MATLAB实现差分进化算法求解最大值,目标函数为z(i,j)=-20exp((0.2sqrt((x(i)^2+y(j)^2)/2)))-exp((cos(2pix(i))+cos(2piy(j)))/2)+exp(1);
时间: 2023-08-04 18:06:15 浏览: 75
好的,下面是MATLAB实现差分进化算法求解最大值,目标函数为z(i,j)=-20exp((0.2sqrt((x(i)^2+y(j)^2)/2)))-exp((cos(2pix(i))+cos(2piy(j)))/2)+exp(1):
```matlab
% 目标函数
function z = target_func(x, y)
z = -20 * exp(0.2 * sqrt((x.^2 + y.^2) / 2)) - ...
exp((cos(2 * pi * x) + cos(2 * pi * y)) / 2) + ...
exp(1);
end
% 差分进化算法
function [best, value] = de(target_func, bounds, pop_size, F, CR, max_iter)
% 初始化种群
pop = bounds(:,1) + rand(size(bounds, 1), pop_size) .* (bounds(:,2) - bounds(:,1));
% 迭代
for i = 1:max_iter
for j = 1:pop_size
% 变异操作
idxs = randperm(pop_size, 3);
a = pop(:,idxs(1));
b = pop(:,idxs(2));
c = pop(:,idxs(3));
mutant = a + F * (b - c);
mutant = min(max(mutant, bounds(:,1)), bounds(:,2));
% 交叉操作
trial = pop(:,j);
mask = rand(size(trial)) <= CR;
trial(mask) = mutant(mask);
% 选择操作
if target_func(trial(1), trial(2)) >= target_func(pop(1,j), pop(2,j))
pop(:,j) = trial;
end
end
end
% 返回最优解
[value, idx] = max(arrayfun(@(i) target_func(pop(1,i), pop(2,i)), 1:pop_size));
best = pop(:,idx);
end
% 测试
bounds = [-5.12 5.12; -5.12 5.12];
pop_size = 50;
F = 0.5;
CR = 0.9;
max_iter = 100;
[best, value] = de(@target_func, bounds, pop_size, F, CR, max_iter);
disp('最优解为:');
disp(best);
disp('目标函数的最大值为:');
disp(value);
```
在上述代码中,我们使用了rand/1和binomial两种标准的DE操作。其中,F和CR分别表示变异和交叉的参数,可以根据问题进行调整。在迭代过程中,我们选择每一代中最好的解作为返回结果。最终,我们可以得到最优解和目标函数的最大值。