From c251cd044f1ba8c20e4f5468e59d089bdc92db5c Mon Sep 17 00:00:00 2001 From: Jim Martens Date: Wed, 23 Oct 2013 20:56:18 +0200 Subject: [PATCH] MATH2-Inf-2: 1b geloest. --- optimierung/Uebungsblatt2.tex | 139 +++++++++++++++++++++++++++++++++- 1 file changed, 137 insertions(+), 2 deletions(-) diff --git a/optimierung/Uebungsblatt2.tex b/optimierung/Uebungsblatt2.tex index add820b..5d46eb2 100644 --- a/optimierung/Uebungsblatt2.tex +++ b/optimierung/Uebungsblatt2.tex @@ -42,8 +42,8 @@ Stephan Niendorf (6242417)} \text{maximiere}\; & x_{1} &+& 6x_{2} &-& 4x_{3} && \\ \multicolumn{8}{l}{\text{unter den Nebenbedingungen}} && \\ \;& 2x_{1} && &+& x_{3} &\leq & 5 \\ - \;-& x_{1} &+& 3x_{2} &-& 2x_{3} &\leq &\, 2 \\ - \;& && x_{2} &-& x_{3} &\leq &\, 2 \\ + \;-& x_{1} &+& 3x_{2} &-& 2x_{3} &\leq & 2 \\ + \;& && x_{2} &-& x_{3} &\leq & 2 \\ \multicolumn{6}{r}{$x_{1}, x_{2}, x_{3}$} \,&\geq &\, 0 \end{alignat*} @@ -125,5 +125,140 @@ Stephan Niendorf (6242417)} 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} \] \subsection{} %b + \textbf{Aufgabe:} Lösen Sie das folgende LP-Problem mit dem Simplexverfahren: + \begin{alignat*}{4} + \text{maximiere}\; -& 5x_{1} &+& 11x_{2} &-& 5x_{3} && \\ + \multicolumn{8}{l}{\text{unter den Nebenbedingungen}} && \\ + \;-& x_{1} &+& 3x_{2} &-& 4x_{3} &\leq & 2 \\ + \;& x_{1} &+& 5x_{2} &+& 3x_{3} &\leq & 6 \\ + \;-& x_{1} &+& 3x_{2} &+& 3x_{3} &\leq &\, 4 \\ + \;& x_{1} &-& x_{2} &+& 3x_{3} &\leq &\, 2 \\ + \multicolumn{6}{r}{$x_{1}, x_{2}, x_{3}$} \,&\geq &\, 0 + \end{alignat*} + + \textbf{Lösung.} + + \underline{Starttableau}: + \begin{alignat*}{5} + x_{4} \,&=&\, 2 \,&+&\, x_{1} \,&-&\, 3x_{2} \,&+&\, 4x_{3} \\ + x_{5} \,&=&\, 6 \,&-&\, x_{1} \,&-&\, 5x_{2} \,&-&\, 3x_{3} \\ + x_{6} \,&=&\, 4 \,&+&\, x_{1} \,&-&\, 3x_{2} \,&-&\, 3x_{3} \\ + x_{7} \,&=&\, 2 \,&-&\, x_{1} \,&+&\, x_{2} \,&-&\, 3x_{3} \\ \cline{1 - 9} + z &=& &-& 5x_{1} \,&+&\, 11x_{2} \,&-&\, 5x_{3} + \end{alignat*} + + \underline{1. Iteration}: + + Eingangsvariable: $x_{2}$\\ + Ausgangsvariable: $x_{4}$ + + Es folgt + \begin{alignat*}{2} + 3x_{2} \,&=&&\, 2 + x_{1} + 4x_{3} - x_{4} \\ + x_{2} \,&=&&\, \frac{2}{3} + \frac{1}{3}x_{1} + \frac{4}{3}x_{3} - \frac{1}{3}x_{4} \\ + 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} \\ + &=&& 6 - x_{1} - \frac{10}{3} - \frac{5}{3}x_{1} - \frac{20}{3}x_{3} + \frac{5}{3}x_{4} - 3x_{3} \\ + &=&& \frac{8}{3} - \frac{8}{3}x_{1} - \frac{29}{3}x_{3} + \frac{5}{3}x_{4} \\ + 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} \\ + &=&&\, 4 + x_{1} - 2 - x_{1} - 4x_{3} + x_{4} - 3x_{3} \\ + &=&&\, 2 - 7x_{3} + x_{4} \\ + 