2022-02-14 13:31:50 +01:00
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\documentclass[a4paper,french,12pt]{article}
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\title{Optimisation et complexité --- DM1}
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\author{}
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\date{Dernière compilation~: \today{} à \currenttime}
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\usepackage{styles}
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\usepackage{xfrac}
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\usepackage{tikz}
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\usetikzlibrary{shapes.multipart}
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\usetikzlibrary{automata, arrows.meta, positioning}
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2022-02-14 15:53:46 +01:00
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\usepackage{xcolor,colortbl}
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\definecolor{Red}{rgb}{0.89,0.45,0.36}
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\newcolumntype{r}{>{\columncolor{Red}}Y}
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\begin{document}
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\maketitle
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2022-02-14 15:53:46 +01:00
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\section{TD3 --- Exercice 3}
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2022-02-14 13:31:50 +01:00
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On considère le programme linéaire suivant~:
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\begin{align*}
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\text{Max} Z = x_1 - x_2 \\
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\left\{
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\begin{array}{l}
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2x_1 - x_2 \geq -4 \\
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2022-02-14 13:31:50 +01:00
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x_1 - x_2 \leq 4 \\
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x_1 + x_2 \leq 10 \\
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x_1, x_2 \geq 0 \\
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\end{array}
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\right.
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\end{align*}
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\subsection{}
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Résoudre le programme par la méthode du simplexe, et déduire la solution optimale $S_1$.
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2022-02-14 13:31:50 +01:00
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2022-02-14 15:53:46 +01:00
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\begin{align*}
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\left\{
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\begin{array}{l}
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-2x_1 + x_2 + x_3 = 4 \\
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x_1 - x_2 + x_4 = 4 \\
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x_1 + x_2 + x_5 = 10 \\
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x_1, x_2, x_3, x_4, x_5 \geq 0 \\
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\end{array}
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\right.
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\end{align*}
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\begin{tabularx}{\linewidth}{|Y|Y|Y|Y|Y|Y|Y|Y|Y|}
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\hline
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Max & \multicolumn{2}{|c|}{$C_i$} & 1 & -1 & 0 & 0 & 0 \\
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\hline
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$C_B$ & B & b & $x_1$ & $x_2$ & $x_3$ & $x_4$ & $x_5$ \\
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\hline
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0 & $x_4$ & 4 & -2 & 1 & 1 & 0 & 0 \\
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0 & $x_5$ & 4 & 1 & -1 & 0 & 1 & 0 \\
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0 & $x_6$ & 10 & 1 & 1 & 0 & 0 & 1 \\
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\hline
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\multicolumn{2}{|c|}{$Z_i$} & 0 & 0 & 0 & 0 & 0 & 0 \\
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\hline
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\multicolumn{3}{|c|}{$C_i - Z_i$} & 1 & -1 & 0 & 0 & 0 \\
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\hline
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\end{tabularx}
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\begin{tabularx}{\linewidth}{|Y|Y|Y|r|Y|Y|Y|Y|Y|}
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\hline
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Max & \multicolumn{2}{|c|}{$C_i$} & 1 & -1 & 0 & 0 & 0 \\
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\hline
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$C_B$ & B & b & $x_1$ & $x_2$ & $x_3$ & $x_4$ & $x_5$ \\
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\hline
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0 & $x_4$ & 12 & 0 & -1 & 1 & 2 & 0 \\
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\rowcolor{Red} 1 & $x_1$ & 4 & 1 & -1 & 0 & 1 & 0 \\
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0 & $x_6$ & 6 & 0 & 2 & 0 & -1 & 1 \\
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\hline
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\multicolumn{2}{|c|}{$Z_i$} & 4 & 1 & -1 & 0 & 1 & 0 \\
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\hline
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\multicolumn{3}{|c|}{$C_i - Z_i$} & 0 & -2 & 0 & -1 & 0 \\
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\hline
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\end{tabularx}
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La solution optimale est donc $S_1 = 4$.
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\end{document}
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