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MATH2-Inf-2: 1a geloest.
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Jim Martens
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@ -15,9 +15,14 @@
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\polyset{style=C, div=:,vars=x}
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\pgfplotsset{compat=1.8}
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\pagenumbering{arabic}
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% ensures that paragraphs are separated by empty lines
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\parskip 12pt plus 1pt minus 1pt
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\parindent 0pt
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% define how the sections are rendered
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\def\thesection{\arabic{section})}
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\def\thesubsection{\alph{subsection})}
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\def\thesubsubsection{(\roman{subsubsection})}
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% some matrix magic
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\makeatletter
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\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
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\hskip -\arraycolsep
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@ -32,6 +37,93 @@ Stephan Niendorf (6242417)}
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\maketitle
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\section{} %1
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\subsection{} %a
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\textbf{Aufgabe:} Lösen Sie das folgende LP-Problem mit dem Simplexverfahren:
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\begin{alignat*}{4}
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\text{maximiere}\; & x_{1} &+& 6x_{2} &-& 4x_{3} && \\
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\multicolumn{8}{l}{\text{unter den Nebenbedingungen}} && \\
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\;& 2x_{1} && &+& x_{3} &\leq & 5 \\
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\;-& x_{1} &+& 3x_{2} &-& 2x_{3} &\leq &\, 2 \\
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\;& && x_{2} &-& x_{3} &\leq &\, 2 \\
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\multicolumn{6}{r}{$x_{1}, x_{2}, x_{3}$} \,&\geq &\, 0
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\end{alignat*}
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\textbf{Lösung.}
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\underline{Starttableau}:
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\begin{alignat*}{5}
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x_{4} \,&=&\, 5 \,&-&\, 2x_{1} && &-&\, x_{3} \\
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x_{5} \,&=&\, 2 \,&+&\, x_{1} \,&-&\, 3x_{2} \,&+&\, 2x_{3} \\
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x_{6} \,&=&\, 2 && &-&\, x_{2} \,&+&\, x_{3} \\ \cline{1 - 9}
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z &=& && x_{1} \,&+&\, 6x_{2} \,&-&\, 4x_{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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Es folgt
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\begin{alignat*}{2}
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3x_{2} \,&=&&\, 2 + x_{1} + 2x_{3} - x_{5} \\
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x_{2} \,&=&&\, \frac{2}{3} + \frac{1}{3}x_{1} + \frac{2}{3}x_{3} - \frac{1}{3}x_{5} \\
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x_{4} \,&=&&\, 5 - 2x_{1} - x_{3} \\
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x_{6} \,&=&&\, 2 - \left(\frac{2}{3} + \frac{1}{3}x_{1} + \frac{2}{3}x_{3} - \frac{1}{3}x_{5}\right) + x_{3} \\
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&=&&\, 2 - \frac{2}{3} - \frac{1}{3}x_{1} - \frac{2}{3}x_{3} + \frac{1}{3}x_{5} + x_{3} \\
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&=&&\, \frac{4}{3} - \frac{1}{3}x_{1} + \frac{1}{3}x_{3} + \frac{1}{3}x_{5} \\
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z \,&=&&\, x_{1} + 6\left(\frac{2}{3} + \frac{1}{3}x_{1} + \frac{2}{3}x_{3} - \frac{1}{3}x_{5}\right) - 4x_{3} \\
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&=&&\, x_{1} + 4 + 2x_{1} + 4x_{3} - 2x_{5} - 4x_{3} \\
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&=&&\, 4 + 3x_{1} - 2x_{5}
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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{2}{3} \,&+&\, \frac{1}{3}x_{1} \,&+&\, \frac{2}{3}x_{3} \,&-&\, \frac{1}{3}x_{5} \\
