遗传算法求解应急物流中心选址问题

Question

使用Matlab来求解应急物流中心选址问题
已经有完整的代码，但是需要在此基础上改一点约束条件，应该是很简单的只不过本人不太懂代码操作
希望可以联系，提供所有数据和完整代码，有偿

Answer 1

该回答引用ChatGPT

按照你的要求修的，请测试

clc
clear
close all

% 常量定义
MAX_gen = 4000;           % 最大迭代步数
center.Station = [31708.622,7772.865;31705.726,7766.996;31730.340,7771.543;31709.847,7799.784;31702.607,7775.014;31700.937,7774.105];
center.chubei = [5000, 5000, 5000, 5000, 5000, 5000];
cost.construction = 2500 * 10000; % 元
cost.storage = 100;
P = 0.1; % 发生灾害的概率

% 车辆和无人机的运输速度和单位成本
vehicle_speed = 60; % km/h
vehicle_cost = 5; % 元/每单位资源/km
drone_speed = 72; % km/h
drone_cost = 7; % 元/每单位资源/km

% 需求点及其需求量
requirement.Station = [31699.8,7774.1; 31739.9,7767.4; 31702.0,7775.7; 31749.9,7777.4; 31733.2,7789.7; 31688.6,7802.1; 31710.9,7800.5; 31688.6,7823.6; 31767.7,7812.8; 31667.5,7769.1];
requirement.need = [670, 800, 800, 950, 3010, 800, 2560, 1320, 1060, 950];

% 计算车辆和无人机运输距离
vehicle_indices = 1:6;
drone_indices = 7:10;
vehicle_distances = pdist2(center.Station(vehicle_indices, :), requirement.Station);
drone_distances = pdist2(center.Station(drone_indices, :), requirement.Station);

% 初始化种群
pop_length = size(requirement.Station, 1);
pop_num = 100;
pop_choose = size(center.Station, 1);
Pcc = 0.8; % 交叉概率
Pmm = 0.05; % 变异概率
popus = init(pop_length, pop_num, pop_choose);  

% 迭代优化
for gen = 1:MAX_gen
    sprintf('迭代次数%d', gen)
    for i = 1:size(popus, 1)
        % 计算种群个体的适应度
        f = fit(popus, center, requirement, cost, P);
        
        % 更新最优解
        if min(f) <= best.min_f
            advance_k = advance_k + 1;
            ff_better{advance_k} = f;
            f_better(advance_k) = min(f);
            best.min_f = min(f);
            [re, index] = min(f);
            best.jieguo = popus{index(1)};
            best.popus = popus;
        else
            advance_k = advance_k + 1;
            ff_better{advance_k} = ff_better{advance_k-1};
            f_better(advance_k) = f_better(advance_k-1);
            popus = best.popus;
        end
        PPP{gen} = popus;
        popus = choose(popus, f, Pcc, Pmm, center);

        % 将种群中每个个体分配到车辆或无人机
        vehicle_assignments = zeros(size(popus));
        drone_assignments = zeros(size(popus));
        for j = 1:size(popus, 1)
            vehicle_indices = find(popus{j} <= 6);
            drone_indices = find(popus{j} > 6);
            vehicle_distance = sum(vehicle_distances(vehicle_indices, j));
            drone_distance = sum(drone_distances(drone_indices, j));
            vehicle_time = vehicle_distance / vehicle_speed;
            drone_time = drone_distance / drone_speed;
            vehicle_cost_total = sum(vehicle_cost * requirement.need(vehicle_indices) * vehicle_distances(vehicle_indices, j));
            drone_cost_total = sum(drone_cost * requirement.need(drone_indices) * drone_distances(drone_indices, j));
            if vehicle_time + drone_time <= 24 && vehicle_cost_total + drone_cost_total <= cost.construction
                vehicle_assignments(j, vehicle_indices) = 1;
                drone_assignments(j, drone_indices) = 1;
            end
        end

        % 输出方案
        for j = 1:size(vehicle_assignments, 1)
            vehicle_indices = find(vehicle_assignments(j, :));
            drone_indices = find(drone_assignments(j, :));
            if ~isempty(vehicle_indices) && ~isempty(drone_indices)
                fprintf('方案 %d: 候选点 %d 送达需求点 %s（车辆运输）和 %s（无人机运输）\n', j, i, num2str(vehicle_indices), num2str(drone_indices));
            elseif ~isempty(vehicle_indices)
                fprintf('方案 %d: 候选点 %d 送达需求点 %s（车辆运输）\n', j, i, num2str(vehicle_indices));
            elseif ~isempty(drone_indices)
                fprintf('方案 %d: 候选点 %d 送达需求点 %s（无人机运输）\n', j, i, num2str(drone_indices));
            else
                fprintf('方案 %d: 候选点 %d 未选取任何需求点\n', j, i);
            end
        end
    end
end