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@@ -1574,10 +1574,13 @@ Suite à ces deux exemples, nous allons essayer de définir de façon assez thé
Il existe comme on vient de le voir deux types disctincts de problèmes d'optimisation:
1. L'optimisation continue.
-2. L'optimisation discrète (ou optimisation combinatoire).
+2. L'optimisation discrète (souvent appelée optimisation combinatoire).
### L'optimisation continue
+L'optimisation continue ou *programme mathématique continu* est un programme d'optimisation soumis à certaines contraintes.
+On peut l'exprimer de la façon suivante.
+
Soit $f:\real^n\rightarrow\real$ une fonction objectif (ou fontion de coût), on cherche $\vec x_0\in\real^n$, tel que $f(\vec x_0)\leq f(\vec x)$ pour $\vec x$ certaines conditions: les contraintes qui sont en général des égalités strictes ou des inégalités qui peuvent s'exprimer de la façon suivante.
Soient $m$ fonctions $g_i:\real^n\rightarrow\real$
\begin{align}
@@ -1602,6 +1605,431 @@ $$
On constate la présence d'un grand nombre de minima locaux qui rendent la recherche du minimum global (se trouvant en $x=0$) particulièrement compliqué à déterminer.
+L'optimisation continue est très communément utilisée en apprentissage automatique (machine learning), en particulier pour
+optimiser les poids des réseaux de neurones.
+
+### Optimisation discrète
+
+L'optimisation discrète, ou *programme mathématique discret*, est très similaire dans la définition mathématique à l'optimisation continue. L'ensemble $\real^n$ est remplcé par un ensemble discret, $\integer^n$. On peut le formuler de la façon suivante. Soit $f:\integer^n\rightarrow \real$ on cherche à résoudre
+\begin{align*}
+&\min_{\vec x\in\integer^n}f(\vec x),\\
+&g_i(\vec x)\leq 0,\quad i=1,...,m,\\
+&\mbox{pour }m\geq 0\mbox{ et }\vec x\in\integer^n.
+\end{align*}
+On voit donc qu'on a une "contrainte" supplémentaire qui est de supposer que les éléments sont tous discrets.
+
+Il existe un grand nombre d'application quotidiennes
+de l'optimisation discrète: le choix du moyen optimal de livraison d'un paquet, le meilleur réseau aérien, dans la logistique, le design de microprocesseurs, ...
+
+L'optimisation discrète est une branche des mathématiques très intéressante. Elle permet de poser des problèmes relativements simples, dont les solutions brute force sont très "simples" conceptuellement. En revanche les solutions "réalisables" dans un temps raisonnables (polynomial) sont souvent très difficiles à obtenir comme on l'a vu avec le problème du voyageur de commerce. Le problème se formule de façon simple, il existe une soltution très simple conceptuellement au problème (l'essai de toutes les possibilités), mais qui n'est pas réalisable en pratique, car déjà à 25 villes il faudrait des milliards d'années pour le résoudre.
+
+Une heuristique possible serait de penser les problèmes d'optimisation discrète comme un cas particulier des problèmes continus.
+En résumé on *relaxe* (on retire la condition que $\vec x\in\natural^n$) le problème discret en le rendant continu,
+on trouve la solution et finalement on tronque la solution pour la rendre entière.
+
+En procédant de la sorte, il est possible d'approximer de trouver une solution approchée de relativement bonne qualité, mais
+on est pas du tout certains de trouver la solution optimale.
+
+On peut s'en convaincre avec le problème suivant. On cherche
+\begin{align}
+&\max 10x_1+11x_2,\\
+&10x_1+12x_2\leq 59,\\
+&x_1,x_2\in \real\mbox{ ou } x_1,x_2\in \natural.
+\end{align}
+
+Dans le cas discret la solution de ce problème est $x_1=1$, $x_2=4$
+\begin{align}
+&10x_1+11x_2=54,\\
+&10x_1+12x_2=58\leq 59.
+\end{align}
+
+1. L'heuristique de la relaxation au cas continu, nous donne $x_1^\ast=5.9$, $x_2^\ast=0$
+
+ \begin{align}
+ &10x_1+11x_2=59,\\
+ &10x_1+12x_2=59\leq 59.
+ \end{align}
+ On aurait donc après avoir tronqué $x_1^\ast$ et $x_2^\ast$, la solution est donc $x_1=5$, $x_2=0$, on obtient
+ $$
+ 10x_1+11x_2=50,
+ $$
+ ce qui est assez différent de $x_1=1$ et $x_4=4$, où 10x_1+11x_2=58 (mais c'est pas non plus la pire solution possible).
