如何解释从Google OR Tools返回的车辆路径问题解决方案？

Question

我有一个使用Google的工具或python库实现的工作车辆路径问题解决方案。我有一个包含9个位置的时间矩阵，每个位置都有时间窗口。所有值均以秒为单位。

（例如，第一个时间窗口是从28800到28800。28800秒相当于上午8:00。我希望在上午8:00访问此位置，即车辆段）

我有意只用一辆车来解决这个问题（本质上是解决一个旅行推销员问题）。我相信我正确地增加了我的维度，但我肯定可能犯了一个错误——我的意图是让车辆在任何位置等待它想要的时间，就像它允许它解决车辆路径问题一样。我将上限最大值设置为86400，因为一天有86400秒，我认为考虑到这些数据，这将是一个足够高的数字。

来源

from ortools.constraint_solver import pywrapcp
from ortools.constraint_solver import routing_enums_pb2

Matrix = [
  [0,557,763,1156,813,618,822,700,112],       # Depot
  [523,0,598,1107,934,607,658,535,589],       # 1 - Location
  [631,480,0,968,960,570,451,135,582],        # 2 - Location
  [1343,1247,1367,0,1270,1289,809,1193,1253], # 3 - Location
  [746,1000,1135,1283,0,1003,1186,1071,776],  # 4 - Location
  [685,627,810,1227,990,0,712,709,550],       # 5 - Location
  [869,718,558,732,1105,650,0,384,821],       # 6 - Location
  [679,528,202,878,1008,618,412,0,630],       # 7 - Location
  [149,626,762,1124,696,532,821,698,0]        # 8 - Location
]

Windows = [
  [ 28800, 28800 ], # Depot
  [ 43200, 43200 ], # 1 - Location
  [ 50400, 50400 ], # 2 - Location
  [ 21600, 79200 ], # 3 - Location
  [ 21600, 79200 ], # 4 - Location
  [ 21600, 79200 ], # 5 - Location
  [ 21600, 79200 ], # 6 - Location
  [ 21600, 79200 ], # 7 - Location
  [ 21600, 79200 ]  # 8 - Location
]

# Create the routing index manager.
manager = pywrapcp.RoutingIndexManager(len(Matrix), 1, 0)

# Create Routing Model.
routing = pywrapcp.RoutingModel(manager)

# Create and register a transit callback.
def time_callback(from_index, to_index):
  # Returns the travel time between the two nodes.
  # Convert from routing variable Index to time matrix NodeIndex.
  from_node = manager.IndexToNode(from_index)
  to_node = manager.IndexToNode(to_index)
  return Matrix[from_node][to_node]

transit_callback_index = routing.RegisterTransitCallback(time_callback)

# Define cost of each arc.
routing.SetArcCostEvaluatorOfAllVehicles(transit_callback_index)

# Add Time Windows constraint.
routing.AddDimension(
    transit_callback_index,
    86400,  # An upper bound for slack (the wait times at the locations).
    86400,  # An upper bound for the total time over each vehicle's route.
    False,  # Determine whether the cumulative variable is set to zero at the start of the vehicle's route.
    'Time')
time_dimension = routing.GetDimensionOrDie('Time')

# Add time window constraints for each location except depot.
for location_idx, time_window in enumerate(Windows):
  if location_idx == 0:
    continue
  index = manager.NodeToIndex(location_idx)
  time_dimension.CumulVar(index).SetRange(time_window[0], time_window[1])

# Add time window constraints for each vehicle start node.
index = routing.Start(0)
time_dimension.CumulVar(index).SetRange(Windows[0][0],Windows[0][1])

# Instantiate route start and end times to produce feasible times.
routing.AddVariableMinimizedByFinalizer(time_dimension.CumulVar(routing.Start(0)))
routing.AddVariableMinimizedByFinalizer(time_dimension.CumulVar(routing.End(0)))

# Setting first solution heuristic. 
search_parameters = pywrapcp.DefaultRoutingSearchParameters()
search_parameters.first_solution_strategy = (routing_enums_pb2.FirstSolutionStrategy.PATH_CHEAPEST_ARC)

# Setting local search metaheuristics:
search_parameters.local_search_metaheuristic = (routing_enums_pb2.LocalSearchMetaheuristic.GUIDED_LOCAL_SEARCH)
search_parameters.time_limit.seconds = 5
search_parameters.log_search = False

# Solve the problem.
solution = routing.SolveWithParameters(search_parameters)

# Return the solution.
time = 0
index = routing.Start(0)
print("Locations:")
while not routing.IsEnd(index):
  time = time_dimension.CumulVar(index)
  print("{0} ({1}, {2})".format(manager.IndexToNode(index),solution.Min(time),solution.Max(time)))
  index = solution.Value(routing.NextVar(index))
print("{0} ({1}, {2})".format(manager.IndexToNode(index),solution.Min(time),solution.Max(time)))

输出

Locations:
0 (28800, 28800)
8 (28912, 42041)
5 (29444, 42573)
1 (43200, 43200)
2 (50400, 50400)
7 (50535, 50535)
6 (50947, 50947)
3 (51679, 51679)
4 (52949, 52949)
0 (52949, 52949)

我的问题是关于解决方案为我计算的输出。我对解决方案中第二个和第三个位置的时间窗口感到困惑。我希望所有的时间窗口都与结果的其余部分一样。解决方案是什么。Min（）和解决方案。Max（）在处理解决方案时，值是否在该问题的范围内？我在使用或工具时是否有任何明显的错误？

Answer 1

我对这个元组的理解是

(Min_time, Max_time)

其中，Min\u time是满足时间窗口要求的最短到达时间。因为最大时间是完全相同的逻辑。

当您可以到达满足约束的节点时，程序会输出一个范围。

Answer 2

Locations:
0 (28800, 28800) // must arrive and leave no later than 28800
8 (28912, 42041) // must arrive at or after 28912 and leave no later than 42041
5 (29444, 42573) // must arrive at or after 29444and leave no later than 42573
1 (43200, 43200) // must arrive and leave no later than 43200
2 (50400, 50400) // must arrive and leave no later than 50400

请参阅我添加的评论。当到达时间是一个范围时，例如节点8或5，基本上意味着到达时间需要落在该时间范围内。只要满足条件，解决方案就仍然可行。

您可以验证如下：

Depot [28800, 28800] -> Travel (0, 8) 112-> Loc 8 [21600, 79200] -> Travel (8, 5) 532 -> Loc 5 [21600, 79200] -> Travel (5, 1) 685 -> Loc 1 [43200, 43200]

在时间28800出发，行程时间为112，您将在时间28912（您的解决方案中的最小值）到达loc 8；立即出发，行程时间为532，您将在时间29444到达loc 5。

现在，loc 1有一个可用的时间段，即43200。因此，如果车辆在时间29444离开，行驶时间为627，它将在时间30071到达位置1，这不是有效的到达时间。但是，如果车辆在43200-627=42573发车，它将准时到达。因此，这意味着车辆需要怠速（松弛）一段时间才能行驶。由于loc 8和loc 5都有一个范围，解决方案是说明这些位置有一些可用的松弛。因此，最小值和最大值真正告诉您的是，只要到达和离开在这些范围内，解决方案是可行的。