【进阶三】Python实现(MD)HFVRPTW常见求解算法——离散粒子群(DPSO)

基于python语言,实现经典离散粒子群算法(DPSO)对多车场(Multi-depot)、异构固定车辆(heterogeneous fixed fleet)、带有服务时间窗(time window)等限制约束的车辆路径规划问题((MD)HFVRPTW)进行求解。

目录

往期资料

1. 适用场景

  • 求解HFVRPTW或MDHFVRPTW
  • 异构固定车队
  • 车辆容量不小于需求节点最大需求
  • 车辆路径长度或运行时间无限制
  • 需求节点服务成本满足三角不等式
  • 节点时间窗至少满足车辆路径只包含一个需求节点的情况
  • 多车辆基地或单一
  • 各车场车辆总数满足实际需求

2. 求解效果

(1)收敛曲线
【进阶三】Python实现(MD)HFVRPTW常见求解算法——离散粒子群(DPSO)

(2)车辆路径
【进阶三】Python实现(MD)HFVRPTW常见求解算法——离散粒子群(DPSO)

3. 代码分析

本算法继续采用将所有需求节点构造为一个有序列表的编码方式,并运用在交叉、变异等寻优过程中。当需要评估染色体质量时需采用split方法,在考虑车场、异构固定车队、服务时间窗等约束条件下,将有序列表分割为多个可行的车辆路径。split过程也是整个算法的核心,这里复现文末的参考文献中的算法3,并做了适当微调。整个算法的函数调用关系如下图(采用PyCallGraph绘制)。
【进阶三】Python实现(MD)HFVRPTW常见求解算法——离散粒子群(DPSO)

4. 数据格式

以csv文件储存数据,其中demand.csv文件记录需求节点数据,共包含需求节点id,需求节点横坐标,需求节点纵坐标,需求量;depot.csv文件记录车场节点数据,共包含车场id,车场横坐标,车场纵坐标,车队类型,车辆容量,车辆速度,车辆数量,车辆固定成本,车辆单位变动成本,车辆最早开始服务时间,车辆最晚结束服务时间。需要注意的是:需求节点id应为整数,从0开始编号,车场节点id任意,但不可与需求节点id重复(建议以 ‘d’+int 形式,便于程序可视化路线)。 可参考github主页相关文件。

5. 分步实现

(1)数据结构
定义Sol()类,Demand()类,Vehicle()类,Model()类,其属性如下表:

  • Sol()类,表示一个可行解
属性 描述
node_id_list 需求节点id有序排列集合
obj 优化目标值
route_list 车辆路径集合,对应MDVRPTW的解
timetable_list 车辆节点访问时间集合,对应MDVRPTW的解
distance_of_routes 总旅行距离
time_of_routes 总时间
  • Demand()类,表示一个需求节点
属性 描述
id 物理节点id,需唯一
x_coord 物理节点x坐标
y_coord 物理节点y坐标
demand 物理节点需求
start_time 最早开始服务(被服务)时间
end_time 最晚结束服务(被服务)时间
service_time 需求节点服务时间
  • Vehicle()类,表示一个车队节点
属性 描述
depot_id 车辆归属的车场节点节点id,需唯一
x_coord 车辆归属车场节点x坐标
y_coord 车辆归属车场节点y坐标
type 车辆类型
capacity 车辆容量
free_speed 车辆运营速度
fixed_cost 车辆固定成本
variable_cost 车辆变动成本
start_time 最早开始服务时间
end_time 最晚结束服务时间
  • Model()类,存储算法参数
属性 描述
best_sol 全局最优解,值类型为Sol()
demand_dict 需求节点集合(字典),值类型为Demand()
vehicle_dict 车队集合(字典),值类型为Vehicle()
vehicle_type_list 车队id集合
demand_id_list 需求节点id集合
sol_list 种群,值类型为Sol()
distance_matrix 节点距离矩阵
number_of_demands 需求节点数量
opt_type 优化目标类型,0:最小旅行距离,1:最小时间成本
popsize 种群规模
v 可行解更新速度
Vmax 最大速度
w 惯性权重
c1 学习因子
c2 学习因子

