Simplex Method Calculator

Maximize $$$Z = 3 x_{1} + 4 x_{2}$$$, subject to $$$\begin{cases} x_{1} + 2 x_{2} \leq 8 \\ x_{1} + x_{2} \leq 6 \\ x_{1} \geq 0 \\ x_{2} \geq 0 \end{cases}$$$.

The problem in the canonical form can be written as follows:

Add variables (slack or surplus) to turn all the inequalities into equalities:

Write down the simplex tableau:

The entering variable is $$$x_{2}$$$, because it has the most negative coefficient $$$-4$$$ in the Z-row.

The leaving variable is $$$S_{1}$$$, because it has the smallest ratio.

Divide row $$$1$$$ by $$$2$$$: $$$R_{1} = \frac{R_{1}}{2}$$$.

Add row $$$2$$$ multiplied by $$$4$$$ to row $$$1$$$: $$$R_{1} = R_{1} + 4 R_{2}$$$.

Subtract row $$$2$$$ from row $$$3$$$: $$$R_{3} = R_{3} - R_{2}$$$.

The entering variable is $$$x_{1}$$$, because it has the most negative coefficient $$$-1$$$ in the Z-row.

The leaving variable is $$$S_{2}$$$, because it has the smallest ratio.

Multiply row $$$2$$$ by $$$2$$$: $$$R_{2} = 2 R_{2}$$$.

Add row $$$3$$$ to row $$$1$$$: $$$R_{1} = R_{1} + R_{3}$$$.

Subtract row $$$3$$$ multiplied by $$$\frac{1}{2}$$$ from row $$$2$$$: $$$R_{2} = R_{2} - \frac{R_{3}}{2}$$$.

None of the Z-row coefficients are negative.

The optimum is reached.

The following solution is obtained: $$$\left(x_{1}, x_{2}, S_{1}, S_{2}\right) = \left(4, 2, 0, 0\right)$$$.

$$$Z = 20$$$A is achieved at $$$\left(x_{1}, x_{2}\right) = \left(4, 2\right)$$$A.