# solving minimum linear programming with simplex method

I want to solve a minimizing linear programming problem with simplex method.

$$min \quad 2x_1+3x_2+x_3 \\ \text{subject to: }x_1+4x_2 \le 3 \\ x_2+4x_3 \le 2 \\ x_1+2x_2+3x_3=5 \\ x_2+x_3=1$$

In order to solve this problem with simplex method it needs to be converted to standard form: $$max \; -2x_1-3x_2-x_3 \\ x_1+4x_2+u=3 \\ x_2+4x_3+v=2 \\ x_1+2x_2+3x_3=5 \\ x_2+x_3=1 \\ x_1, x_2, x_3, u, v \ge 0$$

For simplex tableau, the objective function should be an equation: $$P=-2x_1-3x_2-x_3 \Rightarrow P+2x_1+3x_2+x_3=0$$

The set of basic variables is composed of slack variables $u,v$. Therefore, non-basic variables $x_1, x_2, x_3$ are initialized to $0$. This violates the equality constraints of the original problem and at the same time the objective equation states that the first basic solution is the optimal solution since all coefficients of the objective equation are non-negative. This means the optimal solution is infeasible!

Does it mean simplex method cannot be used to solve this problem or I am missing some important aspects?

There is such aspect. Consider system of equations $$\begin{cases} (1):x_1+4x_2 \le 3 \\ (2):x_2+4x_3 \le 2 \\ (3):x_1+2x_2+3x_3=5 \\ (4):x_2+x_3=1 \end{cases}$$ Make some substitutions: \begin{align} (4)\to (2)&: x_3\le\frac{1}{3}\\ (4)\to (3)&: x_1=3-x_3 \quad(5)\\ (4)&: x_2=1-x_3 \quad(6)\\ (5),(6)\to(1)&: x_3\ge\frac{4}{5} \end{align}
Notice that $x_3\le\frac{1}{3}$ and $x_3\ge\frac{4}{5}$ contradict each other, hence system is incompatible.