单纯形法计算器
使用单纯形法求解优化问题
该计算器将使用单纯形法求解给定的优化问题。必要时,它会加入松弛变量、剩余变量和人工变量。若涉及人工变量,则使用大M法或两阶段法来确定初始解。提供求解步骤。
您的输入
在约束 $$$\begin{cases} x_{1} + 2 x_{2} \leq 8 \\ x_{1} + x_{2} \leq 6 \\ x_{2} \geq 0 \\ x_{1} \geq 0 \end{cases}$$$ 下最大化 $$$Z = 3 x_{1} + 4 x_{2}$$$。
解答
该问题的标准形式可写为:
$$Z = 3 x_{1} + 4 x_{2} \to max$$$$\begin{cases} x_{1} + 2 x_{2} \leq 8 \\ x_{1} + x_{2} \leq 6 \\ x_{1}, x_{2} \geq 0 \end{cases}$$引入(松弛或剩余)变量,将所有不等式转化为等式:
$$Z = 3 x_{1} + 4 x_{2} \to max$$$$\begin{cases} x_{1} + 2 x_{2} + S_{1} = 8 \\ x_{1} + x_{2} + S_{2} = 6 \\ x_{1}, x_{2}, S_{1}, S_{2} \geq 0 \end{cases}$$写出单纯形表:
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-3$$$ | $$$-4$$$ | $$$0$$$ | $$$0$$$ | $$$0$$$ |
| $$$S_{1}$$$ | $$$1$$$ | $$$2$$$ | $$$1$$$ | $$$0$$$ | $$$8$$$ |
| $$$S_{2}$$$ | $$$1$$$ | $$$1$$$ | $$$0$$$ | $$$1$$$ | $$$6$$$ |
入基变量是 $$$x_{2}$$$,因为它在 Z 行中具有最负的系数 $$$-4$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 | Ratio |
| $$$Z$$$ | $$$-3$$$ | $$$-4$$$ | $$$0$$$ | $$$0$$$ | $$$0$$$ | |
| $$$S_{1}$$$ | $$$1$$$ | $$$2$$$ | $$$1$$$ | $$$0$$$ | $$$8$$$ | $$$\frac{8}{2} = 4$$$ |
| $$$S_{2}$$$ | $$$1$$$ | $$$1$$$ | $$$0$$$ | $$$1$$$ | $$$6$$$ | $$$\frac{6}{1} = 6$$$ |
离基变量是$$$S_{1}$$$,因为它的比值最小。
将第$$$1$$$行除以$$$2$$$:$$$R_{1} = \frac{R_{1}}{2}$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-3$$$ | $$$-4$$$ | $$$0$$$ | $$$0$$$ | $$$0$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$S_{2}$$$ | $$$1$$$ | $$$1$$$ | $$$0$$$ | $$$1$$$ | $$$6$$$ |
将第$$$2$$$行的$$$4$$$倍加到第$$$1$$$行: $$$R_{1} = R_{1} + 4 R_{2}$$$.
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-1$$$ | $$$0$$$ | $$$2$$$ | $$$0$$$ | $$$16$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$S_{2}$$$ | $$$1$$$ | $$$1$$$ | $$$0$$$ | $$$1$$$ | $$$6$$$ |
第$$$3$$$行减去第$$$2$$$行:$$$R_{3} = R_{3} - R_{2}$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-1$$$ | $$$0$$$ | $$$2$$$ | $$$0$$$ | $$$16$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$S_{2}$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$- \frac{1}{2}$$$ | $$$1$$$ | $$$2$$$ |
入基变量是 $$$x_{1}$$$,因为它在 Z 行中具有最负的系数 $$$-1$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 | Ratio |
| $$$Z$$$ | $$$-1$$$ | $$$0$$$ | $$$2$$$ | $$$0$$$ | $$$16$$$ | |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ | $$$\frac{4}{\frac{1}{2}} = 8$$$ |
| $$$S_{2}$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$- \frac{1}{2}$$$ | $$$1$$$ | $$$2$$$ | $$$\frac{2}{\frac{1}{2}} = 4$$$ |
离基变量是$$$S_{2}$$$,因为它的比值最小。
将第$$$2$$$行乘以$$$2$$$: $$$R_{2} = 2 R_{2}$$$.
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$-1$$$ | $$$0$$$ | $$$2$$$ | $$$0$$$ | $$$16$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$x_{1}$$$ | $$$1$$$ | $$$0$$$ | $$$-1$$$ | $$$2$$$ | $$$4$$$ |
将第$$$3$$$行加到第$$$1$$$行:$$$R_{1} = R_{1} + R_{3}$$$。
| Basic | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | 解答 |
| $$$Z$$$ | $$$0$$$ | $$$0$$$ | $$$1$$$ | $$$2$$$ | $$$20$$$ |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ |
| $$$x_{1}$$$ | $$$1$$$ | $$$0$$$ | $$$-1$$$ | $$$2$$$ | $$$4$$$ |
从第$$$2$$$行中减去第$$$3$$$行的$$$\frac{1}{2}$$$倍:$$$R_{2} = R_{2} - \frac{R_{3}}{2}$$$。
| 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 行的系数均非负。
已达到最优解。
得到如下解:$$$\left(x_{1}, x_{2}, S_{1}, S_{2}\right) = \left(4, 2, 0, 0\right)$$$。
答案
$$$Z = 20$$$A在$$$\left(x_{1}, x_{2}\right) = \left(4, 2\right)$$$A处取得。