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MATH2-Inf-2: 1b geloest.
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Jim Martens
@ -42,8 +42,8 @@ Stephan Niendorf (6242417)}
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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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\;-& 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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@ -125,5 +125,140 @@ Stephan Niendorf (6242417)}
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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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\textbf{Aufgabe:} Lösen Sie das folgende LP-Problem mit dem Simplexverfahren:
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\begin{alignat*}{4}
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\text{maximiere}\; -& 5x_{1} &+& 11x_{2} &-& 5x_{3} && \\
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\multicolumn{8}{l}{\text{unter den Nebenbedingungen}} && \\
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\;-& x_{1} &+& 3x_{2} &-& 4x_{3} &\leq & 2 \\
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\;& x_{1} &+& 5x_{2} &+& 3x_{3} &\leq & 6 \\
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\;-& x_{1} &+& 3x_{2} &+& 3x_{3} &\leq &\, 4 \\
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\;& x_{1} &-& x_{2} &+& 3x_{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} \,&=&\, 2 \,&+&\, x_{1} \,&-&\, 3x_{2} \,&+&\, 4x_{3} \\
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x_{5} \,&=&\, 6 \,&-&\, x_{1} \,&-&\, 5x_{2} \,&-&\, 3x_{3} \\
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x_{6} \,&=&\, 4 \,&+&\, x_{1} \,&-&\, 3x_{2} \,&-&\, 3x_{3} \\
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x_{7} \,&=&\, 2 \,&-&\, x_{1} \,&+&\, x_{2} \,&-&\, 3x_{3} \\ \cline{1 - 9}
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z &=& &-& 5x_{1} \,&+&\, 11x_{2} \,&-&\, 5x_{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: $x_{4}$
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Es folgt
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\begin{alignat*}{2}
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3x_{2} \,&=&&\, 2 + x_{1} + 4x_{3} - x_{4} \\
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x_{2} \,&=&&\, \frac{2}{3} + \frac{1}{3}x_{1} + \frac{4}{3}x_{3} - \frac{1}{3}x_{4} \\
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x_{5} \,&=&&\, 6 - x_{1} - 5\left(\frac{2}{3} + \frac{1}{3}x_{1} + \frac{4}{3}x_{3} - \frac{1}{3}x_{4}\right) - 3x_{3} \\
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&=&& 6 - x_{1} - \frac{10}{3} - \frac{5}{3}x_{1} - \frac{20}{3}x_{3} + \frac{5}{3}x_{4} - 3x_{3} \\
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&=&& \frac{8}{3} - \frac{8}{3}x_{1} - \frac{29}{3}x_{3} + \frac{5}{3}x_{4} \\
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x_{6} \,&=&&\, 4 + x_{1} - 3\left(\frac{2}{3} + \frac{1}{3}x_{1} + \frac{4}{3}x_{3} - \frac{1}{3}x_{4}\right) - 3x_{3} \\
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&=&&\, 4 + x_{1} - 2 - x_{1} - 4x_{3} + x_{4} - 3x_{3} \\
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&=&&\, 2 - 7x_{3} + x_{4} \\
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x_{7} &=&& 2 - x_{1} + \frac{2}{3} + \frac{1}{3}x_{1} + \frac{4}{3}x_{3} - \frac{1}{3}x_{4} - 3x_{3} \\
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&=&& \frac{8}{3} - \frac{2}{3}x_{1} - \frac{5}{3}x_{3} - \frac{1}{3}x_{4} \\
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z \,&=&&\, - 5x_{1} + 11\left(\frac{2}{3} + \frac{1}{3}x_{1} + \frac{4}{3}x_{3} - \frac{1}{3}x_{4}\right) - 5x_{3} \\
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&=&&\, -5x_{1} + \frac{22}{3} + \frac{11}{3}x_{1} + \frac{44}{3}x_{3} - \frac{11}{3}x_{4} - 5x_{3} \\
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&=&&\, \frac{22}{3} - \frac{4}{3}x_{1} + \frac{29}{3}x_{3} - \frac{11}{3}x_{4}
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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{4}{3}x_{3} \,&-&\, \frac{1}{3}x_{4} \\
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x_{5} \,&=&\, \frac{8}{3} \,&-&\, \frac{8}{3}x_{1} \,&-&\, \frac{29}{3}x_{3} \,&+&\, \frac{5}{3}x_{4} \\
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x_{6} \,&=&\, 2 && \,&-&\, 7x_{3} \,&+&\, x_{4} \\
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x_{7} \,&=&\, \frac{8}{3} \,&-&\, \frac{2}{3}x_{1} \,&-&\, \frac{5}{3}x_{3} \,&-&\, \frac{1}{3}x_{4} \\ \cline{1 - 9}
