Método de la Doble FaseResolver el siguiente problema de Programación Lineal por el método de la doble fase.

Min Z = 0.4X1 + 0.5X2 Forma estándar

Sujeto a: 0.3X1 + 0.1X2 ≤ 2.7 0.3X1 + 0.1X2 + H1 ≤ 2.7

0.5X1 + 0.5X2 = 6 0.5X1 + 0.5X2 + W1 = 6

0.6X1 + 0.4X2 ≥ 6 0.6X1 + 0.4X2 – E2 + W2 ≥ 6

X1 ≥ 0, X2 ≥ 0 X1, X2, H1, W1, E2, W2 ≥ 0

FASE 1: Hasta que W1 = 0, W2 = 0

Min Q = W1 + W2

Q – W1 – W2 = 0

Q X1 X2 H1 W1 E2 W2 Sol. B.Q 1 0 0 0 -1 0 -1 0Q 1 11/10 9/10 0 0 -1 0 12H1 0 3/10 1/10 1 0 0 0 27/10W1 0 1/2 1/2 0 1 0 0 6W1 0 3/5 2/5 0 0 -1 1 6

Q 1 0 8/15 -11/3 0 -1 0 21/10X1 0 1 1/3 10/3 0 0 0 9W1 0 0 1/3 -5/3 1 0 0 3/2W2 0 0 1/5 -2 0 -1 1 3/5

Q 1 0 0 5/3 0 5/3 -5/3 ½X1 0 1 0 20/3 0 5/3 -5/3 8W1 0 0 0 5/3 1 5/3 -5/3 ½X2 0 0 1 -10 0 -5 5 3

Q 1 0 0 0 -1 0 0 0X1 0 1 0 0 -4 -5 5 6H1 0 0 0 1 3/5 1 -1 3/10X2 0 0 1 0 6 5 -5 6

Sol. Básica Factible: Q = O

Donde:

X1, X2: Variables de decisión H1: Variable de holguraW1, W2: Variables artificiales E2: Variable de exceso

X BaseX1 = 6H1 = 3/10X2 = 6

X No BaseW1 = 0 E2 = 0W2 = 0

FASE 2:

Min Z = 0.4X1 + 0.5X2 con W1= 0, W2 = 0

Z X1 X2 H1 E2 Sol. Bas.Z 1 -2/5 -1/2 0 0 0Z 1 0 0 0 1/2 27/5X1 0 1 0 0 -5 6H1 0 0 0 1 1 3/10X2 0 0 1 0 5 6

Z 1 0 0 -1/2 0 21/4X1 0 1 0 5 0 15/2E2 0 0 0 1 1 3/10X2 0 0 1 -5 0 9/2

Sol. Optima Factible: Z = 21/4X Base

X1 = 15/2E2 = 3/10X2 = 9/2

X No BaseW1 = 0 W2 = 0H1 = 0

Min Z=0.4x1 +0.5x2

Sujeto A:

0.3 X1 +0.1 X2 ≤ 2.7 0.3 X1 +0.1 X2+X3 = 2.7

0.5 X1 +0.5 X2 =6 0.5 X1 +0.5 X2 + X4 = 6

0.6 X1 +0.4 X2 ≥6 0.6 X1 +0.4 X2 -X5 + X6=6

X1 ,X2, ≥0 X3,X4,X5,X6 ≥0

Fase 1: Hasta Que : X4=0, X6=0

Min Z = X4 +X6 Max -Z = -X4 -X6

-Z + X4 +X6 =0

Z X1 X2 X3 X4 X5 X6 SOL. B.Z -1 0 0 0 1 0 1 0X3 0 0.3 0.1 1 0 0 0 2.7X4 0 0.5 0.5 0 1 0 0 6X6 0 0.6 0.4 0 0 -1 1 6

Z -1 -0.5 0.5 0 0 0 1 -6Z -1 -1.1 -0.9 0 0 1 0 -12X3 0 0.3 0.1 1 0 0 0 2.7X4 0 0.5 0.5 0 1 0 0 6X6 0 0.6 0.4 0 0 -1 1 6

Z -1 0 -16/30 11/3 0 1 0 -2.1

X1 0 1 1/3 10/3 0 0 0 9X4 0 0 1/3 -5/3 1 0 0 1.5X6 0 0 0.2 -2 0 -1 1 0.6

Z -1 0 0 -5/3 0 -5/3 8/3 -0.5

X1 0 1 0 20/3 0 5/3 -5/3 8X4 0 0 0 5/3 1 5/3 -5/3 0.5X2 0 0 1 -10 0 -5 5 3

-1 0 0 0 1 0 1 0X1 0 1 0 0 -4 -5 5 6X3 0 0 0 1 3/5 1 -1 0.3X2 0 0 1 0 6 5 -5 6

Solución Básica Factible:

Z=0

Básicas

X1 =6, X2 =6, X3 =0.3,

No Básicas

X5 =0,X4 =0, X5 =0

Donde:

X1, X2: Variables de decisiónX3: Variable de holguraX4, X6: Variables artificiales X5: Variable de exceso

Segunda Fase Min Z=0.4x1 +0.5x2 Con x4=0 x6=0

Z X1 X2 X3 X5 SOL. B.Z -1 0.4 0.5 0 0 0X1 0 1 0 0 -5 6X3 0 0 0 1 1 0.3X2 0 0 1 0 5 6

Z -1 0 0.5 0 2 -2.4Z -1 0 0 0 -0.5 -5.4X1 0 1 0 0 -5 6X3 0 0 0 1 1 0.3X2 0 0 1 0 5 6ZZ -1 0 0 0.5 0 -5.25X1 0 1 0 5 0 7.5X5 0 0 0 1 1 0.3X2 0 0 1 -5 0 4.5

Solución óptima Básica Factible:

Z=5.25

Base

X1 =7.5, X2 =4.5, X5 =0.3

No Base

X3 =0, X4 =0, X5 =0