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MATH2-Inf-3: Aufgabe 2a korrigiert.
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Jim Martens
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@ -202,7 +202,7 @@ Stephan Niendorf (6242417)}
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\underline{Starttableau}:
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\begin{alignat*}{5}
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x_{3} \,&=&\, -4 \,&-&\, x_{1} \,&+&\, x_{2} \,&+&\, x_{0} \\
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x_{3} \,&=&\, -4 \,&+&\, x_{1} \,&+&\, x_{2} \,&+&\, x_{0} \\
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x_{4} \,&=&\, 2 \,&-&\, x_{1} \,&-&\, 2x_{2} \,&+&\, x_{0} \\
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x_{5} \,&=&\, -1 \,&+&\, x_{1} \,&-&\, x_{2} \,&+&\, x_{0} \\ \cline{1 - 9}
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w &=& && && \,&-&\, x_{0}
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@ -215,55 +215,75 @@ Stephan Niendorf (6242417)}
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Es folgt
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\begin{alignat*}{2}
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-x_{0} \,&=&&\, -4 - x_{1} + x_{2} - x_{3} \\
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x_{0} \,&=&&\, 4 + x_{1} - x_{2} + x_{3} \\
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x_{4} \,&=&&\, 2 - x_{1} - 2x_{2} + \left(4 + x_{1} - x_{2} + x_{3}\right) \\
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&=&&\, 2 - x_{1} - 2x_{2} + 4 + x_{1} - x_{2} + x_{3} \\
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&=&&\, 6 - 3x_{2} + x_{3} \\
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x_{5} \,&=&&\, -1 + x_{1} - x_{2} + \left(4 + x_{1} - x_{2} + x_{3}\right) \\
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&=&&\, -1 + x_{1} - x_{2} + 4 + x_{1} - x_{2} + x_{3} \\
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&=&&\, 3 + 2x_{1} - 2x_{2} + x_{3} \\
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w \,&=&&\, -\left(4 + x_{1} - x_{2} + x_{3}\right) \\
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&=&&\, -4 - x_{1} + x_{2} - x_{3} \\
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-x_{0} \,&=&&\, -4 + x_{1} + x_{2} - x_{3} \\
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x_{0} \,&=&&\, 4 - x_{1} - x_{2} + x_{3} \\
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x_{4} \,&=&&\, 2 - x_{1} - 2x_{2} + \left(4 - x_{1} - x_{2} + x_{3}\right) \\
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&=&&\, 2 - x_{1} - 2x_{2} + 4 - x_{1} - x_{2} + x_{3} \\
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&=&&\, 6 - 2x_{1} - 3x_{2} + x_{3} \\
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x_{5} \,&=&&\, -1 + x_{1} - x_{2} + \left(4 - x_{1} - x_{2} + x_{3}\right) \\
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&=&&\, -1 + x_{1} - x_{2} + 4 - x_{1} - x_{2} + x_{3} \\
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&=&&\, 3 - 2x_{2} + x_{3} \\
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w \,&=&&\, -\left(4 - x_{1} - x_{2} + x_{3}\right) \\
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&=&&\, -4 + x_{1} + x_{2} - x_{3} \\
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\end{alignat*}
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\underline{Ergebnis des Umwandelns}:
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\begin{alignat*}{5}
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x_{0} \,&=&\, 4 \,&+&\, x_{1} \,&-&\, x_{2} \,&+&\, x_{3} \\
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x_{4} \,&=&\, 6 \,&& &-&\, 3x_{2} \,&+&\, x_{3} \\
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x_{5} \,&=&\, 3 \,&+&\, 2x_{1} \,&-&\, 2x_{2} \,&+&\, x_{3} \\ \cline{1 - 9}
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w &=& -2 \,&-&\, x_{1} \,&+&\, x_{2} \,&-&\, x_{3}
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x_{0} \,&=&\, 4 \,&-&\, x_{1} \,&-&\, x_{2} \,&+&\, x_{3} \\
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x_{4} \,&=&\, 6 \,&-&\, 2x_{1} \,&-&\, 3x_{2} \,&+&\, x_{3} \\
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x_{5} \,&=&\, 3 \,&& &-&\, 2x_{2} \,&+&\, x_{3} \\ \cline{1 - 9}
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w &=& -4 \,&+&\, x_{1} \,&+&\, x_{2} \,&-&\, x_{3}
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\end{alignat*}
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\underline{1. Iteration}:
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Eingangsvariable: $x_{2}$ \\
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Ausgangsvariable: $x_{5}$
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Eingangsvariable: $x_{1}$ \\
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Ausgangsvariable: $x_{0}$
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Es folgt
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\begin{alignat*}{2}
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2x_{2} &=&& 3 + 2x_{1} + x_{3} - x_{5} \\
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x_{2} &=&& \frac{3}{2} + x_{1} + \frac{1}{2}x_{3} - \frac{1}{2}x_{5} \\
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x_{0} &=&& 4 + x_{1} - \left(\frac{3}{2} + x_{1} + \frac{1}{2}x_{3} - \frac{1}{2}x_{5}\right) + x_{3} \\