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z &=& \frac{22}{3} \,&-&\, \frac{4}{3}x_{1} \,&+&\, \frac{29}{3}x_{3} \,&-&\, \frac{11}{3}x_{4}
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\end{alignat*}
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\underline{2. Iteration}:
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Eingangsvariable: $x_{3}$ \\
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Ausgangsvariable: $x_{5}$
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Es folgt
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\begin{alignat*}{2}
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\frac{29}{3}x_{3} &=&& \frac{8}{3} - \frac{8}{3}x_{1} + \frac{5}{3}x_{4} - x_{5} \\
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x_{3} &=&& \frac{8}{29} - \frac{8}{29}x_{1} + \frac{5}{29}x_{4} - \frac{3}{29}x_{5} \\
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x_{2} &=&& \frac{2}{3} + \frac{1}{3}x_{1} + \frac{4}{3}\left(\frac{8}{29} - \frac{8}{29}x_{1} + \frac{5}{29}x_{4} - \frac{3}{29}x_{5}\right) - \frac{1}{3}x_{4} \\
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&=&& \frac{2}{3} + \frac{1}{3}x_{1} + \frac{32}{87} - \frac{32}{87}x_{1} + \frac{20}{87}x_{4} - \frac{4}{29}x_{5} - \frac{1}{3}x_{4} \\
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&=&& \frac{30}{29} - \frac{1}{29}x_{1} - \frac{1}{29}x_{4} - \frac{4}{29}x_{5} \\
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x_{6} &=&& 2 - 7\left(\frac{8}{29} - \frac{8}{29}x_{1} + \frac{5}{29}x_{4} - \frac{3}{29}x_{5}\right) + x_{4} \\
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&=&& 2 - \frac{56}{29} + \frac{56}{29}x_{1} - \frac{35}{29}x_{4} + \frac{21}{29}x_{5} + x_{4} \\
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&=&& \frac{2}{29} + \frac{56}{29}x_{1} - \frac{6}{29}x_{4} + \frac{21}{29}x_{5} \\
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x_{7} &=&& \frac{8}{3} - \frac{2}{3}x_{1} - \frac{5}{3}\left(\frac{8}{29} - \frac{8}{29}x_{1} + \frac{5}{29}x_{4} - \frac{3}{29}x_{5}\right) - \frac{1}{3}x_{4} \\
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&=&& \frac{8}{3} - \frac{2}{3}x_{1} - \frac{40}{87} + \frac{40}{87}x_{1} - \frac{25}{87}x_{4} + \frac{5}{29}x_{5} - \frac{1}{3}x_{4} \\
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&=&& \frac{64}{29} - \frac{6}{29}x_{1} - \frac{18}{29}x_{4} + \frac{5}{29}x_{5} \\
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z &=&& \frac{22}{3} - \frac{4}{3}x_{1} + \frac{29}{3}\left(\frac{8}{29} - \frac{8}{29}x_{1} + \frac{5}{29}x_{4} - \frac{3}{29}x_{5}\right) - \frac{11}{3}x_{4} \\
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&=&& \frac{22}{3} - \frac{4}{3}x_{1} + \frac{8}{3} - \frac{8}{3}x_{1} + \frac{5}{3}x_{4} - x_{5} - \frac{11}{3}x_{4} \\
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&=&& 10 + \frac{4}{3}x_{1} - 2x_{4} - x_{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_{3} \,&=&\, \frac{8}{29} \,&-&\, \frac{8}{29}x_{1} \,&+&\, \frac{5}{29}x_{4} \,&-&\, \frac{3}{29}x_{5} \\
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x_{2} \,&=&\, \frac{30}{29} \,&-&\, \frac{1}{29}x_{1} \,&-&\, \frac{1}{29}x_{4} \,&-&\, \frac{4}{29}x_{5} \\
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x_{6} \,&=&\, \frac{2}{29} \,&+&\, \frac{56}{29}x_{1} \,&-&\, \frac{6}{29}x_{4} \,&+&\, \frac{21}{29}x_{5} \\
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x_{7} \,&=&\, \frac{64}{29} \,&-&\, \frac{6}{29}x_{1} \,&-&\, \frac{18}{29}x_{4} \,&+&\, \frac{5}{29}x_{5} \\ \cline{1 - 9}
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z &=& 10 \,&+&\, \frac{4}{3}x_{1} \,&-&\, 2x_{4} \,&-&\, x_{5}
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\end{alignat*}
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\underline{3. Iteration}:
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Eingangsvariable: $x_{1}$\\
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Ausgangsvariable: $x_{3}$
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Es folgt
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\begin{alignat*}{2}
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\frac{8}{29}x_{1} &=&& \frac{8}{29} + \frac{5}{29}x_{4} - \frac{3}{29}x_{5} - x_{3} \\
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x_{1} &=&& 1 + \frac{5}{8}x_{4} - \frac{3}{8}x_{5} - \frac{29}{8}x_{3} \\
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x_{2} &=&& \frac{30}{29} - \frac{1}{29}\left(1 + \frac{5}{8}x_{4} - \frac{3}{8}x_{5} - \frac{29}{8}x_{3}\right) - \frac{1}{29}x_{4} - \frac{4}{29}x_{5} \\
