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Programmation Linéaireet
Recherche OpérationnelleTravaux Dirigés
Alexandre [email protected]
Université d’OrléansUFR Sciences
Département d’informatique
L3 STIC – année 2006•2007
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 0
Table des matières
1 Modélisation de Problèmes Linéaires
2 Forme Standard/Canonique, Résolution Graphique
3 Programmation Linéaire en Nombres Entiers
4 Algorithme du Simplexe
5 Cas particuliers de l’algorithme du Simplexe
6 Méthode du Simplexe
7 Dualité, Théorème de Dualité
8 Dualité, Écarts Complémentaires et Interprétation Économique
9 Résolution de Problème, Interprétation des Résultats
10 Résolution de Problème, Interprétation des Résultats
11 Algorithme du Simplexe Révisé
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 1 Modélisation de Problèmes Linéaires
(*) Exercice 1.3
min: 3 x1 + 24 x2 + 9 x3 + 20 x4 + 19 x5;calories = 110 x1 + 205 x2 + 160 x3 + 420 x4 + 260 x5;proteines = 4 x1 + 32 x2 + 8 x3 + 4 x4 + 14 x5;calcium = 2 x1 + 12 x2 + 285 x3 + 22 x4 + 80 x5;porridge = 28 x1;poulet = 100 x2;lait = 237 x3;tarte = 170 x4;porc = 260 x5;2000 <= calories;55 <= proteines;800 <= calcium;28 x1 <= 110;100 x2 <= 600;237 x3 <= 2000;170 x4 <= 350;260 x5 <= 500;
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 1 Modélisation de Problèmes Linéaires
(*) Exercice 1.3
Value of objective function: 92.5144x1 3.92857x2 0x3 4.3947x4 2.05882x5 0porridge 110calcium 1305.64tarte 350calories 2000poulet 0lait 1041.54porc 0proteines 59.1071
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 1 Modélisation de Problèmes Linéaires
Exercice 1.4Question 2 (*)
10000pA égale à la moitié de la quantité initiale 10000pA
on a pA = 12 pA
on pose 10000xi = Ti
max −30x1 − 100x2 − 180x3
s.c.
8<:5000 ≥ 6000x1 + 1000x2 + 700x3 + 10000x4
1 = x1 + x2 + x3 + x4
x1, x2, x3, x4 ≥ 0
max −3x1 − 10x2 − 18x3
s.c.
8<:50 ≥ 60x1 + 10x2 + 7x3 + 100x4
1 = x1 + x2 + x3 + x4
x1, x2, x3, x4 ≥ 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 1 Modélisation de Problèmes Linéaires
Exercice 1.4Question 2 (*)
10000pA égale à la moitié de la quantité initiale 10000pA
on a pA = 12 pA
on pose 10000xi = Ti
max −30x1 − 100x2 − 180x3
s.c.
8<:5000 ≥ 6000x1 + 1000x2 + 700x3 + 10000x4
1 = x1 + x2 + x3 + x4
x1, x2, x3, x4 ≥ 0
max −3x1 − 10x2 − 18x3
s.c.
8<:50 ≥ 60x1 + 10x2 + 7x3 + 100x4
1 = x1 + x2 + x3 + x4
x1, x2, x3, x4 ≥ 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 1 Modélisation de Problèmes Linéaires
Exercice 1.4Question 2 (*)
10000pA égale à la moitié de la quantité initiale 10000pA
on a pA = 12 pA
on pose 10000xi = Ti
max −30x1 − 100x2 − 180x3
s.c.
8<:5000 ≥ 6000x1 + 1000x2 + 700x3 + 10000x4
1 = x1 + x2 + x3 + x4
x1, x2, x3, x4 ≥ 0
max −3x1 − 10x2 − 18x3
s.c.
8<:50 ≥ 60x1 + 10x2 + 7x3 + 100x4
1 = x1 + x2 + x3 + x4
x1, x2, x3, x4 ≥ 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 1 Modélisation de Problèmes Linéaires
(*) Exercice 1.6
6 variablesK , S les quantités de kérosène et de super produitesk1, k2 les quantités de PL1 et PL2 intervenant dans le kérosènes1, s2 les quantités de PL1 et PL2 intervenant dans le super
contraintes sur les taux d’octane
(5)104k1 + 94k2
k1 + k2≥ 102
(6)104s1 + 94s2
s1 + s2≥ 96
max: 150 K + 100 S;K = k1 + k2;K <= 20;S = s1 + s2;k1 + s1 <= 30;k2 + s2 <= 70;8 k2 <= 2 k1; /* (5) */2 s2 <= 8 s1; /* (6) */
profit = 325003 k1 = 40/3, k2 = 10/3, s1 = 50/3, s2 = 200/3
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 1 Modélisation de Problèmes Linéaires
(*) Exercice 1.6
6 variablesK , S les quantités de kérosène et de super produitesk1, k2 les quantités de PL1 et PL2 intervenant dans le kérosènes1, s2 les quantités de PL1 et PL2 intervenant dans le super
contraintes sur les taux d’octane
(5)104k1 + 94k2
k1 + k2≥ 102
(6)104s1 + 94s2
s1 + s2≥ 96
max: 150 K + 100 S;K = k1 + k2;K <= 20;S = s1 + s2;k1 + s1 <= 30;k2 + s2 <= 70;8 k2 <= 2 k1; /* (5) */2 s2 <= 8 s1; /* (6) */