+
+2. Il est possible d'appliquer une heuristique un peu meilleure en général. Le problème qui nous intéresse ici peut se formuler de la façon suivante en général:
+
+ \begin{align}
+ &\max m_1x_1+m_2x_2,\\
+ &v_1x_1+v_2x_2\leq V.
+ \end{align}
+ On commence par calculer "la densité" $\rho_i=m_i/v_i$ de chaque indice $i$. Ici on a $\rho_1=1$ et $\rho_2=11/12$.
+ On classe les $\rho_i$ par ordre décroissant, et on prend le maximum possible de chacun.
+ On commence donc par prendre autant de $i=1$ que possible comme il a la plus grande densité.
+ Cela nous donne $x_1=5$ et il ne reste plus la place pour des $x_2$ telle que la contrainte est satisfaire et on a terminé.
+ On a donc $x_1=5$ et $x_2=0$, ce qui est le même résultat que la relaxation continue, mais ce n'est pas toujours le cas.
+ Cette façon de procéder est au moins aussi bonne que la relaxation continue en général (elle peut donc être meilleure).
+
+## Programmation linéaire
+
+Contrairement à ce qu'on pourrait croire, la *programmation linéaire* n'a pas forcément grand chose à avoir avec l'informatique. Il s'agit d'un sous ensemble des problèmes d'optimisation mathématique. Si la fonction objective et les fonctions de contraintes sont toutes **linéaires**, alors le problème en lui est même est dit linéaire. La façon la plus générale d'écrire un programme linéaire discret, est de faire un "mélange" (le programme est dit *mixte*) entre un programme discret et un programme continu
+\begin{align}
+\min_{\vec x_1\in\real^n,\vec x_2\in\integer^n} \vec c_1^T\cdot \vec x_1+ \vec c_2^T\cdot \vec x_2,\\
+\mat{A}_1\cdot \vec x_1+\mat{A}_2\cdot \vec x_2\leq b,
+\end{align}
+où $\vec c_1$ et $\vec c_2$ sont des vecteurs, et $\mat{A}_1$ et $\mat{A}_2$ des matrices des parties continues, $\vec x_1$ et discrfètes, $\vec x_2$.
+
+Les programmes linéaires continus ont tous des solutions polynomiales, en revanche dans le cas discret cela n'est pas le cas comme on a pu le voir avec le cas du voyageur du commerce.
+
+Il existe bien évidemment des programmes non-linéaires (la fonction objectif, ou les fonctions de contraintes ne sont pas linéaires) mais leur résolution est en général beaucoup plus compliquée, car nous ne sommes pas assurés de l'existence d'un minimum global (il existe également des minima locaux).
+
+### Pogrammation linéaire continue: l'algorithme du simplexe
+
+Une des méthodes les plus répandues de résolution de problèmes d'optimisation continue est la méthode du simplexe linéaire.
+Bien que cette méthode soit sous-optimale (elle est a en général un coût exponentiel en fonction du nombre de dimensions)
+elle reste très utilisée, car elle est relativement intuitive et simple à implémenter.
+
+Nous commençons par définir un problème de programmation linéaire continu de façon générale comme
+\begin{align}
+&\min_{\vec x\in\real^n} {\vec c}^T\cdot \vec x,\\
+&\mat{A}\cdot \vec x\leq b,\\
+&\vec x \geq \vec 0,\quad \vec x\in\real^n.
+\end{align}
+Si on suppose que $\mat{A}$ représente une application surjective, on peut représenter
+$\mat{A}$ comme un polyèdre fermé.
+
+---
+
+Exemple (Polygone) +.#
+
+Soit le problème de programmation linéaire suivant
+\begin{align}
+&\max 10x_1+11x_2,\\
+&10x_1+12x_2\leq 59,\\
+&x_1\geq 0,\quad x_2\geq 0.
+\end{align}
+On peut faire correspondre la formulation générale ci-dessus à
+\begin{align}
+\vec x = \vectwo{x_1}{x_2},\quad \vec c=\vectwo{10}{11},\\
+\mat{A}=\left(10\ 12\right),\quad \vec b=(59).
+\end{align}
+Les conditions sur $\vec x$
+\begin{align*}
+&x_2\leq \frac{59-10 x_2}{12},\\
+&x_1\geq 0,\quad x_2\geq 0,
+\end{align*}
+peuvent se représenter graphiquement par un polygone (nous sommes en deux dimensions), comme représenté sur la @fig:polygon.
