使用Pulp解决线性编程python问题

编程问答 2022-06-03

问题描述

基本上，我正在尝试使每个目的地只有一个起点 所有目的地必须只有一个来源 并不一定必须使用所有来源 我希望有人能帮助我，我知道这不是正确的方法，而是我所得到的

.NET 3.5,4,4.52

解决方法

我认为您已经很接近想要的东西，但是您有一些问题。首先，您将Routes定义为所有可能的路线，而您只为某些路线定义了cost_to_send。

假设已定义成本的路线是可行的路线，则最好将Routes定义为：

Routes = [(i,j) for i in origin for j in destination if j in cost_to_send[i]]

我可以看到的另一个问题是第二组约束：

for i in origin:
    prob += lpSum(quantity[i][j] for j in destination) == 1

这就是说，对于每个起点，流向所有目的地的总流量必须为1。但是在问题陈述中，您会说：

所有目的地必须只有一个来源，而不是所有来源 一定要使用

但是由于这种限制，您正在强制使用每个原点。消除此约束即可解决问题：

from pulp import *
import pandas as pd 
origin = ["a","b","c","d","e","f","g","h"]
destination = ["1","2","3","4","5","6","7","8","9","10"]


offer = {"a":3,"b":3,"c":3,"d":3,"e":3,"f":4,"g":3,"h":3}
demand = {"1":1,"2":1,"3":1,"4":1,"5":1,"6":1,"7":1,"8":1,"9":1,"10":1}
cost_to_send = { 
"a":{"1":1,"3":1},"b":{"2":1,"9":1},"c":{"5":1,"7":1},"d":{"7":1,"10":1},"e":{"3":1,"8":1},"f":{"1":1,"g":{"4":1,"h":{"1":1,"8":1}
}
prob = LpProblem("Exercise",LpMinimize)

# Routes = [(i,j) for i in origin for j in destination]
Routes = [(i,j) for i in origin for j in destination if j in cost_to_send[i]]

quantity = LpVariable.dicts("quantity de envio",Routes,0)

prob += lpSum(quantity[(i,j)]*cost_to_send[i][j] for (i,j) in Routes)

for j in destination:
    prob += lpSum(quantity[(i,j)] for i in origin if (i,j) in Routes) == demand[j]

#for i in origin:
#    prob += lpSum(quantity[i][j] for j in destination) == 1

prob.solve()
print("Status: ",LpStatus[prob.status])

for v in prob.variables():
    if v.varValue > 0:
        print(v.name,"=",v.varValue)

print("Answer ",value(prob.objective))


Returns:

Status:  Optimal
quantity_de_envio_('a',_'1') = 1.0
quantity_de_envio_('a',_'2') = 1.0
quantity_de_envio_('a',_'3') = 1.0
quantity_de_envio_('b',_'9') = 1.0
quantity_de_envio_('c',_'5') = 1.0
quantity_de_envio_('c',_'6') = 1.0
quantity_de_envio_('d',_'10') = 1.0
quantity_de_envio_('f',_'4') = 1.0
quantity_de_envio_('f',_'7') = 1.0
quantity_de_envio_('h',_'8') = 1.0
Answer  10.0

请注意，如果您希望给定目的地的需求大于1，但想强制目的地仅接收来自单个来源的输入-那么您将需要添加 binary 每个“路线”的变量来控制每个路线是否开放。然后，您将设置一个约束，即如果这些二进制变量为true，则数量变量只能为非零，然后可以将这些二进制变量的总和限制为== 1。

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