(2)文件读取

def readCSVFile(demand_file,depot_file,model):
    with open(demand_file,'r') as f:
        demand_reader=csv.DictReader(f)
        for row in demand_reader:
            demand = Demand()
            demand.id = int(row['id'])
            demand.x_coord = float(row['x_coord'])
            demand.y_coord = float(row['y_coord'])
            demand.demand = float(row['demand'])
            demand.start_time=float(row['start_time'])
            demand.end_time=float(row['end_time'])
            demand.service_time=float(row['service_time'])
            model.demand_dict[demand.id] = demand
            model.demand_id_list.append(demand.id)
        model.number_of_demands=len(model.demand_id_list)

    with open(depot_file, 'r') as f:
        depot_reader = csv.DictReader(f)
        for row in depot_reader:
            vehicle = Vehicle()
            vehicle.depot_id = row['depot_id']
            vehicle.x_coord = float(row['x_coord'])
            vehicle.y_coord = float(row['y_coord'])
            vehicle.type = row['vehicle_type']
            vehicle.capacity=float(row['vehicle_capacity'])
            vehicle.free_speed=float(row['vehicle_speed'])
            vehicle.numbers=float(row['number_of_vehicle'])
            vehicle.fixed_cost=float(row['fixed_cost'])
            vehicle.variable_cost=float(row['variable_cost'])
            vehicle.start_time=float(row['start_time'])
            vehicle.end_time=float(row['end_time'])
            model.vehicle_dict[vehicle.type] = vehicle
            model.vehicle_type_list.append(vehicle.type)

(3)计算距离矩阵

"计算距离矩阵"
def calDistanceMatrix(model):
    for i in range(len(model.demand_id_list)):
        from_node_id = model.demand_id_list[i]
        for j in range(i + 1, len(model.demand_id_list)):
            to_node_id = model.demand_id_list[j]
            dist = math.sqrt((model.demand_dict[from_node_id].x_coord - model.demand_dict[to_node_id].x_coord) ** 2
                             + (model.demand_dict[from_node_id].y_coord - model.demand_dict[to_node_id].y_coord) ** 2)
            model.distance_matrix[from_node_id, to_node_id] = dist
            model.distance_matrix[to_node_id, from_node_id] = dist
        for _, vehicle in model.vehicle_dict.items():
            dist = math.sqrt((model.demand_dict[from_node_id].x_coord - vehicle.x_coord) ** 2
                             + (model.demand_dict[from_node_id].y_coord - vehicle.y_coord) ** 2)
            model.distance_matrix[from_node_id, vehicle.type] = dist
            model.distance_matrix[vehicle.type, from_node_id] = dist

(4)分割路径
split过程采用标号法最短路思想,为了避免在搜索过程中产生大量劣质节点标签,通过一定规则删除劣质标签:根据帕累托删除被支配的标签;根据剩余容量与剩余需求决定是否生成新标签。
在搜索过程中需要计算可能车辆路径的成本(时间成本或距离成本),如果采用时间成本,这里为了简化,只计算旅行距离成本,忽略了等待时间成本。但在计算适应度部分是严格按照时间成本内容计算的(旅行时间成本+等待时间成本)。

"检查路径是否满足时间要求,不满足要求则不会产生新的标签"
def checkTimeWindow(route,model,vehicle):
    timetable=[]
    departure=0
    for i in range(len(route)):
        if i == 0:
            next_node_id = route[i + 1]
            travel_time = int(model.distance_matrix[vehicle.type, next_node_id] /vehicle.free_speed)
            departure = max(0, model.demand_dict[next_node_id].start_time - travel_time)
            timetable.append((int(departure), int(departure)))
        elif 1 <= i <= len(route) - 2:
            last_node_id = route[i - 1]
            current_node_id = route[i]
            current_node = model.demand_dict[current_node_id]
            travel_time = int(model.distance_matrix[last_node_id, current_node_id] / vehicle.free_speed)
            arrival = max(timetable[-1][1] + travel_time, current_node.start_time)
            departure = arrival + current_node.service_time
            timetable.append((int(arrival), int(departure)))
            if departure > current_node.end_time:
                departure = float('inf')
                break
        else:
            last_node_id = route[i - 1]
            travel_time = int(model.distance_matrix[last_node_id, vehicle.type] / vehicle.free_speed)
            departure = timetable[-1][1] + travel_time
            timetable.append((int(departure), int(departure)))
    if departure<vehicle.end_time:
        return True
    else:
        return False

