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MATH2-Inf-4: Blatt 4 gelöst.
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Jim Martens
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\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
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\usepackage[T1]{fontenc}
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\usepackage[utf8]{inputenc}
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\usepackage[ngerman]{babel}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{paralist}
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\usepackage{gauss}
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\usepackage{pgfplots}
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\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
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\usepackage{tikz}
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\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
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\usepackage{polynom}
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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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\let\@ifnextchar\new@ifnextchar
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\array{#1}}
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\makeatother
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\begin{document}
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\author{Jan Branitz (6326955), Jim Martens (6420323),\\
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Stephan Niendorf (6242417)}
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\title{Hausaufgaben zum 11. November}
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\maketitle
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\section{} %1
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\subsection{} %a
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\subsubsection{} %i
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Regel vom größten Koeffizienten:
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\textbf{Aufgabe:} Lösen Sie das folgende LP-Problem mit dem Simplexverfahren:
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\begin{alignat*}{3}
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\text{maximiere}\; & 2x_{1} \,&+&\, x_{2} && \\
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\multicolumn{6}{l}{\text{unter den Nebenbedingungen}} && \\
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\;& x_{1} \,&+&\, x_{2} \,&\leq & 9 \\
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\;& x_{1} \,&& &\leq & 2 \\
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\;& &&\, x_{2} \,&\leq & 8 \\
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\multicolumn{4}{r}{$x_{1}, x_{2}$} \,&\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*}{4}
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x_{3} \,&=&\, 9 \,&-&\, x_{1} \,&-&\, x_{2} \\
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x_{4} \,&=&\, 2 \,&-&\, x_{1} && \\
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x_{5} \,&=&\, 8 \,&& &-&\, x_{2} \\ \cline{1 - 7}
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z &=& &&\, 2x_{1} \,&+&\, x_{2}
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\end{alignat*}
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\underline{1. 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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x_{1} \,&=&&\, 2 - x_{4} \\
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x_{3} \,&=&&\, 9 - \left(2 - x_{4}\right) - x_{2} \\
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&=&&\, 9 - 2 + x_{4} - x_{2} \\
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&=&&\, 7 - x_{2} + x_{4} \\
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x_{5} \,&=&&\, 8 - x_{2} \\
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z \,&=&&\, 2\left(2 - x_{4}\right) + x_{2} \\
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&=&&\, 4 - 2x_{4} + x_{2} \\
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&=&&\, 4 + x_{2} - 2x_{4}
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\end{alignat*}
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\underline{Ergebnis der 1. Iteration}:
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\begin{alignat*}{4}
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x_{1} \,&=&\, 2 \,&& &-&\, x_{4} \\
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x_{3} \,&=&\, 7 \,&-&\, x_{2} \,&+&\, x_{4} \\
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x_{5} \,&=&\, 8 \,&-&\, x_{2} \,&& \\ \cline{1 - 7}
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z &=& 4 \,&+&\, x_{2} \,&-&\, 2x_{4}
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\end{alignat*}
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\underline{2. Iteration}:
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Eingangsvariable: $x_{2}$ \\
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Ausgangsvariable: $x_{3}$
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Es folgt
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\begin{alignat*}{2}
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x_{2} \,&=&&\, 7 + x_{4} - x_{3} \\
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x_{1} \,&=&&\, 2 - x_{4}\\
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x_{5} \,&=&&\, 8 - \left(7 + x_{4} - x_{3}\right) \\
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&=&&\, 8 - 7 - x_{4} + x_{3} \\
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&=&&\, 1 - x_{4} + x_{3} \\
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z \,&=&&\, 4 + \left(7 + x_{4} - x_{3}\right) - 2x_{4} \\
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&=&&\, 4 + 7 + x_{4} - x_{3} - 2x_{4} \\
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&=&&\, 11 - x_{4} - x_{3}
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\end{alignat*}
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\underline{Ergebnis der 2. Iteration}:
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\begin{alignat*}{4}
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x_{2} \,&=&\, 7 \,&+&\, x_{4} \,&-&\, x_{3} \\
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x_{1} \,&=&\, 2 \,&-&\, x_{4} \,&& \\
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x_{5} \,&=&\, 1 \,&-&\, x_{4} \,&+&\, x_{3} \\ \cline{1 - 7}
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z &=& 11 \,&-&\, x_{4} \,&-&\, x_{3}
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\end{alignat*}
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Dieses Tableau liefert die optimale Lösung $x_{1} = 2, x_{2} = 7$ mit $z = 11$.
