无约束的 VRP 的 OR 工具错误地分组

如何解决无约束的 VRP 的 OR 工具错误地分组

我正在使用 OR 工具来解决 VRP，没有任何限制。这是源代码：

def create_data_model():
    """Stores the data for the problem."""
    data = {}
    data['distance_matrix'] = [
        [0,20079,2613,8005,19277,12468,13701],[0,21285,16012,32574,35394,28806],18233,5392,19965,19650,13064],15013,5639,22883,22570,15982],32991,19256,21815,18414,9112],34348,16976,23122,15678,14647],27652,13917,16476,8043,14820,0]
    ]
    data['time_matrix'] = [
        [0,1955,508,1331,1474,1427,1292],1795,1608,2057,2410,2036],1485,823,1370,1541,1100],1402,924,1533,1637,1263],2308,1663,1853,1766,1104],2231,1373,1660,1441,1554],1998,1353,1543,764,1550,0]
    ]
    data['num_vehicles'] = 6
    data['depot'] = 0
    return data

def print_solution(data,manager,routing,solution):
    """Prints solution on console."""
    max_route_distance = 0
    for vehicle_id in range(data['num_vehicles']):
        index = routing.Start(vehicle_id)
        plan_output = 'Route for vehicle {}:\n'.format(vehicle_id)
        route_distance = 0
        while not routing.IsEnd(index):
            plan_output += ' {} -> '.format(manager.IndexToNode(index))
            prevIoUs_index = index
            index = solution.Value(routing.Nextvar(index))
            route_distance += routing.GetArcCostForVehicle(
                prevIoUs_index,index,vehicle_id)
        plan_output += '{}\n'.format(manager.IndexToNode(index))
        plan_output += 'distance of the route: {}m\n'.format(route_distance)
        print(plan_output)
        max_route_distance = max(route_distance,max_route_distance)
    print('Maximum of the route distances: {}m'.format(max_route_distance))

def test(request):
    # Instantiate the data problem.
    data = create_data_model()
    # Create the routing index manager.
    manager = pywrapcp.RoutingIndexManager(len(data['distance_matrix']),data['num_vehicles'],data['depot'])
    # Create Routing Model.
    routing = pywrapcp.RoutingModel(manager)
    
    def distance_callback(from_index,to_index):
        """Returns the distance between the two nodes."""
        # Convert from routing variable Index to distance matrix NodeIndex.
        from_node = manager.IndexToNode(from_index)
        to_node = manager.IndexToNode(to_index)
        return data['distance_matrix'][from_node][to_node]

    transit_callback_index = routing.RegisterTransitCallback(distance_callback)
    
    routing.SetArcCostEvaluatorOfAllVehicles(transit_callback_index)

    dimension_name = 'distance'
    routing.AddDimension(
        transit_callback_index,# no slack
        1000000000,# vehicle maximum travel distance
        True,# start cumul to zero
        dimension_name)
    distance_dimension = routing.GetDimensionorDie(dimension_name)
    distance_dimension.SetGlobalSpanCostCoefficient(35394)
    
    # Setting first solution heuristic.
    search_parameters = pywrapcp.DefaultRoutingSearchParameters()
    search_parameters.first_solution_strategy = (
        routing_enums_pb2.FirstSolutionStrategy.PATH_CHEApest_ARC)

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

    # Print solution on console.
    if solution:
        print_solution(data,solution)

    return HttpResponse('')

以上所有内容均从 google 示例中复制粘贴而来，但以下内容除外：

我自己的距离矩阵和车辆数量
AddDimension 函数中非常大的车辆最大行驶距离
SetGlobalSpanCostCoefficient(35394)
我按照 OR-Tools solve traveling salesman (TSP) without returning to the home node 将所有节点到 0（仓库）的距离设置为 0。

以上代码的结果如下所示：

Route for vehicle 0:
 0 ->  1 -> 0
distance of the route: 20079m

Route for vehicle 1:
 0 ->  5 -> 0
distance of the route: 12468m

Route for vehicle 2:
 0 ->  4 -> 0
distance of the route: 19277m

Route for vehicle 3:
 0 ->  2 ->  3 -> 0
distance of the route: 8005m

Route for vehicle 4:
 0 ->  6 -> 0
distance of the route: 13701m

Route for vehicle 5:
 0 -> 0
distance of the route: 0m

Maximum of the route distances: 20079m

为了验证上面的输出，我在谷歌地图上标记了点。编号顺序与距离矩阵的顺序相同。

Coords on map

仓库（起点）在 Watthana，可以在标记 B 附近看到。

显然，从 Watthana 出发，单次旅行中最便宜的路径应该是 2-1-3。但是 Google OR 将其返回为两次行程（如车辆 0 和 3 的路线所示）。这也可以通过手动添加距离来验证。

从家到 2 到 1 到 3 的距离 = 2613+5392+15013 = 23018m

车辆 0 和 3 的距离 = 20079+8005 = 28084m

我做错了什么？我怎样才能让谷歌不把第 1 点分开？另请注意，理想情况下，点 E、F、D 也可以分组，但它们没有。

提前致谢！

解决方法

从问题来看，我认为您想要的是减少所有车辆行驶的累计距离。

    distance_dimension.SetGlobalSpanCostCoefficient(35394)

相反，此代码通过将 Span Cost 添加到由 35394 加权的目标函数来确保每辆车行驶的距离最小化。

global_span_cost =
coefficient * (Max(dimension end value) - Min(dimension start value))

在您的情况下，这不是一个非常高的优先级，因此解决方案是注释该行或将系数减少到一个小的值，例如 1 或 2，降低它的唯一重要性。

详细了解GSpanCoeff

现在的解决方案应该是

Route for vehicle 0:
 0 ->  2 ->  3 ->  1 -> 0
Distance of the route: 23018m

Route for vehicle 1:
 0 ->  6 ->  4 -> 0
Distance of the route: 21744m

Route for vehicle 2:
 0 ->  5 -> 0
Distance of the route: 12468m

Maximum of the route distances: 23018m
Sum of the route distances: 57230m