Investigacion Operacional Tarea 4

Variables Enteras:
Xpt: Cantidad de producto p en el periodo t.
Ypt: Cantidad de producto p de sobreproducción en el periodo t.
Zpt: Cantidad de demanda echa del producto p en el periodo t.
Wrt: Cantidad de recursos r extras en el periodo t.
Variables binarias:
Apt: 1 si se almacena el producto p en el periodo t.
0 Otro caso.
El costo mínimo y la mejor solución a nuestroproblema es: 731.1500

min 1.5x11+2.75x12+3x13+1.5x21+1.75x22+2x23+x31+1.1x32+1.2x33+0.5w11+0.5w12+0.5w13+w21+w22+w23+2w31+2w32+2w33+w41+w42+w43+12z11+12z12+12z13+6z21+6z22+6z23+4z31+4z32+4z33+0.2y11+0.2y12+0.2y13+0.25y21+0.25y22+0.25y23+0.5y31+0.5y32+0.5y33

st

0.5x11+0.5y11+0.5x21+0.5y21+x31+y31-w11<=250
0.5x12+0.5y12+0.5x22+0.5y22+x32+y32-w12<=2500.5x13+0.5y13+0.5x23+0.5y23+x33+y33-w13<=250

x11+y11+0.5x21+0.5y21-w21<=300
x12+y12+0.5x22+0.5y22-w22<=300
x13+y13+0.5x23+0.5y23-w23<=300

x11+y11+0.8x21+0.8y21+0.5x31+0.5y31-w31<=200
x12+y12+0.8x22+0.8y22+0.5x32+0.5y32-w32<=200
x13+y13+0.8x23+0.8y23+0.5x33+0.5y33-w33<=200

3x11+3y11+x21+y21+0.25x31+0.25y31-w41<=100
3x12+3y12+x22+y22+0.25x32+0.25y32-w42<=1003x13+3y13+x23+y23+0.25x33+0.25y33-w43<=100

x11+z11-y11=5
x12+z12+y11-y12=5
x13+z13+y12-y13=20

x21+z21-y21=30
x22+z22+y21-y22=30
x23+z23+y22-y23=40

x31+z31-y31=100
x32+z32+y31-y32=100
x33+z33+y32-y33=200

a11+a21+a31<=2
a12+a22+a23<=2
a13+a23+a33<=2

y11-10000a11<=0
y12-10000a12<=0
y13-10000a13<=0
y21-10000a21<=0
y22-10000a22<=0
y23-10000a13<=0
y31-10000a31<=0y32-10000a32<=0
y33-10000a33<=0

end

gin x11
gin x12
gin x13
gin x21
gin x22
gin x23
gin x31
gin x32
gin x33
gin y11
gin y12
gin y13
gin y21
gin y22
gin y23
gin y31
gin y32
gin y33
gin z11
gin z12
gin z13
gin z21
gin z22
gin z23
gin z31
gin z32
gin z33
gin w11
gin w12
gin w13
gin w21
gin w22
gin w23
gin w31
gin w32
gin w33
gin w41
gin w42
gin w43

int a11int a12
int a13
int a21
int a22
int a23
int a31
int a32
int a33
Solución LINDO
LP OPTIMUM FOUND AT STEP 18
OBJECTIVE VALUE = 730.625000

SET A12 TO <= 0 AT 1, BND= -731.0 TWIN= -730.6 36
SET A22 TO <= 0 AT 2, BND= -753.5 TWIN= -731.0 41
SET A11 TO >= 1 AT 3, BND= -753.5 TWIN= -758.8 45NEW INTEGER SOLUTION OF 753.500000 AT BRANCH 3 PIVOT 45
BOUND ON OPTIMUM: 730.6250
DELETE A11 AT LEVEL 3
FLIP A22 TO >= 1 AT 2 WITH BND= -731.00000
SET A11 TO >= 1 AT 3, BND= -731.0 TWIN= -736.2 53
SET Y22 TO >= 23 AT 4, BND= -731.5 TWIN= -731.5 62

NEW INTEGER SOLUTION OF731.500000 AT BRANCH 5 PIVOT 62
BOUND ON OPTIMUM: 730.6250
DELETE Y22 AT LEVEL 4
DELETE A11 AT LEVEL 3
DELETE A22 AT LEVEL 2
FLIP A12 TO >= 1 AT 1 WITH BND= -730.62500
SET Y13 TO <= 0 AT 2, BND= -730.6 TWIN=-0.1000E+31 62
SET Y23 TO <= 0 AT 3, BND= -730.6 TWIN=-0.1000E+31 62
SETY33 TO <= 0 AT 4, BND= -730.6 TWIN=-0.1000E+31 62
SET Z11 TO <= 0 AT 5, BND= -730.6 TWIN=-0.1000E+31 62
SET Z12 TO <= 0 AT 6, BND= -730.6 TWIN=-0.1000E+31 62
SET Z13 TO <= 0 AT 7, BND= -730.6 TWIN=-0.1000E+31 62
SET Z21 TO <= 0 AT 8, BND= -730.6 TWIN=-0.1000E+31 62
SETZ22 TO <= 0 AT 9, BND= -730.6 TWIN=-0.1000E+31 62
SET Z23 TO <= 0 AT 10, BND= -730.6 TWIN=-0.1000E+31 62
SET Z31 TO <= 0 AT 11, BND= -730.6 TWIN=-0.1000E+31 62
SET Z32 TO <= 0 AT 12, BND= -730.6 TWIN=-0.1000E+31 62
SET Z33 TO <= 0 AT 13, BND= -730.6 TWIN=-0.1000E+31 62
SET...