+
+{#fig:polygon width=70%}
+
+---
+
+On peut montrer que si toutes les conditions ci-dessus sont satisfaites, alors la solution optimale se trouve sur un des sommets du polyèdre.
+On résout ce genre de problème en parcourant donc tous les sommets en suivants les arrêtes du polyèdre.
+Comme vous avez déjà vu cet algorithme dans le cours d'algorithmique avancée, on ne va pas s'attarder dessus,
+mais plutôt parler de modélisation d'un problème d'optimisation. En d'autres termes, comment en partant
+d'un problème d'optimisation formulé en phrases on écrit un problème mathématique.
+
+## Modélisation
+
+### Exemple: Le problème du sac à dos
+
+Le problème du sac à dos est un problème classique en optimisation combinatoire. Il n'existe pas de méthode exacte qui
+permette de résoudre ce problème en temps polynomial (pour ceux que ça intéresse, je pense que résdoudre ce problème de façon générale
+vaut beaucoup de gloire et d'argent).
+Il s'énonce de la façon suivante. Soit un ensemble de $n$ objets de poids $p_i$ et de valeur $v_i$. Combien de chaque objet
+pouvons nous mettre dans le sac à dos, tel que la valeur totale des objets dans le sac à dos soit maximisée,
+et que le poids total n'excède pas un poids maximal $P$. Nous simplifions encore le problème en spécifiant
+qu'il peut y avoir au plus un objet de chaque dans le sac.
+
+Nous essayons de maximiser la valeur du sac. Nous représentons la présence ou non de l'objet $i$, par une variable $x_i$:
+$x_i=1$ si l'objet est présent $x_i=0$ si l'objet est absent. On cherche donc un vecteur $\vec x=(x_1,x_2,...,x_n)^T$, avec $x_i\in\{0,1\}$.
+La fonction objectif à maximiser est par conséquent
+$$
+f(\vec x)=\sum_{i=1}^n x_i\cdot v_i.
+$$
+Le poids du sac s'exprime lui avec $p(\vec x)$ comme
+$$
+p(\vec x)=\sum_{i=1}^n x_i\cdot p_i.
+$$
+On formule donc mathématiquement notre problème d'optimisation comme
+\begin{align}
+&\max \sum_{i=1}^n x_i\cdot v_i,\\
+&\sum_{i=1}^n x_i\cdot p_i\leq P,\\
+&x_i\in\{0,1\}.
+\end{align}
+
+### Généralités
+
+L'exercice de transposer un problème d'optimisation en formulation mathématique est appelée **modélisation**. L'exercice en lui même est intéressant, et nous allons en voir quelques exemples que vous devrez proposer et résoudre par vous mêmes. Mais avant d'en arriver là, on peut écrire un certain nombre de règles.
+
+Soient deux événement, $A$ et $B$, et les variables de décision
+binaires, $x_A,\ x_B\in\{0,1\}$
+
+- Si les deux événements ne peuvent pas se produire en même temps, on peut écrire
+$$
+x_A+x_B\leq 1.
+$$
+On a 4 possibilités:
+ 1. $A$ se produit, $x_A=1$, et $B$ se produit $x_B=1$. On a $x_A+x_B=2$ (c'est plus grand que $1$ nous sommes sauvés).
+ 2. $A$ se produit, $x_A=1$, et $B$ ne se produit pas $x_B=0$. On a $x_A+x_B=1$ (c'est égal à $1$ nous sommes satisfaisons la contrainte).
+ 3. $A$ ne se se produit pas, $x_A=0$, et $B$ se produit $x_B=1$. On a $x_A+x_B=1$ (c'est égal à $1$ nous sommes satisfaisons la contrainte).
+ 4. $A$ ne se se produit pas, $x_A=0$, et $B$ ne se produit pas $x_B=0$. On a $x_A+x_B=0$ (c'est égal à $0$ nous sommes satisfaisons la contrainte).
+
+- Si au moins un des deux événements doit se produit
+$$
+x_A+x_B\geq 1.
+$$
+On a 4 possibilités:
+ 1. $A$ se produit, $x_A=1$, et $B$ se produit $x_B=1$. On a $x_A+x_B=2$ (c'est plus grand que $1$ satisfaisons la contrainte).
+ 2. $A$ se produit, $x_A=1$, et $B$ ne se produit pas $x_B=0$. On a $x_A+x_B=1$ (c'est égal à $1$ nous satisfaisons la contrainte).
