如何在python中用纸浆制定和解决优化问题答案

【问题标题】：How to formulate and solve optimization problem with pulp in python如何在python中用纸浆制定和解决优化问题
【发布时间】：2022-11-11 00:34:48
【问题描述】：

输入：

2 个变量列表，list_1 和 list_2。
从每个列表中获取一个元素并返回某个值 >=0 的函数：富（el_list1，el_list2）我们可以假设我们有一个形状为 (len(list_1), len(list_2)) 的矩阵，其值 >=0。

任务：假设每个列表的每个元素只能使用一次，查找 list_1、list_2 的元素对，它们给出 foo(el_list1, el_list2) 的最大总和。

例子：

list_1 = [l1_0，l1_1，l1_2]
list_2 = [l2_0，l2_1，l2_2，l2_3]

A = 
[[0.0, 1.5, 2.4, 0.8],
 [3.1, 0.5, 0.0, 0.0],
 [0.0, 1.3, 2.5, 1.0]
],
where A[i, j] = foo(list_1[i],list_2[j])

对于此示例，结果可能类似于：

最大总和 = 3.1 + 1.5 + 2.5 + 0 = 7.1
给出这个总和的对：
[(l1_1;l2_0),(l1_0;l2_1),(l1_2;l2_2),(无,l2_3)]

【问题讨论】：

标签： python pulp

【解决方案1】：

from pulp import *

model = pulp.LpProblem("_problem", LpMaximize)

A = [[0.0, 1.5, 2.4, 0.8],
 [3.1, 0.5, 0.0, 0.0],
 [0.0, 1.3, 2.5, 1.0]
]
        
# Creates a list of all the supply nodes
list1 = ["I0", "I1", "I2"]

# Creates a list of all demand nodes
list2 = ["J0","J1", "J2", "J3"]

# The cost data is made into a dictionary
Acosts = makeDict([list1,list2],A,0)

Routes = [(w,b) for w in list1 for b in list2]

pv = LpVariable.dicts("Route",(list1,list2),0)

# Objective
model += lpSum([pv[w][b]*Acosts[w][b] for (w,b) in Routes])

# Because you don't have any constraints, I made it up. 
for w in list1:
    model += lpSum([pv[w][b] for b in list2])==1, "Sum_of_Products_out_of_plants_%s"%w

for b in list2:
    model += lpSum([pv[w][b] for w in list1])==1, "Sum_of_Products_into_centers%s"%b
    
model.solve()
print("Status:", LpStatus[model.status])
for v in model.variables():
    print(v.name, "=", v.varValue)
print("Maximize Cost = ", pulp.value(model.objective))

输出：

Status: Infeasible
Route_I0_J0 = 0.0
Route_I0_J1 = 1.0
Route_I0_J2 = 0.0
Route_I0_J3 = 0.0
Route_I1_J0 = 1.0
Route_I1_J1 = 0.0
Route_I1_J2 = 0.0
Route_I1_J3 = 0.0
Route_I2_J0 = 0.0
Route_I2_J1 = 0.0
Route_I2_J2 = 1.0
Route_I2_J3 = 1.0
Maximize Cost =  8.1

【讨论】：