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【运筹优化】带时间窗约束的车辆路径规划问题（VRPTW）详解 + Python 调用 Gurobi 建模求解

admin 阅读： 2024-03-26

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文章目录

一、概述
- 1.1 VRP 问题
- 1.2 CVRP 问题
- 1.3 VRPTW 问题
二、VRPTW 的一般模型
三、Python 调用 Gurobi 建模求解
- 3.1 Solomn 数据集
- 3.2 完整代码
- 3.3 运行结果展示
- - 3.3.1 测试案例：c101.txt
  - 3.3.2 测试案例：r101.txt

一、概述

1.1 VRP 问题

车辆路径规划问题(Vehicle Routing Problem，VRP)一般指的是：对一系列发货点和收货点，组织调用一定的车辆，安排适当的行车路线，使车辆有序地通过它们，在满足指定的约束条件下（例如：货物的需求量与发货量，交发货时间，车辆容量限制，行驶里程限制，行驶时间限制等），力争实现一定的目标（如车辆空驶总里程最短，运输总费用最低，车辆按一定时间到达，使用的车辆数最小等）。

下图给出了一个简单的VRP的例子

在这里插入图片描述

1.2 CVRP 问题

最基本的VRP问题叫做带容量约束的车辆路径规划问题（Capacitated Vehicle Routing Problem，CVRP）。在CVRP中，需要考虑每辆车的容量约束、车辆的路径约束和装载量约束

1.3 VRPTW 问题

为了考虑配送时间要求，带时间窗的车辆路径规划问题（Vehicle Routing Problem with Time Window，VRPTW）应运而生。

VRPTW 不仅考虑CVRP的所有约束，还需要考虑时间窗约束，也就是每个顾客对应一个时间窗 [ e i , l i ] [e_i,l_i] [ei,li]，其中 e i e_i ei 和 l i l_i li 分别代表该点的最早到达时间和最晚到达时间。顾客点 i ∈ V i \in V i∈V 的需求必须要在其时间窗内被送达

VRPTW 已经被证明是 NP-hard 问题，其求解复杂度随着问题规模的增加而急剧增加，求解较为困难。到目前为止，求解 VRPTW 的比较高效的精确算法是分支定价算法和分支定价切割算法。

二、VRPTW 的一般模型

VRPTW 可以建模为一个混合整数规划问题，在给出完整数学模型之前，先引入下面的决策变量：

x i j k = { 1 ，在最优解中，弧 ( i , j ) 被车辆 k 选中 0 ，其他 s i k = 车辆 k 到达 i 的时间模型中涉及的其他参数为 : t i j 表示车辆在弧 ( i , j ) 上的行驶时间 M 为一个足够大的正数 {x_{ijk}}=\begin{cases} 1\text{，在最优解中，弧}\left( i,j \right) \text{被车辆}k\text{选中}\\ 0\text{，其他}\\ \end{cases} \\ {s_{ik}}=\text{车辆}k\text{到达}i\text{的时间} \\ \text{模型中涉及的其他参数为}: \\ {t_{ij}}\text{表示车辆在弧}\left( i,j \right) \text{上的行驶时间} \\ M\text{为一个足够大的正数} xijk={1，在最优解中，弧(i,j)被车辆k选中0，其他sik=车辆k到达i的时间模型中涉及的其他参数为:tij表示车辆在弧(i,j)上的行驶时间M为一个足够大的正数

关于 M M M 的取值，实际上可以直接取非常大的正数，但是为了提高求解效率，拉紧约束。我们可以采用下面的取值方法：

M = m a x { b i + t i j − a j } , ∀ ( i , j ) ∈ A M=max\{b_i+t_{ij}-a_j\} , \forall (i,j)\in A M=max{bi+tij−aj},∀(i,j)∈A

