Rekenmachine voor de simplexmethode

Los optimalisatieproblemen op met behulp van de simplexmethode

De rekenmachine zal het gegeven optimalisatieprobleem oplossen met behulp van het simplex-algoritme. Zo nodig worden slack-, surplus- en kunstmatige variabelen toegevoegd. In het geval van kunstmatige variabelen wordt de Big-M-methode of de tweefasenmethode gebruikt om de startoplossing te bepalen. Stappen zijn beschikbaar.

Door komma's gescheiden.

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Uw invoer

Maximaliseer $$$Z = 3 x_{1} + 4 x_{2}$$$, onderworpen aan $$$\begin{cases} x_{1} + 2 x_{2} \leq 8 \\ x_{1} + x_{2} \leq 6 \\ x_{2} \geq 0 \\ x_{1} \geq 0 \end{cases}$$$.

Oplossing

Het probleem in canonieke vorm kan als volgt worden geschreven:

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

Voeg slack- of surplusvariabelen toe om alle ongelijkheden in gelijkheden om te zetten:

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

Schrijf het simplex-tableau op:

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

De inkomende variabele is $$$x_{2}$$$, omdat deze de meest negatieve coëfficiënt $$$-4$$$ in de Z-rij heeft.

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

De uitgaande variabele is $$$S_{1}$$$, omdat deze het kleinste quotiënt heeft.

Deel rij $$$1$$$ door $$$2$$$: $$$R_{1} = \frac{R_{1}}{2}$$$.

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

Tel $$$4$$$ keer rij $$$2$$$ op bij rij $$$1$$$: $$$R_{1} = R_{1} + 4 R_{2}$$$.

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

Trek rij $$$2$$$ af van rij $$$3$$$: $$$R_{3} = R_{3} - R_{2}$$$.

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

De inkomende variabele is $$$x_{1}$$$, omdat deze de meest negatieve coëfficiënt $$$-1$$$ in de Z-rij heeft.

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

De uitgaande variabele is $$$S_{2}$$$, omdat deze het kleinste quotiënt heeft.

Vermenigvuldig rij $$$2$$$ met $$$2$$$: $$$R_{2} = 2 R_{2}$$$.

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

Tel rij $$$3$$$ op bij rij $$$1$$$: $$$R_{1} = R_{1} + R_{3}$$$.

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

Trek rij $$$3$$$ vermenigvuldigd met $$$\frac{1}{2}$$$ af van rij $$$2$$$: $$$R_{2} = R_{2} - \frac{R_{3}}{2}$$$.

Basic$$$x_{1}$$$$$$x_{2}$$$$$$S_{1}$$$$$$S_{2}$$$Oplossing
$$$Z$$$$$$0$$$$$$0$$$$$$1$$$$$$2$$$$$$20$$$
$$$x_{2}$$$$$$0$$$$$$1$$$$$$1$$$$$$-1$$$$$$2$$$
$$$x_{1}$$$$$$1$$$$$$0$$$$$$-1$$$$$$2$$$$$$4$$$

Geen van de Z-rijcoëfficiënten is negatief.

Het optimum is bereikt.

De volgende oplossing wordt verkregen: $$$\left(x_{1}, x_{2}, S_{1}, S_{2}\right) = \left(4, 2, 0, 0\right)$$$.

Antwoord

$$$Z = 20$$$A wordt bereikt bij $$$\left(x_{1}, x_{2}\right) = \left(4, 2\right)$$$A.