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&=&& \frac{30}{29} - \frac{1}{29} - \frac{5}{232}x_{4} + \frac{3}{232}x_{5} + \frac{1}{8}x_{3} - \frac{1}{29}x_{4} - \frac{4}{29}x_{5} \\
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&=&& 1 - \frac{13}{232}x_{4} - \frac{1}{8}x_{5} + \frac{1}{8}x_{3} \\
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x_{6} &=&& \frac{2}{29} + \frac{56}{29}\left(1 + \frac{5}{8}x_{4} - \frac{3}{8}x_{5} - \frac{29}{8}x_{3}\right) - \frac{6}{29}x_{4} + \frac{21}{29}x_{5} \\
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&=&& \frac{2}{29} + \frac{56}{29} + \frac{35}{29}x_{4} - \frac{21}{29}x_{5} - 7x_{3} - \frac{6}{29}x_{4} + \frac{21}{29}x_{5} \\
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&=&& 2 + x_{4} - 7x_{3} \\
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x_{7} &=&& \frac{64}{29} - \frac{6}{29}\left(1 + \frac{5}{8}x_{4} - \frac{3}{8}x_{5} - \frac{29}{8}x_{3}\right) - \frac{18}{29}x_{4} + \frac{5}{29}x_{5} \\
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&=&& \frac{64}{29} - \frac{6}{29} - \frac{15}{116}x_{4} + \frac{18}{232}x_{5} + \frac{3}{4}x_{3} - \frac{18}{29}x_{4} + \frac{5}{29}x_{5} \\
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&=&& 2 - \frac{3}{4}x_{4} + \frac{1}{4}x_{5} + \frac{3}{4}x_{3} \\
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z &=&& 10 + \frac{4}{3}\left(1 + \frac{5}{8}x_{4} - \frac{3}{8}x_{5} - \frac{29}{8}x_{3}\right) - 2x_{4} - x_{5} \\
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&=&& 10 + \frac{4}{3} + \frac{5}{6}x_{4} - \frac{1}{2}x_{5} - \frac{29}{6}x_{3} - 2x_{4} - x_{5} \\
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&=&& \frac{34}{3} - \frac{7}{6}x_{4} - \frac{3}{2}x_{5} - \frac{29}{6}x_{3}
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\end{alignat*}
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\underline{Ergebnis der 3. Iteration}:
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\begin{alignat*}{5}
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x_{1} \,&=&\, 1 \,&+&\, \frac{5}{8}x_{4} \,&-&\, \frac{3}{8}x_{5} \,&-&\, \frac{29}{8}x_{3} \\
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x_{2} \,&=&\, 1 \,&-&\, \frac{13}{232}x_{4} \,&-&\, \frac{1}{8}x_{5} \,&+&\, \frac{1}{8}x_{3} \\
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x_{6} \,&=&\, 2 \,&+&\, x_{4} && \,&-&\, 7x_{3} \\
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x_{7} \,&=&\, 2 \,&-&\, \frac{3}{4}x_{4} \,&+&\, \frac{1}{4}x_{5} \,&+&\, \frac{3}{4}x_{3} \\ \cline{1 - 9}
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z &=& \frac{34}{3} \,&-&\, \frac{7}{6}x_{4} \,&-&\, \frac{3}{2}x_{5} \,&-&\, \frac{29}{6}x_{3}
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\end{alignat*}
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Dieses Tableau liefert die optimale Lösung $x_{1} = 1, x_{2} = 1, x_{3} = 0$ mit $z = \frac{34}{3}$.
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\underline{Startlösung ("`zulässige Basislösung am Anfang"')}:
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\[
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x_{1} = 0, x_{2} = 0, x_{3} = 0, x_{4} = 2, x_{5} = 6, x_{6} = 4, x_{7} = 2 \text{ mit } z = 0
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\]
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\underline{Zulässige Basislösung nach der 1. Iteration}:
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\[
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x_{1} = 0, x_{2} = \frac{2}{3}, x_{3} = 0, x_{4} = 0, x_{5} = \frac{8}{3}, x_{6} = 2, x_{7} = \frac{8}{3} \text{ mit } z = \frac{22}{3} = 7\frac{1}{3}
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\]
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\underline{Zulässige Basislösung nach der 2. Iteration}:
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\[
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x_{1} = 0, x_{2} = \frac{30}{29}, x_{3} = \frac{8}{29}, x_{4} = 0, x_{5} = 0, x_{6} = 2, x_{7} = \frac{64}{29} \text{ mit } z = 10
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\]
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\underline{Zulässige Basislösung nach der 3. Iteration}:
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\[
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x_{1} = 1, x_{2} = 1, x_{3} = 0, x_{4} = 0, x_{5} = 0, x_{6} = 2, x_{7} = 2 \text{ mit } z = \frac{34}{3} = 11\frac{1}{3}
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\]
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\section{} %2
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\end{document}
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