Tarea de investigación: MAXIMIZAR: 1 X1 + 3 X2 + 2 X3
Enviado por Elperro Elgato • 22 de Octubre de 2015 • Tareas • 765 Palabras (4 Páginas) • 269 Visitas
Presentado por: Israel Antonio Castiblanco C. Código: 31555
MAXIMIZAR: 1 X1 + 3 X2 + 2 X3 | [pic 1] | MAXIMIZAR: 1 X1 + 3 X2 + 2 X3 + 0 X4 + 0 X5 + 0 X6 | ||||||
3 X1 + 1 X2 + 2 X3 ≤ 7 | 3 X1 + 1 X2 + 2 X3 + 1 X4 = 7 | |||||||
X1, X2, X3 ≥ 0 | X1, X2, X3, X4, X5, X6 ≥ 0 | |||||||
Tabla 1 | ||||||||
Base | z | b | X1 | X2 | X3 | X4 | X5 | X6 |
X4 | 0 | 7 | 3 | 1 | 2 | 1 | 0 | 0 |
X5 | 0 | 6 | 1 | 2 | 0 | 0 | 1 | 0 |
X6 | 0 | 40 | 4 | 3 | 8 | 0 | 0 | 1 |
Z |
| 0 | -1 | -3 | -2 | 0 | 0 | 0 |
Tabla 2 | ||||||||
Base | z | b | X1 | X2 | X3 | X4 | X5 | X6 |
X4 | 0 | 4 | 2.5 | 0 | 2 | 1 | -0.5 | 0 |
X2 | 3 | 3 | 0.5 | 1 | 0 | 0 | 0.5 | 0 |
X6 | 0 | 31 | 2.5 | 0 | 8 | 0 | -1.5 | 1 |
Z |
| 9 | 0.5 | 0 | -2 | 0 | 1.5 | 0 |
Tabla 3 | ||||||||
Base | z | b | X1 | X2 | X3 | X4 | X5 | X6 |
X3 | 2 | 2 | 1.25 | 0 | 1 | 0.5 | -0.25 | 0 |
X2 | 3 | 3 | 0.5 | 1 | 0 | 0 | 0.5 | 0 |
X6 | 0 | 15 | -7.5 | 0 | 0 | -4 | 0.5 | 1 |
Z |
| 13 | 3 | 0 | 0 | 1 | 1 | 0 |
La solución óptima es Z = 13
2) El vector de disponibilidad cambia a 4, 5,6
MAXIMIZAR: 1 X1 + 3 X2 + 2 X3 | [pic 2] | MAXIMIZAR: 1 X1 + 3 X2 + 2 X3 + 0 X4 + 0 X5 + 0 X6 |
3 X1 + 1 X2 + 2 X3 ≤ 4 | 3 X1 + 1 X2 + 2 X3 + 1 X4 = 4 | |
X1, X2, X3 ≥ 0 | X1, X2, X3, X4, X5, X6 ≥ 0 |
Tabla 3 | ||||||||
Base | Z | b | X1 | X2 | X3 | X4 | X5 | X6 |
X4 | 0 | 7 / 8 | 15 / 8 | 0 | 0 | 1 | -1 / 8 | -1 / 4 |
X2 | 3 | 5 / 2 | 1 / 2 | 1 | 0 | 0 | 1 / 2 | 0 |
X3 | 2 | 5 / 16 | 5 / 16 | 0 | 1 | 0 | -3 / 16 | 1 / 8 |
Z |
| 65 / 8 | 9 / 8 | 0 | 0 | 0 | 9 / 8 | 1 / 4 |
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