向纸浆添加约束。交通问题

Question

我正在使用纸浆解决运输问题。

我有供应商列表（供应）和客户列表（需求）。供应总是大于需求，在某些情况下，并不是每个供应商都被 PULP 选择。然而，有时我需要选择特定的供应商，无论它是否是最佳选择。我知道应该将其添加为约束，但我不知道如何修改我的脚本。

prob = LpProblem("transportationproblem", LpMaximize)

routes = [(s,c) for s in suppliers for c in customers]

vars = LpVariable.dicts("Route", (suppliers, customers), 0, None, LpInteger)

prob += (
    lpSum([vars[s][c] * costs[s][c] for (s, c) in routes]),
    "sum_of_transp_cost"
)

for s in suppliers:
    prob += (
        lpSum([vars[s][c] for c in customers]) <= supply[s],
        f"sum_of_products_from_suppliers_{s}",
    )

for c in customers:
    prob += (
        lpSum([vars[s][c] for s in suppliers]) == demand[c],
        f"sum_of_products_to_customers{c}"
    )

我尝试添加此约束，但求解器无法正常工作

for s in suppliers:
    if s == "specific supplier": 
        prob += (
            lpSum([vars[s][c] for c in customers]) == supply[s],
            f"sum_of_products_from_suppliers_{s}",
    )

Answer 1

您的问题可能是您对您想要“特定”的供应商行双重应用约束。这有效：

import pulp
from numpy.random import default_rng

rand = default_rng(seed=0)

prob = pulp.LpProblem(name='transportation_problem', sense=pulp.LpMaximize)
suppliers = range(9)
customers = range(7)
costs = rand.random((len(suppliers), len(customers)))
supply = rand.integers(size=len(suppliers), low=10, high=20)
demand = rand.integers(size=len(customers), low=0, high=10)

routes = pulp.LpVariable.matrix(
    name='Route_s%d_c%d',
    indices=(suppliers, customers),
    cat=pulp.LpInteger, lowBound=0,
)

sum_of_transp_cost = pulp.lpDot(routes, costs)
prob.setObjective(sum_of_transp_cost)

specific_supplier_index = 3
specific_supplier_routes = 4

for supplier, supply_avail, row in zip(suppliers, supply, routes):
    if supplier == specific_supplier_index:
        prob.addConstraint(
            constraint=pulp.lpSum(row) == specific_supplier_routes,
            name=f'products_from_specific_supplier_{supplier}',
        )
    else:
        prob.addConstraint(
            constraint=pulp.lpSum(row) <= supply_avail,
            name=f'products_from_supplier_{supplier}',
        )

for customer, cust_demand in zip(customers, demand):
    prob.addConstraint(
        constraint=pulp.lpSum(
            routes[supplier][customer] for supplier in suppliers
        ) == cust_demand,
        name=f'products_to_customer_{customer}',
    )

print(prob)
prob.solve()
assert prob.status == pulp.LpStatusOptimal

for supplier, row in zip(suppliers, routes):
    print(', '.join(f'{pulp.value(route):.0f}' for route in row))

0, 0, 0, 0, 1, 5, 2
0, 0, 3, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 4
0, 0, 0, 0, 0, 0, 0
0, 0, 0, 6, 0, 0, 0
0, 0, 0, 0, 0, 0, 0
9, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0, 0, 0