08-Sept TP7 - Distribucion (Clase 2)

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CLASE PRACTICA: Investigación Operativa
Trabajo Practico Nº 7 – Distribución 
Profesores: JTP. Ing. Néstor O. Cruz
AY1. Ing. Mariela E. Rodríguez
Facultad de Ingeniería
Universidad Nacional de Jujuy
a) Realice la representación grafica de la RED. 
Oferta Demanda
1
2
3
1
2
3
4
50
20
35
25
10
25
45
105 105
MODELO DE TRANSPORTE
1 2 3 4 ai
1 5 2 3 4 50
2 2 4 3 2 20
3 4 3 1 5 35
bj 25 10 25 45
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
10 10
3
4 3 1 5
35
35
bj 25 10 25 45 105
▪ Método Fundamental
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
10 10
3
4 3 1 5
35
35
bj 25 10 25 45 105
▪ Método Fundamental
Z= 5 * 25 + 2 * 10 + 3 * 15 + 3 * 10 + 2 * 10 + 5 * 35
Z= 415 UM
1º Iteración
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
10 10
3
4 3 1 5
35
35
bj 25 10 25 45 105
▪ Método Fundamental
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
10 10
3
4 3 1 5
35
35
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 3 + 3 – 2 = 2
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
10 10
3
4 3 1 5
35
35
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 3 + 3 – 2 = 2
ẟ21= 2 – 3 + 3 – 5 = -3
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
10 10
3
4 3 1 5
35
35
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 3 + 3 – 2 = 2
ẟ21= 2 – 3 + 3 – 5 = -3
ẟ22= 4 – 3 + 3 – 2 = 2
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
10 10
3
4 3 1 5
35
35
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 3 + 3 – 2 = 2
ẟ21= 2 – 3 + 3 – 5 = -3
ẟ22= 4 – 3 + 3 – 2 = 2
ẟ31= 4 – 5 + 3 – 3 +2 - 5 = -4
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
10 10
3
4 3 1 5
35
35
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 3 + 3 – 2 = 2
ẟ21= 2 – 3 + 3 – 5 = -3
ẟ22= 4 – 3 + 3 – 2 = 2
ẟ31= 4 – 5 + 3 – 3 +2 - 5 = -4
ẟ32= 3 – 2 + 3 – 3 + 2 - 5 = -2
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
10 10
3
4 3 1 5
35
35
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 3 + 3 – 2 = 2
ẟ21= 2 – 3 + 3 – 5 = -3
ẟ22= 4 – 3 + 3 – 2 = 2
ẟ31= 4 – 5 + 3 – 3 +2 - 5 = -4
ẟ32= 3 – 2 + 3 – 3 + 2 - 5 = -2
ẟ33= 1 – 3 + 2 – 5 = -5
Sale el 10, La disminución en Z es -5 
=> 10* (-5) = - 50
Z= 415 – 50 = 365
Z= 365 U.M.
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 3 + 3 – 2 = 2
ẟ21= 2 – 3 + 3 – 5 = -3
ẟ22= 4 – 3 + 3 – 2 = 2
ẟ31= 4 – 5 + 3 – 3 +2 - 5 = -4
ẟ32= 3 – 2 + 3 – 3 + 2 - 5 = -2
ẟ33= 1 – 3 + 2 – 5 = -5
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 3 + 3 – 2 = 2
ẟ21= 2 – 3 + 3 – 5 = -3
ẟ22= 4 – 3 + 3 – 2 = 2
ẟ31= 4 – 5 + 3 – 3 +2 - 5 = -4
ẟ32= 3 – 2 + 3 – 3 + 2 - 5 = -2
ẟ33= 1 – 3 + 2 – 5 = -5
Z= 5 * 25 + 2 * 10 + 3* 15 + 2 * 20 + 1 * 10 + 5 * 25 
Z= 365 U.M.
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 5 + 1 – 3 = -3
2º Iteración
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 5 + 1 – 3 = -3
ẟ21= 2 – 5 + 3 – 1 + 5 – 2 = 2
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 5 + 1 – 3 = -3
ẟ21= 2 – 5 + 3 – 1 + 5 – 2 = 2
ẟ22= 4 – 2 + 3 – 1 + 5 – 2 = 7
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 5 + 1 – 3 = -3
ẟ21= 2 – 5 + 3 – 1 + 5 – 2 = 2
ẟ22= 4 – 2 + 3 – 1 + 5 – 2 = 7
ẟ23= 3 – 1 + 5 – 2 = 5
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 5 + 1 – 3 = -3
ẟ21= 2 – 5 + 3 – 1 + 5 – 2 = 2
ẟ22= 4 – 2 + 3 – 1 + 5 – 2 = 7
ẟ23= 3 – 1 + 5 – 2 = 5
ẟ31= 4 – 5 + 3 – 1 = 1
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 5 + 1 – 3 = -3
ẟ21= 2 – 5 + 3 – 1 + 5 – 2 = 2
ẟ22= 4 – 2 + 3 – 1 + 5 – 2 = 7
ẟ23= 3 – 1 + 5 – 2 = 5
ẟ31= 4 – 5 + 3 – 1 = 1
ẟ32= 3 – 2+ 3 – 1 = 3
Sale el 15, La disminución en Z es -15
=> 15* (-3)=-45
Z= 365 – 45 = 320
Z= 320 U.M.
