Pulp 如何处理这些限制？

Question

我使用 Pulp 包解决了具有不等式约束的 MILP 问题。 MILP 问题试图最小化安装设施平台以通过管道为客户提供服务的成本，以便每个平台可以处理一定数量的客户。代码示例如下：

import numpy as np
from pulp import *
import random 
CUSTOMERS = range(1,17) ## generate random Customer Ids
FACILITY =['FAC 1','FAC 2'] # Number and Name of Facilities
randomCosts = random.sample(range(90, 100), 2) ## Generate Random Installation Costs 
actcost = dict(zip(FACILITY, randomCosts)) ## Assign installation cost to each facility
randompipelineCost = random.sample(range(5, 20), 2) ## Generate Random pipeline Costs
pipelineCost = dict(zip(FACILITY, randompipelineCost))## Assign pipeline cost to each facility
sizeOfPlatforms = [10,10] ## Size of Platforms
maxSizeOfPlatforms = dict(zip(FACILITY, sizeOfPlatforms)) ## Assign Size to each Facility
serviceRandom=[] 
serviceCosts = {}
for facility in FACILITY: ## Generate Random Service Costs for each customer
   serviceRandom=[]
   for i in range (16):
     serviceRandom.append(random.randrange(1, 101, 1))
   service = dict(zip(CUSTOMERS, serviceRandom))
   serviceCosts[facility]=service

print 'CUSTOMERS', CUSTOMERS
print 'FACILITY', FACILITY
print 'Facility Cost', actcost 
print 'pipeline Cost',pipelineCost 
print 'service Cost', serviceCosts 

prob = LpProblem("FacilityLocation",LpMinimize)

##Decision Variables

use_facility = LpVariable.dicts("UseFacility", FACILITY, cat=LpBinary)

use_customer = LpVariable.dicts("UseCustomer",[(i,j) for i in CUSTOMERS for j in FACILITY], cat=LpBinary)

## Objective Function 

prob += lpSum(actcost[j]*use_facility[j] for j in FACILITY) + lpSum(pipelineCost[j]*use_facility[j] for j in FACILITY)+ lpSum(serviceCosts[j][i]*use_customer[(i,j)] for i in CUSTOMERS for j in FACILITY)


# Constraints 
for j in FACILITY: 
   prob += lpSum(use_customer[(i,j)] for i in CUSTOMERS) <= maxSizeOfPlatforms[j]

for i in CUSTOMERS: 
   prob += lpSum(use_customer[(i,j)] for j in FACILITY) == 1

for j in FACILITY:
    for i in CUSTOMERS:
        prob += use_facility[j] >= lpSum(use_customer[(i,j)])

我的问题，Pulp 或默认求解器 CPLEX 如何处理这些约束？

Answer 1

如果你在程序的最后添加

prob.solve(pulp.CPLEX_CMD(keepFiles=True))
print(pulp.LpStatus[prob.status])
for variable in prob.variables():
    print ("{} = {}".format(variable.name, variable.varValue))

然后你会看到一些解决方案：

CUSTOMERS range(1, 17)
FACILITY ['FAC 1', 'FAC 2']
Optimal
UseCustomer_(1,_'FAC_1') = 1.0
UseCustomer_(1,_'FAC_2') = 0.0
UseCustomer_(10,_'FAC_1') = 0.0
UseCustomer_(10,_'FAC_2') = 1.0

还有两个文件：FacilityLocation-pulp 和 FacilityLocation-pulp.sol 这将帮助您了解发生了什么。