综合上面引出的决策变量，并参考文献（Desaulniers et al.，2006），给出的 VRPTW 的标准模型如下：

min ⁡ ∑ k ∈ K ∑ i ∈ V ∑ j ∈ V c i j x i j k s . t . ∑ k ∈ K ∑ j ∈ V x i j k = 1 , ∀ i ∈ C ∑ j ∈ V x 0 j k = 1 , ∀ k ∈ K ∑ i ∈ V x i h k − ∑ j ∈ V x h j k = 0 , ∀ h ∈ C , ∀ k ∈ K ∑ i ∈ V x i , n + 1 , k = 1 , ∀ k ∈ K ∑ i ∈ C q i ∑ j ∈ V x i j k ⩽ Q , ∀ k ∈ K s i k + t i j − M ( 1 − x i j k ) ⩽ s j k , ∀ ( i , j ) ∈ A , ∀ k ∈ K e i ⩽ s i k ⩽ l i , ∀ i ∈ V , ∀ k ∈ K x i j k ∈ { 0 , 1 } , ∀ ( i , j ) ∈ A , ∀ k ∈ K \min \sum_{k\in K}{\sum_{i\in V}{\sum_{j\in V}{{c_{ij}}{x_{ijk}}}}} \\ s.t. \sum_{k\in K}{\sum_{j\in V}{{x_{ijk}}=1 , \forall i\in C}} \\ \,\, \sum_{j\in V}{{x_{0jk}}=1 , \forall k\in K} \\ \,\, \sum_{i\in V}{{x_{ihk}}-\sum_{j\in V}{{x_{hjk}}=0 , \forall h\in C,\forall k\in K}} \\ \,\, \sum_{i\in V}{x_{i,n+1,k}=1 , \forall k\in K} \\ \,\, \sum_{i\in C}{q_i\sum_{j\in V}{{x_{ijk}} \leqslant Q , \forall k\in K}} \\ \,\, {s_{ik}}+{t_{ij}}-M\left( 1-{x_{ijk}} \right) \leqslant {s_{jk}}\,\,, \forall \left( i,j \right) \in A,\forall k\in K \\ \,\, e_i\leqslant {s_{ik}}\leqslant l_i\,\,, \forall i\in V,\forall k\in K \\ \,\, {x_{ijk}}\in \left\{ 0,1 \right\} \,\,, \forall \left( i,j \right) \in A,\forall k\in K mink∈K∑i∈V∑j∈V∑cijxijks.t.k∈K∑j∈V∑xijk=1,∀i∈Cj∈V∑x0jk=1,∀k∈Ki∈V∑xihk−j∈V∑xhjk=0,∀h∈C,∀k∈Ki∈V∑xi,n+1,k=1,∀k∈Ki∈C∑qij∈V∑xijk⩽Q,∀k∈Ksik+tij−M(1−xijk)⩽sjk,∀(i,j)∈A,∀k∈Kei⩽sik⩽li,∀i∈V,∀k∈Kxijk∈{0,1},∀(i,j)∈A,∀k∈K

其中， Q Q Q 为车容量， q i q_i qi 为第 i i i 个顾客的需求：

目标函数是为了最小化所有车辆的总行驶成本（距离）
约束1~4保证了每辆车必须从仓库出发，经过一个点就离开那个点，最终返回仓库
约束5为车辆的容量约束
约束6~7是时间窗约束，保证了车辆到达每个顾客点的时间均在时间窗内，点n+1是点o的一个备份，是为了方便实现。

三、Python 调用 Gurobi 建模求解

3.1 Solomn 数据集

Solomn 数据集下载地址

3.2 完整代码

注意，在下面代码中，将弧 i i i 到弧 j j j 所需的时间 t i j t_{ij} tij 和成本 c i j c_{ij} cij 都当作了弧 i i i 到弧 j j j 所需的距离来看待

