INVESTIGACION DE OPERACIONES 1. UNIDAD 2. METODO SIMPLEX
Enviado por luceromariscal • 27 de Abril de 2020 • Tareas • 1.537 Palabras (7 Páginas) • 682 Visitas
INVESTIGACION DE OPERACIONES 1. UNIDAD 2. METODO SIMPLEX.
ALUMNO: ericka osuna molina GRUPO de f
RESOLVER LOS SIGUIENTES EJERCICIOS POR EL METODO SIMPLEX FORMA TABULAR.
Max. Z = -2x1 + 3x2 -2x3
s.a. 4x1 – x2 -5x3 = 10
2x1 + 3x2 + 2x3 = 12
X1>= 0, X2>=0, X3>= 0
MAX Z = 2X1 -3X2 + 2X3
4X1 – X2 -5X3 + S1 = 10
2/3 | 1 | 2/3 | 0 | 1/3 | 4 |
2 | 3 | 2 | 0 | 1 | 12 |
2 | -3 | 2 | 0 | 0 | 0 |
4 | 0 | 4 | 0 | 1 | 12 |
2X1 + 3X2 + 2X3 + S2 =12
EC | VB | X1 | X2 | X3 | S1 | S2 | LD |
0 | Z | 2 | -3 | 2 | 0 | 0 | 0 |
1 | S1 | 4 | -1 | -5 | 1 | 0 | 10 |
2 | S2 | 2 | 3 | 2 | 0 | 1 | 12 |
[pic 1]
10/-1=-10
12/3=4
EC | VB | X1 | X2 | X3 | S1 | S2 | LD |
0 | Z | 4 | 0 | 4 | 0 | 1 | 12 |
1 | S1 | 14/3 | 0 | -13/3 | 1 | 1/3 | 14 |
2 | X2 | 2/3 | 1 | 2/3 | 0 | 1/3 | 4 |
2/3 | 1 | 2/3 | 0 | 1/3 | 4 |
2/3 | 1 | 2/3 | 0 | 1/3 | 4 |
4 | -1 | -5 | 1 | 0 | 10 |
14/3 | 0 | -13/3 | 1 | 1/3 | 14 |
[pic 2]
5/2 / -1/4=-10
7 / 9/2=14/9
Solucion optima Z= 12, x1=0, x2=4.x3=0
2: Max Z = 2x1 + 3x2
s.a. 2x1 +x2 <= 4
x1 + 2x2 <= 5
x1, x2 >= 0
MAX Z: -2X -3X2
EC 1= 2X1 + X2 + S1 = 4
EC 2= X1 + 2X2 + S2 =5
EC | VB | X1 | X2 | S1 | S2 | LD |
0 | Z | -2 | -3 | 0 | 0 | 0 |
1 | S1 | 2 | 1 | 1 | 0 | 4 |
2 | S2 | 1 | 2 | 0 | 1 | 5 |
1/2 | 1 | 0 | 1/2 | 5/2 |
3/2 | 3 | 0 | 3/2 | 15/2 |
-2 | -3 | 0 | 0 | 0 |
-1/2 | 0 | 0 | 3/2 | 15/2 |
[pic 3]
4/1=4
5/2=5/2
EC | VB | X1 | X2 | S1 | S2 | LD |
0 | Z | -1/2 | 0 | 0 | 3/2 | 15/2 |
1 | S1 | 3/2 | 0 | 1 | -1/2 | 3/2 |
2 | X2 | 1/2 | 1 | 0 | 1/2 | 5/2 |
1/2 | 1 | 0 | 1/2 | 5/2 |
-1/2 | -1 | 0 | -1/2 | -5/2 |
2 | 1 | 1 | 0 | 4 |
3/2 | 0 | 1 | -1/2 | 3/2 |
[pic 4]
1.5/1.5=1
2.5/0.5=5
1 | 0 | 2/3 | -1/3 | 1 |
1/2 | 0 | 1/3 | -1/6 | 1/2 |
-1/2 | 0 | 0 | 3/2 | 15/2 |
0 | 0 | 1/3 | 4/3 | 8 |
EC | VB | X1 | X2 | S1 | S2 | LD |
0 | Z | 0 | 0 | 1/3 | 4/3 | 8 |
1 | X1 | 1 | 0 | 2/3 | -1/3 | 1 |
2 | X2 | 0 | 1 | -1/3 | 2/3 | 2 |
[pic 5]
1 | 0 | 2/3 | -1/3 | 1 |
-1/2 | 0 | -1/3 | 1/6 | -1/2 |
1/2 | 1 | 0 | 1/2 | 5/2 |
0 | 1 | -1/3 | 2/3 | 2 |
[pic 6]
Z= 8
X1=1
X2=2
3: Max Z = 2x1 – 4x2 + 5x3 -6x4
S.a. x1 + 4x2 – 2x3 + 8x4 <= 2
- x1 + 2x2 + 3x3 + 4x4 <= 1
X1, ……x4 >=0
...