"当产生新的标签W后,检查剩余的车辆容量之和是否能满足剩余未被检车的检点的总需求,如果总容量<总需求,则舍弃W,表明采用W后会导致解不可行"
"这也是减少无效标签的途径之一"
def checkResidualCapacity(residual_node_id_list,W,model):
    residual_fleet_capacity=0
    residual_demand = 0
    for node_id in residual_node_id_list:
        residual_demand+=model.demand_dict[node_id].demand
    for k,v_type in enumerate(model.vehicle_type_list):
        vehicle=model.vehicle_dict[v_type]
        residual_fleet_capacity+=(vehicle.numbers-W[k+4])*vehicle.capacity
    if residual_demand<=residual_fleet_capacity:
        return True
    else:
        return False

"由于标号法会产生大量标签,为了降标签数量,减少对劣质标签的搜索,在插入新标签时根据帕累托,删除支配解"
def updateNodeLabels(label_list,W,number_of_lables):
    new_label_list=[]
    if len(label_list)==0:
        number_of_lables += 1
        W[0] = number_of_lables
        new_label_list.append(W)
    else:
        for label in label_list:
            if W[3]<=label[3] and sum(W[4:])<=sum(label[4:]):
                if W not in new_label_list:
                    number_of_lables += 1
                    W[0] = number_of_lables
                    new_label_list.append(W)
            elif W[3]<=label[3] and sum(W[4:])>sum(label[4:]):
                new_label_list.append(label)
                if W not in new_label_list:
                    number_of_lables += 1
                    W[0] = number_of_lables
                    new_label_list.append(W)
            elif W[3]>label[3] and sum(W[4:])<sum(label[4:]):
                new_label_list.append(label)
                if W not in new_label_list:
                    number_of_lables += 1
                    W[0] = number_of_lables
                    new_label_list.append(W)
            elif W[3]>label[3] and sum(W[4:])>=sum(label[4:]):
                new_label_list.append(label)
    return new_label_list,number_of_lables

"根据标号法的求解结果,从中提取出各车辆路径"
def extractRoutes(V,node_id_list,model):
    route_list = []
    min_obj=float('inf')
    pred_label_id=None
    v_type=None
    # search the min cost label of last node of the node_id_list
    for label in V[model.number_of_demands-1]:
        if label[3]<=min_obj:
            min_obj=label[3]
            pred_label_id=label[1]
            v_type=label[2]
    # generate routes by pred_label_id
    route=[node_id_list[-1]]
    indexs=list(range(0,model.number_of_demands))[::-1]
    start=1
    while pred_label_id!=1:
        for i in indexs[start:]:
            stop=False
            for label in V[i]:
                if label[0]==pred_label_id:
                    stop=True
                    pred_label_id=label[1]
                    start=i
                    v_type_=label[2]
                    break
            if not stop:
                route.insert(0,node_id_list[i])
            else:
                route.insert(0,v_type)
                route.append(v_type)
                route_list.append(route)
                route=[node_id_list[i]]
                v_type=v_type_
    route.insert(0,v_type)
    route.append(v_type)
    route_list.append(route)
    return route_list

