模拟退火算法实例分析
1. 求解组合优化问题
以TSP问题为例,以TSPLIB的berlin52为例进行求解,berlin52有52座城市,数据下载地址:https://wwwproxy.iwr.uni-heidelberg.de/groups/comopt/software/TSPLIB95/tsp/
clear
clc
a = 0.99; % 温度衰减函数的参数
t0 = 97; tf = 3; t = t0;
Markov_length = 10000; % Markov链长度
coordinates = [
1 565.0 575.0; 2 25.0 185.0; 3 345.0 750.0;
4 945.0 685.0; 5 845.0 655.0; 6 880.0 660.0;
7 25.0 230.0; 8 525.0 1000.0; 9 580.0 1175.0;
10 650.0 1130.0; 11 1605.0 620.0; 12 1220.0 580.0;
13 1465.0 200.0; 14 1530.0 5.0; 15 845.0 680.0;
16 725.0 370.0; 17 145.0 665.0; 18 415.0 635.0;
19 510.0 875.0; 20 560.0 365.0; 21 300.0 465.0;
22 520.0 585.0; 23 480.0 415.0; 24 835.0 625.0;
25 975.0 580.0; 26 1215.0 245.0; 27 1320.0 315.0;
28 1250.0 400.0; 29 660.0 180.0; 30 410.0 250.0;
31 420.0 555.0; 32 575.0 665.0; 33 1150.0 1160.0;
34 700.0 580.0; 35 685.0 595.0; 36 685.0 610.0;
37 770.0 610.0; 38 795.0 645.0; 39 720.0 635.0;
40 760.0 650.0; 41 475.0 960.0; 42 95.0 260.0;
43 875.0 920.0; 44 700.0 500.0; 45 555.0 815.0;
46 830.0 485.0; 47 1170.0 65.0; 48 830.0 610.0;
49 605.0 625.0; 50 595.0 360.0; 51 1340.0 725.0;
52 1740.0 245.0;
];
coordinates(:,1) = []; % 除去第一列的城市编号
amount = size(coordinates,1); % 城市的数目
% 通过向量化的方法计算距离矩阵
dist_matrix = zeros(amount, amount);
coor_x_tmp1 = coordinates(:,1) * ones(1,amount);
coor_x_tmp2 = coor_x_tmp1';
coor_y_tmp1 = coordinates(:,2) * ones(1,amount);
coor_y_tmp2 = coor_y_tmp1';
dist_matrix = sqrt((coor_x_tmp1-coor_x_tmp2).^2 + ...
(coor_y_tmp1-coor_y_tmp2).^2);
sol_new = 1:amount; % 产生初始解
% sol_new是每次产生的新解；sol_current是当前解；sol_best是冷却中的最好解；
E_current = inf;E_best = inf; % E_current是当前解对应的回路距离；
% E_new是新解的回路距离；
% E_best是最优解的
sol_current = sol_new; sol_best = sol_new;
p = 1;
while t>=tf
for r=1:Markov_length % Markov链长度
% 产生随机扰动
if (rand < 0.5) % 随机决定是进行两交换还是三交换
% 两交换
ind1 = 0; ind2 = 0;
while (ind1 == ind2)
ind1 = ceil(rand.*amount);
ind2 = ceil(rand.*amount);
end
tmp1 = sol_new(ind1);
sol_new(ind1) = sol_new(ind2);
sol_new(ind2) = tmp1;
else
% 三交换
ind1 = 0; ind2 = 0; ind3 = 0;
while (ind1 == ind2) || (ind1 == ind3) ...
|| (ind2 == ind3) || (abs(ind1-ind2) == 1)
ind1 = ceil(rand.*amount);
ind2 = ceil(rand.*amount);
ind3 = ceil(rand.*amount);
end
tmp1 = ind1;tmp2 = ind2;tmp3 = ind3;
% 确保ind1 < ind2 < ind3
if (ind1 < ind2) && (ind2 < ind3)
;
elseif (ind1 < ind3) && (ind3 < ind2)
ind2 = tmp3;ind3 = tmp2;
elseif (ind2 < ind1) && (ind1 < ind3)
ind1 = tmp2;ind2 = tmp1;
elseif (ind2 < ind3) && (ind3 < ind1)
ind1 = tmp2;ind2 = tmp3; ind3 = tmp1;
elseif (ind3 < ind1) && (ind1 < ind2)
ind1 = tmp3;ind2 = tmp1; ind3 = tmp2;
elseif (ind3 < ind2) && (ind2 < ind1)
ind1 = tmp3;ind2 = tmp2; ind3 = tmp1;
end
tmplist1 = sol_new((ind1+1):(ind2-1));
sol_new((ind1+1):(ind1+ind3-ind2+1)) = ...
sol_new((ind2):(ind3));
sol_new((ind1+ind3-ind2+2):ind3) = ...
tmplist1;
end
%检查是否满足约束
% 计算目标函数值（即内能）
E_new = 0;
for i = 1 : (amount-1)
E_new = E_new + ...
dist_matrix(sol_new(i),sol_new(i+1));
end
% 再算上从最后一个城市到第一个城市的距离
E_new = E_new + ...
dist_matrix(sol_new(amount),sol_new(1));
if E_new < E_current
E_current = E_new;
sol_current = sol_new;
if E_new < E_best
% 把冷却过程中最好的解保存下来
E_best = E_new;
sol_best = sol_new;
end
else
% 若新解的目标函数值小于当前解的，
% 则仅以一定概率接受新解
if rand < exp(-(E_new-E_current)./t)
E_current = E_new;
sol_current = sol_new;
else
sol_new = sol_current;
end
end
end
t=t.*a; % 控制参数t（温度）减少为原来的a倍
end
disp('最优解为：')
disp(sol_best)
disp('最短距离：')
disp(E_best)
运行结果:
2 求解背包问题
在这个0-1背包的例子中,假设有12件物品,质量分别为2,5,18,3,2,5,10,4,11,7,14,6,价值分别为5,10,13,4,3,11,13,10,8,16,7,4,包的最大允许质量为46.
clear
clc
a = 0.95
k = [5;10;13;4;3;11;13;10;8;16;7;4];
k = -k; % 模拟退火算法是求解最小值，故取负数
d = [2;5;18;3;2;5;10;4;11;7;14;6];
restriction = 46;
num = 12;
sol_new = ones(1,num); % 生成初始解
E_current = inf;E_best = inf;
% E_current是当前解对应的目标函数值（即背包中物品总价值）；
% E_new是新解的目标函数值；
% E_best是最优解的
sol_current = sol_new; sol_best = sol_new;
t0=97; tf=3; t=t0;
p=1;
while t>=tf
for r=1:100
%产生随机扰动
tmp=ceil(rand.*num);
sol_new(1,tmp)=~sol_new(1,tmp);
%检查是否满足约束
while 1
q=(sol_new*d <= restriction);
if ~q
p=~p; %实现交错着逆转头尾的第一个1
tmp=find(sol_new==1);
if p
sol_new(1,tmp)=0;
else
sol_new(1,tmp(end))=0;
end
else
break
end
end
% 计算背包中的物品价值
E_new=sol_new*k;
if E_newif E_new% 把冷却过程中最好的解保存下来
E_best=E_new;
sol_best=sol_new;
end
else
if rand<exp(-(E_new-E_current)./t)
E_current=E_new;
sol_current=sol_new;
else
sol_new=sol_current;
end
end
end
t=t.*a;
end
disp('最优解为：')
sol_best
disp('物品总价值等于：')
val=-E_best;
disp(val)
disp('背包中物品重量是：')
disp(sol_best * d)
运行结果:
