Problem Description
The open shop scheduling problem can be described as: there are n workpieces that need to be processed, each workpiece has m processes, and needs to be processed on m different machines. The processing time of each process is known, but each workpiece The processing order of workpieces is arbitrary; one machine can only process one workpiece at the same time, and one workpiece cannot be processed on two machines at the same time; each workpiece can only be processed on a certain machine at the same time; ultimately It is necessary to find an arrangement and combination of a set of machines and workpieces that will take the shortest time and maximize the efficiency in processing all workpieces.
| artifact | A | B | C | D | AND | F | G | H | I |
|---|---|---|---|---|---|---|---|---|---|
| Workpiece number | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Machine 1 processing time | 4 | 7 | 6 | 5 | 8 | 3 | 5 | 5 | 10 |
| Machine 2 processing time | 7 | 10 | 1 | 5 | 7 | 5 | 8 | 7 | 3 |
| Machine 3 processing time | 7 | 1 | 8 | 9 | 3 | 7 | 8 | 6 | 1 |
| Time of arrival | 3 | 2 | 4 | 5 | 3 | 2 | 1 | 8 | 6 |
| Delivery time | 46 | 35 | 49 | 41 | 40 | 48 | 49 | 37 | 36 |
Number of devices: 3
objective function
Minimize total delivery delay time
Coding instructions
The number of machines is m, the number starting from 0 is 0,1,...,m-1, and the number of workpieces is n, numbering also starts from 0.
I refer to the coding method of genetic algorithmGenetic Algorithms for Open Shop Scheduling and Re-SchedulingThis paper uses A two-digit integer string to represent the corresponding relationship between the workpiece and the assigned machine. For example, the string '43' indicates that the workpiece numbered 3 is assigned to the machine numbered 4 for processing. I think the character The conversion between string and integer is quite troublesome, so it is replaced by tuple, that is, tuple(4,3) means that the workpiece numbered 3 is assigned to the machine numbered 4 for processing (in fact, this I can't see any outstanding effect of this encoding method for the time being. Simply because it is not a mainstream encoding method and it is easier to understand, I gave it a try).
In addition, the cal_start_and_finish_time() function in the genetic algorithm refers to https://github.com/0MarijaS/OpenShopScheduling, it was originally a very simple thing. I couldn’t understand it when I wrote it before. After reading the code, I realized that I had been stuck in an inertial thinking and failed to connect it with the actual situation. It might also be coding. It's a matter of habit.
Operation result
最佳调度顺序和分配： [(0, 1), (0, 8), (1, 5), (2, 0), (2, 3), (1, 6), (0, 7), (2, 6), (1, 2), (1, 1), (0, 5), (2, 5), (0, 3), (2, 1), (1, 4), (0, 0), (0, 6), (1, 8), (2, 8), (2, 7), (0, 2), (0, 4), (2, 2), (2, 4), (1, 3), (1, 0), (1, 7)]
最小交货期延时时间： 36
python code
import random
import numpy as np
import matplotlib.pyplot as plt
import copy
# 定义遗传算法参数
POP_SIZE = 100 # 种群大小
MAX_GEN = 100 # 最大迭代次数
CROSSOVER_RATE = 0.7 # 交叉概率