"采用标号法对node_id_list进行分割,得到车辆路径"
def splitRoutes(node_id_list,model):
    """
    V: dict,key=id,value=[n1,n2,n3,n4,n5,....]
        id:node_id_list的索引
        n1: 当前标签的生成次序
        n2: 生成当前标签的前一个标签的id
        n3: 当前标签对应的车辆类型
        n4: 当前路径的费用,对应与优化目标,当优化目标为旅行时间时,这里为简化计算只考虑节点间的旅行时间,舍去了等待时间
        n5-: 截止到当前标签,各类型车辆的使用数量
    这里采用先搜索车辆集合再搜索标签集合的方法,与原文是相反的"
        假设有a个标签,n个车需要搜索
            若按照原文的搜索顺序:对于任意一个label,都要判断当前路径对于n个车是否满足时间窗要求,搜索次数=a*n;
            若按照本文的搜索顺序。对于任意一个车辆类型,若路径不满足时间窗要求则不进行标签搜索,因此搜索次数应<a*n
    """
    V={i:[] for i in model.demand_id_list}
    V[-1]=[[0]*(len(model.vehicle_type_list)+4)] # -1表示虚拟车场的索引
    V[-1][0][0]=1 # 虚拟车场的标签id为1
    V[-1][0][1]=1 # 虚拟车场的标签的前向标签也为1
    number_of_lables=1
    for i in range(model.number_of_demands):
        n_1=node_id_list[i]
        j=i
        load=0
        distance={v_type:0 for v_type in model.vehicle_type_list}
        while True:
            n_2=node_id_list[j]
            load=load+model.demand_dict[n_2].demand
            stop = False
            for k,v_type in enumerate(model.vehicle_type_list):
                vehicle=model.vehicle_dict[v_type]
                if i == j:
                    distance[v_type]=model.distance_matrix[v_type,n_1]+model.distance_matrix[n_1,v_type]
                else:
                    n_3=node_id_list[j-1]
                    distance[v_type]=distance[v_type]-model.distance_matrix[n_3,v_type]+model.distance_matrix[n_3,n_2]\
                                     +model.distance_matrix[n_2,v_type]
                route=node_id_list[i:j+1]
                route.insert(0,v_type)
                route.append(v_type)
                if not checkTimeWindow(route,model,vehicle): # 检查时间窗,只有满足时间窗才有可能生成新的标签,否则跳过
                    continue
                for id,label in enumerate(V[i-1]):
                    if load<=vehicle.capacity and label[k+4]<vehicle.numbers:
                        stop=True
                        "计算路径成本,这里计算旅行时间成本时,只考虑节点间的旅行时间,暂不考虑等待时间成本"
                        if model.opt_type==0:
                            cost=vehicle.fixed_cost+distance[v_type]*vehicle.variable_cost
                        else:
                            cost=vehicle.fixed_cost+distance[v_type]/vehicle.free_speed*vehicle.variable_cost
                        "由于label是W的前向标签,因此可以在label的基础上生成W"
                        W=copy.deepcopy(label)
                        "将W的前向标签id设置为label的id"
                        W[1]=V[i-1][id][0]
                        "设置W使用的车辆类型"
                        W[2]=v_type
                        "在label的基础上更新W的cost"
                        W[3]=W[3]+cost
                        "在label的基础上更新使用的车辆数"
                        W[k+4]=W[k+4]+1
                        "检车剩余容量约束,判断是否有可能将W作为当前节点的新的标签"
                        if checkResidualCapacity(node_id_list[j+1:],W,model):
                            "根据帕累托将W插入到当前节点的标签列表中,同时删除被支配标签"
                            label_list,number_of_lables=updateNodeLabels(V[j],W,number_of_lables)
                            V[j]=label_list
            j+=1
            if j>=len(node_id_list) or stop==False:
                break
    if len(V[model.number_of_demands-1])>0:
        route_list=extractRoutes(V, node_id_list, model)
        return route_list
    else:
        print("Failed to split the node id list because of the insufficient capacity")
        return None

(5)适应度计算
对于解的评价可采用旅行时间成本或旅行距离成本,关于某一条车辆路径的成本计算如下:

