What to do about equality constraints in the Simplex Tableau method

Question

The question I've got is: Maximise $$2x-y+3z$$ subject to $$2y+z \leq 2$$ $$x+y+z=4$$ $$x-2y+z \geq 3$$ $$x,y,z \geq 0$$

Using the Simplex Tableau method. I know that for $\leq$ constraints you need slack variables, for $\geq$ you need slack and artificial variables, but what are you supposed to do with the equality constraint? I read a lecture online where it said convert it to both a $\leq$ and a $\geq$ constraint, so in this case we would have: $$ x+y+z=4$$ can be written as $x+y+z\leq 4$ and $x+y+z\geq 4$ but then it went onto say that we could just neglect the second constraint which would yield $-x-y-z\leq 4$.

Could anyone give anymore clarification if this is indeed correct and if so, why we get the final result from the two inequalities?

Answer 1

To get an optimal solution, it is necessary to have a basic solution as a starting point. This means, that the number of constraints must be equal to the number of basic variables. Thus every constraint has to have a positive slack variable or a positive artificial variable. If the slack variable is negative or is not needed, then you add an artificial variable.

There are three cases:

$\color{blue}{\leq \texttt{constraint:}}$

If you have a $\leq$-constraint, then you have to add a slack variable for each constraint.

$2y+z \leq 2 \quad \Longrightarrow \quad 2y+z +s_1=2$

$\color{blue}{=\texttt{constraint:}}$

If you have a $=$-constraint, then you do not have to add a slack variable for each constraint. But you have to add an artificial variable for each constraint.

$x+y+z=4 \quad \Longrightarrow \quad x+y+z+a_1=4$

$\color{blue}{\geq \texttt{constraint:}}$

If you have a $\geq$-constraint, then you have to substract a slack variable for each constraint. Additionally you have to add an artificial variable for each constraint.

$x-2y+z \geq 3 \quad \Longrightarrow \quad x-2y+z-s_2+a_2 = 3$

The initial simplex tableau is

$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & y & z & s_1 & s_2 & a_1 & a_2 & RHS \\ \hline -2 & 1 & \color{green}{-3} & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 2 & \color{red} 1 & 1 & 0 & 0 & 0 & \color{orange}{2} \\ \hline 1 & 1 & 1 & 0 & 0 & 1 & 0 & 4 \\ \hline 1 & -2 & 1 & 0 & -1 & 0 & 1 & 3 \\ \hline \end{array} $

The coefficients of the objective function have to be multiplied by (-1), because the objective function has to be maximized.

The initial basic solution is $(x, y, z, s_1, s_2, a_1, a_2)=(0, 0, 0, 2, 0, 4, 3)$

The first pivot element is the red One.