+ 3. $A$ ne se se produit pas, $x_A=0$, et $B$ se produit $x_B=1$. On a $x_A+x_B=1$ (c'est égal à $1$ nous satisfaisons la contrainte).
+ 4. $A$ ne se se produit pas, $x_A=0$, et $B$ ne se produit pas $x_B=0$. On a $x_A+x_B=0$ (c'est plus petit que $1$, nous sommes sauvés).
+
+- Si $A$ se produit, alors $B$ doit se produire
+$$
+x_A\leq x_B.
+$$
+Si $A$ s'est produit, nous avons $x_A=1$, cela signifie que $x_B$ doit se produire également et donc $1\leq x_B$, $x_B$ doit forcément valoir $x_B=1$ sinon la contrainte n'est pas satisfaite. De façon similaire, si $B$ ne se produit pas ($x_B=0$), alors $A$ ne doit pas se produire, car $x_A\leq 0$. On a donc que $x_A=0$ obligatoirement.
+- Soit $C$ un troisième événement et $x_C$ sa variable de décision binaire. Si $A$ se produit, alors $B$ ou/et $C$ doivent se produire:
+$$
+x_A\leq x_B+x_C.
+$$
+
+Souvent, il est utile de rajouter des variables même si cela complexifie *a priori* le problème. Cela permet néanmoins d'être plus expressif.
+
+- Soient un nombre $N$ de fonctions linéaires $c_i x$,
+$i=1, ..., N$. Si l'objectif est de maximiser la valeur minimale prise par cette ensemble de fonction, on peut rajouter une variable $y$ et l'ensemble de contraintes
+$$
+y\leq c_i x.
+$$
+La fonction objectif est elle-même modifiée et devient $\max y$.
+
+- Soient $N$ contraintes
+\begin{align}
+&c_1 x\leq b_1,\\
+&c_2 x\leq b_2,\\
+&\cdots,\nonumber\\
+&c_N x\leq b_N.
+\end{align}
+Si nous souhaitons en imposer un sous ensemble, $K$ en même temps. Nous pouvons définir une variable $M_i$ qui satisfera la $i$-ème contrainte
+quelle que soit $x$
+$$
+c_i x\leq b_i+M_i,\quad\forall x.
+$$
+On définit également $y_i$, une variable binaire, pour chaque contrainte. Nous pouvons dès-lors réécrire les contraintes de la façon suivante
+\begin{align}
+&c_1 x\leq b_1+M_1y_1,\\
+&c_2 x\leq b_2+M_2y_2,\\
+&\cdots,\nonumber\\
+&c_N x\leq b_N+M_Ny_N,\\
+&\sum_{i=1}^N y_i = N-K.
+\end{align}
+
+---
+
+Question +.#
+
+Que se passe-t-il si $N=2$ et $K=1$?
+
+---
+
+Ci-dessous vous trouverez trois problèmes différents qu'il faudra modéliser.
+
+---
+
+Problème (Fabrication) +.#
+
+Une entreprise produit des pizzas de luxe (LU) et "low-cost" (LC).
+Elles sont toutes les deux fabriquées à partir des mêmes produits,
+pâte, fromage, et jambon,
+mais dans des proportions différentes qui sont résumées dans le tableau ci-dessous
+
++----------+--------+-------+
+| | L | LC |
++==========+========+=======+
+| Pâte | 1 | 3 |
++----------+--------+-------+
+| Fromage | 2 | 1 |
++----------+--------+-------+
+| Jambon | 2 | 1 |
++----------+--------+-------+
+
+Chacune de ces matières premières est disponible en quantité limitée:
+$1200\mathrm{kg}$ de pâte, $600\mathrm{kg}$ de fromage, et $400\mathrm{kg}$ de jambom. La pizza de luxe rapporte $10$ francs par kilo et la pizza low-cost rapporte $6$ francs par kilo. Modélisr le problème du calcul de nombre de kilos de pizzas à produire de chaque type pour maximiser le rendement.
+
+Modéliser le problème:
+
+1. Déterminer les variables du problème.
+2. Écrire la fonction objectif.
+3. Ecrire les contraintes.
+
+<!-- Le problème se modélise de la façon suivante:
+
+1. Les variables sont $x_i$ les quantités en kilos de pizzas du type $LU$ et $LC$ (on a donc $x_{LU}$, et $x_{LC}$).
+2. La fonction objectif à maximiser est le rendement des pizzas:
+$$
+\max_{\vec x} f(\vec x)=\max_{\vec x}10\cdot x_{LU}+6\cdot x_{LC}.