profit = 325003 k1 = 40/3, k2 = 10/3, s1 = 50/3, s2 = 200/3
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 1 Modélisation de Problèmes Linéaires
(*) Exercice 1.6
6 variablesK , S les quantités de kérosène et de super produitesk1, k2 les quantités de PL1 et PL2 intervenant dans le kérosènes1, s2 les quantités de PL1 et PL2 intervenant dans le super
contraintes sur les taux d’octane
(5)104k1 + 94k2
k1 + k2≥ 102
(6)104s1 + 94s2
s1 + s2≥ 96
max: 150 K + 100 S;K = k1 + k2;K <= 20;S = s1 + s2;k1 + s1 <= 30;k2 + s2 <= 70;8 k2 <= 2 k1; /* (5) */2 s2 <= 8 s1; /* (6) */
profit = 325003 k1 = 40/3, k2 = 10/3, s1 = 50/3, s2 = 200/3
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = +30x1 +30x2 +40x3
e1 = 2000 −2x1 −x2 −x3e2 = 7000 −x1 −2x2 −4x3
x3 ≤ 2000
x3 ≤ 70004
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, e1 = 2000, e2 = 7000
on substitue x3 = 70004 − 1
4 x1 − 12 x2 − 1
4 e2
z(max) = 70000 +20x1 +10x2 −10e2
e1 = 250 − 74 x1 − 1
2 x2 + 14 e2
x3 = 1750 − 14 x1 − 1
2 x2 − 14 e2
x1 ≤ 10007
x1 ≤ 7000
nouvelle solution
z(max) = 70000, x1 = 0, x2 = 0, x3 = 1750, e1 = 250, e2 = 0
on substitue x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = +30x1 +30x2 +40x3
e1 = 2000 −2x1 −x2 −x3e2 = 7000 −x1 −2x2 −4x3
x3 ≤ 2000
x3 ≤ 70004
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, e1 = 2000, e2 = 7000
on substitue x3 = 70004 − 1
4 x1 − 12 x2 − 1
4 e2
z(max) = 70000 +20x1 +10x2 −10e2
e1 = 250 − 74 x1 − 1
2 x2 + 14 e2
x3 = 1750 − 14 x1 − 1
2 x2 − 14 e2
x1 ≤ 10007
x1 ≤ 7000
nouvelle solution
z(max) = 70000, x1 = 0, x2 = 0, x3 = 1750, e1 = 250, e2 = 0
on substitue x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = +30x1 +30x2 +40x3
e1 = 2000 −2x1 −x2 −x3e2 = 7000 −x1 −2x2 −4x3
x3 ≤ 2000
x3 ≤ 70004
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, e1 = 2000, e2 = 7000
on substitue x3 = 70004 − 1
4 x1 − 12 x2 − 1
4 e2
z(max) = 70000 +20x1 +10x2 −10e2
e1 = 250 − 74 x1 − 1
2 x2 + 14 e2
x3 = 1750 − 14 x1 − 1
2 x2 − 14 e2
x1 ≤ 10007
x1 ≤ 7000
nouvelle solution
z(max) = 70000, x1 = 0, x2 = 0, x3 = 1750, e1 = 250, e2 = 0
on substitue x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = +30x1 +30x2 +40x3
e1 = 2000 −2x1 −x2 −x3e2 = 7000 −x1 −2x2 −4x3
x3 ≤ 2000
x3 ≤ 70004
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, e1 = 2000, e2 = 7000
on substitue x3 = 70004 − 1
4 x1 − 12 x2 − 1
4 e2
z(max) = 70000 +20x1 +10x2 −10e2
e1 = 250 − 74 x1 − 1
2 x2 + 14 e2
x3 = 1750 − 14 x1 − 1
2 x2 − 14 e2
x1 ≤ 10007
x1 ≤ 7000
nouvelle solution
z(max) = 70000, x1 = 0, x2 = 0, x3 = 1750, e1 = 250, e2 = 0
on substitue x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = +30x1 +30x2 +40x3
e1 = 2000 −2x1 −x2 −x3e2 = 7000 −x1 −2x2 −4x3
x3 ≤ 2000
x3 ≤ 70004
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, e1 = 2000, e2 = 7000
on substitue x3 = 70004 − 1
4 x1 − 12 x2 − 1
4 e2
z(max) = 70000 +20x1 +10x2 −10e2
e1 = 250 − 74 x1 − 1
2 x2 + 14 e2
x3 = 1750 − 14 x1 − 1
2 x2 − 14 e2
x1 ≤ 10007
x1 ≤ 7000
nouvelle solution
z(max) = 70000, x1 = 0, x2 = 0, x3 = 1750, e1 = 250, e2 = 0
on substitue x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = +30x1 +30x2 +40x3
e1 = 2000 −2x1 −x2 −x3e2 = 7000 −x1 −2x2 −4x3
x3 ≤ 2000
x3 ≤ 70004
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, e1 = 2000, e2 = 7000
on substitue x3 = 70004 − 1
4 x1 − 12 x2 − 1
4 e2
z(max) = 70000 +20x1 +10x2 −10e2
e1 = 250 − 74 x1 − 1
2 x2 + 14 e2
x3 = 1750 − 14 x1 − 1
2 x2 − 14 e2
x1 ≤ 10007
x1 ≤ 7000
nouvelle solution
z(max) = 70000, x1 = 0, x2 = 0, x3 = 1750, e1 = 250, e2 = 0
on substitue x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = +30x1 +30x2 +40x3
e1 = 2000 −2x1 −x2 −x3e2 = 7000 −x1 −2x2 −4x3
x3 ≤ 2000
x3 ≤ 70004
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, e1 = 2000, e2 = 7000
on substitue x3 = 70004 − 1
4 x1 − 12 x2 − 1
4 e2
z(max) = 70000 +20x1 +10x2 −10e2
e1 = 250 − 74 x1 − 1
2 x2 + 14 e2
x3 = 1750 − 14 x1 − 1
2 x2 − 14 e2
