Big-M Simplex Method Returns Unfeasible Solution Consider the following linear programming problem: 
"Minimize 2x1 + 3x2 subject to the constraints 
2x1 + x2 ≥ 4 
2x1 - x2 ≥ -1
x1,x2 ≥ 0"

Since there are ≥ signs in both constraints, I want to use the big-M method. The original solution in standard form shows that the auxiliary variable has to be added to the first constraint.
So I've got the problem in standard form for the big-M method as:
"Minimize 2x1 + 3x2 + MA 
subject to the constraints 
2x1 + x2 - x3 + A ≥ 4 
2x1 - x2 - x4 ≥ -1
x1,x2 ≥ 0"
I've performed the method on this, with tableau:
| 1 | 2 | 1 | 0 | 0 | -M | 0
|0 | 2 |  1 |-1 | 0 |  1  | 4
|0 | 2 |-1 | 0 |-1 |  0  |-1
(Row 0 at the top is the objective function)
But this gives a tableau with a negative and a zero in the entering column after two iterations, implying that the feasible region is unbounded and I can perform no more iterations despite the objective row not being optimal. 
I can see from drawing the constraints that the optimal solution is at 4, so this cannot be the correct conclusion. What have I done wrong? Or is the method impossible with this problem?
 A: You do not have the right objective function. It is described here, how the objective function has to be created.
$$\begin{array}{|m{cm}|m{1cm}|}
\hline    x_1  &x_2 &s_1 & s_2 & a_1 & RHS  \\ \hline \hline \hline \color{blue}{-2+2M}& -3+M  & M & 0 & 0 & 4M\\ \hline  \color{red}{\boxed{ 2}} & 1 & -1 & 0&1&4 \\  \hline -2 &1&0&1&0&1  \\ \hline  \end{array}$$ 
$s_1$ and $s_2$ are the slack variables. $a_1$ is the artificial variable for the first constraint.
The pivot element is $\color{red}2$, because $\color{blue}{-2+2M}$ is the biggest value  in the first row.
$$\begin{array}{|m{cm}|m{1cm}|}
\hline    x_1  &x_2 &s_1 & s_2 & a_1 & RHS  \\ \hline \hline \hline 0& -2  & -1 & 0 & 1-M & 4\\ \hline  1 & \frac{1}{2} & -\frac{1}{2} & 0&\frac{1}{2}&2 \\  \hline 0 &2&-1&1&1&5  \\ \hline  \end{array}$$ 
All coefficients in the first row are non-positive. $x^*_1=2,x_2^*=0,s_1=0,s_2=5,a_1=0,z^*=4$ 
Remark:
The second constraint has to be transformed, before you can put it in the table. The RHS has to be non-negative.
$2x_1-x_2\geq -1$
The easiest way is to multibly the equation by (-1). The relation sign turns around.
$-2x_1+x_2\leq 1$
Inserting the slack variable
$-2x_1+x_2+s_2= 1$