+$$
+3. Les contraintes sont données par:
+\begin{align}
+1\cdot x_{LU} + 3\cdot x_{LC}\leq 1200,\\
+2\cdot x_{LU} + 1\cdot x_{LC}\leq 600,\\
+2\cdot x_{LU} + 1\cdot x_{LC}\leq 400,\\
+x_{LU}\geq 0,\quad x_{LC}\geq 0.
+\end{align} -->
+
+---
+
+---
+
+Problème (Distribution d'électricité) +.#
+
+Nous souhaitons alimenter en électricité trois villes, Genève, Bâle, et Zürich.
+Nous avons à notre disposition trois centrales (celle de la Grande Dixence, de Leibstadt et
+de Mühleberg). Nous souhaitons à partir du tableau ci-dessous, représentant les coût de production
+et d'acheminement de l'électricité des centrales aux villes, ainsi que leur demande et puissance
+respectives en GWh. Déterminer fournir en électricitié toutes les villes à moindre coût.
+Modéliser le problème d'optimisation des coûts globaux d'élctricité (tous les chiffres donnés ici
+sont choisis au hasard).
+
++----------------+--------+--------+--------+-----------------+
+| | Bâle | Genève | Zürich | Puissance (GWh) |
++================+========+========+========+=================+
+| Grande Dixence | 3 | 5 | 8 | 30 |
++----------------+--------+--------+--------+-----------------+
+| Leibstadt | 8 | 11 | 6 | 50 |
++----------------+--------+--------+--------+-----------------+
+| Mühleberg | 11 | 15 | 12 | 40 |
++----------------+--------+--------+--------+-----------------+
+| Demande (GWh) | 20 | 25 | 40 | |
++----------------+--------+--------+--------+-----------------+
+
+Modéliser le problème:
+
+1. Déterminer les variables du problème.
+2. Écrire la fonction objectif.
+3. Ecrire les contraintes.
+
+<!-- Le problème se modélise de la façon suivante:
+
+1. Les variables du probèmes sont $x_{ij}$, le nombre de GWh acheminés de la centrale $i$ à la ville $j$.
+On peut par exemple numéroter la Grande Dixence, Leibstadt et Mühleberg, 1, 2, et 3 respectivement et
+Bâle, Genève et Zûrich, 1, 2, et 3 respectivement.
+2. Le coût total de production et d'acheminement de l'électricité et la fonction objectif $f(\vec x)$ qui s'écrit
+\begin{align}
+f(\vec x) = &3x_{11} + 5x_{12} + 8x_{13} +\nonumber\\
+ &8x_{21} + 11x_{22} + 6x_{23} +\nonumber\\
+ &11x_{31} + 15x_{32} + 12x_{33}.
+\end{align}
+3. Il y a trois sortes de contraintes.
+
+ a. Les la quantité d'électricité acheminée doit être positive
+ $$
+ x_{ij}\geq 0,\quad\forall{i,j}.
+ $$
+ b. La production maximale de chaque centrale ne doit pas être dépassée
+ \begin{align}
+ &x_{11}+x_{12}+x_{13}\leq 30,\\
+ &x_{21}+x_{22}+x_{23}\leq 50,\\
+ &x_{31}+x_{32}+x_{33}\leq 40.
+ \end{align}
+ c. Chaque ville doit au moins recevoir la quantité d'électricité demandée
+ \begin{align}
+ &x_{11}+x_{21}+x_{31}\geq 20,\\
+ &x_{12}+x_{22}+x_{32}\geq 25,\\
+ &x_{13}+x_{23}+x_{33}\geq 40.
+ \end{align} -->
+
+---
+
+---
+
+Problème (Transport) +.#
+
+Le graphe de la @fig:graph_transport représente le réseau de transport d'un jouet de deux usines $U$, et $V$,
+à deux clients $C$, et $D$ en passant par les entrepôts $E$ et $F$. Sur les arrêtes du graphe se
+situent les coûts de transport par jouet transporté.
+
+{#fig:graph_transport width=100%}
+
+A la sortie de l'usine $A$, il y a une quantité de $90$ jouets disponibles et une quantité de $75$
+jouets à la sortie de l'usine $B$. Les clients $C$ et $D$ ont une demande de respectivement
+$70$ et $80$. Les entrepôts ne sont qu'un point de transit.
+
+Modéliser le problème d'optimisation permettant de déterminer le planning de transport optimal
+qui permet de satisfaire les demandes des clients pour un coût aussi faible que possible.