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\underline{Startlösung ("`zulässige Basislösung am Anfang"')}:
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\[
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x_{1} = 0, x_{2} = 0, x_{3} = 9, x_{4} = 2, x_{5} = 8 \text{ mit } z = 0
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\]
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\underline{Zulässige Basislösung nach der 1. Iteration}:
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\[
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x_{1} = 2, x_{2} = 0, x_{3} = 7, x_{4} = 0, x_{5} = 8 \text{ mit } z = 4
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\]
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\underline{Zulässige Basislösung nach der 2. Iteration}:
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\[
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x_{1} = 2, x_{2} = 7, x_{3} = 0, x_{4} = 0, x_{5} = 1 \text{ mit } z = 11
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\]
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Regel vom größten Zuwachs:
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\textbf{Aufgabe:} Lösen Sie das folgende LP-Problem mit dem Simplexverfahren:
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\begin{alignat*}{3}
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\text{maximiere}\; & 2x_{1} \,&+&\, x_{2} && \\
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\multicolumn{6}{l}{\text{unter den Nebenbedingungen}} && \\
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\;& x_{1} \,&+&\, x_{2} \,&\leq & 9 \\
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\;& x_{1} \,&& &\leq & 2 \\
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\;& &&\, x_{2} \,&\leq & 8 \\
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\multicolumn{4}{r}{$x_{1}, x_{2}$} \,&\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*}{4}
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x_{3} \,&=&\, 9 \,&-&\, x_{1} \,&-&\, x_{2} \\
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x_{4} \,&=&\, 2 \,&-&\, x_{1} && \\
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x_{5} \,&=&\, 8 \,&& &-&\, x_{2} \\ \cline{1 - 7}
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z &=& &&\, 2x_{1} \,&+&\, x_{2}
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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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x_{2} \,&=&&\, 8 - x_{5} \\
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x_{3} \,&=&&\, 9 - x_{1} - \left(8 - x_{5}\right) \\
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&=&&\, 9 - x_{1} - 8 + x_{5} \\
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&=&&\, 1 - x_{1} + x_{5} \\
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x_{4} \,&=&&\, 2 - x_{1} \\
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z \,&=&&\, 2x_{1} + \left(8 - x_{5}\right) \\
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&=&&\, 2x_{1} + 8 - x_{5} \\
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&=&&\, 8 + 2x_{1} - x_{5}
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\end{alignat*}
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\underline{Ergebnis der 1. Iteration}:
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\begin{alignat*}{4}
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x_{2} \,&=&\, 8 \,&& &-&\, x_{5} \\
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x_{3} \,&=&\, 1 \,&-&\, x_{1} \,&+&\, x_{5} \\
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x_{4} \,&=&\, 2 \,&-&\, x_{1} \,&& \\ \cline{1 - 7}
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z &=& 8 \,&+&\, 2x_{1} \,&-&\, x_{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: $x_{3}$
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Es folgt
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\begin{alignat*}{2}
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x_{1} \,&=&&\, 1 + x_{5} - x_{3} \\
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x_{2} \,&=&&\, 8 - x_{5}\\
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x_{4} \,&=&&\, 2 - \left(1 + x_{5} - x_{3}\right) \\
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&=&&\, 2 - 1 - x_{5} + x_{3} \\
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&=&&\, 1 - x_{5} + x_{3} \\
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z \,&=&&\, 8 + 2\left(1 + x_{5} - x_{3}\right) - x_{5} \\
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&=&&\, 8 + 2 + 2x_{5} - 2x_{3} - x_{5} \\
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&=&&\, 10 + x_{5} - 2x_{3}
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\end{alignat*}
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\underline{Ergebnis der 2. Iteration}:
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\begin{alignat*}{4}
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x_{1} \,&=&\, 1 \,&+&\, x_{5} \,&-&\, x_{3} \\
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x_{2} \,&=&\, 8 \,&-&\, x_{5} \,&& \\
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x_{4} \,&=&\, 1 \,&-&\, x_{5} \,&+&\, x_{3} \\ \cline{1 - 7}
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z &=& 10 \,&+&\, x_{5} \,&-&\, 2x_{3}
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\end{alignat*}
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\underline{3. Iteration}:
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Eingangsvariable: $x_{5}$ \\
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Ausgangsvariable: $x_{4}$
