生产线平衡算法matlab,装配生产线任务平衡问题的遗传算法MATLAB源代码
时间: 2024-01-11 22:02:03 浏览: 24
以下是一个简单的基于遗传算法的装配生产线任务平衡问题的MATLAB源代码。这个算法旨在最小化每个工作站的空闲时间,同时保持任务的平衡。
```
% 初始化问题参数
num_jobs = 20; % 任务数量
num_stations = 4; % 工作站数量
processing_times = randi([1, 10], num_jobs, num_stations); % 随机生成处理时间
% 遗传算法参数
num_generations = 100; % 世代数量
population_size = 50; % 种群数量
mutation_rate = 0.01; % 变异率
% 初始化种群
population = zeros(population_size, num_jobs); % 每个染色体代表一个任务序列
for i = 1:population_size
population(i, :) = randperm(num_jobs);
end
% 进化
for gen = 1:num_generations
% 计算适应度
fitness = zeros(population_size, 1); % 每个染色体的适应度
for i = 1:population_size
% 计算每个工作站的空闲时间
station_times = zeros(1, num_stations);
for j = 1:num_jobs
station_times(1) = station_times(1) + processing_times(population(i, j), 1);
for k = 2:num_stations
station_times(k) = max(station_times(k), station_times(k-1)) + processing_times(population(i, j), k);
end
end
% 计算适应度为最大空闲时间
fitness(i) = max(station_times);
end
% 选择
[~, idx] = sort(fitness); % 按适应度排序
elite = population(idx(1:ceil(population_size/10)), :); % 保留前10%的精英
parents = population(randperm(population_size, population_size-ceil(population_size/10)), :); % 随机选择90%的父代
% 交叉
children = zeros(size(parents)); % 子代数量等于父代数量
for i = 1:2:population_size-ceil(population_size/10)
p1 = parents(i, :);
p2 = parents(i+1, :);
pt = randi(num_jobs-1); % 随机选择交叉点
c1 = [p1(1:pt), p2(pt+1:end)]; % 子代1
c2 = [p2(1:pt), p1(pt+1:end)]; % 子代2
children(i, :) = c1;
children(i+1, :) = c2;
end
% 变异
mutation_mask = rand(size(children)) < mutation_rate; % 随机生成变异掩码
mutation_values = randi([1, num_jobs], size(children)); % 随机生成变异值
children(mutation_mask) = mutation_values(mutation_mask); % 变异
% 新种群
population = [elite; children]; % 精英和子代组成新种群
end
% 输出结果
best_fitness = inf;
best_solution = zeros(1, num_jobs);
for i = 1:population_size
% 计算每个工作站的空闲时间
station_times = zeros(1, num_stations);
for j = 1:num_jobs
station_times(1) = station_times(1) + processing_times(population(i, j), 1);
for k = 2:num_stations
station_times(k) = max(station_times(k), station_times(k-1)) + processing_times(population(i, j), k);
end
end
% 记录最佳解
if max(station_times) < best_fitness
best_fitness = max(station_times);
best_solution = population(i, :);
end
end
fprintf('最小化空闲时间为 %d\n', best_fitness);
disp('最佳任务序列为:');
disp(best_solution);
```