+
+1. Déterminer les variables du problème.
+2. Écrire la fonction objectif.
+3. Ecrire les contraintes.
+
+<!-- Le problème se modélise de la façon suivante:
+
+1. Les variables sont les quantités transportées entre les usines et entrepôts,
+$x_{UE}$, $x_{UF}$, $x_{VE}$, $x_{VF}$, et les quantités transportées entre les
+entrepôts et les clients $x_{EC}$, $x_{ED}$, $x_{FC}$, $x_{FD}$.
+2. On cherhce à minimiser la fonction de coût de transport $f(\vec x)$
+$$
+f(x)=x_{UE} + 2x_{UF} + 3x_{VE} + 4x_{VF} + 5x_{EC} + 8x_{ED}
+ + 7x_{FC} + 6x_{FD}.
+$$
+3. Il y a 4 sortes de contraintes.
+
+ a. Les quantités de jouets sont des entiers positifs
+ $$
+ x_{ij}\in\natural.
+ $$
+ b. Ce qui part des usines doit être plus petit que le stock local
+ \begin{align}
+ x_{UE}+x_{UF}&\leq 90,\\
+ x_{VE}+x_{VF}&\leq 75.
+ \end{align}
+ c. Ce qui arrive à chaque entrepôt doit partir de chaque entrepôt
+ \begin{align}
+ x_{UE}+x_{VE}&=x_{EC}+x_{ED},\\
+ x_{UF}+x_{VF}&=x_{FC}+x_{FD}.
+ \end{align}
+ d. Les clients doivent recevoir la quantité demandée
+ \begin{align}
+ x_{EC}+x_{FC}&= 70,\\
+ x_{ED}+x_{FD}&= 80.
+ \end{align} -->
+
+---
+
+## Optimisation continue
+
+Dans cette section, nous allons considérer des problèmes purement continus.
+Nous allons dans un premier temps considérer une fonction opbjectif, $f$,
+$$
+f:D\rightarrow\real,\quad D\subseteq \real,
+$$
+dont nous allons chercher le minimum (pour autant qu'il existe). Nous allons supposer que
+$f$ est une fonction continue et dérivable.
+
+### Minimum local/global
+
+Comme vous le savez, le minimum (ou le maximum) d'une fonction, se situe à un endroit où sa dérivée est nulle.
+On recherche donc, $x$, tel que
+$$
+f'(x)=0.
+$$
+Mais cette contrainte sur $f'(x)$ n'est pas suffisante pour garantir de trouver un minimum.
+En effet, si $f'(x)=0$, peut également vouloir dire qu'on se trouve sur un point d'inflexion
+ou sur un maximum.
+On peut assez facilement, discriminer ces deux cas, en considérant la deuxième dérivée de $f$.
+En effet, nous avons à faire à un minimum seulement si
+$$
+f''(x)>0.
+$$
+Les cas où $f''(x)=0$ est un point d'inflexion et $f''(x)<0$ est un maximum.
+
+Un autre problème beaucoup plus compliqué à résoudre est de déterminer un minimum **global**.
+En effet, comme pour la fonction de @fig:ackley, une fonction peut posséder un grand nombre de minimam **locaux** (où
+$f'(x)=0$ et $f''(x)>0$) mais qui n'est pas un mimumum global.
+
+Mathématiquement un *minimum local* se définit comme $x^\ast$ tel qu'il existe $\delta>0$ et que $f(x^\ast)\leq f(x)$, pour
+$x\in[x^\ast-\dealt,x^\ast+delta]$. Un *minimum global* est un $x^\ast$ tel que $\forall x\in D$, $f(x^\ast)\leq f(x)$.
+
# Remerciements
Je voudrais remercier (par ordre alphabétique) les étudiants du cours
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diff --git a/python/polyedre.py b/python/polyedre.py
new file mode 100644
index 0000000000000000000000000000000000000000..6cb636392f8f719b6d2dcf6be8e9172128f5a8c6
--- /dev/null
+++ b/python/polyedre.py
@@ -0,0 +1,13 @@
+import matplotlib.pyplot as plt
+import numpy as np
+
+x1 = np.arange(0, 6.0, 0.1)
+x2 = (59 - 10 * x1) / 12
+
+x2max = np.max(x2)
+y2 = np.arange(0, x2max, 0.1)
+
+plt.plot(x1,x2,'k-')
+plt.plot(x1,0*x1,'r-')
+plt.plot(0*y2,y2,'b-')
+plt.show()
\ No newline at end of file