MUTATION_RATE = 0.2 # 变异概率
def sort_by_id(id, sequence):
# 根据id对sequence进行排序
new_sequence = sequence[:]
for i in range(len(id)):
sequence[i] = new_sequence[id[i]]
# 随机生成初始种群，这里用一个元组表示工件和机器的分配关系，如元组(0,1)表示编号为1的工件在编号为0的机器上加工，工件和机器编码都是从0开始
def get_init_pop(pop_size):
pop = []
job = [(m, n) for m in range(machine_num) for n in range(len(job_id))]
for _ in range(pop_size):
random.shuffle(job)
pop.append(job[:])
return pop
# 检查工件是否能够插入排程好的机器的间隙
def check_job(start, limit, duration, job_activity):
if len(job_activity) == 0:
if start + duration <= limit:
return True, start
elif len(job_activity) == 1:
if job_activity[0][0] != 0 and job_activity[0][0] >= start + duration:
return True, start
else:
if job_activity[0][1] + duration <= limit:
return True, max(job_activity[0][1], start)
else:
end = start + duration
j = 0
while end <= limit and j < len(job_activity) - 1:
if job_activity[j][0] <= start < job_activity[j][1]:
start = job_activity[j][1]
end = start + duration
if start >= job_activity[j][1] and end <= job_activity[j + 1][0]:
return True, start
j += 1
return False, -1
def cal_start_and_finish_time(operations, pro_times, arr_times, machine_num):
# operations: list[tuple()] 一个列表，列表中元素为二元组(机器编号,工件编号)
jobs = [[] for _ in range(len(operations) // machine_num)]
solution = [[] for _ in range(machine_num)] # 封装一个有效的调度方案，(开始时间, 结束时间, 操作编号)
for operation in operations:
machine, job = operation[0], operation[1]
duration = pro_times[machine][job] # 定位加工时间
machine_activity = solution[machine] # 当前时刻当前机器是否正在加工
job_activity = jobs[job] # 当前时刻当前工件是否正在加工
# print(f'==========operation={operation}===========')
# print("job_activity=",job_activity)
# print("machine_activity=",machine_activity)
fit, start, limit = False, -1, float('inf')
if len(machine_activity) == 0: # 如果机器空闲
start = arr_times[operation[1]]
fit, start = check_job(start, limit, duration, job_activity)
elif len(machine_activity) == 1 and machine_activity[0][
0] > arr_times[operation[1]]: # 如果该机器只有一个操作，且该操作开始时间大于当前操作对应工件的到达时间，则判断一下当前操作是否可以排在该操作的前面
start = arr_times[operation[1]]
limit = machine_activity[0][0]
fit, start = check_job(start, limit, duration, job_activity)
elif len(machine_activity) == 1 and machine_activity[0][0] <= arr_times[
operation[1]]: # 如果该机器只有一个操作，且该操作开始时间小于等于当前操作对应工件的到达时间，则将当前操作排在该操作后边
start = max(arr_times[operation[1]], machine_activity[0][1])
fit, start = check_job(start, limit, duration, job_activity)
else: # 如果该机器有两个或两个以上的操作，则看下操作与操作之间是否存在足够的间隙可容纳当前操作
for i in range(len(machine_activity) - 1):
start = max(arr_times[operation[1]], machine_activity[i][1])
limit = machine_activity[i + 1][0]
options = []
if limit - start >= duration:
fit, start = check_job(start, limit, duration, job_activity)
if fit:
options.append((start, limit - (start + duration)))
if len(options) > 0:
options.sort(key=lambda x: x[1])
start = options[0][0]
fit = True
if fit and start + duration <= limit:
jobs[job].append((start, start + duration)) # (开始时间,结束时间)
solution[machine].append((start, start + duration, operation))
jobs[job].sort() # 升序排序（时间是从小到大排列）
solution[machine].sort()