# -*- coding: utf-8 -*-# # Author: WSKH # Blog: wskh0929.blog.csdn.net # Time: 2023/2/8 11:14 # Description: Python 调用 Gurobi 建模求解 VRPTW 问题 import time import matplotlib.pyplot as plt import numpy as np from gurobipy import * class Data: customerNum = 0 nodeNum = 0 vehicleNum = 0 capacity = 0 corX = [] corY = [] demand = [] serviceTime = [] readyTime = [] dueTime = [] distanceMatrix = [[]] def readData(path, customerNum): data = Data() data.customerNum = customerNum if customerNum is not None: data.nodeNum = customerNum + 2 with open(path, 'r') as f: lines = f.readlines() count = 0 for line in lines: count += 1 if count == 5: line = line[:-1] s = re.split(r" +", line) data.vehicleNum = int(s[1]) data.capacity = float(s[2]) elif count >= 10 and (customerNum is None or count <= 10 + customerNum): line = line[:-1] s = re.split(r" +", line) data.corX.append(float(s[2])) data.corY.append(float(s[3])) data.demand.append(float(s[4])) data.readyTime.append(float(s[5])) data.dueTime.append(float(s[6])) data.serviceTime.append(float(s[7])) data.nodeNum = len(data.corX) + 1 data.customerNum = data.nodeNum - 2 # 回路 data.corX.append(data.corX[0]) data.corY.append(data.corY[0]) data.demand.append(data.demand[0]) data.readyTime.append(data.readyTime[0]) data.dueTime.append(data.dueTime[0]) data.serviceTime.append(data.serviceTime[0]) # 计算距离矩阵 data.distanceMatrix = np.zeros((data.nodeNum, data.nodeNum)) for i in range(data.nodeNum): for j in range(i + 1, data.nodeNum): distance = math.sqrt((data.corX[i] - data.corX[j]) ** 2 + (data.corY[i] - data.corY[j]) ** 2) data.distanceMatrix[i][j] = data.distanceMatrix[j][i] = distance return data class Solution: ObjVal = 0 X = [[]] S = [[]] routes = [[]] routeNum = 0 def __init__(self, data, model): self.ObjVal = model.ObjVal # X_ijk self.X = [[([0] * data.vehicleNum) for _ in range(data.nodeNum)] for _ in range(data.nodeNum)] # S_ik self.S = [([0] * data.vehicleNum) for _ in range(data.nodeNum)] # routes self.routes = [] def getSolution(data, model): solution = Solution(data, model) for m in model.getVars(): split_arr = re.split(r"_", m.VarName) if split_arr[0] == 'X' and m.x > 0.5: solution.X[int(split_arr[1])][int(split_arr[2])][int(split_arr[3])] = m.x elif split_arr[0] == 'S' and m.x > 0.5: solution.S[int(split_arr[1])][int(split_arr[2])] = m.x for k in range(data.vehicleNum): i = 0 subRoute = [] subRoute.append(i) finish = False while not finish: for j in range(data.nodeNum): if solution.X[i][j][k] > 0.5: subRoute.append(j) i = j if j == data.nodeNum - 1: finish = True if len(subRoute) >= 3: subRoute[-1] = 0 solution.routes.append(subRoute) solution.routeNum += 1 return solution def plot_solution(solution, customer_num): plt.xlabel("x") plt.ylabel("y") plt.title(f"{data_type} : {customer_num} Customers") plt.scatter(data.corX[0], data.corY[0], c='blue', alpha=1, marker=',', linewidths=3, label='depot') # 起点 plt.scatter(data.corX[1:-1], data.corY[1:-1], c='black', alpha=1, marker='o', linewidths=3, label='customer') # 普通站点 for k in range(solution.routeNum): for i in range(len(solution.routes[k]) - 1): a = solution.routes[k][i] b = solution.routes[k][i + 1] x = [data.corX[a], data.corX[b]] y = [data.corY[a], data.corY[b]] plt.plot(x, y, 'k', linewidth=1) plt.grid(False) plt.legend(loc='best') plt.show() def print_solution(solution, data): for index, subRoute