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compile and add consignes exos 26 a 31 -- RO2
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src/HAI820I_Complements_RO/assets/tikz/td1_ex04_ftest.tex

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\ffigbox[\FBwidth]{%
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\caption{\centering Un graphe non orienté}\label{fig:td4_ex31_f1}
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}{
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\fbox{
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\begin{tikzpicture}[scale=1, main node/.style={circle, draw, fill=blue!20, inner sep=1pt, font=\scriptsize, minimum size=6mm, text=black}]
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% les sommets initiaux
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\node[main node] (v1) at (0,0) {\(v_1\)};
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\node[main node] (v2) at (-1,-1) {\(v_2\)};
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\node[main node] (v3) at (0,-2) {\(v_3\)};
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\node[main node] (v4) at (1,-1) {\(v_4\)};
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% les arcs avec capacités
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\draw (v1) to node[above left] {\(e_1\)} (v2);
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\draw (v2) to node[below left] {\(e_2\)} (v3);
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\draw (v2) to node[above] {\(e_3\)} (v4);
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\draw (v3) to node[below right] {\(e_4\)} (v4);
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\end{tikzpicture}
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}
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}

src/HAI820I_Complements_RO/exos/exo_26.tex

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% ----- Consignes exo 26 ----- %
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\begin{td-exo}[title]\, % 26
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\begin{td-exo}[Génération de colonnes]\,\\ % 26
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On considère le problème \(PL\) suivant:
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\begin{equation*}
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PL =
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\begin{cases}
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\min(z) = 2x_1 + 3x_2 \\
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4x_1 + x_2 - x_3 \geq 5 \\
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3x_1 + 2x_2 - x_4 \geq 6 \\
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x_1 + 2x_2 - x_5 \geq 3 \\
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x_i \geq 0, \forall i \in \{1, \ldots, 5\}
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\end{cases}
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\end{equation*}
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Commencer la méthode de génération de colonne avec les variables \(x_1, x_3\) et \(x_4\).
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Ecrire le problème restreint \(PL_R\).
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Calculer les coûts réduits.
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\end{td-exo}
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% ----- Solutions exo 26 ----- %

src/HAI820I_Complements_RO/exos/exo_27.tex

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% ----- Consignes exo 27 ----- %
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\begin{td-exo}[title]\, % 27
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\begin{td-exo}[Encadrement de la valeur optimale]\,\\ % 27
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Montrer qu'à chaque itération de la génération de colonnes on a
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\begin{equation*}
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\text{valeur}(PL) \leq \text{valeur}(PL_R)
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\end{equation*}
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et
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\begin{equation*}
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\text{score}(PL_R) + k \cdot \text{cred} \leq \text{score}(PL)
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\end{equation*}
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\(k\) est la valeur d'une solution optimale \(\sum_{j=1}^{n} x_j^\ast \leq k\) et \(\text{cred}\) est le cout reduit minimum à une itération donnée.
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\end{td-exo}
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% ----- Solutions exo 27 ----- %

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