  • 距离成本=车辆固定成本+旅行距离*变动成本
  • 时间成本=车辆固定成本+(旅行时间+等待时间)*变动成本

这里认为单位距离成本=单位旅行时间成本=单位等待时间成本

"计算解的成本,这里对于时间成本包含了节点间旅行时间以及节点处的等待时间"
def calTravelCost(route_list,model):
    timetable_list=[]
    distance_of_routes=0
    time_of_routes=0
    obj=0
    for route in route_list:
        timetable=[]
        vehicle=model.vehicle_dict[route[0]]
        travel_distance=0
        travel_time=0
        v_type = route[0]
        free_speed=vehicle.free_speed
        fixed_cost=vehicle.fixed_cost
        variable_cost=vehicle.variable_cost
        for i in range(len(route)):
            if i == 0:
                next_node_id=route[i+1]
                travel_time_between_nodes=model.distance_matrix[v_type,next_node_id]/free_speed
                departure=max(0,model.demand_dict[next_node_id].start_time-travel_time_between_nodes)
                timetable.append((int(departure),int(departure)))
            elif 1<= i <= len(route)-2:
                last_node_id=route[i-1]
                current_node_id=route[i]
                current_node = model.demand_dict[current_node_id]
                travel_time_between_nodes=model.distance_matrix[last_node_id,current_node_id]/free_speed
                arrival=max(timetable[-1][1]+travel_time_between_nodes,current_node.start_time)
                departure=arrival+current_node.service_time
                timetable.append((int(arrival),int(departure)))
                travel_distance += model.distance_matrix[last_node_id, current_node_id]
                travel_time += model.distance_matrix[last_node_id, current_node_id]/free_speed+\
                                + max(current_node.start_time - arrival, 0)
            else:
                last_node_id = route[i - 1]
                travel_time_between_nodes = model.distance_matrix[last_node_id,v_type]/free_speed
                departure = timetable[-1][1]+travel_time_between_nodes
                timetable.append((int(departure),int(departure)))
                travel_distance += model.distance_matrix[last_node_id,v_type]
                travel_time += model.distance_matrix[last_node_id,v_type]/free_speed
        distance_of_routes+=travel_distance
        time_of_routes+=travel_time
        if model.opt_type==0:
            obj+=fixed_cost+travel_distance*variable_cost
        else:
            obj += fixed_cost + travel_time *variable_cost
        timetable_list.append(timetable)
    return timetable_list,time_of_routes,distance_of_routes,obj

def calObj(sol,model):
    # calculate travel distance and travel time
    ret = splitRoutes(sol.node_id_list, model)
    if ret is not None:
        sol.route_list = ret
        sol.timetable_list, sol.time_of_routes, sol.distance_of_routes, sol.obj = calTravelCost(sol.route_list, model)
    else:
        sol.obj = 10**8

(6)生成初始粒子群

def generateInitialSol(model):
    demand_id_list=copy.deepcopy(model.demand_id_list)
    best_sol=Sol()
    best_sol.obj=float('inf')
    for i in range(model.popsize):
        seed = int(random.randint(0, 10))
        random.seed(seed)
        random.shuffle(demand_id_list)
        sol=Sol()
        sol.node_id_list= copy.deepcopy(demand_id_list)
        calObj(sol,model)
        model.sol_list.append(sol)
        model.v.append([model.Vmax]*len(model.demand_id_list))
        model.pl.append(sol.node_id_list)
        if sol.obj<best_sol.obj:
            best_sol=copy.deepcopy(sol)
    model.best_sol=best_sol
    model.pg=best_sol.node_id_list

(7)速度及位置更新

def updatePosition(model):
    w=model.w
    c1=model.c1
    c2=model.c2
    pg = model.pg
    for id,sol in enumerate(model.sol_list):
        x=sol.node_id_list
        v=model.v[id]
        pl=model.pl[id]
        r1=random.random()
        r2=random.random()
        new_v=[]
        for i in range(len(model.demand_id_list)):
            v_=w*v[i]+c1*r1*(pl[i]-x[i])+c2*r2*(pg[i]-x[i])
            if v_>0:
                new_v.append(min(v_,model.Vmax))
            else:
                new_v.append(max(v_,-model.Vmax))
        new_x=[min(int(x[i]+new_v[i]),len(model.demand_id_list)-1) for i in range(len(model.demand_id_list)) ]
        new_x=adjustRoutes(new_x,model)
        model.v[id]=new_v
        new_sol=Sol()
        new_sol.node_id_list=new_x
        calObj(new_sol,model)
        if new_sol.obj<sol.obj:
            model.pl[id]=copy.deepcopy(new_x)
        if new_sol.obj<model.best_sol.obj:
            model.best_sol=copy.deepcopy(new_sol)
            model.pg=copy.deepcopy(new_x)
        model.sol_list[id]= copy.deepcopy(new_sol)

def adjustRoutes(node_id_list,model):
    all_node_list=copy.deepcopy(model.demand_id_list)
    repeat_node=[]
    for id,node_id in enumerate(node_id_list):
        if node_id in all_node_list:
            all_node_list.remove(node_id)
        else:
            repeat_node.append(id)
    for i in range(len(repeat_node)):
        node_id_list[repeat_node[i]]=all_node_list[i]
    return node_id_list

(8)绘制收敛曲线

def plotObj(obj_list):
    plt.rcParams['font.sans-serif'] = ['SimHei'] #show chinese
    plt.rcParams['axes.unicode_minus'] = False  # Show minus sign
    plt.plot(np.arange(1,len(obj_list)+1),obj_list)
    plt.xlabel('Iterations')
    plt.ylabel('Obj Value')
    plt.grid()
    plt.xlim(1,len(obj_list)+1)
    plt.show()

