求具有11,390,625个变量组合的函数的最小值

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我正在编写一个代码来解决管道数量的直径大小的最佳组合。目标函数是求出6条管道压降之和最小。

由于我有15个离散直径大小的选择，它们是2,4,6,8,12,16,20,24,30,36,40,42,50,60,80，可以用于系统中六个管道中的任何一个，因此可能的解决方案列表变为15^6，等于11,390,625

为了解决这个问题，我使用了使用纸浆包的混合整数线性规划。我能够找到相同直径组合的解决方案(例如2，2，2，2，2或4,4,4,4,4,4)，但我需要的是通过所有组合(例如2,4,2,2,4,2或4,2,4,2,4,2 )找到最小值。我试图这样做，但这个过程需要很长时间才能完成所有组合。有没有更快的方法来做这件事？

请注意，我无法计算每条管道的压降，因为直径的选择将影响系统的总压降。因此，在任何时候，我都需要计算系统中每种组合的压降。

我还需要限制问题，使流水线区域的速率/横截面> 2。

非常感谢您的帮助。

我的代码的第一次尝试如下：

from pulp import * 
import random 
import itertools
import numpy

rate = 5000
numberOfPipelines = 15 

def pressure(diameter):
    diameterList = numpy.tile(diameter,numberOfPipelines)
    pressure = 0.0
    for pipeline in range(numberOfPipelines):
        pressure +=  rate/diameterList[pipeline]
    return pressure 
diameterList = [2,4,6,8,12,16,20,24,30,36,40,42,50,60,80]

pipelineIds = range(0,numberOfPipelines)
pipelinePressures = {} 

for diameter in diameterList: 
   pressures = [] 
   for pipeline in range(numberOfPipelines): 
      pressures.append(pressure(diameter))
   pressureList = dict(zip(pipelineIds,pressures))
   pipelinePressures[diameter] = pressureList 
   print 'pipepressure', pipelinePressures 
prob = LpProblem("Warehouse Allocation",LpMinimize)

use_diameter = LpVariable.dicts("UseDiameter", diameterList, cat=LpBinary) 
use_pipeline = LpVariable.dicts("UsePipeline", [(i,j) for i in pipelineIds for j in diameterList], cat = LpBinary)

## Objective Function: 
prob += lpSum(pipelinePressures[j][i] * use_pipeline[(i,j)] for i in pipelineIds for j in diameterList)

## At least each pipeline must be connected to a diameter: 
for i in pipelineIds: 
   prob += lpSum(use_pipeline[(i,j)] for j in diameterList) ==1 

## The diameter is activiated if at least one pipelines is assigned to it: 
for j in diameterList: 
  for i in pipelineIds: 
     prob += use_diameter[j] >= lpSum(use_pipeline[(i,j)])


## run the solution

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

for i in diameterList:
    if use_diameter[i].varValue> pressureTest:
        print("Diameter Size",i)

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

这就是我为组合部分所做的，这花了很长时间。

xList = np.array(list(itertools.product(diameterList,repeat = numberOfPipelines)))
        print len(xList)
        for combination in xList:
            pressures = [] 
            for pipeline in range(numberOfPipelines):
               pressures.append(pressure(combination))
            pressureList = dict(zip(pipelineIds,pressures))
            pipelinePressures[combination] = pressureList
            print 'pipelinePressures',pipelinePressures

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