x1 ≤ 10007
x1 ≤ 7000
nouvelle solution
z(max) = 70000, x1 = 0, x2 = 0, x3 = 1750, e1 = 250, e2 = 0
on substitue x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = 5100007 + 30
7 x2 − 507 e2 − 80
7 e1
x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
x3 = 120007 − 3
7 x2 − 27 e2 + 1
7 e1
x2 ≤ 500x2 ≤ 4000
z(max) = 5100007 , x1 = 1000
7 , x2 = 0, x3 = 120007 , e1 = 0, e2 = 0
on substitue x2 = 500 + 12 e2 − 2e1 − 7
2 x1
z(max) = 75000 −5e2 −20e1 −15x1
x2 = 500 + 12 e2 −2e1 − 7
2 x1
x3 = 1500 − 12 e2 +e1 + 3
2 x1
solution optimale
z(max) = 75000, x1 = 0, x2 = 500, x3 = 1500, e1 = 0, e2 = 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = 5100007 + 30
7 x2 − 507 e2 − 80
7 e1
x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
x3 = 120007 − 3
7 x2 − 27 e2 + 1
7 e1
x2 ≤ 500x2 ≤ 4000
z(max) = 5100007 , x1 = 1000
7 , x2 = 0, x3 = 120007 , e1 = 0, e2 = 0
on substitue x2 = 500 + 12 e2 − 2e1 − 7
2 x1
z(max) = 75000 −5e2 −20e1 −15x1
x2 = 500 + 12 e2 −2e1 − 7
2 x1
x3 = 1500 − 12 e2 +e1 + 3
2 x1
solution optimale
z(max) = 75000, x1 = 0, x2 = 500, x3 = 1500, e1 = 0, e2 = 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = 5100007 + 30
7 x2 − 507 e2 − 80
7 e1
x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
x3 = 120007 − 3
7 x2 − 27 e2 + 1
7 e1
x2 ≤ 500x2 ≤ 4000
z(max) = 5100007 , x1 = 1000
7 , x2 = 0, x3 = 120007 , e1 = 0, e2 = 0
on substitue x2 = 500 + 12 e2 − 2e1 − 7
2 x1
z(max) = 75000 −5e2 −20e1 −15x1
x2 = 500 + 12 e2 −2e1 − 7
2 x1
x3 = 1500 − 12 e2 +e1 + 3
2 x1
solution optimale
z(max) = 75000, x1 = 0, x2 = 500, x3 = 1500, e1 = 0, e2 = 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = 5100007 + 30
7 x2 − 507 e2 − 80
7 e1
x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
x3 = 120007 − 3
7 x2 − 27 e2 + 1
7 e1
x2 ≤ 500x2 ≤ 4000
z(max) = 5100007 , x1 = 1000
7 , x2 = 0, x3 = 120007 , e1 = 0, e2 = 0
on substitue x2 = 500 + 12 e2 − 2e1 − 7
2 x1
z(max) = 75000 −5e2 −20e1 −15x1
x2 = 500 + 12 e2 −2e1 − 7
2 x1
x3 = 1500 − 12 e2 +e1 + 3
2 x1
solution optimale
z(max) = 75000, x1 = 0, x2 = 500, x3 = 1500, e1 = 0, e2 = 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.2
z(max) = 5100007 + 30
7 x2 − 507 e2 − 80
7 e1
x1 = 10007 − 2
7 x2 + 17 e2 − 4
7 e1
x3 = 120007 − 3
7 x2 − 27 e2 + 1
7 e1
x2 ≤ 500x2 ≤ 4000
z(max) = 5100007 , x1 = 1000
7 , x2 = 0, x3 = 120007 , e1 = 0, e2 = 0
on substitue x2 = 500 + 12 e2 − 2e1 − 7
2 x1
z(max) = 75000 −5e2 −20e1 −15x1
x2 = 500 + 12 e2 −2e1 − 7
2 x1
x3 = 1500 − 12 e2 +e1 + 3
2 x1
solution optimale
z(max) = 75000, x1 = 0, x2 = 500, x3 = 1500, e1 = 0, e2 = 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.4
z(max) = + 135 t1 +4t2 + 18
5 t3x1 = 4000 − 3
8 t1 − 18 t2 − 1
10 t3x2 = 800 − 1
2 t1 − 120 t2 − 1
5 t3x3 = 1500 − 1
2 t1 − 15 t2 − 3
20 t3x4 = 8000 −t1 −t2 −t3
t2 ≤ 32000t2 ≤ 16000t2 ≤ 7500t2 ≤ 8000
on substitue t2 = 7500− 52 t1 − 3
4 t3 − 5x3
z(max) = 30000 − 375 t1 + 3
5 t3 −20x3
x1 = 61252 − 1
16 t1 − 1160 t3 + 5
8 x3
x2 = 425 − 38 t1 − 13
80 t3 + 14 x3
t2 = 7500 − 52 t1 − 3
4 t3 −5x3
x4 = 500 + 32 t1 − 1
4 t3 +5x3
t3 ≤ 490000t3 ≤ 34000
13 ' 2615t3 ≤ 10000t3 ≤ 2000
on substitue t3 = 2000 + 6t1 + 20x3 − 4x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.4
z(max) = + 135 t1 +4t2 + 18
5 t3x1 = 4000 − 3
8 t1 − 18 t2 − 1
10 t3x2 = 800 − 1
2 t1 − 120 t2 − 1
5 t3x3 = 1500 − 1
2 t1 − 15 t2 − 3
20 t3x4 = 8000 −t1 −t2 −t3
t2 ≤ 32000t2 ≤ 16000t2 ≤ 7500t2 ≤ 8000
on substitue t2 = 7500− 52 t1 − 3
4 t3 − 5x3
z(max) = 30000 − 375 t1 + 3
5 t3 −20x3
x1 = 61252 − 1
16 t1 − 1160 t3 + 5
8 x3
x2 = 425 − 38 t1 − 13
80 t3 + 14 x3
t2 = 7500 − 52 t1 − 3
4 t3 −5x3
x4 = 500 + 32 t1 − 1
4 t3 +5x3
t3 ≤ 490000t3 ≤ 34000
13 ' 2615t3 ≤ 10000t3 ≤ 2000
on substitue t3 = 2000 + 6t1 + 20x3 − 4x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.4
z(max) = + 135 t1 +4t2 + 18
5 t3x1 = 4000 − 3
8 t1 − 18 t2 − 1
10 t3x2 = 800 − 1
2 t1 − 120 t2 − 1
5 t3x3 = 1500 − 1
2 t1 − 15 t2 − 3
20 t3x4 = 8000 −t1 −t2 −t3
