(Pulp)如何创建约束，以便如果值大于 0，则它也必须大于 5000

Question

我正在解决纸浆运输优化问题，我有一定的工厂和订单，并且根据每个工厂的运输成本，我必须生成旨在成本最小化的最佳优化。 我的算法有效，但我需要创建一个约束，例如，如果有工厂移动，则最小移动必须为 5000。

from pulp import *

capacityFactory = {
    "Factory01": 56000, 
    "Factory02": 50000, 
    "Factory03": 40000, 
    "Factory04": 30000,
    "Factory05": 37000,
    "Factory06": 42000,
    "Factory07": 45000,
    "Factory08": 52000,
    "Factory09": 57000
}
factories = capacityFactory.keys()

orderBalance = {
    "Order01": 14000,
    "Order02": 35000,
    "Order03": 76000,
    "Order04": 36000,
    "Order05": 72000,
    "Order06": 41000,
    "Order07": 12000,
    "Order08": 56000,
    "Order09": 10000,
    "Order10": 7000
}
orders = orderBalance.keys()

# Creates a list of cost of each transportation path
cost = [  # orders
    [3.26, 9.31, 8.49, 5.96, 8.61, 5.45, 6.2, 7.31, 4.89, 8.34], # A   factories
    [5.5, 8.75, 9.67, 5.05, 8.54, 5.5, 6.3, 7.36, 4.35, 8.51],  # B
    [4.66, 8.36, 8.13, 4.33, 8.59, 5.6, 6.4, 7.42, 4.91, 9.71], # C
    [1.66, 1.36, 1.88, 2.77, 100, 5.65, 6.5, 7.89, 5.21, 7.01],   # D
    [2.456, 6.54, 7.4, 3.54, 7.98, 5.44, 6.4, 7.89, 5.21, 7.01],   # E
    [3.54, 6.98, 8.24, 3.12, 7.77, 5.12, 6.52, 7.12, 5.16, 7.5],   # F
    [2.15, 4.54, 7.87, 3.15, 7.45, 5.396, 6.98, 7.02, 5, 7.4],   # G
    [2.48, 5.55, 8.341, 3.19, 7.48, 5.9, 6.21, 7.23, 5.08, 7.36],   # H
    [3.41, 5.69, 8.14, 3.22, 7.66, 5.87, 6.35, 8.23, 5.39, 7.99],   # I
]
cost = makeDict([factories, orders], cost, 0)
prob = LpProblem("Material capacityFactory Problem", LpMinimize)
Routes = [(w, b) for w in factories for b in orders]
vars = LpVariable.dicts("Receive", (factories, orders), 0, None, LpInteger)

prob += (
    lpSum([vars[w][b] * cost[w][b] for (w, b) in Routes]),
    "Sum_of_Transporting_cost",
)

# The capacityFactory maximum constraints are added to prob for each capacityFactory node (factories)
for w in factories:
    prob += (
        lpSum([vars[w][b] for b in orders]) <= capacityFactory[w],
        "Sum_of_Products_out_of_factories_%s" % w,
    )

# The orderBalance minimum constraints are added to prob for each orderBalance node (project)
for b in orders:
    prob += (
        lpSum([vars[w][b] for w in factories]) >= orderBalance[b],
        "Sum_of_Products_into_orders%s" % b,
    )

# Add non-negativity constraints for all variables
for w in factories:
    for b in orders:
        prob += (
            vars[w][b] >= 0,
            "Non_Negative_%s_to_%s" % (w, b),
        )

prob.solve()

for v in prob.variables():
    print(v.name, "=", v.varValue)
    
print("Value of Objective Function = ", value(prob.objective))

如何使用 Pulp 来实现此约束？

pythonpulp线性规划优化

Answer 1

您正在寻找的概念是“半连续”变量的实现，其中变量为 0 或在某个范围内。这个网站上有很多例子。

我假设一家工厂要么根本不使用，要么必须用于至少 5000 个任何类型的订单，所以...

本质上你想要：

引入另一个二进制指示变量来指示工厂是否用于任何订单

use[factory]   ∈ {0, 1}   1 if used, 0 else

然后您需要将每个工厂的工厂活动限制为零或相关范围

∑ receive[factory, order] for all orders ≤ use[factory] * max_orders[factory]

和

∑ receive[factory, order] for all orders ≥ use[factory] * min_orders[factory]

如果您的程序开始变慢，我还鼓励您考虑将

receive

设置为连续变量而不是整数。使用连续变量求解总是更快，并且当您的情况下数字“很大”达到 10,000 时，通常连续近似就可以了。