Investigacion Operativa

Universidad Nacional Tecnológica del Cono Sur de Lima - UNTECS
Universidad Nacional Tecnológica del Cono Sur de Lima - UNTECS

MÉTODO SIMPLEX Y GRAN M


MÉTODO SIMPLEX Y GRAN M



INTEGRANTES:
Gallardo Salazar, Mariella
Peña Berrú, Verónica
PROFESOR:
Mario Ninaquispe Soto

INTEGRANTES:
Gallardo Salazar, Mariella
Peña Berrú, Verónica
PROFESOR:
Mario NinaquispeSoto

Hallar la solución de los siguientes problemas:

1. Maximizar:
Maximizar: z= -x1 + 2 x2
Dependiendo de: 6x1 - 2x2 ≤ 3
-2x1+ 3x2 ≤ 6
x1+ x2 ≤ 3
x1 , x2 ≥ 0
Forma estándar:
Maximizar: z= -x1 + 2 x2 +0S1+ 0S2+ 0S3
Dependiendo de: 6x1 - 2x2 + S1= 3
-2x1+ 3x2+ S2= 6

x1+ x2 + S3= 3
x1, x2 ≥ 0

| Z | X1 | X2 | S1 | S2 | S3 | SOLUCIÓN |Z | 1 | 1 | -2 | 0 | 0 | 0 | 0 |
S1 | 0 | 6 | -2 | 1 | 0 | 0 | 3 |
S2 | 0 | -2 | 3 | 0 | 1 | 0 | 6 |
S3 | 0 | 1 | 1 | 0 | 0 | 1 | 3 |
Z | 1 | -1/3 | 0 | 0 | 2/3 | 0 | 4 |
S1 | 0 | 4.667 | 0 | 1 | 2/3 | 0 | 7 |
X2 | 0 | -2/3 | 1 | 0 | 1/3 | 0 | 2 |
S3 | 0 | 1.667 | 0 | 0 | -1/3 | 1 | 1 |
Z | 1 | 0 | 0 | 0 | 3/5 | 1/5 | 4.2 |
S1 | 0 | 0 | 0 | 1 | 1.6 | -14/5 | 4.2 |
X2 | 0 | 0| 1 | 0 | 1/5 | 2/5 | 2.4 |
X1 | 0 | 1 | 0 | 0 | -1/5 | 3/5 | 0.6 |