t2 ≤ 32000t2 ≤ 16000t2 ≤ 7500t2 ≤ 8000
on substitue t2 = 7500− 52 t1 − 3
4 t3 − 5x3
z(max) = 30000 − 375 t1 + 3
5 t3 −20x3
x1 = 61252 − 1
16 t1 − 1160 t3 + 5
8 x3
x2 = 425 − 38 t1 − 13
80 t3 + 14 x3
t2 = 7500 − 52 t1 − 3
4 t3 −5x3
x4 = 500 + 32 t1 − 1
4 t3 +5x3
t3 ≤ 490000t3 ≤ 34000
13 ' 2615t3 ≤ 10000t3 ≤ 2000
on substitue t3 = 2000 + 6t1 + 20x3 − 4x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.4
z(max) = + 135 t1 +4t2 + 18
5 t3x1 = 4000 − 3
8 t1 − 18 t2 − 1
10 t3x2 = 800 − 1
2 t1 − 120 t2 − 1
5 t3x3 = 1500 − 1
2 t1 − 15 t2 − 3
20 t3x4 = 8000 −t1 −t2 −t3
t2 ≤ 32000t2 ≤ 16000t2 ≤ 7500t2 ≤ 8000
on substitue t2 = 7500− 52 t1 − 3
4 t3 − 5x3
z(max) = 30000 − 375 t1 + 3
5 t3 −20x3
x1 = 61252 − 1
16 t1 − 1160 t3 + 5
8 x3
x2 = 425 − 38 t1 − 13
80 t3 + 14 x3
t2 = 7500 − 52 t1 − 3
4 t3 −5x3
x4 = 500 + 32 t1 − 1
4 t3 +5x3
t3 ≤ 490000t3 ≤ 34000
13 ' 2615t3 ≤ 10000t3 ≤ 2000
on substitue t3 = 2000 + 6t1 + 20x3 − 4x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.4
z(max) = + 135 t1 +4t2 + 18
5 t3x1 = 4000 − 3
8 t1 − 18 t2 − 1
10 t3x2 = 800 − 1
2 t1 − 120 t2 − 1
5 t3x3 = 1500 − 1
2 t1 − 15 t2 − 3
20 t3x4 = 8000 −t1 −t2 −t3
t2 ≤ 32000t2 ≤ 16000t2 ≤ 7500t2 ≤ 8000
on substitue t2 = 7500− 52 t1 − 3
4 t3 − 5x3
z(max) = 30000 − 375 t1 + 3
5 t3 −20x3
x1 = 61252 − 1
16 t1 − 1160 t3 + 5
8 x3
x2 = 425 − 38 t1 − 13
80 t3 + 14 x3
t2 = 7500 − 52 t1 − 3
4 t3 −5x3
x4 = 500 + 32 t1 − 1
4 t3 +5x3
t3 ≤ 490000t3 ≤ 34000
13 ' 2615t3 ≤ 10000t3 ≤ 2000
on substitue t3 = 2000 + 6t1 + 20x3 − 4x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.4
z(max) = + 135 t1 +4t2 + 18
5 t3x1 = 4000 − 3
8 t1 − 18 t2 − 1
10 t3x2 = 800 − 1
2 t1 − 120 t2 − 1
5 t3x3 = 1500 − 1
2 t1 − 15 t2 − 3
20 t3x4 = 8000 −t1 −t2 −t3
t2 ≤ 32000t2 ≤ 16000t2 ≤ 7500t2 ≤ 8000
on substitue t2 = 7500− 52 t1 − 3
4 t3 − 5x3
z(max) = 30000 − 375 t1 + 3
5 t3 −20x3
x1 = 61252 − 1
16 t1 − 1160 t3 + 5
8 x3
x2 = 425 − 38 t1 − 13
80 t3 + 14 x3
t2 = 7500 − 52 t1 − 3
4 t3 −5x3
x4 = 500 + 32 t1 − 1
4 t3 +5x3
t3 ≤ 490000t3 ≤ 34000
13 ' 2615t3 ≤ 10000t3 ≤ 2000
on substitue t3 = 2000 + 6t1 + 20x3 − 4x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.4
z(max) = 31200 − 195 t1 −8x3 − 12
5 x4
x1 = 3050 + . . .x2 = 100 + . . .t2 = 6000 + . . .t3 = 2000 + . . .
z(max) = 31200t1 = 0, t2 = 6000, t3 = 2000
x1 = 3050, x2 = 100, x3 = 0, x4 = 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 4 Algorithme du Simplexe
(*) Exercice 4.4
z(max) = 31200 − 195 t1 −8x3 − 12
5 x4
x1 = 3050 + . . .x2 = 100 + . . .t2 = 6000 + . . .t3 = 2000 + . . .
z(max) = 31200t1 = 0, t2 = 6000, t3 = 2000
x1 = 3050, x2 = 100, x3 = 0, x4 = 0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on substitue x1 = 11x2 + 5x3 − 18x4 − 2x5
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +53x2 +41x3 −204x4 −20x5
x1 = +11x2 +5x3 −18x4 −2x5
x6 = −4x2 −2x3 +8x4 +x5
x7 = 1 −11x2 −5x3 +18x4 +2x5
x2 ≤ 0x2 ≤ 1
11
on substitue x2 = −0.5x3 + 2x4 + 0.25x5 − 0.25x6 retour
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on substitue x1 = 11x2 + 5x3 − 18x4 − 2x5
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +53x2 +41x3 −204x4 −20x5
x1 = +11x2 +5x3 −18x4 −2x5
x6 = −4x2 −2x3 +8x4 +x5
x7 = 1 −11x2 −5x3 +18x4 +2x5
x2 ≤ 0x2 ≤ 1
11
on substitue x2 = −0.5x3 + 2x4 + 0.25x5 − 0.25x6 retour
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on substitue x1 = 11x2 + 5x3 − 18x4 − 2x5
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +53x2 +41x3 −204x4 −20x5
x1 = +11x2 +5x3 −18x4 −2x5
x6 = −4x2 −2x3 +8x4 +x5
x7 = 1 −11x2 −5x3 +18x4 +2x5
x2 ≤ 0x2 ≤ 1
11
on substitue x2 = −0.5x3 + 2x4 + 0.25x5 − 0.25x6 retour
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on substitue x1 = 11x2 + 5x3 − 18x4 − 2x5
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +53x2 +41x3 −204x4 −20x5
x1 = +11x2 +5x3 −18x4 −2x5
x6 = −4x2 −2x3 +8x4 +x5
x7 = 1 −11x2 −5x3 +18x4 +2x5
x2 ≤ 0x2 ≤ 1
11
on substitue x2 = −0.5x3 + 2x4 + 0.25x5 − 0.25x6 retour
