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.
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 | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | Oplossing | Ratio |
| $$$Z$$$ | $$$-3$$$ | $$$-4$$$ | $$$0$$$ | $$$0$$$ | $$$0$$$ | |
| $$$S_{1}$$$ | $$$1$$$ | $$$2$$$ | $$$1$$$ | $$$0$$$ | $$$8$$$ | $$$\frac{8}{2} = 4$$$ |
| $$$S_{2}$$$ | $$$1$$$ | $$$1$$$ | $$$0$$$ | $$$1$$$ | $$$6$$$ | $$$\frac{6}{1} = 6$$$ |
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 | $$$x_{1}$$$ | $$$x_{2}$$$ | $$$S_{1}$$$ | $$$S_{2}$$$ | Oplossing | Ratio |
| $$$Z$$$ | $$$-1$$$ | $$$0$$$ | $$$2$$$ | $$$0$$$ | $$$16$$$ | |
| $$$x_{2}$$$ | $$$\frac{1}{2}$$$ | $$$1$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$4$$$ | $$$\frac{4}{\frac{1}{2}} = 8$$$ |
| $$$S_{2}$$$ | $$$\frac{1}{2}$$$ | $$$0$$$ | $$$- \frac{1}{2}$$$ | $$$1$$$ | $$$2$$$ | $$$\frac{2}{\frac{1}{2}} = 4$$$ |
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.