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Es folgt
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\begin{alignat*}{2}
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x_{5} \,&=&&\, 1 + x_{3} - x_{4} \\
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x_{1} \,&=&&\, 1 + \left(1 + x_{3} - x_{4}\right) - x_{3} \\
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&=&&\, 1 + 1 + x_{3} - x_{4} - x_{3} \\
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&=&&\, 2 - x_{4} \\
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x_{2} \,&=&&\, 8 - \left(1 + x_{3} - x_{4}\right) \\
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&=&&\, 8 - 1 - x_{3} + x_{4} \\
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&=&&\, 7 - x_{3} + x_{4} \\
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z \,&=&&\, 10 + \left(1 + x_{3} - x_{4}\right) - 2x_{3} \\
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&=&&\, 10 + 1 + x_{3} - x_{4} - 2x_{3} \\
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&=&&\, 11 - x_{3} - x_{4}
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\end{alignat*}
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\underline{Ergebnis der 3. Iteration}:
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\begin{alignat*}{4}
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x_{5} \,&=&\, 1 \,&+&\, x_{3} \,&-&\, x_{4} \\
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x_{1} \,&=&\, 2 \,&& &-&\, x_{4} \\
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x_{2} \,&=&\, 7 \,&-&\, x_{3} \,&+&\, x_{4} \\ \cline{1 - 7}
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z &=& 11 \,&-&\, x_{3} \,&-&\, x_{4}
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\end{alignat*}
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Dieses Tableau liefert die optimale Lösung $x_{1} = 2, x_{2} = 7$ mit $z = 11$.
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\underline{Startlösung ("`zulässige Basislösung am Anfang"')}:
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\[
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x_{1} = 0, x_{2} = 0, x_{3} = 9, x_{4} = 2, x_{5} = 8 \text{ mit } z = 0
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\]
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\underline{Zulässige Basislösung nach der 1. Iteration}:
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\[
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x_{1} = 0, x_{2} = 8, x_{3} = 1, x_{4} = 2, x_{5} = 0 \text{ mit } z = 8
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\]
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\underline{Zulässige Basislösung nach der 2. Iteration}:
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\[
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x_{1} = 1, x_{2} = 8, x_{3} = 0, x_{4} = 1, x_{5} = 0 \text{ mit } z = 10
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\]
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\underline{Zulässige Basislösung nach der 3. Iteration}:
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\[
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x_{1} = 2, x_{2} = 7, x_{3} = 0, x_{4} = 0, x_{5} = 1 \text{ mit } z = 11
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\]
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Die Regel vom größten Koeffizienten schneidet besser ab und ist eine Iteration früher fertig.
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\subsubsection{} %ii
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\begin{tikzpicture}[>=stealth]
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\begin{axis}[
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ymin=0,ymax=10,
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x=1cm,
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y=1cm,
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axis x line=middle,
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axis y line=middle,
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axis line style=->,
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xlabel={$x_{1}$},
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ylabel={$x_{2}$},
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xmin=0,xmax=4
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]
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\addplot[no marks, black, -] expression[domain=0:3,samples=100]{-1*x + 9} node[pos=0.65,anchor=north]{};
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\addplot[no marks, black, -] expression[domain=0:3,samples=100]{8} node[pos=0.65,anchor=north]{};
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\node at (axis cs: 0.25,8.25) {P};
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\node at (axis cs: 1,8.25) {Q};
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\node at (axis cs: 2.25,7.25) {R};
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\node at (axis cs: 2.25,0.25) {S};
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\node at (axis cs: 0.25,0.25) {T};
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\draw[>=stealth] (axis cs:2,0) -- (axis cs:2,10) node [pos=0.65,anchor=north]{};
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\end{axis}
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\end{tikzpicture}
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Bei der Regel des größten Koeffizienten wurden die Eckpunkte in der Reihenfolge T, S und R durchlaufen. Bei der Regel des größten Zuwachses wurden die Eckpunkte in der Reihenfolge T, P, Q und R durchlaufen.