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on substitue x1 = 11x2 + 5x3 − 18x4 − 2x5
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +53x2 +41x3 −204x4 −20x5
x1 = +11x2 +5x3 −18x4 −2x5
x6 = −4x2 −2x3 +8x4 +x5
x7 = 1 −11x2 −5x3 +18x4 +2x5
x2 ≤ 0x2 ≤ 1
11
on substitue x2 = −0.5x3 + 2x4 + 0.25x5 − 0.25x6 retour
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on substitue x1 = 11x2 + 5x3 − 18x4 − 2x5
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +53x2 +41x3 −204x4 −20x5
x1 = +11x2 +5x3 −18x4 −2x5
x6 = −4x2 −2x3 +8x4 +x5
x7 = 1 −11x2 −5x3 +18x4 +2x5
x2 ≤ 0x2 ≤ 1
11
on substitue x2 = −0.5x3 + 2x4 + 0.25x5 − 0.25x6 retour
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +14.5x3 −98x4 −6.75x5 −13.25x6
x1 = −0.5x3 +4x4 +0.75x5 −2.75x6
x2 = −0.5x3 +2x4 +0.25x5 −0.25x6
x7 = 1 +0.5x3 −4x4 −0.75x5 +2.75x6
x3 ≤ 0x3 ≤ 0
on substitue x3 = −2x1 + 8x4 + 1.5x5 − 5.5x6
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +18x4 +15x5 −93x6 −29x1
x3 = +8x4 +1.5x5 −5.5x6 −2x1
x2 = −2x4 −0.5x5 +2.5x6 +x1
x7 = 1 −x1
x4 ≤ 0
on substitue x4 = −0.25x5 + 1.75x6 + 0.5x1− 0.5x2
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +14.5x3 −98x4 −6.75x5 −13.25x6
x1 = −0.5x3 +4x4 +0.75x5 −2.75x6
x2 = −0.5x3 +2x4 +0.25x5 −0.25x6
x7 = 1 +0.5x3 −4x4 −0.75x5 +2.75x6
x3 ≤ 0x3 ≤ 0
on substitue x3 = −2x1 + 8x4 + 1.5x5 − 5.5x6
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +18x4 +15x5 −93x6 −29x1
x3 = +8x4 +1.5x5 −5.5x6 −2x1
x2 = −2x4 −0.5x5 +2.5x6 +x1
x7 = 1 −x1
x4 ≤ 0
on substitue x4 = −0.25x5 + 1.75x6 + 0.5x1− 0.5x2
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +14.5x3 −98x4 −6.75x5 −13.25x6
x1 = −0.5x3 +4x4 +0.75x5 −2.75x6
x2 = −0.5x3 +2x4 +0.25x5 −0.25x6
x7 = 1 +0.5x3 −4x4 −0.75x5 +2.75x6
x3 ≤ 0x3 ≤ 0
on substitue x3 = −2x1 + 8x4 + 1.5x5 − 5.5x6
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +18x4 +15x5 −93x6 −29x1
x3 = +8x4 +1.5x5 −5.5x6 −2x1
x2 = −2x4 −0.5x5 +2.5x6 +x1
x7 = 1 −x1
x4 ≤ 0
on substitue x4 = −0.25x5 + 1.75x6 + 0.5x1− 0.5x2
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +14.5x3 −98x4 −6.75x5 −13.25x6
x1 = −0.5x3 +4x4 +0.75x5 −2.75x6
x2 = −0.5x3 +2x4 +0.25x5 −0.25x6
x7 = 1 +0.5x3 −4x4 −0.75x5 +2.75x6
x3 ≤ 0x3 ≤ 0
on substitue x3 = −2x1 + 8x4 + 1.5x5 − 5.5x6
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +18x4 +15x5 −93x6 −29x1
x3 = +8x4 +1.5x5 −5.5x6 −2x1
x2 = −2x4 −0.5x5 +2.5x6 +x1
x7 = 1 −x1
x4 ≤ 0
on substitue x4 = −0.25x5 + 1.75x6 + 0.5x1− 0.5x2
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +14.5x3 −98x4 −6.75x5 −13.25x6
x1 = −0.5x3 +4x4 +0.75x5 −2.75x6
x2 = −0.5x3 +2x4 +0.25x5 −0.25x6
x7 = 1 +0.5x3 −4x4 −0.75x5 +2.75x6
x3 ≤ 0x3 ≤ 0
on substitue x3 = −2x1 + 8x4 + 1.5x5 − 5.5x6
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +18x4 +15x5 −93x6 −29x1
x3 = +8x4 +1.5x5 −5.5x6 −2x1
x2 = −2x4 −0.5x5 +2.5x6 +x1
x7 = 1 −x1
x4 ≤ 0
on substitue x4 = −0.25x5 + 1.75x6 + 0.5x1− 0.5x2
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +14.5x3 −98x4 −6.75x5 −13.25x6
x1 = −0.5x3 +4x4 +0.75x5 −2.75x6
x2 = −0.5x3 +2x4 +0.25x5 −0.25x6
x7 = 1 +0.5x3 −4x4 −0.75x5 +2.75x6
x3 ≤ 0x3 ≤ 0
on substitue x3 = −2x1 + 8x4 + 1.5x5 − 5.5x6
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +18x4 +15x5 −93x6 −29x1
x3 = +8x4 +1.5x5 −5.5x6 −2x1
x2 = −2x4 −0.5x5 +2.5x6 +x1
x7 = 1 −x1
x4 ≤ 0
on substitue x4 = −0.25x5 + 1.75x6 + 0.5x1− 0.5x2
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10.5x5 −70.5x6 −20x1 −9x2
x3 = −0.5x5 +4.5x6 +2x1 −4x2
x4 = −0.25x5 +1.75x6 +0.5x1 −0.5x2
x7 = 1 −x1
x5 ≤ 0x5 ≤ 0
on substitue x5 = 9x6 + 4x1 − 8x2 − 2x3
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +24x6 +22x1 −93x2 −21x3
x5 = +9x6 +4x1 −8x2 −2x3
x4 = −x6 −0.5x1 +1.5x2 +0.5x3
x7 = 1 −x1
x6 ≤ 0
on substitue x6 = −0.5x1 + 1.5x2 + 0.5x3 − x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10.5x5 −70.5x6 −20x1 −9x2
x3 = −0.5x5 +4.5x6 +2x1 −4x2
x4 = −0.25x5 +1.75x6 +0.5x1 −0.5x2
x7 = 1 −x1
x5 ≤ 0x5 ≤ 0
on substitue x5 = 9x6 + 4x1 − 8x2 − 2x3