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x_{4} \,&=&\, 5 \,&-&\, 2x_{1} \,&-&\, x_{3} && \\
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x_{6} \,&=&\, \frac{4}{3} \,&-&\, \frac{1}{3}x_{1} \,&+&\, \frac{1}{3}x_{3} \,&+&\, \frac{1}{3}x_{5} \\ \cline{1 - 9}
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z &=& 4 \,&+&\, 3x_{1} && &-& 2x_{5}
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\end{alignat*}
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\underline{2. Iteration}:
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Eingangsvariable: $x_{1}$ \\
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Ausgangsvariable: $x_{4}$
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Es folgt
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\begin{alignat*}{2}
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2x_{1} &=&& 5 - x_{3} - x_{4} \\
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x_{1} &=&& \frac{5}{2} - \frac{1}{2}x_{3} - \frac{1}{2}x_{4} \\
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x_{2} &=&& \frac{2}{3} + \frac{1}{3}\left(\frac{5}{2} - \frac{1}{2}x_{3} - \frac{1}{2}x_{4}\right) + \frac{2}{3}x_{3} - \frac{1}{3}x_{5} \\
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&=&& \frac{3}{2} - \frac{1}{6}x_{3} - \frac{1}{6}x_{4} + \frac{2}{3}x_{3} - \frac{1}{3}x_{5} \\
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&=&& \frac{3}{2} + \frac{1}{2}x_{3} - \frac{1}{6}x_{4} - \frac{1}{3}x_{5} \\
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x_{6} &=&& \frac{4}{3} - \frac{1}{3}\left(\frac{5}{2} - \frac{1}{2}x_{3} - \frac{1}{2}x_{4}\right) + \frac{1}{3}x_{3} + \frac{1}{3}x_{5} \\
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&=&& \frac{1}{2} + \frac{1}{6}x_{3} + \frac{1}{6}x_{4} + \frac{1}{3}x_{3} + \frac{1}{3}x_{5} \\
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&=&& \frac{1}{2} + \frac{1}{2}x_{3} + \frac{1}{6}x_{4} + + \frac{1}{3}x_{5} \\
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z &=&& 4 + 3\left(\frac{5}{2} - \frac{1}{2}x_{3} - \frac{1}{2}x_{4}\right) - 2x_{5} \\
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&=&& \frac{23}{2} - \frac{3}{2}x_{3} - \frac{3}{2}x_{4} - 2x_{5} \\
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&=&& \frac{23}{2} - \frac{3}{2}x_{3} - \frac{3}{2}x_{4} - 2x_{5}
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\end{alignat*}
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\underline{Ergebnis der 2. Iteration}:
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\begin{alignat*}{5}
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x_{1} \,&=&\, \frac{5}{2} \,&-&\, \frac{1}{2}x_{3} \,&-&\, \frac{1}{2}x_{4} && \\
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x_{2} \,&=&\, \frac{3}{2} \,&+&\, \frac{1}{2}x_{3} \,&-&\, \frac{1}{6}x_{4} \,&-&\, \frac{1}{3}x_{5} \\
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x_{6} \,&=&\, \frac{1}{2} \,&+&\, \frac{1}{2}x_{3} \,&+&\, \frac{1}{6}x_{4} \,&+&\, \frac{1}{3}x_{5} \\ \cline{1 - 9}
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z &=& \frac{23}{2} \,&-&\, \frac{3}{2}x_{3} \,&-&\, \frac{3}{2}x_{4} \,&-&\, 2x_{5}
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\end{alignat*}
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Dieses Tableau liefert die optimale Lösung $x_{1} = \frac{5}{2}, x_{2} = \frac{3}{2}, x_{3} = 0$ mit $z = \frac{23}{2}$.
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\underline{Startlösung ("`zulässige Basislösung am Anfang"')}:
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\[
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x_{1} = 0, x_{2} = 0, x_{3} = 0, x_{4} = 5, x_{5} = 2, x_{6} = 2 \text{ mit } z = 0
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\]
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\underline{Zulässige Basislösung nach der 1. Iteration}:
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\[
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x_{1} = 0, x_{2} = \frac{2}{3}, x_{3} = 0, x_{4} = 5, x_{5} = 0, x_{6} = \frac{4}{3} \text{ mit } z = 4
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\]
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\underline{Zulässige Basislösung nach der 2. Iteration}:
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\[
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x_{1} = \frac{5}{2}, x_{2} = \frac{3}{2}, x_{3} = 0, x_{4} = 0, x_{5} = 0, x_{6} = \frac{1}{2} \text{ mit } z = \frac{23}{2}
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\]
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\subsection{} %b
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\section{} %2
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\end{document}
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