metodo

Desarrollo

ART
PROC X Y CAPACIDAD
HORAS 4 3 48
TELA 3 5 60
Z 40 20 X



Z = 40X + 20Y 4x + 3y + S1 = 48
4x + 3y ≤ 48 3x + 5y + S2 = 603x +5y ≤ 60 x + 0 + S3 = 9
X ≤ 9 Z -40x – 20y +0S1+0 S2+0+ 0S3=0
X, Y ≥ 0


X Y S1 S2 S3 I
4 3 1 0 0 48
3 5 0 1 0 60
1 0 0 0 1 9
-40 -20 0 00 0



Tabla I (X=0, Y=0, Z=0)
VNB
VB X Y S1 S2 S3 I
S1 4 3 1 0 0 48
S2 3 5 0 1 0 60
S3 1 1 0 0 1 9
Z -40 -20 0 0 0 0




F/N


4 48/412
C.P:
3
60/3 20
1 9/1 9
VE: X
VS: S3
NP: 1

FS1-4FX
F S1 4 3 1 0 0 48
4 FX 4 0 0 0 4 36
______________________________
0 3 1 0 -412
S2-3 FX =
F S2 3 5 0 1 0 60
3 FX 3 0 0 0 3 27
_____________________________
0 5 0 1 -3 33

FZ + 40FX =
FZ -40 -20 0 0 0 0
40 FX 40 0 00 40 360
____________________________
0 -20 0 0 40 360


Tabla II (X=9, Y=0, Z=360)
VNB
VB X Y S1 S2 S3 I
S1 0 3 10 -4 12
S2 0 5 0 1 -3 33
X 1 0 0 0 1 9
Z 0 -20 0 0 40 360


FN





3 12/3 4
C.P:
5
33/5 20
0 9/0 ∞

VE: Y
VS: S1
NP: 3Tabla III(X=9, Y=4, Z=440)

VNB
VB X Y S1 S2 S3 I
Y 0 1 0.333 0 -1.333 4
S2 0 0 -1.665 1 3.665 3
X 1 0 0 0 1 9
Z 0 0 6.66 0 13.34 440


FNS2 - 5 FY =
F S2 0 5 0 1 -3 33
5 FY 0 5 1.665 0 -6.665 20
__________________________________
0 0 -1.665 1 3.665 13

FZ + 20FY =
FZ 0 -20 0 040 360
20 FY 0 20 6.66 0 -26.66 80
___________________________________
0 0 6.66 0 13.34 440