else:
start = 0
if len(machine_activity) >= 1:
start = machine_activity[-1][1]
if len(job_activity) >= 1:
start = max(start, job_activity[-1][1])
jobs[job].append((start, start + duration))
solution[machine].append((start, start + duration, operation))
# print(jobs)
# print(solution)
return solution
# TODO: 开始时间和结束时间计算有问题，这个函数废了
# 将输入的一维job进行二维转换，计算start_time和finish_time
def cal_start_and_finish_time_failed(job):
# 按照输入的工序进行排序
sorted_job = sorted(job, key=lambda x: x[0]) # 按照设备排序
sorted_pro_time = [pro_times[j[0]][j[1]] for j in sorted_job] # 对应的加工时间
# 转换成二维列表
sorted_job = [sorted_job[i:i + len(job_id)] for i in range(0, len(sorted_job), len(job_id))]
sorted_pro_time = [sorted_pro_time[i:i + len(job_id)] for i in range(0, len(sorted_pro_time), len(job_id))]
# 对每台机器的总加工时间进行加和，用于排序
# 这里直接规定当各机器无法同时开工时，选择总加工时间最长的机器最先开工
start_seq = [i[0] for i in sorted(enumerate(pro_times), key=lambda x: sum(x[1]), reverse=True)] # 机器开始加工的顺序
start_time = [[] for _ in range(len(sorted_job))]
finish_time = [[] for _ in range(len(sorted_job))]
# 第一台开工的机器的第一个工件的start_time和finish_time
start_time[start_seq[0]].append(arr_times[sorted_job[start_seq[0]][0][1]])
finish_time[start_seq[0]].append(start_time[start_seq[0]][0] + sorted_pro_time[start_seq[0]][0])
# 单独先把第一个开工的机器的start_time和finish_time计算出来
for j in range(1, len(pro_times[0])):
start_time[start_seq[0]].append(
max(finish_time[start_seq[0]][-1], arr_times[sorted_job[start_seq[0]][j][1]]))
finish_time[start_seq[0]].append(start_time[start_seq[0]][j] + sorted_pro_time[start_seq[0]][j])
# 计算每台机器的第一个工件的开始加工时间
for i in range(1, len(start_seq)):
# 一个工件不能同时在两台机器加工
start_time[start_seq[i]].append(-1)
for k in range(i):
if sorted_job[start_seq[i]][0][1] == sorted_job[start_seq[k]][0][1]:
start_time[start_seq[i]][-1] = finish_time[start_seq[k]][0]
elif k == i - 1 and start_time[start_seq[i]][-1] == -1:
start_time[start_seq[i]][-1] = arr_times[sorted_job[start_seq[i]][0][1]]
finish_time[start_seq[i]].append(start_time[start_seq[i]][0] + sorted_pro_time[start_seq[i]][0])
# 计算其他行列
for i in range(1, len(start_seq)):
for j in range(1, len(pro_times[0])):
start_time[start_seq[i]].append(-1)
for k in range(i):
# 写着写着突然发现不对，还是需要判断前边两台机器上正在加工的工件是不是当前机器即将要加工的工件
# 感觉这里把自己绕进去了，实际上，当某个工件在当前机器上的加工时间特别大，可能是其在别的机器上加工时间的几倍或者是别的工件加工时间之和的时候，这个if判断还要向前或者向后搜索很多轮
# 但是如果真有这样一道工序存在，应该不需要什么算法直接可以看出结果，所以这里只搜索相邻的两轮
# TODO: but，逻辑还是欠严谨，这是一段失败的代码，或许证明这思路就不行？先到这里吧，麻了
if finish_time[start_seq[i]][j - 1] <= finish_time[start_seq[k]][j - 1]:
if sorted_job[start_seq[i]][j][1] == sorted_job[start_seq[k]][j - 1][1]:
start_time[start_seq[i]][-1] = max(finish_time[start_seq[i]][-1],
finish_time[start_seq[k]][j - 1])
elif k == i - 1 and start_time[start_seq[i]][-1] == -1:
start_time[start_seq[i]][-1] = max(finish_time[start_seq[i]][-1],
arr_times[sorted_job[start_seq[i]][j][1]])
elif finish_time[start_seq[k]][j - 1] < finish_time[start_seq[i]][j - 1] <= finish_time[start_seq[k]][
j]:
if sorted_job[start_seq[i]][j][1] == sorted_job[start_seq[k]][j][1]:
start_time[start_seq[i]][-1] = max(finish_time[start_seq[i]][-1], finish_time[start_seq[k]][j])