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +24x6 +22x1 −93x2 −21x3
x5 = +9x6 +4x1 −8x2 −2x3
x4 = −x6 −0.5x1 +1.5x2 +0.5x3
x7 = 1 −x1
x6 ≤ 0
on substitue x6 = −0.5x1 + 1.5x2 + 0.5x3 − x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10.5x5 −70.5x6 −20x1 −9x2
x3 = −0.5x5 +4.5x6 +2x1 −4x2
x4 = −0.25x5 +1.75x6 +0.5x1 −0.5x2
x7 = 1 −x1
x5 ≤ 0x5 ≤ 0
on substitue x5 = 9x6 + 4x1 − 8x2 − 2x3
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +24x6 +22x1 −93x2 −21x3
x5 = +9x6 +4x1 −8x2 −2x3
x4 = −x6 −0.5x1 +1.5x2 +0.5x3
x7 = 1 −x1
x6 ≤ 0
on substitue x6 = −0.5x1 + 1.5x2 + 0.5x3 − x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10.5x5 −70.5x6 −20x1 −9x2
x3 = −0.5x5 +4.5x6 +2x1 −4x2
x4 = −0.25x5 +1.75x6 +0.5x1 −0.5x2
x7 = 1 −x1
x5 ≤ 0x5 ≤ 0
on substitue x5 = 9x6 + 4x1 − 8x2 − 2x3
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +24x6 +22x1 −93x2 −21x3
x5 = +9x6 +4x1 −8x2 −2x3
x4 = −x6 −0.5x1 +1.5x2 +0.5x3
x7 = 1 −x1
x6 ≤ 0
on substitue x6 = −0.5x1 + 1.5x2 + 0.5x3 − x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10.5x5 −70.5x6 −20x1 −9x2
x3 = −0.5x5 +4.5x6 +2x1 −4x2
x4 = −0.25x5 +1.75x6 +0.5x1 −0.5x2
x7 = 1 −x1
x5 ≤ 0x5 ≤ 0
on substitue x5 = 9x6 + 4x1 − 8x2 − 2x3
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +24x6 +22x1 −93x2 −21x3
x5 = +9x6 +4x1 −8x2 −2x3
x4 = −x6 −0.5x1 +1.5x2 +0.5x3
x7 = 1 −x1
x6 ≤ 0
on substitue x6 = −0.5x1 + 1.5x2 + 0.5x3 − x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10.5x5 −70.5x6 −20x1 −9x2
x3 = −0.5x5 +4.5x6 +2x1 −4x2
x4 = −0.25x5 +1.75x6 +0.5x1 −0.5x2
x7 = 1 −x1
x5 ≤ 0x5 ≤ 0
on substitue x5 = 9x6 + 4x1 − 8x2 − 2x3
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +24x6 +22x1 −93x2 −21x3
x5 = +9x6 +4x1 −8x2 −2x3
x4 = −x6 −0.5x1 +1.5x2 +0.5x3
x7 = 1 −x1
x6 ≤ 0
on substitue x6 = −0.5x1 + 1.5x2 + 0.5x3 − x4
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on entre dans une boucle ! 1er dictionnaire
lp_solve boucle aussi sur ce programme, mais trouve la solutionquand on échange les lignes x5 et x6 au départ :
max: 10 x1 - 57 x2 - 9 x3 - 24 x4;x6 = - 0.5 x1 + 1.5 x2 + 0.5 x3 - x4;x5 = - 0.5 x1 + 5.5 x2 + 2.5 x3 - 9 x4;x7 = 1 - x1 ;
on trouve :z(max)=1, x1=1, x2=0, x3=1, x4=0, x5=2, x6=0, x7=0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on entre dans une boucle ! 1er dictionnaire
lp_solve boucle aussi sur ce programme, mais trouve la solutionquand on échange les lignes x5 et x6 au départ :
max: 10 x1 - 57 x2 - 9 x3 - 24 x4;x6 = - 0.5 x1 + 1.5 x2 + 0.5 x3 - x4;x5 = - 0.5 x1 + 5.5 x2 + 2.5 x3 - 9 x4;x7 = 1 - x1 ;
on trouve :z(max)=1, x1=1, x2=0, x3=1, x4=0, x5=2, x6=0, x7=0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on entre dans une boucle ! 1er dictionnaire
lp_solve boucle aussi sur ce programme, mais trouve la solutionquand on échange les lignes x5 et x6 au départ :
max: 10 x1 - 57 x2 - 9 x3 - 24 x4;x6 = - 0.5 x1 + 1.5 x2 + 0.5 x3 - x4;x5 = - 0.5 x1 + 5.5 x2 + 2.5 x3 - 9 x4;x7 = 1 - x1 ;
on trouve :z(max)=1, x1=1, x2=0, x3=1, x4=0, x5=2, x6=0, x7=0
L3 STIC info/miage Programmation Linéaire (PLRO)
Feuille 5 Cas particuliers de l’algorithme du Simplexe
Exercice 5.2
z(max) = 0, x1 = 0, x2 = 0, x3 = 0, x4 = 0, x5 = 0, x6 = 0, x7 = 1
z(max) = +10x1 −57x2 −9x3 −24x4
x5 = −0.5x1 +5.5x2 +2.5x3 −9x4
x6 = −0.5x1 +1.5x2 +0.5x3 −x4
x7 = 1 −x1
x1 ≤ 0x1 ≤ 0x1 ≤ 1
on entre dans une boucle ! 1er dictionnaire
lp_solve boucle aussi sur ce programme, mais trouve la solutionquand on échange les lignes x5 et x6 au départ :
max: 10 x1 - 57 x2 - 9 x3 - 24 x4;x6 = - 0.5 x1 + 1.5 x2 + 0.5 x3 - x4;x5 = - 0.5 x1 + 5.5 x2 + 2.5 x3 - 9 x4;x7 = 1 - x1 ;
on trouve :z(max)=1, x1=1, x2=0, x3=1, x4=0, x5=2, x6=0, x7=0
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Feuille 5 Cas particuliers de l’algorithme du Simplexe
(*) Exercice 5.4
z(max) = +400m1 +600m2 +800m3 +1000m4
x1 = 6000 −8m1 −10m2 −12m3 −15m4
x2 = 1500 −2m1 −2m2 −4m3 −5m4
x3 = 450 −m1 −m2 −m3 −m4
x4 = 300 −m3 −m4
