使用混合整数规划限制最大分配大小

Question

我有2个设施，每个都有管道和安装费用。我还有16个客户要服务，每个客户都有服务成本。我想为每个设施分配最多 10 个客户，以使管道成本、安装和服务成本最小化。

我已经实现了以下代码，但它不能正常工作。它应该将每个设施分配给它应该服务的客户数量和 returns 最低成本。但是，输出将水井分配给所有设施（我认为）。

非常感谢您的帮助：

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,0,1,LpBinary)

use_customer = LpVariable.dicts("UseCustomer",[(i,j) for i in CUSTOMERS for j in FACILITY],1)

## 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 j in FACILITY: 
   prob += lpSum(use_facility[j] for j in FACILITY) <=1.0 ##Constraint 1 

##Solution 

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

TOL = 0.00001 


## print Decision Variables
for i in FACILITY: 
   if use_facility[i].varValue > TOL:
     print("Establish Facility at Site",i)

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


##optimal Solution
print ("The cost of production in dollars for one year=", value(prob.objective))

Answer 1

有 4 个不同的问题。

1)

use_customer = LpVariable.dicts("UseCustomer",[(i,j) for i in CUSTOMERS for j in FACILITY],1)

这相当于说 每个变量都是小数并且应该大于一个。也就是说，您正在设置 lowBound=1。我想你想在这里说变量是二进制的：

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

2)

第二个问题是您不限制客户至少被分配到一个设施（至少因为问题是最小化问题并且您的成本严格为正，客户永远不会被分配到更多不止一个设施，但在下文中我将假设正好是一个)

for i in CUSTOMERS: 
     prob += lpSum(use_customer[(i,j)] for j in FACILITY) == 1.0

3)

我不确定你想在这里说什么：

for j in FACILITY: 
   prob += lpSum(use_facility[j] for j in FACILITY) <=1.0 ##Constraint 1

每次您对所有设施点求和并将它们的值之和限制为最多为 1。我想这是一个错误。

4)

最后，您不要 link 变量 use_facility 和 use_customer 在一起。也就是说，变量 use_facility 的值永远不会大于 0。因为它是二进制的，所以我假设 use_facility[j] 代表激活成本。因此，您需要添加以激活设施，即 如果至少有一个客户使用它，设施 j 将被激活:

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

放在一起：

##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)])

使用混合整数规划限制最大分配大小

Constrain the maximum allocation size using Mixed Integer Programming

python

mathematical-optimization

linear-programming

pulp

mixed-integer-programming