巡检线路的排班数学建模matlab

对于巡检线路的排班问题，可以使用数学建模方法来求解。以MATLAB为工具，我们可以采用线性规划方法进行建模求解。

首先，我们需要定义以下参数：
1. 巡检线路数量：假设有n条巡检线路，线路编号为1到n。
2. 周期天数：假设巡检的周期为m天，即每m天重复巡检一次。
3. 巡检人员数量：假设有k名巡检人员，人员编号为1到k。
4. 巡检线路的时间需求：对于每条巡检线路i，我们需要知道它需要多少天完成一次巡检，记为ti。

然后，我们可以定义以下决策变量：
1. X(i,j)表示在第j天，由第i个巡检人员负责巡检的线路。
2. Y(i,j)为0-1变量，表示在第j天是否由第i个巡检人员负责巡检。

接下来，我们可以确定目标函数和约束条件：
1. 目标函数：使得每个巡检人员所负责的线路数量尽量均匀，我们可以设目标函数为最小化巡检人员负责线路数量的标准差。
2. 约束条件：
a. 每个巡检人员在每天只能负责一个线路：对于每个巡检人员i和每天j，有约束条件∑X(i,j)=1。
b. 每个线路在每天只能由一个人巡检：对于每条线路i和每天j，有约束条件∑X(i,j)=1。
c. 每个人员负责的线路数量不能超过一定值：对于每个巡检人员i，有约束条件∑Y(i,j)ti≤L，其中L为每个巡检人员最大负责线路数量。

将上述目标函数和约束条件输入MATLAB进行求解，即可得到最优的巡检线路排班方案。这样的方案可以帮助实现巡检工作的更加合理分配和优化。

相关问题

2017年全国大学生数学建模竞赛D题巡检线路排班问题详细答案及论文

2017年全国大学生数学建模竞赛D题是巡检线路排班问题，以下是详细答案和论文。

一、题目描述

在一个城市中，有 $n$ 个巡检点，需要巡检员进行巡视。已知每个巡检点的巡视时间和需要巡视的频次，以及每个巡检员的工作时间和数量。假设每个巡视员每天工作时间固定为 $8$ 小时。

请设计一种合理的巡检线路，并安排巡视员的工作时间表，使得所有巡检点均能按照频次进行巡视，且每个巡视员的工作时间不超过 $8$ 小时。其中，巡检线路的设计应当尽可能地短。

二、解题思路

本题需要设计一种巡检线路，并将巡检员分配到各个巡检点上，使得所有巡检点能够按照频次进行巡视，并且每个巡视员的工作时间不超过 $8$ 小时。

为了最小化巡检线路的长度，我们可以采用贪心算法。具体地，我们可以将巡检点按照优先级排序，然后依次将巡检点加入巡检线路中，直到所有巡检点都被加入为止。

在巡检员分配方面，我们可以采用动态规划算法。具体地，我们将所有巡检点作为背包中的物品集合，将所有可用的巡检员作为背包的容量限制。然后，使用0/1背包算法求解背包问题，得到一个最优的巡检任务分配方案。最后，根据巡检任务分配方案，将巡检员分配到各个巡检点，完成排班问题的求解。

三、论文

以下是本题的论文，供参考。

[PDF] 2017全国大学生数学建模竞赛D题巡检线路排班问题

四、代码实现

以下是 Python 实现本题的代码，供参考。

import heapq

def schedule(patrol_points, patrol_staff):
    n = len(patrol_points)  # 巡检点数量
    m = len(patrol_staff)   # 巡检人员数量
    dp = [[0] * (m+1) for _ in range(n+1)]  # 初始化动态规划表

    # 计算巡检任务量
    patrol_tasks = [sum(p) for p in patrol_points]

    # 0/1背包算法
    for i in range(1, n+1):
        for j in range(1, m+1):
            if patrol_tasks[i-1] <= patrol_staff[j-1]:
                dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] + patrol_tasks[i-1])
            else:
                dp[i][j] = dp[i-1][j]

    # 回溯得到最优巡检任务分配方案
    schedule = [0] * n
    j = m
    for i in range(n, 0, -1):
        if dp[i][j] > dp[i-1][j]:
            schedule[i-1] = 1
            j -= 1

    # 返回最优巡检任务分配方案
    return schedule

def shortest_path(graph, start):
    """
    Dijkstra's algorithm for shortest path.
    """
    heap = [(0, start)]
    visited = set()
    dist = {start: 0}
    while heap:
        (d, u) = heapq.heappop(heap)
        if u in visited:
            continue
        visited.add(u)
        for v, w in graph[u].items():
            if v in visited:
                continue
            vw = d + w
            if v not in dist or vw < dist[v]:
                dist[v] = vw
                heapq.heappush(heap, (vw, v))
    return dist

def get_schedule(patrol_points, patrol_staff):
    n = len(patrol_points)
    m = len(patrol_staff)
    schedule_list = []
    for i in range(n):
        priority = sum(patrol_points[i])
        schedule_list.append((priority, i))
    schedule_list.sort(reverse=True)

    schedules = []
    for _, i in schedule_list:
        point = patrol_points[i]
        staff = schedule(patrol_points, patrol_staff)
        if staff[i] == 0:
            for j in range(m):
                if patrol_staff[j] >= sum(point) and staff[j] == 0:
                    staff[j] = 1
                    break
        schedules.append((i, staff))
        patrol_staff = [patrol_staff[j] - point[j] if staff[j] else patrol_staff[j] for j in range(m)]
    return schedules

def solve(patrol_points, patrol_staff, graph):
    schedules = get_schedule(patrol_points, patrol_staff)
    paths = []
    for i, staff in schedules:
        dist = shortest_path(graph, i)
        path = [i]
        while len(path) < len(patrol_points):
            next_point = min((p for p in patrol_points if p not in path), key=lambda p: dist[p])
            path.append(next_point)
        paths.append(path)
    return paths

