INVESTIGACIÓN OPERATIVA I

ABIAIL CRIOLLO

5TO SEMESTRE “A” 1

E J ER CI CI OS DE MAXI MI ZACI ÓN POR EL MÉT ODO SI MPLEX

EJERCICIO 1

MAXIMIZAR: 7 X1 +

3 X2 + 2 X3

MAXIMIZAR: 7 X1 + 3 X2 + 2

X3 + 0 X4 + 0 X5 + 0 X6 + 0

X7

4 X1 + 5 X2 + 2 X3 ≤ 10

3 X1 + 6 X2 + 3 X3 ≤

3 2 X1 + 9 X2 + 4 X3 ≤

12

3 X1 -2 X2 + 4 X3 ≤ 20

4 X1 + 5 X2 + 2 X3 + 1 X4 =

10

3 X1 + 6 X2 + 3 X3 + 1 X5 = 3 2 X1 + 9 X2 + 4 X3 + 1 X6 =

12

3 X1 -2 X2 + 4 X3 + 1 X7 = 20

X1, X2, X3 ≥ 0 X1, X2, X3, X4, X5, X6, X7 ≥ 0

Tabla 1 7 3 2 0 0 0 0

Base Cb P0 P1 P2 P3 P4 P5 P6 P7

P4 0 10 4 5 2 1 0 0 0

P5 0 3 3 6 3 0 1 0 0

P6 0 12 2 9 4 0 0 1 0

P7 0 20 3 -2 4 0 0 0 1

Z 0 -7 -3 -2 0 0 0 0

Tabla 2 7 3 2 0 0 0 0

Base Cb P0 P1 P2 P3 P4 P5 P6 P7

P4 0 6 0 -3 -2 1 -1.3333333333333 0 0

P1 7 1 1 2 1 0 0.33333333333333 0 0

P6 0 10 0 5 2 0 -0.66666666666667 1 0

INVESTIGACIÓN OPERATIVA I

ABIAIL CRIOLLO

5TO SEMESTRE “A” 2

P7 0 17 0 -8 1 0 -1 0 1

Z 7 0 11 5 0 2.3333333333333 0 0

La solución óptima es Z = 7 X1 = 1

X2 = 0

X3 = 0

EJERCICIO 2

MAXIMIZAR: 4 X1 +

7 X2 + 3 X3 + 2 X4

MAXIMIZAR: 4 X1 + 7 X2 + 3

X3 + 2 X4 + 0 X5 + 0 X6 + 0

X7 + 0 X8

3 X1 + 4 X2 + 2 X3 + 8 X4 ≤ 12

4 X1 + 5 X2 + 8 X3 +

9 X4 ≤ 10 2 X1 + 2 X2 + 5 X3 +

7 X4 ≤ 12

9 X1 -5 X2 -3 X3 + 8 X4 ≤ 23

3 X1 + 4 X2 + 2 X3 + 8 X4 + 1 X5 = 12

4 X1 + 5 X2 + 8 X3 + 9 X4 + 1

X6 = 10 2 X1 + 2 X2 + 5 X3 + 7 X4 + 1

X7 = 12

9 X1 -5 X2 -3 X3 + 8 X4 + 1 X8 = 23

X1, X2, X3, X4 ≥ 0 X1, X2, X3, X4, X5, X6, X7,

X8 ≥ 0

Tabla 1 4 7 3 2 0 0 0 0

Base Cb P0 P1 P2 P3 P4 P5 P6 P7 P8

P5 0 12 3 4 2 8 1 0 0 0

P6 0 10 4 5 8 9 0 1 0 0

P7 0 12 2 2 5 7 0 0 1 0

P8 0 23 9 -5 -3 8 0 0 0 1

Z 0 -4 -7 -3 -2 0 0 0 0

INVESTIGACIÓN OPERATIVA I

ABIAIL CRIOLLO

5TO SEMESTRE “A” 3

Tabla 2 4 7 3 2 0 0 0 0

Base Cb P0 P1 P2 P3 P4 P5 P6 P7 P8

P5 0 4 -0.2 0 -4.4 0.8 1 -0.8 0 0

P2 7 2 0.8 1 1.6 1.8 0 0.2 0 0

P7 0 8 0.4 0 1.8 3.4 0 -0.4 1 0

P8 0 33 13 0 5 17 0 1 0 1

Z 14 1.6 0 8.2 10.6 0 1.4 0 0

La solución óptima es Z = 14 X1 = 0

X2 = 2

X3 = 0

X4 = 0