Fiscalité marocaine
967 mots
4 pages
Exercice 1.2.1.R´soudre par le simplexe e Max x1 + 2x2
−3x1 + 2x2 ≤ 2
sous
−x1 + 2x2 ≤ 4
x1 + x 2 ≤ 5
xi ≥ 0 i = 1, 2
1) Forme standard
Min z = −(x1 + 2x2)
−3x1 + 2x2 + x3
=2
sous −x1 + 2x2
+ x4
=4
x1 + x 2
+ x5 = 5
xi ≥ 0 i = 1, . . . , 5
1
2) Tableau du simplexe (forme canonique !) x1 -1
-3
-1
1
x2
-2
2
2
1
x3
0
1
0
0
x4
0
0
1
0
x5
0
0
0
1
z
-1
0
0
0
b
0
2
4
5
3) Si SBR, alors phase II (sinon phase I)
Ici, ´vident e
x1
x
3
x4
x5
= x2 = 0
=2≥0
=4≥0
=5≥0
4) sol pas optimale car ∃ cj ≤ 0
5) Changement de base : c2 + n´gatif que c1 → x2 rentre dans la base. e ? Variable xs sortant de la base b 5
2
t = arg mini{ a i }|ai2≥0 = min{ 2 , 4 , 1 } = 2
2 2 i2 ⇒t=1
1
xs tq B −1 as = et = 0 → s = 3
0
2
6) Tableau canonique de la nouvelle base l2 l1 l3 l4
x1
-4
-3
2
2
5
2
x2
0
1
0
0
=
=
=
=
x3
1
1
2
-1
-1
2
l2/2 l1 + l2 l3 − l2 l4 − l2/2
x4
0
0
1
0
x5
0
0
0
1
z
-1
0
0
0
b
2
1
2
4
7) seul c1 < 0 → x1 entre en base
4
2 min{ 2 , 5/2 } = 2 → x4 sort de la base
2
l3 l1 l2 l4 =
=
=
=
l3/2 l1 + 2l3 l2 + 3l3/4 l4 − 5l3/4
3
x1
0
0
1
0
x2
0
1
0
0
x3
-1
-1
4
1
-2
3
4
x4
2
3
4
1
2
-5
4
x5
0
0
0
1
z
-1
0
0
0
b
6
5
2
1
3
2
8) seul c3 < 0 → x3 entre en base
3/2
min{ 3/4 } → x5 sort de la base l4 l1 l2 l3
x1
0
0
1
0
x2
0
1
0
0
=
=
=
=
x3
0
0
0
1
4l4 /3 l1 + 4l4 /3 l2 + l4/3 l3 + 2l4 /3
x4
1
3
1
3
-1
3
5
-3
x5
4
3
1
3
2
3
4
3
z
-1
0
0
0
b
8
3
2
2
• sol : x1 = 2; x2 = 3; x3 = 2; x4 = x5 = 0
• coˆt = -8 u • sol optimale car tous les cj ≥ 0
4
Exercice 1.2.2. x1 0
0
1
0
x2
6
5
4
7
x3
0
1
0
0
x4
0
0
0
1
z
-1
0
0
0
b
31
7
5
12