シンプレックス法計算機

シンプレックス法を用いて最適化問題を解く

この計算機は、シンプレックス法(単体法)を用いて与えられた最適化問題を解きます。必要に応じて、スラック変数、余剰変数、人工変数を導入します。人工変数を用いる場合は、初期解を求めるためにビッグM法または二段階法を使用します。解法の手順も表示されます。

カンマ区切り。

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入力内容

制約 $$$\begin{cases} x_{1} + 2 x_{2} \leq 8 \\ x_{1} + x_{2} \leq 6 \\ x_{2} \geq 0 \\ x_{1} \geq 0 \end{cases}$$$ のもとで $$$Z = 3 x_{1} + 4 x_{2}$$$ を最大化する。

解答

標準形の問題は次のように書けます:

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

不等式をすべて等式にするために、スラック変数または余剰変数を追加せよ:

$$Z = 3 x_{1} + 4 x_{2} \to max$$$$\begin{cases} x_{1} + 2 x_{2} + S_{1} = 8 \\ x_{1} + x_{2} + S_{2} = 6 \\ x_{1}, x_{2}, S_{1}, S_{2} \geq 0 \end{cases}$$

シンプレックス表を作成しなさい:

Basic$$$x_{1}$$$$$$x_{2}$$$$$$S_{1}$$$$$$S_{2}$$$解答
$$$Z$$$$$$-3$$$$$$-4$$$$$$0$$$$$$0$$$$$$0$$$
$$$S_{1}$$$$$$1$$$$$$2$$$$$$1$$$$$$0$$$$$$8$$$
$$$S_{2}$$$$$$1$$$$$$1$$$$$$0$$$$$$1$$$$$$6$$$

入基変数は $$$x_{2}$$$ で、Z 行における係数 $$$-4$$$ が最も負であるためです。

Basic$$$x_{1}$$$$$$x_{2}$$$$$$S_{1}$$$$$$S_{2}$$$解答Ratio
$$$Z$$$$$$-3$$$$$$-4$$$$$$0$$$$$$0$$$$$$0$$$
$$$S_{1}$$$$$$1$$$$$$2$$$$$$1$$$$$$0$$$$$$8$$$$$$\frac{8}{2} = 4$$$
$$$S_{2}$$$$$$1$$$$$$1$$$$$$0$$$$$$1$$$$$$6$$$$$$\frac{6}{1} = 6$$$

離基変数は$$$S_{1}$$$で、最小比だからです。

$$$1$$$行を$$$2$$$で割る: $$$R_{1} = \frac{R_{1}}{2}$$$

Basic$$$x_{1}$$$$$$x_{2}$$$$$$S_{1}$$$$$$S_{2}$$$解答
$$$Z$$$$$$-3$$$$$$-4$$$$$$0$$$$$$0$$$$$$0$$$
$$$x_{2}$$$$$$\frac{1}{2}$$$$$$1$$$$$$\frac{1}{2}$$$$$$0$$$$$$4$$$
$$$S_{2}$$$$$$1$$$$$$1$$$$$$0$$$$$$1$$$$$$6$$$

$$$1$$$に行$$$2$$$$$$4$$$倍を加える: $$$R_{1} = R_{1} + 4 R_{2}$$$.

Basic$$$x_{1}$$$$$$x_{2}$$$$$$S_{1}$$$$$$S_{2}$$$解答
$$$Z$$$$$$-1$$$$$$0$$$$$$2$$$$$$0$$$$$$16$$$
$$$x_{2}$$$$$$\frac{1}{2}$$$$$$1$$$$$$\frac{1}{2}$$$$$$0$$$$$$4$$$
$$$S_{2}$$$$$$1$$$$$$1$$$$$$0$$$$$$1$$$$$$6$$$

$$$3$$$行から第$$$2$$$行を引く: $$$R_{3} = R_{3} - R_{2}$$$

Basic$$$x_{1}$$$$$$x_{2}$$$$$$S_{1}$$$$$$S_{2}$$$解答
$$$Z$$$$$$-1$$$$$$0$$$$$$2$$$$$$0$$$$$$16$$$
$$$x_{2}$$$$$$\frac{1}{2}$$$$$$1$$$$$$\frac{1}{2}$$$$$$0$$$$$$4$$$
$$$S_{2}$$$$$$\frac{1}{2}$$$$$$0$$$$$$- \frac{1}{2}$$$$$$1$$$$$$2$$$

入基変数は $$$x_{1}$$$ で、Z 行における係数 $$$-1$$$ が最も負であるためです。

Basic$$$x_{1}$$$$$$x_{2}$$$$$$S_{1}$$$$$$S_{2}$$$解答Ratio
$$$Z$$$$$$-1$$$$$$0$$$$$$2$$$$$$0$$$$$$16$$$
$$$x_{2}$$$$$$\frac{1}{2}$$$$$$1$$$$$$\frac{1}{2}$$$$$$0$$$$$$4$$$$$$\frac{4}{\frac{1}{2}} = 8$$$
$$$S_{2}$$$$$$\frac{1}{2}$$$$$$0$$$$$$- \frac{1}{2}$$$$$$1$$$$$$2$$$$$$\frac{2}{\frac{1}{2}} = 4$$$

離基変数は$$$S_{2}$$$で、最小比だからです。

$$$2$$$行を$$$2$$$倍する: $$$R_{2} = 2 R_{2}$$$.

Basic$$$x_{1}$$$$$$x_{2}$$$$$$S_{1}$$$$$$S_{2}$$$解答
$$$Z$$$$$$-1$$$$$$0$$$$$$2$$$$$$0$$$$$$16$$$
$$$x_{2}$$$$$$\frac{1}{2}$$$$$$1$$$$$$\frac{1}{2}$$$$$$0$$$$$$4$$$
$$$x_{1}$$$$$$1$$$$$$0$$$$$$-1$$$$$$2$$$$$$4$$$

$$$3$$$を行$$$1$$$に加える: $$$R_{1} = R_{1} + R_{3}$$$.

Basic$$$x_{1}$$$$$$x_{2}$$$$$$S_{1}$$$$$$S_{2}$$$解答
$$$Z$$$$$$0$$$$$$0$$$$$$1$$$$$$2$$$$$$20$$$
$$$x_{2}$$$$$$\frac{1}{2}$$$$$$1$$$$$$\frac{1}{2}$$$$$$0$$$$$$4$$$
$$$x_{1}$$$$$$1$$$$$$0$$$$$$-1$$$$$$2$$$$$$4$$$

$$$2$$$行から$$$3$$$行の$$$\frac{1}{2}$$$倍を引く: $$$R_{2} = R_{2} - \frac{R_{3}}{2}$$$

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行の係数はすべて非負です。

最適解に到達しました。

次の解が得られます: $$$\left(x_{1}, x_{2}, S_{1}, S_{2}\right) = \left(4, 2, 0, 0\right)$$$.

解答

$$$Z = 20$$$A$$$\left(x_{1}, x_{2}\right) = \left(4, 2\right)$$$A において達成される。