elif k == i - 1 and start_time[start_seq[i]][-1] == -1:
start_time[start_seq[i]][-1] = max(finish_time[start_seq[i]][-1],
arr_times[sorted_job[start_seq[i]][j][1]])
else:
start_time[start_seq[i]][-1] = max(finish_time[start_seq[i]][-1],
arr_times[sorted_job[start_seq[i]][j][1]])
finish_time[start_seq[i]].append(start_time[start_seq[i]][j] + sorted_pro_time[start_seq[i]][j])
return sorted_job, sorted_pro_time, start_time, finish_time
# 计算染色体的适应度（makespan） 以最小化交货期延时为目标函数，这里计算的是交货期总延时时间
def fitness(job):
solution = cal_start_and_finish_time(job, pro_times, arr_times,
machine_num)
finish_time = [[ft[1] for ft in sub] for sub in solution]
# 按照输入的工序进行排序
sorted_job = sorted(job, key=lambda x: x[0]) # 按照设备排序
# 转换成二维列表
sorted_job = [sorted_job[i:i + len(job_id)] for i in range(0, len(sorted_job), len(job_id))]
# 计算适应度，目标函数是最小化总延货期
max_finish_time = [] # 记录每个工件在各机器上的最后的完工时间
for k in range(len(job_id)):
# 定位，定位到sorted_job中所有元组的第二个元素为i的索引
indices = [(i, j) for i, row in enumerate(sorted_job) for j, (x, y) in enumerate(row) if y == k]
max_ft = 0
for id in indices:
max_ft = finish_time[id[0]][id[1]] if finish_time[id[0]][id[1]] > max_ft else max_ft
max_finish_time.append(max_ft)
delay_time = [max(mf - d, 0) for mf, d in zip(max_finish_time, deadlines)] # 总延货期
# print("sorted_job=",sorted_job)
# print("sorted_pro_time=",sorted_pro_time)
# print("finish_time=",finish_time)
# print("delay_time=",delay_time)
# print("sum(delay_time)=",sum(delay_time))
return sum(delay_time)
# 选择父代，这里选择POP_SIZE/2个作为父代
def selection(pop):
fitness_values = [1 / fitness(job) for job in pop] # 以最小化交货期总延时为目标函数，这里把最小化问题转变为最大化问题
total_fitness = sum(fitness_values)
prob = [fitness_value / total_fitness for fitness_value in fitness_values] # 轮盘赌，这里是每个适应度值被选中的概率
# 按概率分布prob从区间[0,len(pop))中随机抽取size个元素，不允许重复抽取，即轮盘赌选择
selected_indices = np.random.choice(len(pop), size=POP_SIZE // 2, p=prob, replace=False)
return [pop[i] for i in selected_indices]
# 交叉操作 这里是单点交叉
def crossover(job_p1, job_p2):
cross_point = random.randint(1, len(job_p1) - 1)
job_c1 = job_p1[:cross_point] + [gene for gene in job_p2 if gene not in job_p1[:cross_point]]
job_c2 = job_p2[:cross_point] + [gene for gene in job_p1 if gene not in job_p2[:cross_point]]
return job_c1, job_c2
# 变异操作，因为元组是不可变类型，这里采取的变异方式是随机交换两个点
def mutation(job):
while True:
index1 = random.randint(0, len(job) - 1)
index2 = random.randint(0, len(job) - 1)
if index1 != index2:
break
temp = job[index1]
job[index1] = job[index2]
job[index2] = temp
return job
# 主遗传算法循环
# 以最小化延迟交货时间为目标函数
def GA(): # 工件加工顺序是否为无序
best_job = [(m, n) for m in range(machine_num) for n in range(len(job_id))] # 初始化最佳个体
best_makespan = fitness(best_job) # 获得最佳个体的适应度值
# 创建一个空列表来存储每代的适应度值
fitness_history = [best_makespan]
pop = get_init_pop(POP_SIZE)
for _ in range(1, MAX_GEN + 1):
pop = selection(pop) # 选择
new_population = []
while len(new_population) < POP_SIZE:
parent1, parent2 = random.sample(pop, 2) # 不重复抽样2个
if random.random() < CROSSOVER_RATE:
child1, child2 = crossover(parent1, parent2) # 交叉
new_population.extend([child1, child2])