m4 ≤ 400m4 ≤ 300m4 ≤ 450m4 ≤ 300
on substitue m4 = 300− 0.4m1 − 0.4m2 − 0.8m3 − 0.2x2
z(max) 300000 +200m2 −200x2
x1 = 1500 −2m1 −4m2 +3x2
m4 = 300 −0.4m1 −0.4m2 −0.8m3 −0.2x2
x3 = 150 −0.6m1 −0.6m2 −0.2m3 +0.2x2
x4 = +0.4m1 +0.4m2 −0.2m3 +0.2x2
m2 ≤ 375m2 ≤ 750m2 ≤ 250
on substitue m2 = 250−m1 − 13 m3 + 1
3 x2 − 53 x3
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Feuille 5 Cas particuliers de l’algorithme du Simplexe
(*) Exercice 5.4
z(max) = +400m1 +600m2 +800m3 +1000m4
x1 = 6000 −8m1 −10m2 −12m3 −15m4
x2 = 1500 −2m1 −2m2 −4m3 −5m4
x3 = 450 −m1 −m2 −m3 −m4
x4 = 300 −m3 −m4
m4 ≤ 400m4 ≤ 300m4 ≤ 450m4 ≤ 300
on substitue m4 = 300− 0.4m1 − 0.4m2 − 0.8m3 − 0.2x2
z(max) 300000 +200m2 −200x2
x1 = 1500 −2m1 −4m2 +3x2
m4 = 300 −0.4m1 −0.4m2 −0.8m3 −0.2x2
x3 = 150 −0.6m1 −0.6m2 −0.2m3 +0.2x2
x4 = +0.4m1 +0.4m2 −0.2m3 +0.2x2
m2 ≤ 375m2 ≤ 750m2 ≤ 250
on substitue m2 = 250−m1 − 13 m3 + 1
3 x2 − 53 x3
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Feuille 5 Cas particuliers de l’algorithme du Simplexe
(*) Exercice 5.4
z(max) = +400m1 +600m2 +800m3 +1000m4
x1 = 6000 −8m1 −10m2 −12m3 −15m4
x2 = 1500 −2m1 −2m2 −4m3 −5m4
x3 = 450 −m1 −m2 −m3 −m4
x4 = 300 −m3 −m4
m4 ≤ 400m4 ≤ 300m4 ≤ 450m4 ≤ 300
on substitue m4 = 300− 0.4m1 − 0.4m2 − 0.8m3 − 0.2x2
z(max) 300000 +200m2 −200x2
x1 = 1500 −2m1 −4m2 +3x2
m4 = 300 −0.4m1 −0.4m2 −0.8m3 −0.2x2
x3 = 150 −0.6m1 −0.6m2 −0.2m3 +0.2x2
x4 = +0.4m1 +0.4m2 −0.2m3 +0.2x2
m2 ≤ 375m2 ≤ 750m2 ≤ 250
on substitue m2 = 250−m1 − 13 m3 + 1
3 x2 − 53 x3
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Feuille 5 Cas particuliers de l’algorithme du Simplexe
(*) Exercice 5.4
z(max) = +400m1 +600m2 +800m3 +1000m4
x1 = 6000 −8m1 −10m2 −12m3 −15m4
x2 = 1500 −2m1 −2m2 −4m3 −5m4
x3 = 450 −m1 −m2 −m3 −m4
x4 = 300 −m3 −m4
m4 ≤ 400m4 ≤ 300m4 ≤ 450m4 ≤ 300
on substitue m4 = 300− 0.4m1 − 0.4m2 − 0.8m3 − 0.2x2
z(max) 300000 +200m2 −200x2
x1 = 1500 −2m1 −4m2 +3x2
m4 = 300 −0.4m1 −0.4m2 −0.8m3 −0.2x2
x3 = 150 −0.6m1 −0.6m2 −0.2m3 +0.2x2
x4 = +0.4m1 +0.4m2 −0.2m3 +0.2x2
m2 ≤ 375m2 ≤ 750m2 ≤ 250
on substitue m2 = 250−m1 − 13 m3 + 1
3 x2 − 53 x3
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Feuille 5 Cas particuliers de l’algorithme du Simplexe
(*) Exercice 5.4
z(max) = +400m1 +600m2 +800m3 +1000m4
x1 = 6000 −8m1 −10m2 −12m3 −15m4
x2 = 1500 −2m1 −2m2 −4m3 −5m4
x3 = 450 −m1 −m2 −m3 −m4
x4 = 300 −m3 −m4
m4 ≤ 400m4 ≤ 300m4 ≤ 450m4 ≤ 300
on substitue m4 = 300− 0.4m1 − 0.4m2 − 0.8m3 − 0.2x2
z(max) 300000 +200m2 −200x2
x1 = 1500 −2m1 −4m2 +3x2
m4 = 300 −0.4m1 −0.4m2 −0.8m3 −0.2x2
x3 = 150 −0.6m1 −0.6m2 −0.2m3 +0.2x2
x4 = +0.4m1 +0.4m2 −0.2m3 +0.2x2
m2 ≤ 375m2 ≤ 750m2 ≤ 250
on substitue m2 = 250−m1 − 13 m3 + 1
3 x2 − 53 x3
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Feuille 5 Cas particuliers de l’algorithme du Simplexe
(*) Exercice 5.4
z(max) = +400m1 +600m2 +800m3 +1000m4
x1 = 6000 −8m1 −10m2 −12m3 −15m4
x2 = 1500 −2m1 −2m2 −4m3 −5m4
x3 = 450 −m1 −m2 −m3 −m4
x4 = 300 −m3 −m4
m4 ≤ 400m4 ≤ 300m4 ≤ 450m4 ≤ 300
on substitue m4 = 300− 0.4m1 − 0.4m2 − 0.8m3 − 0.2x2
z(max) 300000 +200m2 −200x2
x1 = 1500 −2m1 −4m2 +3x2
m4 = 300 −0.4m1 −0.4m2 −0.8m3 −0.2x2
x3 = 150 −0.6m1 −0.6m2 −0.2m3 +0.2x2
x4 = +0.4m1 +0.4m2 −0.2m3 +0.2x2
m2 ≤ 375m2 ≤ 750m2 ≤ 250
on substitue m2 = 250−m1 − 13 m3 + 1
3 x2 − 53 x3
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Feuille 5 Cas particuliers de l’algorithme du Simplexe
(*) Exercice 5.4
z(max) = 350000 −200m1 − 10003 x3 − 200
3 m3 − 4003 x2
x1 = 500 +2m1 + 203 x3 + 5
3 x2
m4 = 200 + 23 x3 − 2
3 m3 − 13 x2
m2 = 250 −m1 − 53 x3 − 1
3 m3 + 13 x2
x4 = 100 − 23 x3 − 1
3 m3 + 13 x2
solution optimale : z(max) = 350000x1 = 500, x2 = 0, x3 = 0, x4 = 100
m1 = 0, m2 = 250, m3 = 0, m4 = 200
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Feuille 5 Cas particuliers de l’algorithme du Simplexe
(*) Exercice 5.4
z(max) = 350000 −200m1 − 10003 x3 − 200
3 m3 − 4003 x2
x1 = 500 +2m1 + 203 x3 + 5
3 x2