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\subsection{} %b
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\subsubsection{} %i
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Regel vom größten Koeffizienten:
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\textbf{Aufgabe:} Lösen Sie das folgende LP-Problem mit dem Simplexverfahren:
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\begin{alignat*}{3}
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\text{maximiere}\; & x_{1} \,&+&\, 2x_{2} && \\
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\multicolumn{6}{l}{\text{unter den Nebenbedingungen}} && \\
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\;& x_{1} \,&+&\, 4x_{2} \,&\leq & 4 \\
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\multicolumn{4}{r}{$x_{1}, x_{2}$} \,&\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*}{4}
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x_{3} \,&=&\, 4 \,&-&\, x_{1} \,&-&\, 4x_{2} \\ \cline{1 - 7}
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z &=& &&\, x_{1} \,&+&\, 4x_{2}
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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: $x_{3}$
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Es folgt
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\begin{alignat*}{2}
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4x_{2} \,&=&&\, 4 - x_{1} - x_{3} \\
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x_{2} \,&=&&\, 1 - \frac{1}{4}x_{1} - \frac{1}{4}x_{3} \\
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z \,&=&&\, x_{1} + 4\left(1 - \frac{1}{4}x_{1} - \frac{1}{4}x_{3}\right) \\
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&=&&\, x_{1} + 4 - x_{1} - x_{3} \\
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&=&&\, 4 - x_{3}
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\end{alignat*}
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\underline{Ergebnis der 1. Iteration}:
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\begin{alignat*}{4}
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x_{2} \,&=&\, 1 \,&-&\, \frac{1}{4}x_{1} &-&\, \frac{1}{4}x_{3} \\ \cline{1 - 7}
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z &=& 4 \,&& &-&\, x_{3}
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\end{alignat*}
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Dieses Tableau liefert die optimale Lösung $x_{1} = 0, x_{2} = 1$ mit $z = 4$.
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\underline{Startlösung ("`zulässige Basislösung am Anfang"')}:
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\[
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x_{1} = 0, x_{2} = 0, x_{3} = 4 \text{ mit } z = 0
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\]
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\underline{Zulässige Basislösung nach der 1. Iteration}:
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\[
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x_{1} = 0, x_{2} = 1, x_{3} = 0 \text{ mit } z = 4
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\]
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Regel vom größten Zuwachs:
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\textbf{Aufgabe:} Lösen Sie das folgende LP-Problem mit dem Simplexverfahren:
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\begin{alignat*}{3}
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\text{maximiere}\; & x_{1} \,&+&\, 2x_{2} && \\
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\multicolumn{6}{l}{\text{unter den Nebenbedingungen}} && \\
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\;& x_{1} \,&+&\, 4x_{2} \,&\leq & 4 \\
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\multicolumn{4}{r}{$x_{1}, x_{2}$} \,&\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*}{4}
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x_{3} \,&=&\, 4 \,&-&\, x_{1} \,&-&\, 4x_{2} \\ \cline{1 - 7}
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z &=& &&\, x_{1} \,&+&\, 4x_{2}
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\end{alignat*}
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\underline{1. 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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x_{1} \,&=&&\, 4 - 4x_{2} - x_{3} \\
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z \,&=&&\, \left(4 - 4x_{2} - x_{3}\right) + 4x_{2} \\
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&=&&\, 4 - 4x_{2} - x_{3} + 4x_{2} \\
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&=&&\, 4 - x_{3}
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\end{alignat*}
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\underline{Ergebnis der 1. Iteration}:
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\begin{alignat*}{4}
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x_{1} \,&=&\, 4 \,&-&\, 4x_{2} \,&-&\, x_{3} \\ \cline{1 - 7}
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z &=& 4 \,&& &-&\, x_{3}
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\end{alignat*}
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Dieses Tableau liefert die optimale Lösung $x_{1} = 4, x_{2} = 0$ mit $z = 4$.
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\underline{Startlösung ("`zulässige Basislösung am Anfang"')}:
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\[
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x_{1} = 0, x_{2} = 0, x_{3} = 4 \text{ mit } z = 0
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\]
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\underline{Zulässige Basislösung nach der 1. Iteration}:
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\[
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x_{1} = 4, x_{2} = 0, x_{3} = 0 \text{ mit } z = 4
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\]
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Beide Regeln sind nach der gleichen Anzahl Iterationen fertig, womit es keine bessere Regel gibt.
