文章目录

• 题目
• 完整代码
• 模拟退火(SA)算法
• 遗传算法(GA)
• 结果
• 模拟退火(SA)算法
• 遗传算法(GA)

题目

• Suppose there are n facilities and m customers. We wish to choose:
  • which of the n facilities to open
  • the assignment of customers to facilities
• The objective is to minimize the sum of the opening cost and the assignment cost.
• The total demand assigned to a facility must not exceed its capacity.

Capacitated Facility Location Problem 是一个 NP-hard 的问。 随机搜索算法是解决这类问题的一类比较好的解法。
在本篇博客中，使用 模拟退火(SA)算法 和 遗传算法(GA) 求解该问题。

完整代码

完整代码见github

模拟退火(SA)算法

无论是模拟退火算法，还是遗传算法，都是在初始解的基础上通过相应的操作，得到较好的解的方法。
因为模拟退火算法与初始解无关，所以初始解可以随便给一个。

• 首先将顾客按顺序依次分配给各个设施，如果分配失败（分配后设备超出容量），则将顾客分配给下一个。
  采用这样的分配方式，我们可以得到一个不确定好坏的初始解。

代码实现如下：

  bool solutionGenerator() {
    for (int i = 0; i < numOfCustomer; ++i) {
      if (!assignCustomerToFicility(i, 0)) {
        return false;
      }
    }
    return true;
  }

模拟退火算法的核心就是在初始解的基础上添加扰动产生新解，然后接受更优的解，对于差解，产生一个 [ 0 , 1 ) [0, 1) [0,1)的随机数 R R R，如果 R &lt; e − Δ E / t R \lt e^{-\Delta E / t} R<e−ΔE/t , 则接受该解。

代码实现如下：

//初始温度为1000.0， 退火系数为 0.99
double t= 1000.0, a = 0.99;
int MapkobChainLength = numOfCustomer * numOfFacility;
int count = 0, lastCost = totalCost;
while (t > 0.01 && count < 20) {
    for (int i = 0; i < MapkobChainLength; ++i) {
        // 对解进行扰动
        disturbance(t);
    }
    if (lastCost == solutionCost) {
    	count = count + 1;
    } else {
    	count = 0;
    }
    t = a * t;
}

void disturbance(double t) {
    int i = rand() % numOfCustomer;
    int j = rand() % numOfFacility;
    int temp = customerAssignedTable[i];
    if (assignCustomerToFicility(i, j)) {
        int dE = calculationCost() - solutionCost;
        int k = customerAssignedTable[i];
        if (dE <= 0 || rand() / (RAND_MAX + 1.0) < exp(-dE / t)) {
            openingSolution[k] = actualCapacity[k];
            openingSolution[temp] = actualCapacity[temp];
            assignSolution[i] = customerAssignedTable[i];
            solutionCost = totalCost;
        } else {
            actualCapacity[k] = openingSolution[k];
            actualCapacity[temp] = openingSolution[temp];
            customerAssignedTable[i] = assignSolution[i];
        }
    }
}

遗传算法(GA)

遗传算法的初始解和退火算法的初始解的产生方式大致相同，但是在分配顾客时采用的随机分配，产生100个初始个体

bool initSolutionGenerator() {
    bool isSuccess = true;
    int count = 0, maxCount = 3;
    for (int i = 0; i < MAX_INDIVIDUAL_NUM; ++i) {
        isSuccess = true;
        for (int j = 0; j < numOfCustomer; ++j) {
            if (!assignCustomerToFicility(population[i], j, rand() % numOfFacility)) {
                isSuccess = false;
                break;
            }
        }
        if (!isSuccess) {
            --i;
            count = count + 1;
        } else {
            count = 0;
            calculationCost(population[i]);
        }
        if (count >= 3) {
            return false;
        }
    }
    return true;
}

采用锦标赛的选择方式进行自然选择，然后交叉，变异。

void select() {
    Individual temp[MAX_INDIVIDUAL_NUM];
    memcpy(temp, population, sizeof(population));
    for (int i = 0; i < MAX_INDIVIDUAL_NUM; ++i) {
        int x = rand() % MAX_INDIVIDUAL_NUM;
        int y = rand() % MAX_INDIVIDUAL_NUM;
        x = population[x].totalCost < population[y].totalCost ? x : y;
        memcpy(&population[i], &temp[x], sizeof(Individual));
    }
}
void crossover() {
    int x = rand() % (numOfCustomer - 1);
    int y = rand() % (numOfCustomer - x - 1) + x + 1;
    // 交叉概率
    double Pc = 0.8;
    for (int i = 0; i < MAX_INDIVIDUAL_NUM; i = i + 2) {
        if (rand() / (RAND_MAX + 1.0) < Pc) {
            for (int j = x; j < y; ++j) {
                assignCustomerToFicility(population[i], j, population[i + 1].assignTable[j]);
            }
            for (int j = x; j < y; ++j) {
                assignCustomerToFicility(population[i + 1], j, population[i].assignTable[j]);
            }
        }
    }
}
void mutation() {
    // 变异率
    double Pm = 0.1;
    for (int i = 0; i < MAX_INDIVIDUAL_NUM; ++i) {
        if (rand() / (RAND_MAX + 1.0) < Pm) {
            int k = rand() % numOfCustomer;
            int kk = rand() % numOfFacility;
            assignCustomerToFicility(population[i], k, kk);
        }
        calculationCost(population[i]);
    }
}

累积5000代后退出循环

结果

模拟退火(SA)算法

遗传算法(GA)