else:
new_population.extend([parent1, parent2])
pop = [mutation(job) if random.random() < MUTATION_RATE else job for job in new_population] # 变异
best_gen_job = min(pop, key=lambda x: fitness(x))
best_gen_makespan = fitness(best_gen_job) # 每一次迭代获得最佳个体的适应度值
# print(job_id)
# print(best_gen_job)
# print(best_gen_makespan,best_makespan,best_gen_makespan < best_makespan)
if best_gen_makespan < best_makespan: # 更新最小fitness值
best_makespan = best_gen_makespan
best_job = copy.deepcopy(best_gen_job)
fitness_history.append(best_makespan) # 把本次迭代结果保存到fitness_history中（用于绘迭代曲线）
# print(best_job)
# print(best_makespan)
# print('==========================')
# 绘制迭代曲线图
plt.plot(range(MAX_GEN + 1), fitness_history)
plt.xlabel('Generation')
plt.ylabel('Fitness Value')
plt.title('Genetic Algorithm Convergence')
plt.show()
return best_job, best_makespan
def plot_gantt(operations):
solution = cal_start_and_finish_time(operations, pro_times, arr_times, machine_num)
# 准备一系列颜色
colors = ['blue', 'yellow', 'orange', 'green', 'palegoldenrod', 'purple', 'pink', 'Thistle', 'Magenta', 'SlateBlue', 'RoyalBlue', 'Cyan', 'Aqua', 'floralwhite', 'ghostwhite', 'goldenrod', 'mediumslateblue', 'navajowhite', 'moccasin', 'white', 'navy', 'sandybrown', 'moccasin']
# colors = ['r', 'g', 'b', 'm', 'c', 'y', 'w', 'o', 'p']
job_colors = random.sample(colors, len(job_id))
start_times = [[-1 for _ in range(len(solution[0]))] for _ in range(len(solution))]
end_times = [[-1 for _ in range(len(solution[0]))] for _ in range(len(solution))]
job_decodes = [[-1 for _ in range(len(solution[0]))] for _ in range(len(solution))]
for i in range(len(solution)):
for j in range(len(solution[0])):
start_times[i][j] = solution[i][j][0]
end_times[i][j] = solution[i][j][1]
job_decodes[i][j] = solution[i][j][2] # (机器编号, 工件编号)
# 创建图表和子图
plt.figure(figsize=(12, 6))
# 绘制工序的甘特图
for i in range(len(start_times)):
for j in range(len(start_times[i])):
plt.barh(i, end_times[i][j] - start_times[i][j], height=0.5, left=start_times[i][j],
color=job_colors[job_decodes[i][j][1]], edgecolor='black')
plt.text(x=(start_times[i][j] + end_times[i][j]) / 2, y=i, s=job_decodes[i][j], fontsize=14)
# 设置纵坐标轴刻度为机器编号
machines = [f'Machine {
i}' for i in range(len(start_times))]
plt.yticks(range(len(machines)), machines)
# 设置横坐标轴刻度为时间
# start = min([min(row) for row in start_times])
start = 0
end = max([max(row) for row in end_times])
plt.xticks(range(start, end + 1))
plt.xlabel('Time')
# 图表样式设置
plt.ylabel('Machines')
plt.title('Gantt Chart')
# plt.grid(axis='x')
# 自动调整图表布局
plt.tight_layout()
# 显示图表
plt.show()
if __name__ == '__main__':
# n个工件，每个工件都需要到m台不同机器上各加工一次，加工顺序任意，每台机器上的加工时间已知
job_id = [0, 1, 2, 3, 4, 5, 6, 7, 8] # 工件编号
pro_times = [[4, 7, 6, 5, 8, 3, 5, 5, 10],
[7, 10, 1, 5, 7, 5, 8, 7, 3],
[7, 1, 8, 9, 3, 7, 8, 6, 1]] # 加工时间
arr_times = [3, 2, 4, 5, 3, 2, 1, 8, 6] # 到达时间
deadlines = [46, 35, 49, 41, 40, 48, 49, 37, 36] # 交货期
machine_num = 3 # 3台完全相同的并行机，编号为0,1,2
best_job, best_makespan = GA()
print("最佳调度顺序和分配：", best_job)
print("最小交货期延时时间：", best_makespan)
plot_gantt(best_job)