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\subsubsection{} %ii
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\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=2,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x_{1}$},
|
||||
ylabel={$x_{2}$},
|
||||
xmin=0,xmax=5
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=0:4,samples=100]{-0.25*x + 1} node[pos=0.65,anchor=north]{};
|
||||
%\addplot[no marks, black, -] expression[domain=0:3,samples=100]{8} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 0.25,1.25) {P};
|
||||
\node at (axis cs: 4.25,0.25) {Q};
|
||||
\node at (axis cs: 0.25,0.25) {R};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
|
||||
Bei der Regel mit dem größten Koeffizienten wurden die Eckpunkte in der Reihenfolge R und P durchlaufen. Bei der Regel mit dem größten Zuwachs wurden die Eckpunkte in der Reihenfolge R und Q durchlaufen.
|
||||
\section{} %2
|
||||
|
||||
\textbf{Aufgabe:} Lösen Sie das folgende LP-Problem mit dem Simplexverfahren:
|
||||
\begin{alignat*}{4}
|
||||
\text{maximiere}\; & 8x_{1} \,&-&\, x_{2} \,&+&\, 2x_{3} && \\
|
||||
\multicolumn{8}{l}{\text{unter den Nebenbedingungen}} && \\
|
||||
\;& x_{1} \,&& && &\leq & 2 \\
|
||||
\;& 2x_{1} \,&+&\, 3x_{2} \,&-&\, x_{3} \,&\leq & 4 \\
|
||||
\;& 3x_{1} \,&-&\, 4x_{2} \,&+&\, 2x_{3} \,&\leq & 6 \\
|
||||
\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} && && \\
|
||||
x_{5} \,&=&\, 4 \,&-&\, 2x_{1} \,&-&\, 3x_{2} \,&+&\, x_{3} \\
|
||||
x_{6} \,&=&\, 6 \,&-&\, 3x_{1} \,&+&\, 4x_{2} \,&-&\, 2x_{3} \\ \cline{1 - 9}
|
||||
z &=& &&\, 8x_{1} \,&-&\, x_{2} \,&+&\, 2x_{3}
|
||||
\end{alignat*}
|
||||
|
||||
\underline{1. Iteration}:
|
||||
|
||||
Eingangsvariable: $x_{1}$\\
|
||||
Ausgangsvariable: $x_{4}$
|
||||
|
||||
Es folgt
|
||||
\begin{alignat*}{2}
|
||||
x_{1} \,&=&&\, 2 - x_{4} \\
|
||||
x_{5} \,&=&&\, 4 - 2\left(2 - x_{4}\right) - 3x_{2} + x_{3} \\
|
||||
&=&&\, 4 - 4 + 2x_{4} - 3x_{2} + x_{3} \\
|
||||
&=&&\, - 3x_{2} + x_{3} + 2x_{4} \\
|
||||
x_{6} \,&=&&\, 6 - 3\left(2 - x_{4}\right) + 4x_{2} - 2x_{3} \\
|
||||
&=&&\, 6 - 6 + 3x_{4} + 4x_{2} - 2x_{3} \\
|
||||
&=&&\, 4x_{2} - 2x_{3} + 3x_{4} \\
|
||||
z \,&=&&\, 8\left(2 - x_{4}\right) - x_{2} + 2x_{3} \\
|
||||
&=&&\, 16 - 8x_{4} - x_{2} + 2x_{3} \\
|
||||
&=&&\, 16 - x_{2} + 2x_{3} - 8x_{4}
|
||||
\end{alignat*}
|
||||
|
||||
\underline{Ergebnis der 1. Iteration}:
|
||||
\begin{alignat*}{5}
|
||||
x_{1} \,&=&\, 2 \,&& && &-&\, x_{4} \\
|
||||
x_{5} \,&=&\, 0 \,&-&\, 3x_{2} \,&+&\, x_{3} \,&+&\, 2x_{4} \\
|
||||