in enumerate(solution.routes): distance = 0 load = 0 for i in range(len(subRoute) - 1): distance += data.distanceMatrix[subRoute[i]][subRoute[i + 1]] load += data.demand[subRoute[i]] print(f"Route-{index + 1} : {subRoute} , distance: {distance} , load: {load}") def solve(data): # 声明模型 model = Model("VRPTW") # 模型设置 # 关闭输出 model.setParam('OutputFlag', 0) # 定义变量 X = [[[[] for _ in range(data.vehicleNum)] for _ in range(data.nodeNum)] for _ in range(data.nodeNum)] S = [[[] for _ in range(data.vehicleNum)] for _ in range(data.nodeNum)] for i in range(data.nodeNum): for k in range(data.vehicleNum): S[i][k] = model.addVar(data.readyTime[i], data.dueTime[i], vtype=GRB.CONTINUOUS, name=f'S_{i}_{k}') for j in range(data.nodeNum): X[i][j][k] = model.addVar(vtype=GRB.BINARY, name=f"X_{i}_{j}_{k}") # 目标函数 obj = LinExpr(0) for i in range(data.nodeNum): for j in range(data.nodeNum): if i != j: for k in range(data.vehicleNum): obj.addTerms(data.distanceMatrix[i][j], X[i][j][k]) model.setObjective(obj, GRB.MINIMIZE) # 约束1：车辆只能从一个点到另一个点 for i in range(1, data.nodeNum - 1): expr = LinExpr(0) for j in range(data.nodeNum): if i != j: for k in range(data.vehicleNum): if i != 0 and i != data.nodeNum - 1: expr.addTerms(1, X[i][j][k]) model.addConstr(expr == 1) # 约束2：车辆必须从仓库出发 for k in range(data.vehicleNum): expr = LinExpr(0) for j in range(1, data.nodeNum): expr.addTerms(1, X[0][j][k]) model.addConstr(expr == 1) # 约束3：车辆经过一个点就必须离开一个点 for k in range(data.vehicleNum): for h in range(1, data.nodeNum - 1): expr1 = LinExpr(0) expr2 = LinExpr(0) for i in range(data.nodeNum): if h != i: expr1.addTerms(1, X[i][h][k]) for j in range(data.nodeNum): if h != j: expr2.addTerms(1, X[h][j][k]) model.addConstr(expr1 == expr2) # 约束4：车辆最终返回仓库 for k in range(data.vehicleNum): expr = LinExpr(0) for i in range(data.nodeNum - 1): expr.addTerms(1, X[i][data.nodeNum - 1][k]) model.addConstr(expr == 1) # 约束5：车辆容量约束 for k in range(data.vehicleNum): expr = LinExpr(0) for i in range(1, data.nodeNum - 1): for j in range(data.nodeNum): if i != 0 and i != data.nodeNum - 1 and i != j: expr.addTerms(data.demand[i], X[i][j][k]) model.addConstr(expr <= data.capacity) # 约束6：时间窗约束 for k in range(data.vehicleNum): for i in range(data.nodeNum): for j in range(data.nodeNum): if i != j: model.addConstr(S[i][k] + data.distanceMatrix[i][j] - S[j][k] <= M - M * X[i][j][k]) # 记录求解开始时间 start_time = time.time() # 求解 model.optimize() if model.status == GRB.OPTIMAL: print("-" * 20, "Solved Successfully", '-' * 20) # 输出求解总用时 print(f"Solve Time: {time.time() - start_time} s") print(f"Total Travel Distance: {model.ObjVal}") solution = getSolution(data, model) plot_solution(solution, data.customerNum) print_solution(solution, data) else: print("此题无解") if __name__ == '__main__': # 哪个数据集 data_type = "c101" # 数据集路径 data_path = f'../../data/solomn_data/{data_type}.txt' # 顾客个数设置（从上往下读取完 customerNum 个顾客为止，例如c101文件中有100个顾客点， # 但是跑100个顾客点太耗时了，设置这个数是为了只选取一部分顾客点进行计算，用来快速测试算法） # 如果想用完整的顾客点进行计算，设置为None即可 customerNum = 50 # 一个很大的正数 M = 10000000 # 读取数据 data = readData(data_path, customerNum) # 输出相关数据 print("-" * 20, "Problem Information", '-' * 20) print(f'Data Type: {data_type}') print(f'Node Num: {data.nodeNum}') print(f'Customer Num: {data.customerNum}') print(f'Vehicle Num: {data.vehicleNum}') print(f'Vehicle Capacity: {data.capacity}') # 建模求解 solve(data)

