1. 問題描述

給定n個工廠和m個顧客，開工廠需要一定的費用，一個工廠有一定的容量限制，每個顧客也有一定的需求，而每個顧客要選取某個工廠也需要一定的分配費用，現在要求找出一個分配方案，把顧客分配給不同的工廠，然後在可以滿足所有顧客需求的前提下讓所有的花費（開工廠的花費和分配的花費）最小。

問題分析：

2. 解決思路

設集合 I  = { 1, … , m } 是所有Facilities

集合 J = { 1, …, n } 是所有Customers

對於每個Customer 都有一個 demand, ，只能被一個facility提供

對於每個Facility 都有一個 capacity,

對於每個Facility 都有一個 fixed cost, ，當該facility開啟的時候固定的開銷

定義 是facility j 滿足 customer i 的要求的開銷

對於每個facility i ∈ I ，定義變數

對於每個facility i ∈ I ，每個customer j ∈J， 定義變數 

對於Single Source Capacitated Facility Location Problemk可以描述為如下

因為customer一般都比facility多，所以只關注customer的cost，可以制定出如下策略：

1. 遍歷所有customer，對於每一個customer，將該customer安排到所有facility的cost進行從小到大排序
2.對排序好的cost進行遍歷，假如該facility剩餘的容量滿足customer的要求，則將該customer安排到該facility

3. 程式碼實現

import random
import math
import matplotlib.pyplot as plt

CAPACITY = []
OPENCOST = []
ASSIGNCOST = []
OPENSTATUS = []
DEMAND = []
FACNUM = 0
CUSNUM = 0
Num = 0

#讀取檔案，獲取各項變數
def init(filepath):
    with open(filepath, 'r') as f:
        global FACNUM
        global CUSNUM
        global CAPACITY
        global OPENCOST
        global ASSIGNCOST
        global OPENSTATUS
        FACNUM, CUSNUM = f.readline().strip("\n").split()
        FACNUM = int(FACNUM)
        CUSNUM = int(CUSNUM)
        OPENSTATUS = [0] * FACNUM
        for i in range(FACNUM):
            line = f.readline().strip("\n").split()
            CAPACITY.append(int(line[0]))
            OPENCOST.append(int(line[1]))

        for i in range(int(CUSNUM / 10)):
            line = f.readline().strip("\n").replace(' ', '').split(".")
            for j in range(10):
                DEMAND.append(int(line[j]))

        for i in range(CUSNUM):
            linetoread = int(FACNUM / 10)
            temp = []
            for j in range(linetoread):
                line = f.readline().strip("\n").replace(' ', '').split(".")
                for k in range(10):
                    temp.append(int(line[k]))

            ASSIGNCOST.append(temp)

    print(ASSIGNCOST)

def greedy():
    global CAPACITY

    capacity = CAPACITY.copy()
    openstatus = [0] * FACNUM
    totalcost = 0
    assign = [-1] * CUSNUM
    assigncost = 0
    opencost = 0

    for cus in range(CUSNUM):
    
        facIndex = []       # the facilities may be chosen
        for i in range(FACNUM):
            if capacity[i] >= DEMAND[cus]:
                facIndex.append(i)

        # ASSIGNCOST for customer i
        asforeach = ASSIGNCOST[cus]

        # ASSIGNCOST + OPENCOST for customer i
        temp = [sum(x) for x in zip(asforeach, OPENCOST)]

        # select the min cost index
        sindex = facIndex[0]
        for i in facIndex:
            if temp[i] < temp[sindex]:
                sindex = i

        openstatus[sindex] = 1
        capacity[sindex] = capacity[sindex] - DEMAND[cus]
        assign[cus] = sindex

        assigncost += asforeach[sindex]

    for i in range(FACNUM):
        opencost += openstatus[i] * OPENCOST[i]

    totalcost = assigncost + opencost

    print(assign)
    print("assignment cost: %d"%assigncost)
    print("opencost: %d"%opencost)
    print(totalcost)
    print("openstatus: {}".format(openstatus))

    return assign

 

執行結果