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Copy pathObjectFunc.py
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120 lines (108 loc) · 4.7 KB
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import numpy as np
class ObjFunc():
def __init__(self,datapre):
self.data = datapre.data
self.car_type_exchange_matrix = datapre.car_type_exchange_matrix
self.color_self_exchange_matrix = datapre.color_self_exchange_matrix
self.color_exchange_matrix = datapre.color_exchange_matrix
self.transmission_matrix = datapre.transmission_matrix
self.colorContinutiyM = datapre.colorContinutiyM
self.numOfCar = datapre.numOfCar
self.weldingTime = 80 #焊装耗时80s/车
self.minimumServiceTime = 1800
self.cleaningNozzleTime = 80 #清洗喷头耗时80s/次
# f1:焊装车间装备切换次数
def cal_f1(self,sol:list):
f1 = 0
for i in range(self.numOfCar-1):
f1 += self.car_type_exchange_matrix[sol[i],sol[i+1]]
# print('焊装车间装备切换次数:',f1)
return f1/self.numOfCar
def cal_f2(self,sol:list):
f2 = 0
record = 1 #连续次数
for i in range(1,self.numOfCar):
if self.colorContinutiyM[sol[i-1],sol[i]] and record < 5:
record += 1
else:
f2 += record ** 2
record = 1
return 1/f2
# f3:总装车间，四驱车尽量连放，但连放次数<4，即最小化>=4的四驱车连放数
# 连续二两或者连续三辆四驱车被认为是最优的,和示例完全一样(被示例误导了)
# 满足条件的连续四驱车和总的四驱车比例
def cal_f3(self,sol:list):
# 记录四驱车辆数
fourNum = 0
for i in range(self.numOfCar):
if self.transmission_matrix[sol[i]]:
fourNum += 1
#不妨从后往前退
f3 = 0 #记录不满足连续条件的四驱车数
newsol = sol.copy()
temp = [] #临时存储连续的四驱
while len(newsol) > 0:
carId = newsol.pop()
# 如果是四驱车则添加
if self.transmission_matrix[carId]:
temp.append(carId)
else:
if len(temp) not in [2,3]:
f3 += len(temp)
temp = []
return f3/fourNum
def cal_t1(self,sol:list):
'''计算焊装车间等待时间'''
beginTime = 0 #开始使用时间
cumulativeWeldingTime = self.weldingTime
for i in range(1,self.numOfCar):
# 车辆类型发生改变
if self.car_type_exchange_matrix[sol[i-1],sol[i]]:
# 30min中内发生设备切换，需要等到第30min才可切换
if cumulativeWeldingTime - beginTime < self.minimumServiceTime:
cumulativeWeldingTime = beginTime + self.minimumServiceTime
beginTime = cumulativeWeldingTime
# 车辆类型未发生改变
cumulativeWeldingTime += self.weldingTime
cumulativeWeldingTime -= self.weldingTime * self.numOfCar #焊装设备等待时间
return cumulativeWeldingTime
def cal_t2(self,sol:list):
cumulativeCleaningNozzleTime = 0.0 #额外的清洗时间
i = 0
# 这里只计算5的倍数
while i < self.numOfCar-1:
# 可能会和5的倍数重叠
index = list(range(i,i+5)) #5的倍数生产
count = 1
for j in index[1:]:
if j < self.numOfCar:
if self.color_exchange_matrix[sol[i],sol[j]] == 1:
cumulativeCleaningNozzleTime += self.cleaningNozzleTime
# 颜色连续结束
if self.colorContinutiyM[sol[i],sol[j]] == 0:
i = j
break
else:
count += 1
i = j
else:
break
if count == 5 and i < self.numOfCar-1:
# 下一辆车也连续，不论是否连续都要清洗喷头，所以需要清洗喷头
# if self.color_exchange_matrix[sol[i],sol[i+1]] == 0:
# cumulativeCleaningNozzleTime += self.cleaningNozzleTime
cumulativeCleaningNozzleTime += self.cleaningNozzleTime
return cumulativeCleaningNozzleTime
# f3:总装车间，四驱车尽量连放，但连放次数<4，即最小化>=4的四驱车连放数
def cal_f4(self,sol:list):
f4 = self.cal_t1(sol) + self.cal_t2(sol)
# 换成小时数
return f4 / 3600
def wObjFunc(self,sol:list):
'''加权目标函数'''
f = np.zeros(4)
f[0] = self.cal_f1(sol)
f[1] = self.cal_f2(sol)
f[2] = self.cal_f3(sol)
f[3] = self.cal_f4(sol)
return f