3.3 运行结果展示

3.3.1 测试案例：c101.txt

设置 customerNum = 20

-------------------- Problem Information -------------------- Data Type: c101 Node Num: 22 Customer Num: 20 Vehicle Num: 25 Vehicle Capacity: 200.0 -------------------- Solved Successfully -------------------- Solve Time: 0.2966279983520508 s Total Travel Distance: 160.81590595966603 Route-1 : [0, 20, 13, 17, 18, 19, 15, 16, 14, 12, 0] , distance: 101.32767502613292 , load: 200.0 Route-2 : [0, 5, 3, 7, 8, 10, 11, 9, 6, 4, 2, 1, 0] , distance: 59.48823093353308 , load: 160.0

在这里插入图片描述

设置 customerNum = 50

Data Type: c101 Node Num: 52 Customer Num: 50 Vehicle Num: 25 Vehicle Capacity: 200.0 -------------------- Solved Successfully -------------------- Solve Time: 4.383494138717651 s Total Travel Distance: 363.2468004115909 Route-1 : [0, 5, 3, 7, 8, 10, 11, 9, 6, 4, 2, 1, 0] , distance: 59.48823093353308 , load: 160.0 Route-2 : [0, 32, 33, 31, 35, 37, 38, 39, 36, 34, 0] , distance: 97.2271627850669 , load: 200.0 Route-3 : [0, 43, 42, 41, 40, 44, 46, 45, 48, 50, 49, 47, 0] , distance: 59.843107259523165 , load: 140.0 Route-4 : [0, 20, 24, 25, 27, 29, 30, 28, 26, 23, 22, 21, 0] , distance: 50.80359030264955 , load: 170.0 Route-5 : [0, 13, 17, 18, 19, 15, 16, 14, 12, 0] , distance: 95.88470913081827 , load: 190.0

在这里插入图片描述

设置 customerNum = None

-------------------- Problem Information -------------------- Data Type: c101 Node Num: 102 Customer Num: 100 Vehicle Num: 25 Vehicle Capacity: 200.0 -------------------- Solved Successfully -------------------- Solve Time: 272.5895857810974 s Total Travel Distance: 828.9368669428341 Route-1 : [0, 20, 24, 25, 27, 29, 30, 28, 26, 23, 22, 21, 0] , distance: 50.80359030264955 , load: 170.0 Route-2 : [0, 57, 55, 54, 53, 56, 58, 60, 59, 0] , distance: 101.88256760196126 , load: 200.0 Route-3 : [0, 5, 3, 7, 8, 10, 11, 9, 6, 4, 2, 1, 75, 0] , distance: 59.618077542105574 , load: 180.0 Route-4 : [0, 98, 96, 95, 94, 92, 93, 97, 100, 99, 0] , distance: 95.94313062205805 , load: 190.0 Route-5 : [0, 81, 78, 76, 71, 70, 73, 77, 79, 80, 0] , distance: 127.29748041459519 , load: 150.0 Route-6 : [0, 32, 33, 31, 35, 37, 38, 39, 36, 34, 0] , distance: 97.2271627850669 , load: 200.0 Route-7 : [0, 43, 42, 41, 40, 44, 46, 45, 48, 51, 50, 52, 49, 47, 0] , distance: 64.80747449698114 , load: 160.0 Route-8 : [0, 90, 87, 86, 83, 82, 84, 85, 88, 89, 91, 0] , distance: 76.06956532288787 , load: 170.0 Route-9 : [0, 13, 17, 18, 19, 15, 16, 14, 12, 0] , distance: 95.88470913081827 , load: 190.0 Route-10 : [0, 67, 65, 63, 62, 74, 72, 61, 64, 68, 66, 69, 0] , distance: 59.403108723710105 , load: 200.0