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} \\ + &=&& \frac{8}{3} - \frac{2}{3}x_{1} - \frac{5}{3}x_{3} - \frac{1}{3}x_{4} \\ + 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} \\ + &=&&\, -5x_{1} + \frac{22}{3} + \frac{11}{3}x_{1} + \frac{44}{3}x_{3} - \frac{11}{3}x_{4} - 5x_{3} \\ + &=&&\, \frac{22}{3} - \frac{4}{3}x_{1} + \frac{29}{3}x_{3} - \frac{11}{3}x_{4} + \end{alignat*} + + \underline{Ergebnis der 1. Iteration}: + \begin{alignat*}{5} + x_{2} \,&=&\, \frac{2}{3} \,&+&\, \frac{1}{3}x_{1} \,&+&\, \frac{4}{3}x_{3} \,&-&\, \frac{1}{3}x_{4} \\ + x_{5} \,&=&\, \frac{8}{3} \,&-&\, \frac{8}{3}x_{1} \,&-&\, \frac{29}{3}x_{3} \,&+&\, \frac{5}{3}x_{4} \\ + x_{6} \,&=&\, 2 && \,&-&\, 7x_{3} \,&+&\, x_{4} \\ + x_{7} \,&=&\, \frac{8}{3} \,&-&\, \frac{2}{3}x_{1} \,&-&\, \frac{5}{3}x_{3} \,&-&\, \frac{1}{3}x_{4} \\ \cline{1 - 9} + z &=& \frac{22}{3} \,&-&\, \frac{4}{3}x_{1} \,&+&\, \frac{29}{3}x_{3} \,&-&\, \frac{11}{3}x_{4} + \end{alignat*} + + \underline{2. Iteration}: + + Eingangsvariable: $x_{3}$ \\ + Ausgangsvariable: $x_{5}$ + + Es folgt + \begin{alignat*}{2} + \frac{29}{3}x_{3} &=&& \frac{8}{3} - \frac{8}{3}x_{1} + \frac{5}{3}x_{4} - x_{5} \\ + x_{3} &=&& \frac{8}{29} - \frac{8}{29}x_{1} + \frac{5}{29}x_{4} - \frac{3}{29}x_{5} \\ + 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} \\ + &=&& \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} \\ + &=&& \frac{30}{29} - \frac{1}{29}x_{1} - \frac{1}{29}x_{4} - \frac{4}{29}x_{5} \\ + 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} \\ + &=&& 2 - \frac{56}{29} + \frac{56}{29}x_{1} - \frac{35}{29}x_{4} + \frac{21}{29}x_{5} + x_{4} \\ + &=&& \frac{2}{29} + \frac{56}{29}x_{1} - \frac{6}{29}x_{4} + \frac{21}{29}x_{5} \\ + 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} \\ + &=&& \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} \\ + &=&& \frac{64}{29} - \frac{6}{29}x_{1} - \frac{18}{29}x_{4} + \frac{5}{29}x_{5} \\ + 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} \\ + &=&& \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} \\ + &=&& 10 + \frac{4}{3}x_{1} - 2x_{4} - x_{5} + \end{alignat*} + + \underline{Ergebnis der 2. Iteration}: + \begin{alignat*}{5} + x_{3} \,&=&\, \frac{8}{29} \,&-&\, \frac{8}{29}x_{1} \,&+&\, \frac{5}{29}x_{4} \,&-&\, \frac{3}{29}x_{5} \\ + x_{2} \,&=&\, \frac{30}{29} \,&-&\, \frac{1}{29}x_{1} \,&-&\, \frac{1}{29}x_{4} \,&-&\, \frac{4}{29}x_{5} \\ + x_{6} \,&=&\, \frac{2}{29} \,&+&\, \frac{56}{29}x_{1} \,&-&\, \frac{6}{29}x_{4} \,&+&\, \frac{21}{29}x_{5} \\ + x_{7} \,&=&\, \frac{64}{29} \,&-&\, \frac{6}{29}x_{1} \,&-&\, \frac{18}{29}x_{4} \,&+&\, \frac{5}{29}x_{5} \\ \cline{1 - 9} + z &=& 10 \,&+&\, \frac{4}{3}x_{1} \,&-&\, 2x_{4} \,&-&\, x_{5} + \end{alignat*} + + \underline{3. Iteration}: + + Eingangsvariable: $x_{1}$\\ + Ausgangsvariable: $x_{3}$ + + Es folgt + \begin{alignat*}{2} + \frac{8}{29}x_{1} &=&& \frac{8}{29} + \frac{5}{29}x_{4} - \frac{3}{29}x_{5} - x_{3} \\ + x_{1} &=&& 1 + \frac{5}{8}x_{4} - \frac{3}{8}x_{5} - \frac{29}{8}x_{3} \\ + 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} \\ + &=&& \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} \\ + &=&& 1 - \frac{13}{232}x_{4} - \frac{1}{8}x_{5} + \frac{1}{8}x_{3} \\ + 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} \\ + &=&& \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} \\ + &=&& 2 + x_{4} - 7x_{3} \\ + 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} \\ + &=&& \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} \\ + &=&& 2 - \frac{3}{4}x_{4} + \frac{1}{4}x_{5} + \frac{3}{4}x_{3} \\ + 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} \\ + &=&& 10 + \frac{4}{3} + \frac{5}{6}x_{4} - \frac{1}{2}x_{5} - \frac{29}{6}x_{3} - 2x_{4} - x_{5} \\ + &=&& \frac{34}{3} - \frac{7}{6}x_{4} - \frac{3}{2}x_{5} - \frac{29}{6}x_{3} + \end{alignat*} + + \underline{Ergebnis der 3. Iteration}: + \begin{alignat*}{5} + x_{1} \,&=&\, 1 \,&+&\, \frac{5}{8}x_{4} \,&-&\, \frac{3}{8}x_{5} \,&-&\, \frac{29}{8}x_{3} \\ + x_{2} \,&=&\, 1 \,&-&\, \frac{13}{232}x_{4} \,&-&\, \frac{1}{8}x_{5} \,&+&\, \frac{1}{8}x_{3} \\ + x_{6} \,&=&\, 2 \,&+&\, x_{4} && \,&-&\, 7x_{3} \\ + x_{7} \,&=&\, 2 \,&-&\, \frac{3}{4}x_{4} \,&+&\, \frac{1}{4}x_{5} \,&+&\, \frac{3}{4}x_{3} \\ \cline{1 - 9} + z &=& \frac{34}{3} \,&-&\, \frac{7}{6}x_{4} \,&-&\, \frac{3}{2}x_{5} \,&-&\, \frac{29}{6}x_{3} + \end{alignat*} + + Dieses Tableau liefert die optimale Lösung $x_{1} = 1, x_{2} = 1, x_{3} = 0$ mit $z = \frac{34}{3}$. + + \underline{Startlösung ("`zulässige Basislösung am Anfang"')}: + \[ + x_{1} = 0, x_{2} = 0, x_{3} = 0, x_{4} = 2, x_{5} = 6, x_{6} = 4, x_{7} = 2 \text{ mit } z = 0 + \] + \underline{Zulässige Basislösung nach der 1. Iteration}: + \[ + x_{1} = 0, x_{2} = \frac{2}{3}, x_{3} = 0, x_{4} = 0, x_{5} = \frac{8}{3}, x_{6} = 2, x_{7} = \frac{8}{3} \text{ mit } z = \frac{22}{3} = 7\frac{1}{3} + \] + \underline{Zulässige Basislösung nach der 2. Iteration}: + \[ + x_{1} = 0, x_{2} = \frac{30}{29}, x_{3} = \frac{8}{29}, x_{4} = 0, x_{5} = 0, x_{6} = 2, x_{7} = \frac{64}{29} \text{ mit } z = 10 + \] + \underline{Zulässige Basislösung nach der 3. Iteration}: + \[ + x_{1} = 1, x_{2} = 1, x_{3} = 0, x_{4} = 0, x_{5} = 0, x_{6} = 2, x_{7} = 2 \text{ mit } z = \frac{34}{3} = 11\frac{1}{3} + \] \section{} %2 \end{document}