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CONTINUACION DEL 2DO PARCIAL
Min Z= 2x1 + 4x2 + x3 Z - 2x1 - 4x2 - x3 - w1 - w2 = 0
S.A. x1 + x2 ð 5 x1 + x2 + x4 = 5
2x1 - x2 + 2x3 = 2 2x1 - x2 + 2x3 + w1 = 2
- x1 - 2x2 + 2x3 ≥ 1 - x1 - 2x2 + 2x3 -x5 + w2 = 1
ðx1 ≥ 0
Tableau inicial
| Z | X1 | X2 | X3 | X4 | X5 | W1 | W2 | Solución |
F.O. | 1 | 0 | 0 | 0 | 0 | 0 | -1 | -1 | 0 |
| | 2 | -1 | 2 | 0 | 0 | 0 | -1 | 2 |
| | 1 | -3 | 4 | 0 | -1 | 0 | 0 | 3 |
X4 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 5 |
W1 | 0 | 2 | -1 | 2 | 0 | 0 | 1 | 0 | 2 |
W2 | 0 | -1 | -2 | 2 | 0 | -1 | 0 | 1 | 1 |
W1 2 -1 2 0 0 1 0 2 * Para hacer
+ F.O. 0 0 0 0 0 -1 -1 0 W1 = 0
2 -1 2 0 0 0 -1 2
* Para hacer
+W2 -1 -2 2 0 -1 0 1 1 W2 = 0
1 -3 4 0 -1 0 0 3
Las dos funciones artificiales son cero, así ya se puede hacer el método simplex
2do tableau
| Z | X1 | X2 | X3 | X4 | X5 | W1 | W2 |
Descargar
Enviado por: | Colaboradora |
Idioma: | castellano |
País: | México |