m4 = 200 + 23 x3 − 2
3 m3 − 13 x2
m2 = 250 −m1 − 53 x3 − 1
3 m3 + 13 x2
x4 = 100 − 23 x3 − 1
3 m3 + 13 x2
solution optimale : z(max) = 350000x1 = 500, x2 = 0, x3 = 0, x4 = 100
m1 = 0, m2 = 250, m3 = 0, m4 = 200
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Feuille 6 Méthode du Simplexe
Exercice 6.2Problème 3. (*)
introduction des variables d’écarts
8<:max: x1 + x2
e1 = −4− x1 + 4x2
e2 = 6− x1 − x2
phase 1
8><>:min: 4 + x1 − 4x2 + e1
a1 = 4 + x1 −4x2 + e1
e2 = 6− x1 − x2
8<:min: a1
x2 = 1 + 14 x1 + 1
4 e1 − 14 a1
e2 = 5− 54 x1 − 1
4 e1 + 14 a1
phase 28><>:max: 1 + 5
4 x1 + 14 e1
x2 = 1 + 14 x1 + 1
4 e1
e2 = 5 − 54 x1 − 1
4 e1
8<:max: 6− e2
x2 = 2 + 15 e1 − 1
5 e2
x1 = 4− 15 e1 − 4
5 e2
x1
x2
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Feuille 6 Méthode du Simplexe
Exercice 6.2Problème 4. (*)
phase 18<:min: ae1 = −1− x1 + 2x2 + ae2 = −1− 2x1 + x2 +a
8<:min: 1 + 2x1 − x2 + e2
e1 = x1 + x2 + e2
a = 1 + 2x1 −x2 + e2
8<:min: ae1 = 1 + 3x1 + 2e2 − ax2 = 1 + 2x1 + e2 − a
phase 28><>:max: 1 +3x1 + e2
e1 = 1 + 3x1 + 2e2
x2 = 1 + 2x1 + e2
=⇒ problème non borné
x1
x2
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Feuille 6 Méthode du Simplexe
Exercice 6.2Problème 6. (*)
x2
x1
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Feuille 10 Résolution de Problème, Interprétation des Résultats
(*) Exercice 10.2
Correction en
salle machines
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Feuille 11 Algorithme du Simplexe Révisé
(*) Exercice 11.2
max : CNxN + CBxBs.c. ANxN + ABxB = b
CN =(
3 2 4)
CB =(
0 0 0)
AN =
1 1 22 0 32 1 3
AB =
1 0 00 1 00 0 1
b =
457
= x∗B
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Feuille 11 Algorithme du Simplexe Révisé
(*) Exercice 11.2
Itération 1 :
1 yAB = CB ⇐⇒ y
0@ 1 0 00 1 00 0 1
1A =`
0 0 0´
=⇒ y =`
0 0 0´
2 on peut choisir la troisième colonne :quatrième : a =
0@ 233
1A3 ABd = a ⇐⇒
0@ 1 0 00 1 00 0 1
1A d =
0@ 233
1A =⇒ d =
0@ 233
1A
4 x∗B − td ≥ 0 ⇐⇒
0@ 4− 2t5− 3t7− 3t
1A ≥
0@ 000
1A → t ≤ 2
→ t ≤ 53
→ t ≤ 73
5 variable sortante : x5, colonne sortante : deuxième colonne de AB ,variable entrante : x3, colonne entrante : troisième colonne de AN .
Solution : x∗B =
0@ x∗4x∗3x∗6
1A =
0@ 23532
1ACN =
`3 2 0
´CB =
`0 4 0
´AN =
0@ 1 1 02 0 12 1 0
1A AB =
0@ 1 2 00 3 00 3 1
1A
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Feuille 11 Algorithme du Simplexe Révisé
(*) Exercice 11.2
Itération 2 :
1 yAB = CB ⇐⇒ y
0@ 1 2 00 3 00 3 1
1A =`
0 4 0´
=⇒ y =`
0 43 0
´
2 on choisit la première colonne : a =
0@ 122
1A3 ABd = a ⇐⇒
0@ 1 2 00 3 00 3 1
1A d =
0@ 122
1A =⇒ d =
0@ − 13
230
1A
4 x∗B − td ≥ 0 ⇐⇒
0@ 23 + 1
3 t53 −
23 t
2
1A ≥
0@ 000
1A → t ≥ −2
→ t ≤ 52
→ t ∈ R5 variable sortante : x3, colonne sortante : deuxième colonne de AB ,
variable entrante : x1, colonne entrante : première colonne de AN .
Solution : x∗B =
0@ x∗4x∗1x∗6
1A =
0@ 32522
1ACN =
`4 2 0
´CB =
`0 3 0
´AN =
0@ 2 1 03 0 13 1 0
1A AB =
0@ 1 1 00 2 00 2 1
1A
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Feuille 11 Algorithme du Simplexe Révisé
(*) Exercice 11.2
Itération 3 :
1 yAB = CB ⇐⇒ y
0@ 1 1 00 2 00 2 1
1A =`
0 3 0´
=⇒ y =`
0 32 0
´
2 on choisit la deuxière colonne : a =
0@ 101
1A3 ABd = a ⇐⇒
0@ 1 1 00 2 00 2 1
1A d =
0@ 101
1A =⇒ d =
0@ 101
1A
4 x∗B − td ≥ 0 ⇐⇒
0@ 32 − t
52
2− t
1A ≥
0@ 000
1A → t ≤ 32
→ t ∈ R→ t ≤ 2
5 variable sortante : x4, colonne sortante : première colonne de AB ,variable entrante : x2, colonne entrante : deuxième colonne de AN .
Solution : x∗B =
0@ x∗2x∗1x∗6
1A =
0@ 325212
1ACN =
`4 0 0
´CB =
`2 3 0
´AN =
0@ 2 1 03 0 13 0 0
1A AB =
0@ 1 1 00 2 01 2 1
1A
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Feuille 11 Algorithme du Simplexe Révisé
(*) Exercice 11.2
Itération 4 :
1 yAB = CB ⇐⇒ y
1 1 00 2 01 2 1
=(
2 3 0)
=⇒ y =(2 1
2 0)
2 aucune colonne entrante donc solution optimale.
z = CNx∗N + CBx∗B = CBx∗B =(
2 3 0) 3
25212
= 212
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