Resuelve el modelo de PuLP
Analítica de la cadena de suministro en Python
Aaren Stubberfield
Supply Chain Analytics Mgr.
Proceso común de modelado en PuLP
Inicializar el modeloDefinir variables de decisiónDefinir la función objetivoDefinir las restricciones- Resolver el modelo
- llamar al método
solve() - comprobar el estado de la solución
- imprimir variables de decisión óptimas
- imprimir la función objetivo óptima
- llamar al método
Resolver modelo: método solve
.solve(solver=None)
solver= Opcional: solver específico a usar; por defecto, el predeterminado.
# Initialize, Define Decision Vars., Objective Function, and Constraints
from pulp import *
import pandas as pd
model = LpProblem("Minimize Transportation Costs", LpMinimize)
cust = ['A','B','C']
warehouse = ['W1','W2']
demand = {'A': 1500, 'B': 900, 'C': 800}
costs = {('W1','A'): 232, ('W1','B'): 255, ('W1','C'): 264,
('W2','A'): 255, ('W2','B'): 233, ('W2','C'): 250}
ship = LpVariable.dicts("s_", [(w,c) for w in warehouse for c in cust],
lowBound=0, cat='Integer')
model += lpSum([costs[(w, c)] * ship[(w, c)] for w in warehouse for c in cust])
for c in cust: model += lpSum([ship[(w, c)] for w in warehouse]) == demand[c]
# Solve Model
model.solve()
Resolver modelo: estado de la solución
LpStatus[model.status]
- Not Solved: Estado previo a resolver el problema.
- Optimal: Se encontró una solución óptima.
- Infeasible: No hay soluciones factibles (p. ej., si pones las restricciones x ≤ 1 y x ≥ 2).
- Unbounded: La función objetivo no está acotada; maximizar o minimizar tiende a infinito (p. ej., si la única restricción fuera x ≥ 3).
- Undefined: Puede existir una solución óptima, pero no se ha encontrado.
1 Keen, Ben Alex. “Linear Programming with Python and PuLP 2 Part 2.” _Ben Alex Keen_, 1 Apr. 2016, benalexkeen.com/linear-programming-with-python-and-pulp-part-2/._{{5}}
# Initialize, Define Decision Vars., Objective Function, and Constraints
from pulp import *
import pandas as pd
model = LpProblem("Minimize Transportation Costs", LpMinimize)
cust = ['A','B','C']
warehouse = ['W1','W2']
demand = {'A': 1500, 'B': 900, 'C': 800}
costs = {('W1','A'): 232, ('W1','B'): 255, ('W1','C'): 264,
('W2','A'): 255, ('W2','B'): 233, ('W2','C'): 250}
ship = LpVariable.dicts("s_", [(w,c) for w in warehouse for c in cust], lowBound=0, cat='Integer')
model += lpSum([costs[(w, c)] * ship[(w, c)] for w in warehouse for c in cust])
for c in cust: model += lpSum([ship[(w, c)] for w in warehouse]) == demand[c]
# Solve Model
model.solve()
print("Status:", LpStatus[model.status])
Status: Optimal
Imprime variables en la salida estándar:
for v in model.variables():
print(v.name, "=", v.varValue)
Estructura de datos de pandas:
o = [{A:ship[(w,'A')].varValue, B:ship[(w,'B')].varValue, C:ship[(w,'C')].varValue}
for w in warehouse]
print(pd.DataFrame(o, index=warehouse))
- recorre variables del modelo
- guarda valores en un DataFrame de pandas
# Solve Model
model.solve()
print(LpStatus[model.status])
o = [{A:ship[(w,'A')].varValue, B:ship[(w,'B')].varValue, C:ship[(w,'C')].varValue}
for w in warehouse]
print(pd.DataFrame(o, index=warehouse))
Salida:
Status: Optimal
| |A |B |C |
|:------|:------|:------|:------|
|W1 |1500.0 |0.0 |0.0 |
|W2 |0.0 |900.0 |800.0 |
Resolver modelo: función objetivo óptima
Imprime el valor de la función objetivo óptima:
print("Objective = ", value(model.objective))
# Solve Model
model.solve()
print(LpStatus[model.status])
output = []
for w in warehouse: t = [ship[(w,c)].varValue for c in cust] output.append(t)
opd = pd.DataFrame.from_records(output, index=warehouse, columns=cust)
print(opd)
print("Objective = ", value(model.objective))
Status: Optimal
| |A |B |C |
|:------|:------|:------|:------|
|W1 |1500.0 |0.0 |0.0 |
|W2 |0.0 |900.0 |800.0 |
Objective = 757700.0
Resumen
Resolver el modelo
- Llamar al método
solve() - Comprobar el estado de la solución
- Imprimir valores de las variables de decisión
- Imprimir valor de la función objetivo
¡Vamos a practicar!
Analítica de la cadena de suministro en Python
