使用 Mi-26 型运输直升机进行医疗物资配送的最佳路线规划

使用 Mi-26 型运输直升机进行医疗物资配送的最佳路线规划/n/n假设基地位于经纬度坐标为 (30.127692, 104.628690)，需要同时前往四川省 21 个市州配送药物。/n/nMi-26 型运输直升机最大航程为 2000 公里，最大载重 12000 公斤，飞行速度为 255 公里/小时。/n/n每个地方所需货物如下：/n/n| 城市名称 | 所需要的医疗物资 | /n|---|---| /n| 成都市 | 2000 | /n| 自贡市 | 800 | /n| 攀枝花市 | 500 | /n| 泸州市 | 500 | /n| 德阳市 | 500 | /n| 绵阳市 | 800 | /n| 广元市 | 500 | /n| 遂宁市 | 500 | /n| 内江市 | 800 | /n| 乐山市 | 500 | /n| 南充市 | 500 | /n| 眉山市 | 500 | /n| 宜宾市 | 500 | /n| 广安市 | 500 | /n| 达州市 | 500 | /n| 雅安市 | 500 | /n| 巴中市 | 500 | /n| 资阳市 | 500 | /n| 阿坝州 | 200 | /n| 甘孜州 | 200 | /n| 凉山州 | 200 | /n/n基地拥有总共 10 架直升机。/n/n直升机派送完所载的全部货物后需要返回基地。/n/n问题： 基地应该同时派遣几架 Mi-26 型运输直升机运送医疗物资，使得所有直升机飞行总距离之和最短？/n/n解决方案：/n/n这个问题可以用优化模型 TSP (Traveling Salesman Problem) 来解决，即旅行商问题。/n/n数学建模：/n/n1. 定义决策变量：/n/n* $x_{i,j}$ 表示第 $i$ 架直升机从城市 $j$ 出发的标志变量，若该直升机从城市 $j$ 出发则 $x_{i,j}=1$，否则 $x_{i,j}=0$。/n* $y_{i,j,k}$ 表示第 $i$ 架直升机从城市 $j$ 到城市 $k$ 的路径上是否经过城市 $k$，若经过则 $y_{i,j,k}=1$，否则 $y_{i,j,k}=0$。/n/n2. 定义目标函数：/n/n目标是最小化所有直升机的总飞行距离，即/n/n$$/min /sum_{i=1}^{10}/sum_{j=1}^{21}/sum_{k=1}^{21} y_{i,j,k} d_{j,k}$$ /n/n其中，$d_{j,k}$ 是城市 $j$ 到城市 $k$ 之间的距离。/n/n3. 定义约束条件：/n/n* 每个城市的需求必须被满足，即/n/n$$/sum_{i=1}^{10}/sum_{j=1}^{21} x_{i,j} w_j /geq d_k, /forall k=1,2,/cdots,21$$ /n/n其中，$w_j$ 是城市 $j$ 需要的药物数量。/n/n* 每架直升机只能从一个城市出发，即/n/n$$/sum_{j=1}^{21} x_{i,j} = 1, /forall i=1,2,/cdots,10$$ /n/n* 每架直升机只能到达一个城市，即/n/n$$/sum_{k=1}^{21} y_{i,j,k} = x_{i,j}, /forall i=1,2,/cdots,10, j=1,2,/cdots,21$$ /n/n* 使用 Miller-Tucker-Zemlin (MTZ) 约束条件防止出现子回路：/n/n$$u_i - u_j + ny_{i,j} /leq n-1, /forall i,j=1,2,/cdots,21, i/neq j$$ /n/n其中，$u_i$ 表示城市 $i$ 在路径上的位置，$n$ 表示城市的数量。/n/nMATLAB 实现：/n/nmatlab/n% 定义城市数量和直升机数量/nn_city = 21;/nn_helicopter = 10;/n/n% 定义城市坐标和需求/nlocations = [30.127692, 104.628690;/n 29.339030, 104.778442;/n 26.584704, 101.718637;/n 28.895929, 105.443359;/n 31.131302, 104.397894;/n 31.467450, 104.679114;/n 32.433668, 105.829757;/n 30.532487, 105.582146;/n 29.587080, 105.062408;/n 28.874057, 105.441574;/n 29.034117, 103.768123;/n 30.075440, 103.848538;/n 28.769675, 104.776071;/n 30.455960, 106.635720;/n 31.211217, 107.494973;/n 29.980537, 103.013261;/n 31.899413, 106.752711;/n 30.122211, 104.641930;/n 31.905762, 102.222791;/n 30.055144, 101.963815;/n 27.886763, 102.267335];/ndemands = [2000;/n 800;/n 500;/n 500;/n 500;/n 800;/n 500;/n 500;/n 800;/n 500;/n 500;/n 500;/n 500;/n 500;/n 500;/n 500;/n 500;/n 500;/n 200;/n 200;/n 200];/n/n% 计算城市之间的距离/ndistances = zeros(n_city, n_city);/nfor i = 1:n_city/n for j = 1:n_city/n distances(i,j) = distance(locations(i,:), locations(j,:));/n end/nend/n/n% 定义决策变量/nx = binvar(n_helicopter, n_city, 'full');/ny = binvar(n_helicopter, n_city, n_city, 'full');/n/n% 定义目标函数/nobj = sum(sum(sum(y .