單純形法計算器
使用單純形法求解最佳化問題
該計算器將使用單純形法求解給定的最佳化問題。必要時,它會加入鬆弛變數、盈餘變數與人工變數。若有人工變數,將採用大M法或兩階段法來確定初始解。提供步驟。
您的輸入
最大化 $$$Z = 3 x_{1} + 4 x_{2}$$$,受限於 $$$\begin{cases} x_{1} + 2 x_{2} \leq 8 \\ x_{1} + x_{2} \leq 6 \\ x_{2} \geq 0 \\ x_{1} \geq 0 \end{cases}$$$。
解答
問題的標準形可以表示如下:
$$Z = 3 x_{1} + 4 x_{2} \to max$$$$\begin{cases} x_{1} + 2 x_{2} \leq 8 \\ x_{1} + x_{2} \leq 6 \\ x_{1}, x_{2} \geq 0 \end{cases}$$加入(鬆弛或剩餘)變數,使所有不等式轉為等式:
$$Z = 3 x_{1} + 4 x_{2} \to max$$$$\begin{cases} x_{1} + 2 x_{2} + S_{1} = 8 \\ x_{1} + x_{2} + S_{2} = 6 \\ x_{1}, x_{2}, S_{1}, S_{2} \geq 0 \end{cases}$$寫出單純形表:
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-3$$$ | $$$-4$$$ | $$$0$$$ | $$$0$$$ | $$$0$$$ |
| $$$S_{1}$$$ | $$$1$$$ | $$$2$$$ | $$$1$$$ | $$$0$$$ | $$$8$$$ |
| $$$S_{2}$$$ | $$$1$$$ | $$$1$$$ | $$$0$$$ | $$$1$$$ | $$$6$$$ |
進入變數為 $$$x_{2}$$$,因為它在 Z 行中具有最負的係數 $$$-4$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 | Ratio |
| $$$Z$$$ | $$$-3$$$ | $$$-4$$$ | $$$0$$$ | $$$0$$$ | $$$0$$$ | |
| $$$S_{1}$$$ | $$$1$$$ | $$$2$$$ | $$$1$$$ | $$$0$$$ | $$$8$$$ | $$$\frac{8}{2} = 4$$$ |
| $$$S_{2}$$$ | $$$1$$$ | $$$1$$$ | $$$0$$$ | $$$1$$$ | $$$6$$$ | $$$\frac{6}{1} = 6$$$ |
離基變數為 $$$S_{1}$$$,因為它具有最小比值。
將第 $$$1$$$ 行除以 $$$2$$$:$$$R_{1} = \frac{R_{1}}{2}$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-3$$$ | $$$-4$$$ | $$$0$$$ | $$$0$$$ | $$$0$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$S_{2}$$$ | $$$1$$$ | $$$1$$$ | $$$0$$$ | $$$1$$$ | $$$6$$$ |
將第$$$2$$$行乘以$$$4$$$後加到第$$$1$$$行:$$$R_{1} = R_{1} + 4 R_{2}$$$
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-1$$$ | $$$0$$$ | $$$2$$$ | $$$0$$$ | $$$16$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$S_{2}$$$ | $$$1$$$ | $$$1$$$ | $$$0$$$ | $$$1$$$ | $$$6$$$ |
從第$$$3$$$行減去第$$$2$$$行: $$$R_{3} = R_{3} - R_{2}$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-1$$$ | $$$0$$$ | $$$2$$$ | $$$0$$$ | $$$16$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$S_{2}$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$- \frac{1}{2}$$$ | $$$1$$$ | $$$2$$$ |
進入變數為 $$$x_{1}$$$,因為它在 Z 行中具有最負的係數 $$$-1$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 | Ratio |
| $$$Z$$$ | $$$-1$$$ | $$$0$$$ | $$$2$$$ | $$$0$$$ | $$$16$$$ | |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ | $$$\frac{4}{\frac{1}{2}} = 8$$$ |
| $$$S_{2}$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$- \frac{1}{2}$$$ | $$$1$$$ | $$$2$$$ | $$$\frac{2}{\frac{1}{2}} = 4$$$ |
離基變數為 $$$S_{2}$$$,因為它具有最小比值。
將第$$$2$$$行乘以$$$2$$$: $$$R_{2} = 2 R_{2}$$$.
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-1$$$ | $$$0$$$ | $$$2$$$ | $$$0$$$ | $$$16$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$x_{1}$$$ | $$$1$$$ | $$$0$$$ | $$$-1$$$ | $$$2$$$ | $$$4$$$ |
將第$$$3$$$行加到第$$$1$$$行:$$$R_{1} = R_{1} + R_{3}$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$0$$$ | $$$0$$$ | $$$1$$$ | $$$2$$$ | $$$20$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$x_{1}$$$ | $$$1$$$ | $$$0$$$ | $$$-1$$$ | $$$2$$$ | $$$4$$$ |
從第$$$2$$$行減去$$$\frac{1}{2}$$$倍的第$$$3$$$行:$$$R_{2} = R_{2} - \frac{R_{3}}{2}$$$.
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$0$$$ | $$$0$$$ | $$$1$$$ | $$$2$$$ | $$$20$$$ |
| $$$x_{2}$$$ | $$$0$$$ | $$$1$$$ | $$$1$$$ | $$$-1$$$ | $$$2$$$ |
| $$$x_{1}$$$ | $$$1$$$ | $$$0$$$ | $$$-1$$$ | $$$2$$$ | $$$4$$$ |
Z 行的係數皆非負。
已達到最優解。
可得如下解:$$$\left(x_{1}, x_{2}, S_{1}, S_{2}\right) = \left(4, 2, 0, 0\right)$$$。
答案
在 $$$\left(x_{1}, x_{2}\right) = \left(4, 2\right)$$$A 取得 $$$Z = 20$$$A。