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。

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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

$$$\left(x_{1}, x_{2}\right) = \left(4, 2\right)$$$A 取得 $$$Z = 20$$$A