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 5 + 1 – 3 = -3
ẟ21= 2 – 5 + 3 – 1 + 5 – 2 = 2
ẟ22= 4 – 2 + 3 – 1 + 5 – 2 = 7
ẟ23= 3 – 1 + 5 – 2 = 5
ẟ31= 4 – 5 + 3 – 1 = 1
ẟ32= 3 – 2+ 3 – 1 = 3
Sale el 15, La disminución en Z es -15
=> 15* (-3)=-45
Z= 365 – 45 = 320
Z= 320 U.M.
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
25 10
bj 25 10 25 45 105
▪ Método Fundamental
ẟ14= 4 – 5 + 1 – 3 = -3
ẟ21= 2 – 5 + 3 – 1 + 5 – 2 = 2
ẟ22= 4 – 2 + 3 – 1 + 5 – 2 = 7
ẟ23= 3 – 1 + 5 – 2 = 5
ẟ31= 4 – 5 + 3 – 1 = 1
ẟ32= 3 – 2+ 3 – 1 = 3
Sale el 15, La disminución en Z es -15
=> 15* (-3)=-45
Z= 365 – 45 = 320
Z= 320 U.M.Z= 5 * 25 + 2 * 10 + 4 * 15 + 2 * 20 + 1 * 25 + 5 * 10
Z= 320 UM
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
25 10
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 4 + 5 – 1 = 3
ẟ21= 2 – 5 + 4 – 2 = -1
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
25 10
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 4 + 5 – 1 = 3
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22= 4 – 2 + 4 – 2 = 4
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
25 10
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 4 + 5 – 1 = 3
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 5 – 2 = 5
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
25 10
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 4 + 5 – 1 = 3
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 5 – 2 = 5
ẟ31= 4 – 5 + 4 – 5 = -2
1 2 3 4 ai
1
5 2 3 4
50
25 10 15
2
2 4 3 2
20
20
3
4 3 1 5
35
25 10
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 4 + 5 – 1 = 3
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 5 – 2 = 5
ẟ31= 4 – 5 + 4 – 5 = -2
ẟ32= 3 – 2 + 4 – 5 = 0
Sale el 10, por tener el menor ahorro
La disminución en Z es – 20
 10* (-2)=- 20
Z= 320 – 20 = 300
Z= 300 U.M.
▪ Método Fundamental
ẟ13= 3 – 4 + 5 – 1 = 3
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 5 – 2 = 5
ẟ31= 4 – 5 + 4 – 5 = -2
ẟ32= 3 – 2 + 4 – 5 = 0
Sale el 10, por tener el menor ahorro
La disminución en Z es – 20
 10* (-2)=- 20
Z= 320 – 20 = 300
Z= 300 U.M.
1 2 3 4 ai
1
5 2 3 4
50
15 10 25
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
Z= 5 * 15 + 2 * 10 + 4 * 25 +2 * 20 + 4 * 10 + 1 * 25
Z= 320 UM.
1 2 3 4 ai
1
5 2 3 4
50
15 10 25
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 1 + 4 – 5 = 1
4º Iteración
1 2 3 4 ai
1
5 2 3 4
50
15 10 25
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 1 + 4 – 5 = 1
ẟ21= 2 – 5 + 4 – 2 = -1
1 2 3 4 ai
1
5 2 3 4
50
15 10 25
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 1 + 4 – 5 = 1
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22= 4 – 2 + 4 – 2 = 4
1 2 3 4 ai
1
5 2 3 4
50
15 10 25
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 1 + 4 – 5 = 1
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 4 - 5 + 4 – 2 = 3
1 2 3 4 ai
1
5 2 3 4
50
15 10 25
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 1 + 4 – 5 = 1
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 4 - 5 + 4 – 2 = 3
ẟ32= 3 – 2 + 5 – 4 = 2
1 2 3 4 ai
1
5 2 3 4
50
15 10 25
2
2 4 3 2
20
20
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ13= 3 – 1 + 4 – 5 = 1
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 4 - 5 + 4 – 2 = 3
ẟ32= 3 – 2 + 5 – 4 = 2
ẟ34= 5 – 4 + 5 – 4 = 2
Sale el 15, por tener el menor ahorro
La disminución en Z es – 1
 15* (-1)= - 15
Z= 300 – 15 = 285
Z= 285 U.M.
