S simpleksno metodo rešite linearni program:
\[\begin{aligned} \max \quad 2 x_1 + 5 x_2 \\[1ex] 2 x_1 - x_2 &\le 3 \\ x_1 + x_2 &\le 3 \\ -2 x_1 + x_2 &\le 3 \\ -2 x_1 + 3 x_2 &\le 6 \\[1ex] x_1, x_2 &\ge 0 \end{aligned}\]Začetni slovar:
\[\begin{aligned} x_3 &= 3 - 2x_1 + x_2 & \text{ne omejuje $x_2$} \\ x_4 &= 3 - x_1 - x_2 & x_2 \le 3\\ x_5 &= 3 + 2x_1 - x_2 & x_2 \le 3 \\ x_6 &= 6 + 2x_1 - 3x_2 & x_2 \le 2 \\ z &= 2x_1 + 5x_2 \end{aligned}\]Drugi slovar:
\[\begin{aligned} x_2 &= 2 + 2/3 x_1 - 1/3 x_6 \\ x_3 &= 3 - 2x_1 + (2 + 2/3 x_1 - 1/3 x_6) \\ &= 5 - 4/3 x_1 - 1/3 x_6 \\ x_4 &= 3 - x_1 - (2 + 2/3 x_1 - 1/3 x_6) \\ &= 1 - 5/3 x_1 + 1/3 x_6 \\ x_5 &= 3 + 2x_1 - (2 + 2/3 x_1 - 1/3 x_6) \\ &= 1 + 4/3 x_1 + 1/3 x_6 \\ z &= 2x_1 + 5(2 + 2/3 x_1 - 1/3 x_6) \\ &= 10 + 16/3 x_1 - 5/3 x_6 \end{aligned}\]Tretji slovar:
\[\begin{aligned} x_1 &= 3/5 - 3/5 x_4 + 1/5 x_6 \\ x_2 &= 12/5 - 2/5 x_4 - 1/5 x_6 \\ x_3 &= 21/5 + 4/5 x_4 - 3/5 x_6 \\ x_5 &= 9/5 - 4/5 x_4 + 3/5 x_6 \\ z &= 66/5 - 16/5 x_4 - 3/5 x_6 \end{aligned}\]Ni več kandidatov za vstop, imamo torej zadnji slovar!
Optimalna rešitev:
V standardni obliki zapišite linearni program:
\[\begin{aligned} \min \quad 2 x + y \\[1ex] 3 x + 2y - z &\ge 10 \\ x + z &\le 30 \\ y + z &= 10 \\ x &\ge 0 \\ y &\le 0 \end{aligned}\]