在这里插入图片描述

3.3.2 测试案例：r101.txt

设置 customerNum = 20

-------------------- Problem Information -------------------- Data Type: r101 Node Num: 22 Customer Num: 20 Vehicle Num: 25 Vehicle Capacity: 200.0 -------------------- Solved Successfully -------------------- Solve Time: 0.9535932540893555 s Total Travel Distance: 463.69270291007086 Route-1 : [0, 9, 20, 1, 0] , distance: 74.91992978886165 , load: 35.0 Route-2 : [0, 12, 3, 4, 0] , distance: 76.18033988749895 , load: 51.0 Route-3 : [0, 2, 15, 13, 0] , distance: 62.180339887498945 , load: 38.0 Route-4 : [0, 5, 18, 8, 17, 0] , distance: 86.57837545317302 , load: 49.0 Route-5 : [0, 14, 16, 6, 0] , distance: 72.40405733948208 , load: 42.0 Route-6 : [0, 11, 19, 7, 10, 0] , distance: 91.42966055355615 , load: 50.0

在这里插入图片描述

设置 customerNum = 50

-------------------- Problem Information -------------------- Data Type: r101 Node Num: 52 Customer Num: 50 Vehicle Num: 25 Vehicle Capacity: 200.0 -------------------- Solved Successfully -------------------- Solve Time: 4.6791017055511475 s Total Travel Distance: 946.6603871872358 Route-1 : [0, 21, 40, 26, 0] , distance: 43.35023188854984 , load: 37.0 Route-2 : [0, 33, 29, 9, 34, 24, 25, 0] , distance: 139.4708769010923 , load: 59.0 Route-3 : [0, 39, 23, 41, 22, 4, 0] , distance: 99.11062351878482 , load: 102.0 Route-4 : [0, 28, 12, 3, 50, 0] , distance: 51.94121366484106 , load: 61.0 Route-5 : [0, 36, 47, 11, 19, 49, 10, 32, 1, 0] , distance: 154.4302586824376 , load: 140.0 Route-6 : [0, 42, 14, 44, 16, 38, 37, 17, 0] , distance: 131.9204195702968 , load: 88.0 Route-7 : [0, 2, 15, 43, 13, 0] , distance: 72.54724253800985 , load: 45.0 Route-8 : [0, 45, 8, 46, 48, 0] , distance: 84.49944230335126 , load: 62.0 Route-9 : [0, 5, 7, 18, 6, 0] , distance: 73.5917360311745 , load: 46.0 Route-10 : [0, 27, 31, 30, 20, 35, 0] , distance: 95.79834208869767 , load: 81.0

在这里插入图片描述

设置 customerNum = 70

-------------------- Problem Information -------------------- Data Type: r101 Node Num: 72 Customer Num: 70 Vehicle Num: 25 Vehicle Capacity: 200.0 -------------------- Solved Successfully -------------------- Solve Time: 189.01783299446106 s Total Travel Distance: 1182.9787814963945 Route-1 : [0, 63, 62, 11, 64, 49, 48, 0] , distance: 125.38755919928242 , load: 116.0 Route-2 : [0, 65, 66, 20, 32, 70, 0] , distance: 117.49399251197822 , load: 82.0 Route-3 : [0, 28, 12, 26, 0] , distance: 33.795507476994075 , load: 52.0 Route-4 : [0, 33, 29, 3, 50, 68, 0] , distance: 90.77710269056311 , load: 82.0 Route-5 : [0, 2, 15, 41, 22, 56, 4, 0] , distance: 88.90058825018636 , load: 63.0 Route-6 : [0, 27, 69, 31, 30, 51, 9, 34, 35, 1, 0] , distance: 111.48892006549234 , load: 128.0 Route-7 : [0, 45, 8, 46, 17, 60, 0] , distance: 93.91701945260407 , load: 31.0 Route-8 : [0, 59, 42, 14, 44, 38, 57, 43, 58, 0] , distance: 131.96251141349887 , load: 119.0 Route-9 : [0, 39, 23, 67, 55, 54, 24, 25, 0] , distance: 140.03829072128988 , load: 114.0 Route-10 : [0, 52, 18, 6, 0] , distance: 41.290161379846566 , load: 24.0 Route-11 : [0, 36, 47, 19, 7, 10, 0] , distance: 107.49141646738926 , load: 70.0 Route-12 : [0, 21, 40, 53, 0] , distance: 36.27916407668437 , load: 34.0 Route-13 : [0, 5, 61, 16, 37, 13, 0] , distance: 64.15654779058515 , load: 89.0

在这里插入图片描述

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