▪ Método Fundamental
ẟ13= 3 – 1 + 4 – 5 = 1
ẟ21= 2 – 5 + 4 – 2 = -1
ẟ22=4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 4 - 5 + 4 – 2 = 3
ẟ32= 3 – 2 + 5 – 4 = 2
ẟ34= 5 – 4 + 5 – 4 = 2
Sale el 15, por tener el menor ahorro
La disminución en Z es – 1
 15* (-1)= - 15
Z= 300 – 15 = 285
Z= 285 U.M.
1 2 3 4 ai
1
5 2 3 4
50
10 40
2
2 4 3 2
20
15 5
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
Z= 2 * 10 + 4 * 40 + 2 * 15 + 2 * 5 + 4 * 10 + 1 * 25
Z= 285 UM.
1 2 3 4 ai
1
5 2 3 4
50
10 40
2
2 4 3 2
20
15 5
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ11= 5 – 2 + 2 – 4 = 1
5º Iteración
1 2 3 4 ai
1
5 2 3 4
50
10 40
2
2 4 3 2
20
15 5
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ11= 5 – 2 + 2 – 4 = 1
ẟ13= 3 – 4 + 2 – 2 + 4 – 1 = 2
1 2 3 4 ai
1
5 2 3 4
50
10 40
2
2 4 3 2
20
15 5
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ11= 5 – 2 + 2 – 4 = 1
ẟ13= 3 – 4 + 2 – 2 + 4 – 1 = 2
ẟ22= 4 – 2 + 4 – 2 = 4
1 2 3 4 ai
1
5 2 3 4
50
10 40
2
2 4 3 2
20
15 5
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ11= 5 – 2 + 2 – 4 = 1
ẟ13= 3 – 4 + 2 – 2 + 4 – 1 = 2
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 4 - 2 = 4
1 2 3 4 ai
1
5 2 3 4
50
10 40
2
2 4 3 2
20
15 5
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ11= 5 – 2 + 2 – 4 = 1
ẟ13= 3 – 4 + 2 – 2 + 4 – 1 = 2
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 4 - 2 = 4
ẟ32= 3 – 4 + 2 – 2 + 4 – 2 = 1
1 2 3 4 ai
1
5 2 3 4
50
10 40
2
2 4 3 2
20
15 5
3
4 3 1 5
35
10 25
bj 25 10 25 45 105
▪ Método Fundamental
ẟ11= 5 – 2 + 2 – 4 = 1
ẟ13= 3 – 4 + 2 – 2 + 4 – 1 = 2
ẟ22= 4 – 2 + 4 – 2 = 4
ẟ23= 3 – 1 + 4 - 2 = 4
ẟ32= 3 – 4 + 2 – 2 + 4 – 2 = 1
ẟ34= 5 – 4 + 2 – 2 = 1
Todos los delta son positivos, la solución no se puede mejorar mas => estamos en el optimo.
Z= 285 UM.
1 2 3 4
1
5 2 3 4
10 40
2
2 4 3 2
20
3
4 3 1 5
5 25 5
Z= 2*20+4*5+2*10+1*25+4*40+5*5
Z=290 U.M.
v) Método de Vogel
1 2 3 4
1
5 2 3 4
10 40
2
2 4 3 2
20
3
4 3 1 5
5 25 5
Z= 2*20+4*5+2*10+1*25+4*40+5*5
Z=290 U.M.
v) Método de Vogel
Resolver en Clase aplicando el Método Fundamental: Subir el ejercicio al Aula Virtual. 
ẟ11= 5 – 4 + 5 – 4 = 2
ẟ13= 3 – 4 + 5 – 1 = 3
ẟ22= 4 – 2 + 4 – 5 + 4 - 2 = 3
ẟ23= 3 – 1 + 4 - 2 = 4
ẟ24= 2 – 5 + 4 - 2 = - 1
ẟ32= 3 – 2 + 4 - 5 = 0
1 2 3 4
1
5 2 3 4
10 40
2
2 4 3 2
15 5
3
4 3 1 5
10 25
Z= 2 * 10 + 4 * 40 + 2 * 15 + 2 * 5 + 4 * 10 + 1* 25
Z=285 U.M.
ẟ11= 5 – 4 + 5 – 4 = 2
ẟ13= 3 – 4 + 5 – 1 = 3
ẟ22= 4 – 2 + 4 – 5 + 4 - 2 = 3
ẟ23= 3 – 1 + 4 - 2 = 4
ẟ24= 2 – 5 + 4 - 2 = - 1
ẟ32= 3 – 2 + 4 - 5 = 0
Resolver aplicando el Método Fundamental
25 10 25 45 105
ai
50
20
35
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