Programación Lineal

1.- Resuelve el siguiente problema de Programación Lineal.

MAX Z = 3x1 + 2x2 + 5x3

S.a. x1 + 2x2 + x3 ≤ 430 a) Calcule el nuevo valor de Z,

3x1 + 2x3 ≤ 460 si el nuevo vector b es b^T=

x1 + 4x2 ≤ 420 (500, 700, 200)

x1, x2, x3 ≥ 0 b) Calcule el nuevo vector de Z,

Si el nuevo valor de C^T= 4

5

5

Tabla óptima:

Bas Z X1 X2 X3 S1 S2 S3 Sol

Z 1 4 0 0 1 2 0 1350

X2 0 -1/4 1 0 ½ -1/4 0 100

X3 0 3/2 0 1 0 ½ 0 230

S3 0 2 0 0 -2 1 1 20

XB=B^-1(b+∆B) = (■(1/2&-1/4&0@0&1/2&0@-2&1&1))[■(500@700@200)] = (■(250&-175&+0@0&+350&+0@-1000&+700&+200)) = [■(75@350@-100)] ≥/ 0 (Dif.)

∴ aplicar simplex.

Bas Z X1 X2 X3 S1 S2 S3 Sol

Z 1 4 0 0 1 2 0 1350

X2 0 -1/4 1 0 ½ -1/4 0 75

X3 0 3/2 0 1 0 ½ 0 350

S3 0 2 0 0 -2 1 1 -100

Bas Z X1 X2 X3 S1 S2 S3 Sol

Z 1 5 0 0 0 5/2 ½

X2 0 1/4 1 0 0 0 1/4 50

X3 0 3/2 0 1 0 ½ 0 350

S3 0 -1 0 0 1 -1/2 -1/2 50

S1 / -1/2 ( ½ 0 0 -1/2 ¼ ¼ -25) Piv * -1 ( 1 0 0 -1 ½ ½ )

+ x2 (-1/4 1 0 ½ -1/4 0 75) ( 4 0 0 1 2 0 )

¼ 1 0 0 0 ¼ 50 5 0 0 0 5/2 ½

Z0 = (C2, C3, S1) ^x2 = (2, 5, 0) 50 = 100 + 1750 = $1850

^x3 350

^S1 50

b) C^T= 4 MaxZ= 4x1 + 5x2 + 5x3

5 S.a. x1 + 2x2 + x3 ≤ 430

5 3x1 + 2x3 ≤ 460

x1 + 4x2 ≤ 420

x1, x2, x3 ≥ 0

1

Z1 – (C1 + ∆C1) π a1- (C1 + ∆C1) = (1, 2, 0) 3 – 1 = [1 + 6 + 0] - 1= 7 – 1 = 6 1

2

Z2 – (C2 + ∆C2) π a2- (C2 + ∆C2) = (2, 0, 0) 0 – 1 = [2 + 0 + 0] - 1= 2 – 1 = 1

4

1

Z3 – (C3 + ∆C3) π a3- (C3 + ∆C3) = (1, 2, 0) 2 – 1 = [1 + 4 + 0] - 1= 5 – 1 = 4

0

Bas Z X1 X2 X3 S1 S2 S3 Sol

Z 1 1 2 0 1350

X2 0 -1/4 1 0 ½ -1/4 0 100

X3 0 3/2 0 1 0 ½ 0 230

S3 0 2 0 0 -2 1 1 20

Bas Z X1 X2 X3 S1 S2 S3 Sol

Z 1 ¼ 0 0 ½ ¼ 0 330

X2 0 -1/4 1 0 ½ ¼ 0 100

X3 0 3/2 0 1 0 ½ 0 250

S3 0 -1 0 0 -2 1 1 20

EcZ = (1 6 1 4 1 2 0 1350)

Ecx2(-1) = (0 ¼ -1 0 -1/2 ¼ 0

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