模拟退火算法matalab解决旅行商问题
时间: 2024-01-01 12:21:28 浏览: 29
以下是使用模拟退火算法在Matlab中解决旅行商问题的示例代码:
```matlab
% 旅行商问题的模拟退火算法解决方案
% 假设有5个城市,它们的坐标分别为(0,0)、(1,2)、(3,1)、(2,3)和(4,2)
% 问题是:从第一个城市出发,如何才能经过每个城市一次,最后回到第一个城市,并且总路程最短?
% 初始化城市坐标
city = [0,0;1,2;3,1;2,3;4,2];
% 计算城市之间的距离
n = size(city,1);
distance = zeros(n,n);
for i = 1:n
for j = 1:n
distance(i,j) = sqrt((city(i,1)-city(j,1))^2 + (city(i,2)-city(j,2))^2);
end
end
% 初始化参数
T = 100; % 初始温度
T_min = 1; % 最小温度
alpha = 0.99; % 降温系数
iter = 1000; % 每个温度下的迭代次数
% 随机生成一个初始解
cur_solution = randperm(n);
% 计算初始解的总路程
cur_distance = 0;
for i = 1:n-1
cur_distance = cur_distance + distance(cur_solution(i),cur_solution(i+1));
end
cur_distance = cur_distance + distance(cur_solution(n),cur_solution(1));
% 模拟退火算法主循环
while T > T_min
for i = 1:iter
% 随机交换两个城市的位置,生成一个新解
new_solution = cur_solution;
p = randperm(n,2);
new_solution(p(1)) = cur_solution(p(2));
new_solution(p(2)) = cur_solution(p(1));
% 计算新解的总路程
new_distance = 0;
for j = 1:n-1
new_distance = new_distance + distance(new_solution(j),new_solution(j+1));
end
new_distance = new_distance + distance(new_solution(n),new_solution(1));
% 判断是否接受新解
if new_distance < cur_distance
cur_solution = new_solution;
cur_distance = new_distance;
else
delta = new_distance - cur_distance;
p = exp(-delta/T);
if rand < p
cur_solution = new_solution;
cur_distance = new_distance;
end
end
end
T = T * alpha; % 降温
end
% 输出结果
disp('最优解:');
disp(cur_solution);
disp('最短路程:');
disp(cur_distance);
```