x_{6} \,&=&\, 0 \,&+&\, 4x_{2} \,&-&\, 2x_{3} \,&+&\, 3x_{4} \\ \cline{1 - 9}
|
||||
z &=& 16 \,&-&\, x_{2} \,&+&\, 2x_{3} \,&-&\, 8x_{4}
|
||||
\end{alignat*}
|
||||
|
||||
\underline{2. Iteration}:
|
||||
|
||||
Eingangsvariable: $x_{3}$ \\
|
||||
Ausgangsvariable: $x_{6}$
|
||||
|
||||
Es folgt
|
||||
\begin{alignat*}{2}
|
||||
2x_{3} \,&=&&\, 4x_{2} + 3x_{4} - x_{6} \\
|
||||
x_{3} \,&=&&\, 2x_{2} + \frac{3}{2}x_{4} - \frac{1}{2}x_{6} \\
|
||||
x_{1} \,&=&&\, 2 - x_{4}\\
|
||||
x_{5} \,&=&&\, - 3x_{2} + \left(2x_{2} + \frac{3}{2}x_{4} - \frac{1}{2}x_{6}\right) + 2x_{4} \\
|
||||
&=&&\, - 3x_{2} + 2x_{2} + \frac{3}{2}x_{4} - \frac{1}{2}x_{6} + 2x_{4} \\
|
||||
&=&&\, - x_{2} + \frac{7}{2}x_{4} - \frac{1}{2}x_{6} \\
|
||||
z \,&=&&\, 16 - x_{2} + 2\left(2x_{2} + \frac{3}{2}x_{4} - \frac{1}{2}x_{6}\right) - 8x_{4} \\
|
||||
&=&&\, 16 - x_{2} + 4x_{2} + 3x_{4} - x_{6} - 8x_{4} \\
|
||||
&=&&\, 16 + 3x_{2} - 5x_{4} - x_{6}
|
||||
\end{alignat*}
|
||||
|
||||
\underline{Ergebnis der 2. Iteration}:
|
||||
\begin{alignat*}{5}
|
||||
x_{3} \,&=&\, 0 \,&+&\, 2x_{2} \,&+&\, \frac{3}{2}x_{4} \,&-&\, \frac{1}{2}x_{6} \\
|
||||
x_{1} \,&=&\, 2 \,&& &-&\, x_{4} && \\
|
||||
x_{5} \,&=&\, 0 \,&-&\, x_{2} \,&+&\, \frac{7}{2}x_{4} \,&-&\, \frac{1}{2}x_{6} \\ \cline{1 - 9}
|
||||
z &=& 16 \,&+&\, 3x_{2} \,&-&\, 5x_{4} \,&-&\, x_{6}
|
||||
\end{alignat*}
|
||||
|
||||
\underline{3. Iteration}:
|
||||
|
||||
Eingangsvariable: $x_{2}$ \\
|
||||
Ausgangsvariable: $x_{5}$
|
||||
|
||||
Es folgt
|
||||
\begin{alignat*}{2}
|
||||
x_{2} \,&=&&\, \frac{7}{2}x_{4} - \frac{1}{2}x_{6} - x_{5} \\
|
||||
x_{3} \,&=&&\, 2\left(\frac{7}{2}x_{4} - \frac{1}{2}x_{6} - x_{5}\right) + \frac{3}{2}x_{4} - \frac{1}{2}x_{6} \\
|
||||
&=&&\, 7x_{4} - x_{6} - 2x_{5} + \frac{3}{2}x_{4} - \frac{1}{2}x_{6} \\
|
||||
&=&&\, \frac{17}{2}x_{4} - \frac{3}{2}x_{6} - 2x_{5} \\
|
||||
x_{1} \,&=&&\, 2 - x_{4} \\
|
||||
z \,&=&&\, 16 + 3\left(\frac{7}{2}x_{4} - \frac{1}{2}x_{6} - x_{5}\right) - 5x_{4} - x_{6} \\
|
||||
&=&&\, 16 + \frac{21}{2}x_{4} - \frac{3}{2}x_{6} - 3x_{5} - 5x_{4} - x_{6} \\
|
||||
&=&&\, 16 + \frac{11}{2}x_{4} - \frac{5}{2}x_{6} - 3x_{5}
|
||||
\end{alignat*}
|
||||
|
||||
\underline{Ergebnis der 3. Iteration}:
|
||||
\begin{alignat*}{5}
|
||||
x_{2} \,&=&\, 0 \,&+&\, \frac{7}{2}x_{4} \,&-&\, \frac{1}{2}x_{6} \,&-&\, x_{5} \\
|
||||
x_{3} \,&=&\, 0 \,&+&\, \frac{17}{2}x_{4} \,&-&\, \frac{3}{2}x_{6} \,&-&\, 2x_{5} \\
|
||||
x_{1} \,&=&\, 2 \,&-&\, x_{4} && && \\ \cline{1 - 9}
|
||||
z &=& 16 \,&+&\, \frac{11}{2}x_{4} \,&-&\, \frac{5}{2}x_{6} \,&-&\, 3x_{5}
|
||||
\end{alignat*}
|
||||
|
||||