* distances)));/n/n% 定义约束条件/nconstraints = [sum(x,2) == 1;/n sum(y,3) == x;/n sum(x .* repmat(demands', n_helicopter, 1), 2) >= demands;/n y(1,:,:) == 0;/n y(:,1,:) == 0;/n y(:,:,1) == 0;/n y(n_helicopter,:,:) == 0;/n y(:,n_city,:) == 0;/n y(:,:,n_city) == 0;/n x >= 0;/n y >= 0];/nfor i = 2:n_city/n for j = 2:n_city/n if i ~= j/n constraints = [constraints, sum(y(:,i,j)) <= n_helicopter - 1];/n constraints = [constraints, sum(y(:,i,j)) <= sum(x(:,i))];/n constraints = [constraints, sum(y(:,i,j)) >= sum(x(:,j))];/n constraints = [constraints, sum(y(:,i,j)) >= 1 - n_helicopter + sum(x(:,i)) + sum(x(:,j))];/n end/n end/nend/n/n% 求解问题/nops = sdpsettings('verbose', 1, 'solver', 'intlinprog');/nresult = optimize(constraints, obj, ops);/n/n% 输出结果/nif result.problem == 0/n fprintf('Total distance: %f//n', value(obj));/n for i = 1:n_helicopter/n fprintf('Helicopter %d path://n', i);/n current_city = find(value(x(i,:)));/n while ~isempty(current_city)/n fprintf('%d -> ', current_city(1));/n next_city = find(value(y(i,current_city(1),:)));/n next_city = next_city(next_city ~= current_city(1));/n current_city = next_city;/n end/n fprintf('1//n');/n end/nelse/n fprintf('Problem with code %d//n', result.problem);/nend/n/n% 计算两个坐标之间的距离/nfunction d = distance(a, b)/n r = 6371; % 地球半径/n lat1 = deg2rad(a(1));/n lat2 = deg2rad(b(1));/n dlat = deg2rad(b(1) - a(1));/n dlon = deg2rad(b(2) - a(2));/n h = sin(dlat/2)^2 + cos(lat1)*cos(lat2)*sin(dlon/2)^2;/n d = 2*r*atan2(sqrt(h), sqrt(1-h));/nend/n/n/n画图：/n/nmatlab/n% 画出城市和直升机的位置/nfigure;/nhold on;/nplot(locations(:,2), locations(:,1), 'ro', 'MarkerSize', 10);/nfor i = 1:n_helicopter/n plot(locations(1,2), locations(1,1), 'bx', 'MarkerSize', 10);/nend/nxlabel('Longitude');/nylabel('Latitude');/ntitle('Locations');/n/n% 画出每个直升机的路线/ncolors = {'b', 'g', 'r', 'c', 'm', 'y', 'k', 'b', 'g', 'r'};/nfigure;/nhold on;/nfor i = 1:n_helicopter/n current_city = find(value(x(i,:)));/n x_path = [locations(current_city,2); locations(1,2)];/n y_path = [locations(current_city,1); locations(1,1)];/n plot(x_path, y_path, '-', 'Color', colors{i}, 'LineWidth', 2);/n plot(locations(current_city,2), locations(current_city,1), 'o', 'Color', colors{i}, 'MarkerSize', 10);/nend/nxlabel('Longitude');/nylabel('Latitude');/ntitle('Paths');/n/n/n参考文献：/n/n1. https://en.wikipedia.org/wiki/Travelling_salesman_problem/n2. https://www.mathworks.com/help/optim/ug/travelling-salesman-problem.html/n3. https://www.mathworks.com/help/optim/ug/intlinprog.html/n4. [https://www.mathworks.com/help/optim/ug/sdpsettings.html](https://www.mathworks.com/help/optim/ug/sdpsettings.htm