(9)绘制车辆路线

def plotRoutes(model):
    for route in model.best_sol.route_list:
        x_coord=[model.vehicle_dict[route[0]].x_coord]
        y_coord=[model.vehicle_dict[route[0]].y_coord]
        for node_id in route[1:-1]:
            x_coord.append(model.demand_dict[node_id].x_coord)
            y_coord.append(model.demand_dict[node_id].y_coord)
        x_coord.append(model.vehicle_dict[route[-1]].x_coord)
        y_coord.append(model.vehicle_dict[route[-1]].y_coord)
        plt.grid()
        if route[0]=='v1':
            plt.plot(x_coord,y_coord,marker='o',color='black',linewidth=0.5,markersize=5)
        elif route[0]=='v2':
            plt.plot(x_coord,y_coord,marker='o',color='orange',linewidth=0.5,markersize=5)
        elif route[0]=='v3':
            plt.plot(x_coord,y_coord,marker='o',color='r',linewidth=0.5,markersize=5)
        else:
            plt.plot(x_coord, y_coord, marker='o', color='b', linewidth=0.5, markersize=5)
    plt.xlabel('x_coord')
    plt.ylabel('y_coord')
    plt.show()

(10)输出结果

def outPut(model):
    work=xlsxwriter.Workbook('result.xlsx')
    worksheet=work.add_worksheet()
    worksheet.write(0, 0, 'time_of_routes')
    worksheet.write(0, 1, 'distance_of_routes')
    worksheet.write(0, 2, 'opt_type')
    worksheet.write(0, 3, 'obj')
    worksheet.write(1,0,model.best_sol.time_of_routes)
    worksheet.write(1,1,model.best_sol.distance_of_routes)
    worksheet.write(1,2,model.opt_type)
    worksheet.write(1,3,model.best_sol.obj)
    worksheet.write(2, 0,'vehicleID')
    worksheet.write(2, 1,'depotID')
    worksheet.write(2, 2, 'vehicleType')
    worksheet.write(2, 3,'route')
    worksheet.write(2, 4,'timetable')
    for row,route in enumerate(model.best_sol.route_list):
        worksheet.write(row+3,0,str(row+1))
        depot_id=model.vehicle_dict[route[0]].depot_id
        worksheet.write(row+3,1,depot_id)
        worksheet.write(row+3,2,route[0])
        r=[str(i)for i in route]
        worksheet.write(row+3,3, '-'.join(r))
        r=[str(i)for i in model.best_sol.timetable_list[row]]
        worksheet.write(row+3,4, '-'.join(r))
    work.close()

(11)主函数

def run(demand_file,depot_file,epochs,popsize,Vmax,opt_type,w,c1,c2):
    """
    :param demand_file: demand file path
    :param depot_file: depot file path
    :param epochs: 迭代次数
    :param popsize: 邻域规模
    :param v_cap: 车辆容量
    :param Vmax :最大速度
    :param opt_type: 优化类型:0:最小化车辆数,1:最小化行驶距离
    :param w: 惯性权重
    :param c1:学习因子
    :param c2:学习因子
    :return:
    """
    model=Model()
    model.Vmax=Vmax
    model.opt_type=opt_type
    model.w=w
    model.c1=c1
    model.c2=c2
    model.popsize=popsize
    readCSVFile(demand_file,depot_file,model)
    calDistanceMatrix(model)
    history_best_obj=[]
    generateInitialSol(model)
    history_best_obj.append(model.best_sol.obj)
    start_time=time.time()
    for ep in range(epochs):
        updatePosition(model)
        history_best_obj.append(model.best_sol.obj)
        print("%s/%s, best obj: %s, runtime: %s" % (ep + 1, epochs, model.best_sol.obj, time.time() - start_time))
    plotObj(history_best_obj)
    plotRoutes(model)
    outPut(model)

6. 完整代码

如有错误,欢迎交流。
代码和数据文件可从github主页免费获取:

https://github.com/PariseC/Algorithms_for_solving_VRP

参考

  1. Order-first split-second methods for vehicle routing problems: A review
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