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MyProblem.py
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MyProblem.py
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import numpy as np
import geatpy as ea
import random
'''
1. 工序链中每个工件的工序总数有限制
2. 若要用多染色体，则使用多个种群、并把每个种群对应个体关联起来即可
'''
class MyProblem(ea.Problem):
def __init__(self, steps=[3, 1, 4, 2, 3, 2, 2, 2, 2, 1]): # steps=[2,3,3,4]工件的工序数
self.a = np.array([
[3, 4, 5, 14],
[4, 5, 7, 14],
[5, 6, 7, 9],
[5, 7, 14, 14],
[4, 14, 14, 14],
[5, 7, 14, 14],
[3, 4, 6, 14],
[5, 6, 8, 14],
[4, 5, 14, 14],
[6, 7, 8, 9],
[6, 8, 10, 14],
[3, 5, 6, 7],
[5, 6, 8, 9],
[3, 5, 7, 14],
[4, 6, 6, 7],
[4, 14, 14, 14],
[5, 7, 8, 14],
[4, 6, 14, 14],
[5, 6, 7, 7],
[6, 7, 7, 8]
])
self.n = 10 # 工件数
self.m = 4 # 机器数
self.H = np.array([3, 1, 4, 2, 3, 2, 2, 2, 2, 1]) # 工件的工序数
self.H_1 = np.cumsum(self.H) - self.H # [1 3 6 9]
self.sumH = sum(self.H)
self.xianbiankucun = 15 # 线边库存
self.yunshufeiyong = 400 # 运输费用
self.lilv = 0.0035
self.tiaozhengshijian = np.array([0.2, 0.3, 0.15, 0.2]) # 调整时间
self.yuancailiaofei = np.array([2400, 3000, 2800, 4000]) # 原材料费
self.jiaohuoqi = np.array([15, 24, 25, 30]) # 交货期 小时
self.tiqianchengfa = np.array([30, 30, 30, 30]) # 提前惩罚 元/小时
self.tuoqichengfa = np.array([60, 70, 50, 50]) # 拖期惩罚 元/小时
self.yunshushijian = np.array([0.15, 0.2, 0.25, 0.3]) # 运输时间 小时
self.gongshifei = np.array([60, 30, 50, 40, 30]) # 机器工时费
self.gongrenfei = np.array([25, 30, 25, 25, 25]) # 工人费
name = 'MyProblem'
M = 2
maxormins = [1] * M # f1成本最低 f2时间最短
Dim = 2 * sum(steps) # 工序编码 机器编码
varTypes = [1] * Dim
lb = [1] * Dim
ub = [self.n] * (self.sumH) + [self.m] * (self.sumH)
lbin = [1] * Dim
ubin = [1] * Dim
ea.Problem.__init__(self, name, M, maxormins, Dim, varTypes, lb, ub, lbin, ubin)
def aimFunc(self, pop): # 目标函数
x = pop.Phen.astype(int)
init_step = [1, 1, 1, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10]
for i in range(40):
np.random.shuffle(init_step)
x[i][:self.sumH] = init_step
process = x[:, :self.sumH]
machine = x[:, self.sumH:]
f1 = np.zeros(np.size(process, 0))
f2 = np.zeros(np.size(process, 0))
for population in range(np.size(process, 0)):
try:
step_1 = self.H_1
time_array = np.zeros((self.sumH, self.m), dtype=int)
for i in range(np.size(process, 1)):
time_array[step_1[process[population][i] - 1], machine[population][i]] = 1
step_1[process[population][i] - 1] += 1
# 计算时间
f1[population] = max(np.sum(time_array * self.a, axis=0))
# 计算成本
f2[population] = np.sum((time_array * self.a) * (self.gongshifei + self.gongrenfei))
except:
f1[population] = 10000000000
f2[population] = 10000000000
# 可行性法则约束工序数 行代表个体 列代表约束条件 CV[1]>0 代表1号个体不满足约束条件
pop.CV = np.abs(np.abs(self.sumH - np.count_nonzero(process - 1, axis=1) - self.H[0]) +
np.abs(self.sumH - np.count_nonzero(process - 2, axis=1) - self.H[1]) +
np.abs(self.sumH - np.count_nonzero(process - 3, axis=1) - self.H[2]) +
np.abs(self.sumH - np.count_nonzero(process - 4, axis=1) - self.H[3]) +
np.abs(self.sumH - np.count_nonzero(process - 5, axis=1) - self.H[4]) +
np.abs(self.sumH - np.count_nonzero(process - 6, axis=1) - self.H[5]) +
np.abs(self.sumH - np.count_nonzero(process - 7, axis=1) - self.H[6]) +
np.abs(self.sumH - np.count_nonzero(process - 8, axis=1) - self.H[7]) +
np.abs(self.sumH - np.count_nonzero(process - 9, axis=1) - self.H[8]) +
np.abs(self.sumH - np.count_nonzero(process - 10, axis=1) - self.H[9]))
pop.CV = np.reshape(pop.CV, (np.size(process, 0), 1))
pop.ObjV = np.vstack([f1, f2]).T