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&=&& 4 + x_{1} - \frac{3}{2} - x_{1} - \frac{1}{2}x_{3} + \frac{1}{2}x_{5} + x_{3}\\
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&=&& \frac{5}{2} + \frac{1}{2}x_{3} + \frac{1}{2}x_{5} \\
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x_{4} &=&& 6 - 3\left(\frac{3}{2} + x_{1} + \frac{1}{2}x_{3} - \frac{1}{2}x_{5}\right) + x_{3} \\
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&=&& 6 - \frac{9}{2} - 3x_{1} + \frac{3}{2}x_{3} - \frac{3}{2}x_{5} + x_{3}\\
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&=&& \frac{3}{2} - 3x_{1} + \frac{5}{2}x_{3} - \frac{3}{2}x_{5} \\
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w &=&& -2 - x_{1} + \left(\frac{3}{2} + x_{1} + \frac{1}{2}x_{3} - \frac{1}{2}x_{5}\right) - x_{3} \\
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&=&& -2 - x_{1} + \frac{3}{2} + x_{1} + \frac{1}{2}x_{3} - \frac{1}{2}x_{5} - x_{3} \\
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&=&& \frac{1}{2} - \frac{1}{2}x_{3} - \frac{1}{2}x_{5}
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x_{1} \,&=&&\, 4 - x_{2} + x_{3} - x_{0} \\
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x_{4} \,&=&&\, 6 - 2\left(4 - x_{2} + x_{3} - x_{0}\right) - 3x_{2} + x_{3} \\
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&=&&\, 6 - 8 + 2x_{2} - 2x_{3} + 2x_{0} - 3x_{2} + x_{3}\\
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&=&&\, -2 - x_{2} - x_{3} + 2x_{0} \\
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x_{5} \,&=&&\, 3 - 2x_{2} + x_{3} \\
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w \,&=&&\, -4 + \left(4 - x_{2} + x_{3} - x_{0}\right) + x_{2} - x_{3} \\
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&=&&\, -4 + 4 - x_{2} + x_{3} - x_{0} + x_{2} - x_{3} \\
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&=&&\, 0 - x_{0}
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\end{alignat*}
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\underline{Ergebnis der 1. Iteration}:
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\begin{alignat*}{5}
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x_{2} \,&=&\, \frac{3}{2} \,&+&\, x_{1} \,&+&\, \frac{1}{2}x_{3} \,&-&\, \frac{1}{2}x_{5} \\
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x_{0} \,&=&\, \frac{5}{2} \,&& &+&\, \frac{1}{2}x_{3} \,&+&\, \frac{1}{2}x_{5} \\
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x_{4} \,&=&\, \frac{3}{2} \,&-&\, 3x_{1} \,&+&\, \frac{5}{2}x_{3} \,&-&\, \frac{3}{2}x_{5} \\ \cline{1 - 9}
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w &=& \frac{1}{2} \,&& &-&\, \frac{1}{2}x_{3} \,&-&\, \frac{1}{2}x_{5}
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x_{1} \,&=&\, 4 \,&-&\, x_{2} \,&+&\, x_{3} \,&-&\, x_{0} \\
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x_{4} \,&=&\, -2 \,&-&\, x_{2} \,&-&\, x_{3} \,&+&\, 2x_{0} \\
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x_{5} \,&=&\, 3 \,&-&\, 2x_{2} \,&+&\, x_{3} \,&& \\ \cline{1 - 9}
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w &=& && && &-&\, x_{0}
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\end{alignat*}
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Da das Hilfsproblem keine optimale Lösung besitzt, besitzt das ursprüngliche Problem keine zulässige Lösung und ist damit unlösbar.
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Das Tableau ist optimal. Als optimale Lösung des Hilfsproblem erhält man:
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\[
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x_{0} = 0, x_{1} = 4, x_{2} = 0
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\]
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Als zulässige Lösung für das ursprüngliche Problem ergibt sich:
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\[
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x_{1} = 4, x_{2} = 0
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\]
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Die ursprüngliche Zielfunktion lautet $z = x_{1} + 3x_{2}$. Setzt man für $x_{1}$ die rechte Seite der Gleichung im obigen Tableau ein, erhält man:
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\[
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z = 4 - x_{2} + x_{3} + 3x_{2}
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\]
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Daraus ergibt sich dieses Starttableau:
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\begin{alignat*}{4}
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x_{1} \,&=&\, 4 \,&-&\, x_{2} \,&+&\, x_{3} \\
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x_{4} \,&=&\, -2 \,&-&\, x_{2} \,&-&\, x_{3} \\
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x_{5} \,&=&\, 3 \,&-&\, 2x_{2} \,&+&\, x_{3} \\ \cline{1 - 7}
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z \,&=&\, 4 \,&+&\, 2x_{2} \,&+&\, x_{3}
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\end{alignat*}
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\subsection{} %b
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\textbf{Aufgabe:} Lösen Sie das folgende LP-Hilfsproblem mit dem Simplexverfahren:
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