2. Maximizar: z= 3x1 + 5 x2
Dependiendo de: -3x1+ 2x2 ≤ 6
-x1+ x2 ≤ 5
3x1+ 8x2 ≥ 12
3x1+ 2x2 ≥ 18
x1 , x2 ≥ 0

Forma estándar:
Maximizar: z= 3x1 + 5 x2 + 0S1 + 0S2 + 0S3 + 0S4 - MR1 - MR2
Dependiendo de: -3x1+ 2x2 + S1 = 6
-x1+ x2 + S2= 5
3x1+ 8x2 - S3+ R1= 12
3x1+ 2x2 -S4+ R2= 18
| Z | X1 | X2 | S3 | S4 | R1 | R2 | S1 | S2 | SOLUCIÓN |
Z | 1 | -3 | -5 | 0 | 0 | 0 | 0 | M | M | 0 |
S1 | 0 | -3 | 2 | 0 | 0 | 1 | 0 | 0 | 0 | 6 |
S2 | 0 | -1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 5 |
R1 | 0 | 3 | 8 | -1 | 0 | 0 | 0 | 1 | 0 | 12 |
R2 | 0 | 3 | 2 | 0 | -1 | 0 | 0 | 0 | 1 | 18 |
Z | 1 | -6M-3 | -10M+5 | M | M | 0 | 0 | 0 | 0 | -30M |
S1 | 0 | -3 | 2 | 0 | 0 | 1| 0 | 0 | 0 | 6 |
S2 | 0 | -1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 5 |
R1 | 0 | 3 | 8 | -1 | 0 | 0 | 0 | 1 | 0 | 12 |
R2 | 0 | 3 | 2 | 0 | -1 | 0 | 0 | 0 | 1 | 18 |
Z | 1 | -2.25M -1.125 | 0 | -0.25M -0.625 | M | 0 | 0 | 1.25M +0.625 | 0 | -15M 7.5 |
S1 | 0 | -3.75 | 0 | 0.25 | 0 | 1 | 0 | -0.25 | 0 | 3 |
S2 | 0 | -1.375 | 0 | 0.125 | 0 | 0 | 1 | -0.125 | 0 | 3.5 |
X2 | 0 | 0.375 | 1 |-1/8 | 0 | 0 | 0 | 1/8 | 0 | 1.5 |
R2 | 0 | 2.25 | 0 | 0.25 | -1 | 0 | 0 | -0.25 | 1 | 15 |
Z | 1 | 0 | 6M+3 | -M-1 | M | 0 | 0 | 2M+1 | 0 | -6M+12 |
S1 | 0 | 0 | 10 | -1 | 0 | 1 | 0 | 1 | 0 | 18 |
S2 | 0 | 0 | 3.67 | -0.3289 | 0 | 0 | 1 | 0.4538 | 0 | 9 |
X1 | 0 | 1 | 8/3 | -1/3 | 0 | 0 | 0 | 1/3 | 0 | 4 |
R2 | 0 | 0 | -6 | 1 | -1 | 0 | 0 | -1 | 1 | 6 |
Z | 1 | 0 | -3 | 0 | -1 | 0 | 0| M | M+1 | 18 |
S1 | 0 | 0 | 4 | 0 | -1 | 1 | 0 | 0 | 1 | 24 |
S2 | 0 | 0 | 1.7 | 0 | -0.3289 | 0 | 1 | 0.1249 | 0.3289 | 11 |
X1 | 0 | 1 | 2/3 | 0 | -1/3 | 0 | 0 | 0 | 1/3 | 6 |
S3 | 0 | 0 | -6 | 1 | -1 | 0 | 0 | -1 | 1 | 6 |
Z | 1 | 0 | 0 | 0 | -1.75 | 3/4 | 0 | M | 3/4M+3/4 | 36 |
X2 | 0 | 0 | 1 | 0 | -1/4 | ¼ | 0 | 0 | ¼ | 6 |
S2 | 0 | 0 | 0 | 0 | 0.08 | -0.425 | 1 | 0.125 |-0.0961 | 1 |
X1 | 0 | 1 | 0 | 0 | -1/6 | -1/6 | 0 | 0 | ½ | 2 |
S3 | 0 | 0 | 0 | 1 | -2.5 | 3/2 | 0 | -1 | 3/2 | 42 |
Z | 1 | 0 | 0 | 0 | 0 | -8 | 21 | M | M | 57 |
X2 | 0 | 0 | 1 | 0 | 0 | -1 | 3 | 0 | 0 | 9 |
S4 | 0 | 0 | 0 | 0 | 1 | -5 | 12 | 0 | -1 | 12 |
X1 | 0 | 1 | 0 | 0 | 0 | -1 | 2 | 0 | 0 | 4 |
S3 | 0 | 0 | 0 | 1 | 0 | -11 | 30 | -1 | 0 | 72 |

3. Maximizar: z= 3x1 + 7x2

Dependiendo de: x1- x2 ≥ 4
x1 - 2x2 ≤ 10
-2x1 - x2 ≥ 12
x1 , x2 ≥ 0

Forma estándar:
Maximizar: z= 3x1 + 7 x2 + 0S1 + 0S2 + 0S3 - MR1 - MR2
Dependiendo de: x1- x2 - S1 + R1 = 4
x1- 2x2 + S2= 10
-2x1- x2 - S3+ R2= 2
x1 , x2 ≥ 0

| Z | X1 | X2 | S1 | S3 | R1 | R2 | S2 | SOLUCIÓN |
Z | 1 | -3 | -7 | 0 | 0 | M | M | 0 | 0 |
R1 | 0 | 1 |...