+1) El siguiente modelo de programacin linealMax: Z . 3x1+ 2x2 +5 x2 X1+2x2+x3430 3x1 + 2x3460 X1+ 4x2 420Tiene la tabla ptima BasZX1X2X3S1S2S3SOL

Z14001201350

X21-1/4101/2-1/40100

X303/20101/20230

S30200-21120

CALCULAR NUEVO VALOR DE Z= b= (500, 700,200)XB =B-1 (b+b) = 1/2 -1/4 0 500 250 -175 0 75 0 1/2 0 700 = 0 350 0 = 350 -2 1 1 200 -1000 700 200 -100

Como no se cumple 0 hacemos el dual simplex NO ES FACTIBLE SIMPLEX VARIABLE ENTRADA Z=SI

BasZX1X2X3S1S2S3SOL

Z14001201

X20-1/4101/2-1/4075

X303/20101/20350

S30200-211-1OO

Z150005/21/2.

X2010001/450

X303/20101/20350

SI0-100+1-1/2-1/250

SALIDA. (MS NEGATIVA) Vieja e.c Z=(1 4 0 0 1 2 0 )Zi(-1) = (0 2 0 0 -1 1/2 1/2 ) Z= 1 5 0 0 0 5/2 1/2 ) X2Vieja c.c = (0 -1/4 1 0 1/2 - 1/4 0 75) X3 Zi(-1/2 =( 0 1/2 0 0 -1/2 1/4 0 -75) Zo= ( C2,C3,C4) S1 0 1/4 1 0 0 0 50 Zo= (Z.5 0 ) 50 350 Zo= (100 ,1750,0) = 18 50

B) calcula el Nuevo valor de Z= si c= (4,5,5)Max Z= 4x1 + 5x2 + 5x3 X1 + 2x2 + x3 430 3x1 + +2x3460 X1+ 4x2 420 X1,X2,X30 a1 a2 a3Z1= - (C1+ C1)= a1,- (C1 +c1) 1(1 2 0 ) 3 -4 =( 1, 6, 0) -4 = 3

Z2=- ( C2+ C2) = a2 (C2+ C2) 2( 1 2 0 ) 0 -5 = (2,0.0)-5= -3 4Z3=-(C3+C3)=a3 (C3+ C3)= 1( 1 2 3) 2 -5 = ( 1 4 0) -5 = 0 0

Z3 ( C3+C3) = a3 ( C3 +C3) 1( 1 2 3 ) 2 -5 = ( 1 4 0 ) -5=0 0

TABLA OPTMABAS Z1/11/21/3S1S2S3SOL

Z13-301201350

X20-1/4101/2-1/40100

X303/20101/20230

S20200-21120

Z19/4005/25/201650

X20 -3/4101/2-1/40100

X303/20101/20230

S30200-21120

Viej Ec Z= ( 1 3 -3 0 1 2 0 1350)X2 (3) = (0 -3/4 3 3/2 -3/4 0 300) 1 9/4 0 0 5/2 5/4 0 1650)

EJERCICIO 22) La siguiente es una tabla ptima de un modelo de programacin lineal donde. S1 ,S2, S3 son holguras de la primera, segunda y tercera restriccin.a) Complete la tablab) Calcule el modelo original

BasZX1X2X3S1S2S3SOL

Z1400120

X21-1/4101/2-1/40100

X303/20101/20230

S30200-21120

12 -1/4 0 1 0 0 120 0 0 1 0 1/2 -2 1 1 0 0 1

1 -1/4 0 12 0 0 0 1 0 0 2 0 (1/48)-2 1 1 0 0 11 0 0 12 1/24 00 1 0 0 2 0 (-1)-2 1 1 0 0 11 0 0 12 1/24 0 1 0 0 12 1/24 00 1 0 0 2 0 0 1 0 0 2 0 = B-2 0 1 0 -2 1 0 0 1 24 1/12 1

A= B 12 1/24 0 -1/4 1 0 0 2 0 3/2 0 1 24 1/12 1 2 0 03.06 12 1/24 A= 3 0 1/12 -13/4 24 1/12

CB= B 12 1/24 00 2 0 ( 1 2 0) = 12.08 4 29824 1/12 1

B= BXd 12 1/24 0100 129.6 0 2 0230 = 460 24 1/12 1 20 615Zj-Cj= a-c = ( 1 2 0) 3.06 12 1/24 3 0 1/12 - C -13/424 1/12= ( 9.06 , 12 ,.20) - C C= (9.06 12 .20 - ( 4 , 0, 0 ) = 5.06 , 12 , .20 MAX =5.06X1 +12X2 + .20 X3 3.06X1 + 12X2 + 1/24 X3 129.6 3 X1 +1/12 X3 460 -13/4X1 + 24 X2 + 1/12 X3615