
Then we would have to solve for our basic variable $a_1$:
\begin{array} {|c|c|}\hline BV & z & x_1 & x_2 & x_3 & a_1 & RHS & RT \\ \hline z & 1 & M-c_1 & M-c_2 & M-c_3 & 0 & M & - \\ \hline a_1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}
Since $c_2$ is the smallest number to take away from $M$, the $x_2$ column is the largest $C^\pi_j$ greater than or equal to zero in the objective function row. Thus, the $x_2$ column is where we would pivot to get:
\begin{array} {|c|c|}\hline BV & z & x_1 & x_2 & x_3 & a_1 & RHS & RT \\ \hline z & 1 & -c_1+c_2 & 0 & -c_3+c_2 & -M+c_2& c_2 & - \\ \hline x_2 & 0 & 1 & 1 & 1 & 1 & 1 & - \\ \hline \end{array}
Since $c_2$ is smaller than $c_1$ and $c_3$, there are no more pivots available for Simplex to take. Thus, we are done.