Investigacion operativa

Problema 2
Z= -x1+4x2+6x3
sa
x1+2x2+3x3 ≤ 6
-2x1-5x2+x3 ≤10
X1,X2 ≤ 0
DESARROLLO
X1-4x2-6x3+Z = 0X1+2X2+3X3+S1=6
-2X1-5X2+X3+S2=10
| X1 | X2 | X3 | S1 | S2 | VS |
Z | 1 | -4 | -6 | 0 | 0 | 0 |
S1 | 1 | 2 | 3| 1 | 0 | 6 |
S2 | -2 | -5 | 1 | 0 | 1 | 10 |

| X1 | X2 | X3 | S1 | S2 | VS |
Z | 3 | 0 | 0 | 2 | 0 | 12 |X3 | 1/3 | 2/3 | 1 | 1/3 | 0 | 2 |
S2 | -7/3 | -17/3 | 0 | -1/3 | 1 | 8 |



Problema 1
Z= 5x1+4x2
Sa.X1+x2 ≤ 20
2x1+x2 ≤ 35
-3x1+x2 ≤ 12
DESARROLLO
-S1-5X1-4x2+Z = 0
X1+X2+S1=20
2X1+X2+S2=35-3x1+x2+S3=12

| X1 | X2 | S1 | S2 | S3 | VS |
Z | -5 | -4 | 0 | 0 | 0 | 0 |
S1 | 1 | 1 | 1 | 0 | 0 | 20 |
S2 | 2 |1 | 0 | 1 | 0 | 35 |
S3 | -3 | 1 | 1 | 0 | 0 | 12 |

| X1 | X2 | S1 | S2 | S3 | VS |
Z | 0 | -3/2 | 0 | 5/2| 0 | 87,5 |
S1 | 0 | 1/2 | 1 | -1/2 | 0 | 2,5 |
X1 | 1 | 1/2 | 0 | 1/2 | 0 | 17,5 |
S3 | 0 | 5/2 | 1 | 3/2 |0 | 64,5 |

| X1 | X2 | S1 | S2 | S3 | VS |
Z | 0 | 0 | 3 | 1 | 0 | 95 |
X2 | 0 | 1 | 2 | -1 | 0 | 5 |
X1| 1 | 0 | -1 | 1 | 0 | 15 |
S3 | 0 | 0 | -4 | 4 | 0 | 52 |

Z= 95
X2=5
X1=15
S3=52
investigacion operativa