问题描述
我成功地实施了一个计划，我为一周中的每一天分配 N truck drivers 到 M gathering hubs。我实施的约束是：
-
- 司机不能工作超过 6 天，即休息 1 天
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- 一名司机每天不能分配到超过 1 个集线器
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- 每个枢纽都必须满足一周中每一天的司机要求
在程序运行平稳，满足总体目标，并输出对于每个集线器驱动程序对以下形式的时间表：
Monday Tuesday Wednesday Thursday Friday Saturday Sunday
Hub Driver
Hub 1 Driver_20 1 0 0 0 0 0 0
Hub 2 Driver_20 0 0 0 0 0 0 0
Hub 3 Driver_20 0 0 0 0 0 0 0
Hub 4 Driver_20 0 0 0 0 0 0 0
Hub 5 Driver_20 0 1 0 0 0 0 0
Hub 6 Driver_20 0 0 0 0 1 0 0
Hub 7 Driver_20 0 0 0 1 0 1 1
但是，我想添加一个额外约束，如果可能的话，强制司机在一个中心工作，而不是将他们的工作日分散在许多中心，即在将驱动程序分配到不同的集线器之前，最大限度地提高一个集线器的工作量。
例如，在上面的输出中，我们看到驱动程序在不同的集线器工作 3 天，在集线器 7 工作 3 天。我们如何编写约束来使驱动程序被分配（如果可能的话）在一个集线器工作如果可能？
请在下面找到我的代码。
谢谢
import pulp
import pandas as pd
import numpy as np
pd.set_option('display.max_rows',None)
pd.set_option('display.max_columns',None)
pd.set_option('display.width',2000)
pd.set_option('display.float_format','{:20,.2f}'.format)
pd.set_option('display.max_colwidth',None)
day_requirement = [[2,2,3,5,2],[2,[3,3],[4,4,4],]
total_day_requirements = ([sum(x) for x in zip(*day_requirement)])
hub_names = {0: 'Hub 1',1: 'Hub 2',2: 'Hub 3',3: 'Hub 4',4: 'Hub 5',5: 'Hub 6',6: 'Hub 7'}
total_drivers = max(total_day_requirements) # number of drivers
total_days = 7 # The number of days in week
total_hubs = len(day_requirement) # number of hubs
def schedule(drivers,days,hubs):
driver_names = ['Driver_{}'.format(i) for i in range(drivers)]
var = pulp.LpVariable.dicts('VAR',(range(hubs),range(drivers),range(days)),1,'Binary')
problem = pulp.LpProblem('shift',pulp.LpMinimize)
obj = None
for h in range(hubs):
for driver in range(drivers):
for day in range(days):
obj += var[h][driver][day]
problem += obj
# schedule must satisfy daily requirements of each hub
for day in range(days):
for h in range(hubs):
problem += pulp.lpSum(var[h][driver][day] for driver in range(drivers)) == \
day_requirement[h][day]
# a driver cannot work more than 6 days
for driver in range(drivers):
problem += pulp.lpSum([var[h][driver][day] for day in range(days) for h in range(hubs)]) <= 6
# if a driver works one day at a hub,he cannot work that day in a different hub obvIoUsly
for driver in range(drivers):
for day in range(days):
problem += pulp.lpSum([var[h][driver][day] for h in range(hubs)]) <= 1
# Solve problem.
status = problem.solve(pulp.PULP_CBC_CMD(msg=0))
idx = pd.MultiIndex.from_product([hub_names.values(),driver_names],names=['Hub','Driver'])
col = ['Monday','Tuesday','Wednesday','Thursday','Friday','Saturday','Sunday']
dashboard = pd.DataFrame(0,idx,col)
for h in range(hubs):
for driver in range(drivers):
for day in range(days):
if var[h][driver][day].value() > 0.0:
dashboard.loc[hub_names[h],driver_names[driver]][col[day]] = 1
driver_table = dashboard.groupby('Driver').sum()
driver_sums = driver_table.sum(axis=1)
# print(driver_sums)
day_sums = driver_table.sum(axis=0)
# print(day_sums)
print("Status",pulp.LpStatus[status])
if (driver_sums > 6).any():
print('One or more drivers have been allocated more than 6 days of work so we must add one '
'driver: {}->{}'.format(len(driver_names),len(driver_names) + 1))
schedule(len(driver_names) + 1,hubs)
else:
print(dashboard)
print(driver_sums)
print(day_sums)
for driver in range(drivers):
driver_name = 'Driver_{}'.format(driver)
print(dashboard[np.in1d(dashboard.index.get_level_values(1),[driver_name])])
schedule(total_drivers,total_days,total_hubs)
解决方法
您可以添加二进制变量 z，指示驱动程序是否在集线器上处于活动状态：
z = pulp.LpVariable.dicts('Z',(range(hubs),range(drivers)),1,'Binary')
然后将您的目标更改为（最小化集线器上活动的驱动程序总数）：
for h in range(hubs):
for driver in range(drivers):
obj += z[h][driver]
problem += obj
添加约束以将 z 与 var 连接起来：
for driver in range(drivers):
for h in range(hubs):
problem += z[h][driver] <= pulp.lpSum(var[h][driver][day] for day in range(days))
problem += total_days*z[h][driver] >= pulp.lpSum(var[h][driver][day] for day in range(days))
然而，这个模型更复杂，找到最佳解决方案似乎需要一段时间。您可以设置超时（此处为 10 秒）以获取解决方案：
status = problem.solve(pulp.PULP_CBC_CMD(msg=0,timeLimit=10))