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Maximizar: 7 X Maximizar: 7 X

This document describes solving a linear programming problem with 11 variables (X1-X11) and 6 constraints using the two-phase simplex method. The problem is infeasible as shown in the final simplex table where there is a negative value in the RHS column.

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33 views3 pages

Maximizar: 7 X Maximizar: 7 X

This document describes solving a linear programming problem with 11 variables (X1-X11) and 6 constraints using the two-phase simplex method. The problem is infeasible as shown in the final simplex table where there is a negative value in the RHS column.

Uploaded by

Angela
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Ejercicio No.

13

MAXIMIZAR: 7 X1 + 8 MAXIMIZAR: 7 X1 + 8 X2 + 5 X3 + 4
X2 + 5 X3 + 4 X4 + 6 X5 + 2 X4 + 6 X5 + 2 X6 + 0 X7 + 0 X8 + 0 X9 +
X6 0 X10 + 0 X11
.5 X1 + .45 X2 + .6 X3 + 0
X4 + 0 X5 + 0 X6 2400
1 X1 + 1 X2 + 1 X3 + 0 X4 + 0
0.5 X1 + 0.45 X2 + 0.6 X3 + 1 X7 = 2400
X5 + 0 X6 5500
1 X1 + 1 X2 + 1 X3 + 1 X8 = 5500
1 X1 + 1 X2 + 0 X3 + 0 X4 + 0
1 X1 + 1 X2 + 1 X11 = 2000
X5 + 0 X6 = 2000
1 X1 + 1 X2 + 1 X10 = 4000
1 X1 + 1 X2 + 0 X3 + 0 X4 + 0
1 X1 + 1 X2 + 1 X9 = 2000
X5 + 0 X6 = 4000
1 X1 + 1 X2 + 0 X3 + 0 X4 + 0
X5 + 0 X6 = 2000
X1, X2, X3, X4, X5, X6, X7, X8, X9, X10,
X1, X2, X3, X4, X5, X6 0
X11 0

Pasamos a construir la primera tabla de la Fase I del mtodo de las Dos Fases.

Tabla 1 0 0 0 0 0 0 0 0 -1 -1 -1
Base Cb P0 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11
P7 0 2400 0.5 0.45 0.6 0 0 0 1 0 0 0 0
P8 0 5500 1 1 1 0 0 0 0 1 0 0 0
P11 -1 2000 1 1 0 0 0 0 0 0 0 0 1
P10 -1 4000 1 1 0 0 0 0 0 0 0 1 0
P9 -1 2000 1 1 0 0 0 0 0 0 1 0 0
Z -8000 -3 -3 0 0 0 0 0 0 0 0 0

La variable que sale de la base es P9 y la que entra es P1.


Fila pivote (Fila 5):
2000 / 1 = 2000
1/1=1
1/1=1
0/1=0
0/1=0
0/1=0
0/1=0
0/1=0
0/1=0
1/1=1
0/1=0
0/1=0

Fila 1:
2400 - (0.5 * 2000) = 1400
0.5 - (0.5 * 1) = 0
0.45 - (0.5 * 1) = -0.05
0.6 - (0.5 * 0) = 0.6
0 - (0.5 * 0) = 0
0 - (0.5 * 0) = 0
0 - (0.5 * 0) = 0
1 - (0.5 * 0) = 1
0 - (0.5 * 0) = 0
0 - (0.5 * 1) = -0.5
0 - (0.5 * 0) = 0
0 - (0.5 * 0) = 0

Fila 2:
5500 - (1 * 2000) = 3500
1 - (1 * 1) = 0
1 - (1 * 1) = 0
1 - (1 * 0) = 1
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
1 - (1 * 0) = 1
0 - (1 * 1) = -1
0 - (1 * 0) = 0
0 - (1 * 0) = 0

Fila 3:
2000 - (1 * 2000) = 0
1 - (1 * 1) = 0
1 - (1 * 1) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 1) = -1
0 - (1 * 0) = 0
1 - (1 * 0) = 1

Fila 4:
4000 - (1 * 2000) = 2000
1 - (1 * 1) = 0
1 - (1 * 1) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 0) = 0
0 - (1 * 1) = -1
1 - (1 * 0) = 1
0 - (1 * 0) = 0

Fila Z:
-8000 - (-3 * 2000) = -2000
-3 - (-3 * 1) = 0
-3 - (-3 * 1) = 0
0 - (-3 * 0) = 0
0 - (-3 * 0) = 0
0 - (-3 * 0) = 0
0 - (-3 * 0) = 0
0 - (-3 * 0) = 0
0 - (-3 * 0) = 0
0 - (-3 * 1) = 3
0 - (-3 * 0) = 0
0 - (-3 * 0) = 0

Tabla 2 0 0 0 0 0 0 0 0 -1 -1 -1
Base Cb P0 P1 P2 P3 P4 P5 P6 P7 P8 P9 P10 P11
P7 0 1400 0 -0.05 0.6 0 0 0 1 0 -0.5 0 0
P8 0 3500 0 0 1 0 0 0 0 1 -1 0 0
P11 -1 0 0 0 0 0 0 0 0 0 -1 0 1
P10 -1 2000 0 0 0 0 0 0 0 0 -1 1 0
P1 0 2000 1 1 0 0 0 0 0 0 1 0 0
Z -2000 0 0 0 0 0 0 0 0 3 0 0

No existe ninguna solucin posible para el problema.

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