Comercio Internacional

Páginas: 2 (406 palabras) Publicado: 29 de octubre de 2012
10. Maximizar:
Z=-x1+2x2
Sujeta a:
x1+x2≤1 x1-x2≤1 x1-x1≥-2 (-1) x1≤2 x1,x2≥0
x1+x2≤1 x1-x2≤1 -x1+x1≤2 x1≤2 x1,x2≥0

| x1 | x2 | s1 | s2 | s3 | s4| z | b | |
s1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1÷1=1 |
s2 | 1 | -1 | 0 | 1 | 0 | 0 | 0 | 1 | 1÷-1= -1 |
s3 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 2 | 2÷0=∞ |
s4 | 1 | 0 | 0 | 0 | 0 | 1 | 0| 2 | 2÷0=∞ |
z | 1 | -2 | 0 | 0 | 0 | 0 | 1 | 0 | |
Columna
Pivote
Columna
Pivote

F2= -1F2
| x1 | x2 | s1 | s2 | s3 | s4 | z | b | |
s1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | |s2 | -1 | 1 | 0 | -1 | 0 | 0 | 0 | -1 | |
s3 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 2 | |
s4 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 2 | |
z | # Pivote
# Pivote
1 | -2 | 0 | 0 | 0 | 0 | 1 | 0 ||
F1= F1 – F2
F1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
-F2 | 1 | -1 | 0 | 1 | 0 | 0 | 0 | 1 |
F1 | 2 | 0 | 1 | 1 | 0 | 0 | 0 | 2 |

F5= 2F2 + F5
2F2 | -2 | 2 | 0 | -2 | 0 | 0 | 0 | -2|
F5 | 1 | -2 | 0 | 0 | 0 | 0 | 1 | 0 |
F5 | -1 | 0 | 0 | -2 | 0 | 0 | 1 | -2 |

Nueva matriz
| x1 | x2 | s1 | s2 | s3 | s4 | z | b | |
s1 | 2 | 0 | 1 | 1 | 0 | 0 | 0 | 2 |2÷1=2 |
x2 | -1 | 1 | 0 | -1 | 0 | 0 | 0 | -1 | |
s3 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 2 | 2÷0=∞ |
s4 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 2 | 2÷0=∞ |
z | -1 | 0 | 0 | -2 | 0 | 0 | 1 | -2 | |F2= F1 + F2
F1 | 2 | 0 | 1 | 1 | 0 | 0 | 0 | 2 |
F2 | -1 | 1 | 0 | -1 | 0 | 0 | 0 | -1 |
F2 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |

F5= 2F1 + F5
2F1 | 4 | 0 | 2 | 2 | 0 | 0 | 0 | 4 |F5 | -1 | 0 | 0 | -2 | 0 | 0 | 1 | -2 |
F5 | 3 | 0 | 2 | 0 | 0 | 0 | 1 | 2 |
Nueva matriz
| x1 | x2 | s1 | s2 | s3 | s4 | z | b |
s2 | 2 | 0 | 1 | 1 | 0 | 0 | 0 | 2 |
x2 | 1 | 1 | 1| 0 | 0 | 0 | 0 | 1 |
s3 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 2 |
s4 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 2 |
z | 3 | 0 | 2 | 0 | 0 | 0 | 1 | 2 |

conjunto soluciónx1=0x2=1s1=0s2=2s3=2s4=2Z=2
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