1. 问题描述

给定n个工厂和m个顾客，开工厂需要一定的费用，一个工厂有一定的容量限制，每个顾客也有一定的需求，而每个顾客要选取某个工厂也需要一定的分配费用，现在要求找出一个分配方案，把顾客分配给不同的工厂，然后在可以满足所有顾客需求的前提下让所有的花费（开工厂的花费和分配的花费）最小。

问题分析：

2. 解决思路

设集合 I = { 1, … , m } 是所有Facilities

集合 J = { 1, …, n } 是所有Customers

对于每个Customer 都有一个 demand, $d_{j}$ ，只能被一个facility提供

对于每个Facility 都有一个 capacity, $b_{i}$ ，是该facility最多能提供的容量

对于每个Facility 都有一个 fixed cost, $f_{i}$ ，当该facility开启的时候固定的开销

定义 $c_{ij}$ 是facility j 满足 customer i 的要求的开销

对于每个facility i ∈ I ，定义变量 $y_{i}$

$y_{i} = \left\{\begin{matrix} 1 , if facility open & \\ 0 , if facility close & \end{matrix}\right.$

对于每个facility i ∈ I ，每个customer j ∈J，定义变量 $x_{ij}$

$x_{ij} = \left\{\begin{matrix} 1 , if.facility.i.is.assigned.for.customer.j & \\ 0 , otherwise& \end{matrix}\right.$
对于Single Source Capacitated Facility Location Problemk可以描述为如下

因为customer一般都比facility多，所以只关注customer的cost，可以制定出如下策略：

1. 遍历所有customer，对于每一个customer，将该customer安排到所有facility的cost进行从小到大排序
2.对排序好的cost进行遍历，假如该facility剩余的容量满足customer的要求，则将该customer安排到该facility

3. 代码实现

import random
import math
import matplotlib.pyplot as plt

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

#读取文件，获取各项变量
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

运行结果

	Result	Time(s)
p1	9440	0.000475
p2	8126	0.000641
p3	10126	0.000518
p4	12126	0.000599
p5	9375	0.000423
p6	8061	0.00029
p7	10061	0.001217
p8	12061	0.000298
p9	9040	0.000337
p10	7726	0.000411
p11	9726	0.000441
p12	11726	0.000364
p13	12032	0.001095
p14	9180	0.00073
p15	13180	0.000643
p16	17180	0.000682
p17	12032	0.000952
p18	9180	0.00051
p19	13180	0.000724
p20	17180	0.001145
p21	12032	0.001239
p22	9180	0.000643
p23	13180	0.000572
p24	17180	0.000571
p25	18753	0.003807
p26	15831	0.003425
p27	21031	0.002265
p28	26231	0.002439
p29	20007	0.002902
p30	16812	0.003054
p31	22212	0.00184
p32	27612	0.004533
p33	18611	0.002598
p34	15689	0.002882
p35	20889	0.003181
p36	26089	0.002619
p37	18611	0.004204
p38	15689	0.003696
p39	20889	0.00272
p40	26089	0.002611
p41	7226	0.000569
p42	9957	0.001687
p43	12448	0.001289
p44	7585	0.000704
p45	9848	0.000852
p46	12639	0.001293
p47	6634	0.000446
p48	9044	0.000963
p49	12867	0.001634
p50	10062	0.000485
p51	11175	0.000991
p52	10364	0.000545
p53	12876	0.001308
p54	10351	0.000907
p55	12383	0.001636
p56	23882	0.003144
p57	32882	0.004378
p58	53882	0.004073
p59	39121	0.003088
p60	23882	0.00407
p61	32882	0.003173
p62	53882	0.004647
p63	39121	0.003963
p64	23882	0.004642
p65	32882	0.002274
p66	53882	0.003318
p67	39671	0.005819
p68	23882	0.003292
p69	32882	0.003722
p70	53882	0.003989
p71	39121	0.00351

Capacitated Facility Location Problem贪心算法实现

1. 问题描述

2. 解决思路

3. 代码实现

运行结果

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