\underline{4. Iteration}:
|
||||
|
||||
Eingangsvariable: $x_{4}$ \\
|
||||
Ausgangsvariable: $x_{1}$
|
||||
|
||||
Es folgt
|
||||
\begin{alignat*}{2}
|
||||
x_{4} \,&=&&\, 2 - x_{1} \\
|
||||
x_{2} \,&=&&\, \frac{7}{2}\left(2 - x_{1}\right) - \frac{1}{2}x_{6} - x_{5} \\
|
||||
&=&&\, 7 - \frac{7}{2}x_{1} - \frac{1}{2}x_{6} - x_{5} \\
|
||||
&=&&\, 7 - \frac{1}{2}x_{6} - x_{5} - \frac{7}{2}x_{1} \\
|
||||
x_{3} \,&=&&\, \frac{17}{2}\left(2 - x_{1}\right) - \frac{3}{2}x_{6} - 2x_{5} \\
|
||||
&=&&\, 17 - \frac{17}{2}x_{1} - \frac{3}{2}x_{6} - 2x_{5} \\
|
||||
&=&&\, 17 - \frac{3}{2}x_{6} - 2x_{5} - \frac{17}{2}x_{1} \\
|
||||
z \,&=&&\, 16 + \frac{11}{2}\left(2 - x_{1}\right) - \frac{5}{2}x_{6} - 3x_{5} \\
|
||||
&=&&\, 16 + 11 - \frac{11}{2}x_{1} - \frac{5}{2}x_{6} - 3x_{5} \\
|
||||
&=&&\, 27 - \frac{5}{2}x_{6} - 3x_{5} - \frac{11}{2}x_{1}
|
||||
\end{alignat*}
|
||||
|
||||
\underline{Ergebnis der 4. Iteration}:
|
||||
\begin{alignat*}{5}
|
||||
x_{4} \,&=&\, 2 \,&& && &-&\, x_{1} \\
|
||||
x_{2} \,&=&\, 7 \,&-&\, \frac{1}{2}x_{6} \,&-&\, x_{5} \,&-&\, \frac{7}{2}x_{1} \\
|
||||
x_{3} \,&=&\, 17 \,&-&\, \frac{3}{2}x_{6} \,&-&\, 2x_{5} \,&-&\, \frac{17}{2}x_{1} \\ \cline{1 - 9}
|
||||
z &=& 27 \,&-&\, \frac{5}{2}x_{6} \,&-&\, 3x_{5} \,&-&\, \frac{11}{2}x_{1}
|
||||
\end{alignat*}
|
||||
Dieses Tableau liefert die optimale Lösung $x_{1} = 0, x_{2} = 7, x_{3} = 17$ mit $z = 27$.
|
||||
|
||||
\underline{Startlösung ("`zulässige Basislösung am Anfang"')}:
|
||||
\[
|
||||
x_{1} = 0, x_{2} = 0, x_{3} = 0, x_{4} = 2, x_{5} = 4, x_{6} = 6 \text{ mit } z = 0
|
||||
\]
|
||||
\underline{Zulässige Basislösung nach der 1. Iteration}:
|
||||
\[
|
||||
x_{1} = 2, x_{2} = 0, x_{3} = 0, x_{4} = 0, x_{5} = 0, x_{6} = 0 \text{ mit } z = 16
|
||||
\]
|
||||
\underline{Zulässige Basislösung nach der 2. Iteration}:
|
||||
\[
|
||||
x_{1} = 2, x_{2} = 0, x_{3} = 0, x_{4} = 0, x_{5} = 0, x_{6} = 0 \text{ mit } z = 16
|
||||
\]
|
||||
\underline{Zulässige Basislösung nach der 3. Iteration}:
|
||||
\[
|
||||
x_{1} = 2, x_{2} = 0, x_{3} = 0, x_{4} = 0, x_{5} = 0, x_{6} = 0 \text{ mit } z = 16
|
||||
\]
|
||||
\underline{Zulässige Basislösung nach der 4. Iteration}:
|
||||
\[
|
||||
x_{1} = 0, x_{2} = 7, x_{3} = 17, x_{4} = 2, x_{5} = 0, x_{6} = 0 \text{ mit } z = 27
|
||||
\]
|
||||
Die erste Iteration ist kein degenerierter Schritt. Aber die zweite und dritte Iteration sind degenerierte Schritte, da sich die zulässige Basislösung nicht geändert hat. Die vierte Iteration ist kein degenerierter Schritt.
|
||||
|
||||
\end{document}
|
||||
Loading…
Reference in New Issue