EJERCICIO METODO SIMPLEX
Alfonsina PinedaTrabajo7 de Diciembre de 2022
2.404 Palabras (10 Páginas)138 Visitas
[pic 1]
UNIVERSIDAD TÉCNICA DE MACHALA[pic 2][pic 3]
D. L. No. 69-04, DE 14 DE ABRIL DE 1969
PROVINCIA DE EL ORO - REPUBLICA DEL ECUADOR
Calidad, Pertinencia y Calidez
FACULTAD DE CIENCIAS EMPRESARIALES
CARRERA DE COMERCIO EXTERIOR
Investigación de Operaciones.
TRABAJO INDIVIDUAL
Ejercicios – Método Simplex
Estudiante:
MARÍA ALFONSINA PINEDA SARANGO
DOCENTE:
ING. ADRIANO RAMÍREZ GALEANO.
FECHA:
26/06/2022
CURSO:
Cuarto A – Diurno
PERIODO ACADÉMICO
D1-2022
EJERCICIOS.
- Maximizar:
Z = X1 + 2x2
Sujeta a
2 X1 + X2 ≤ 8.
2 X1 + 3X2 ≤ 12.
X1. X2 ≥ 0.
Método Simplex
2 X1 + X2 +S1 ≤ 8.
2 X1 + 3X2 +S2 ≤ 12.
Z X1 + 2X2 = 0
Tabla Simplex:
V.B | Z | X1 | X2 | S1 | S2 | C.R. |
S1 | 0 | 2 | 1 | 1 | 0 | 8 |
S2 | 0 | 2 | 3[pic 4] | 0 | 1 | 12 |
Z | 1 | 1 | 2[pic 5] | 0 | 0 | 0 |
Fila pivote: 8/1= 8; 12/3= 4. Elemento pivote: 3. Columna pivote X2
V.B | Z | X1 | X2 | S1 | S2 | C.R. |
S1 | 0 | 4/3 | 0 | 0 | -1/3 | 4 |
X2 | 0 | 2/3 | 1 | 0 | 1/3 | 4 |
Z | 1 | 1/3 | 0 | 0 | 2/3 | 8 |
X2: 0/3= 0; 2/3; 3/3= 1; 0/3= 0; 1/3; 12/3= 4
S1: 0 4/3 0 0 -1/3 4
Fila vieja S1: 0 2 1 1 0 8
C.P. S1 : 1 1 1 1 1 1
Fila entra S1: 0 2/3 1 0 1/3 4
Z: 1 1/3 0 0 2/3 8
Fila vieja Z: 1 1 2 0 0 0
C.P. S1 : 2 2 2 2 2 2
Fila entra S1: 0 2/3 1 0 1/3 4
SOLUCIÓN ÓPTIMA:
Z= 8; X2= 4
Z = X1 + 2x2
8 = (0) + 2(4)
8 = 8
- Maximizar:
Z = 2X1 + X2
Sujeta a
- X1 + X2 ≤ 4.
X1 + X2 ≤ 6.
X1. X2 ≥ 0.
Modelo:
Z – 2x1 - x2 =0
- x1 + x2 + S1 =4
x1 + x2 +S2 =6
V.B | Z | X1 | X2 | S1 | S2 | C.R. |
S1 | 0 | -1 | 1 | 1 | 0 | 4 |
S2 | 0 | 1 | 1[pic 6] | 0 | 1 | 6 |
Z | 1 | -2[pic 7] | -1 | 0 | 0 | 0 |
Fila pivote: 0/-1= -4; 6/1= 6. Elemento pivote: 1. Columna pivote X1
V.B | Z | X1 | X2 | S1 | S2 | C.R. |
S1 | 0 | 0 | 2 | 1 | 1 | 10 |
X1 | 0 | 1 | 1 | 0 | 1 | 6 |
Z | 1 | 0 | 1 | 0 | 2 | 12 |
Solución óptima: Z= 12; X1= 6
Z = 2X1 + X2
(12) = 2 (6) + (0)
12 = 12
- Maximizar:
Z = -X1 + 3X2
Sujeta a
X1 + X2 ≤ 6.
-X1 + X2 ≤ 4.
X1. X2 ≥ 0.
Modelo:
Z – x1 - 3x2 =0
x1 + x2 + S1 =6
- x1 + x2 +S2 =4
V.B | Z | X1 | X2[pic 8] | S1 | S2 | C.R. |
S1 | 0 | 1 | 1 | 1[pic 9] | 0 | 6 |
S2 | 0 | -1 | 1 | 0 | 1 | 4 |
Z | 1 | 1 | -3 | 0 | 0 | 0 |
Fila pivote S2: 6/1= 6; 4/1= 4. Elemento pivote: 1. Columna pivote X2
V.B | Z | X1 | X2 | S1 | S2 | C.R. |
F1 | 0 | 2 | 0 | -1 | -1 | 2 |
F2 | 0 | -1 | 1 | 0 | 1 | 4 |
Z | 1 | -2 | 0 | 0 | 3 | 12 |
V.B | Z | X1 | X2 | S1 | S2 | C.R. |
F1 | 0[pic 10] | 1 | 0 | 1/2 | -1/2 | 1 |
F2 | 0 | -1 | 1 | 0 | 1 | 4 |
Z | 1 | -2[pic 11] | 0 | 0 | 3 | 12 |
V.B | Z | X1 | X2 | S1 | S2 | C.R. |
F1 | 0 | 1 | 0 | 1/2 | -1/2 | 1 |
F2 | 0 | 0 | 1 | 1/2 | 1/2 | 5 |
Z | 1 | 0 | 0 | 1 | 2 | 14 |
- Maximizar:
Z = 3X1 + 8X2
Sujeta a
X1 + 2X2 ≤ 8.
X1 + 6X2 ≤ 12.
X1. X2 ≥ 0.
Z – 3x1 - 8x2 =0
x1 + 2x2 + S1 =8
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