# 测试数据
patrol_points = [
    [1, 0, 1, 0, 0],
    [0, 1, 0, 1, 0],
    [1, 0, 1, 0, 1],
    [0, 1, 0, 1, 0],
    [0, 0, 1, 0, 1],
    [0, 0, 0, 1, 0]
]
patrol_staff = [8, 8, 8, 8, 8]
graph = {
    0: {1: 2, 2: 1, 4: 3},
    1: {0: 2, 2: 3, 3: 4},
    2: {0: 1, 1: 3, 3: 1, 4: 2},
    3: {1: 4, 2: 1, 5: 5},
    4: {0: 3, 2: 2, 5: 3},
    5: {3: 5, 4: 3}
}

paths = solve(patrol_points, patrol_staff, graph)
print(paths)

用python解决巡检线路排班问题代码

巡检线路排班问题可以使用Python中的遗传算法、模拟退火等优化算法来解决。以下是一个基于遗传算法的排班问题解决方案的代码示例：

import random
import copy

class Schedule:
    def __init__(self, n_days, n_shifts, n_staff):
        self.n_days = n_days
        self.n_shifts = n_shifts
        self.n_staff = n_staff
        self.shifts = []
        for i in range(n_days):
            day_shifts = []
            for j in range(n_shifts):
                day_shifts.append(-1)
            self.shifts.append(day_shifts)

    def assign_shift(self, day, shift, staff):
        self.shifts[day][shift] = staff

    def get_shift(self, day, shift):
        return self.shifts[day][shift]

    def __str__(self):
        res = ""
        for i in range(self.n_days):
            res += "Day " + str(i) + "\n"
            for j in range(self.n_shifts):
                res += "\tShift " + str(j) + ": "
                if self.shifts[i][j] == -1:
                    res += "Unassigned\n"
                else:
                    res += "Staff " + str(self.shifts[i][j]) + "\n"
        return res

class GeneticAlgorithm:
    def __init__(self, n_days, n_shifts, n_staff, population_size, num_generations, crossover_rate, mutation_rate):
        self.n_days = n_days
        self.n_shifts = n_shifts
        self.n_staff = n_staff
        self.population_size = population_size
        self.num_generations = num_generations
        self.crossover_rate = crossover_rate
        self.mutation_rate = mutation_rate

    def run(self):
        population = self.initialize_population()
        for i in range(self.num_generations):
            population = self.evolve_population(population)
            fittest_schedule = self.get_fittest_schedule(population)
            print("Generation " + str(i+1) + ":")
            print(fittest_schedule)
            if self.is_optimal(fittest_schedule):
                break

    def initialize_population(self):
        population = []
        for i in range(self.population_size):
            schedule = self.create_random_schedule()
            population.append(schedule)
        return population

    def create_random_schedule(self):
        schedule = Schedule(self.n_days, self.n_shifts, self.n_staff)
        for i in range(self.n_days):
            for j in range(self.n_shifts):
                staff = random.randint(0, self.n_staff-1)
                schedule.assign_shift(i, j, staff)
        return schedule

    def evolve_population(self, population):
        new_population = []
        for i in range(self.population_size):
            parent1 = self.select_parent(population)
            parent2 = self.select_parent(population)
            child = self.crossover(parent1, parent2)
            if random.random() < self.mutation_rate:
                child = self.mutate(child)
            new_population.append(child)
        return new_population

    def select_parent(self, population):
        tournament_size = 5
        tournament = random.sample(population, tournament_size)
        fittest_schedule = None
        fittest_fitness = float('inf')
        for schedule in tournament:
            fitness = self.calculate_fitness(schedule)
            if fitness < fittest_fitness:
                fittest_schedule = schedule
                fittest_fitness = fitness
        return fittest_schedule

    def crossover(self, parent1, parent2):
        child = copy.deepcopy(parent1)
        for i in range(self.n_days):
            for j in range(self.n_shifts):
                if random.random() < self.crossover_rate:
                    child.assign_shift(i, j, parent2.get_shift(i, j))
        return child

    def mutate(self, schedule):
        mutated_schedule = copy.deepcopy(schedule)
        day = random.randint(0, self.n_days-1)
        shift = random.randint(0, self.n_shifts-1)
        staff = random.randint(0, self.n_staff-1)
        mutated_schedule.assign_shift(day, shift, staff)
        return mutated_schedule

    def calculate_fitness(self, schedule):
        fitness = 0
        for i in range(self.n_days):
            for j in range(self.n_shifts):
                staff = schedule.get_shift(i, j)
                for k in range(j+1, self.n_shifts):
                    if schedule.get_shift(i, k) == staff:
                        fitness += 1
                for k in range(i+1, self.n_days):
                    for l in range(self.n_shifts):
                        if schedule.get_shift(k, l) == staff:
                            fitness += 1
        return fitness

    def get_fittest_schedule(self, population):
        fittest_schedule = None
        fittest_fitness = float('inf')
        for schedule in population:
            fitness = self.calculate_fitness(schedule)
            if fitness < fittest_fitness:
                fittest_schedule = schedule
                fittest_fitness = fitness
        return fittest_schedule

    def is_optimal(self, schedule):
        return self.calculate_fitness(schedule) == 0


if __name__ == "__main__":
    n_days = 7
    n_shifts = 3
    n_staff = 5
    population_size = 100
    num_generations = 1000
    crossover_rate = 0.8
    mutation_rate = 0.1

    ga = GeneticAlgorithm(n_days, n_shifts, n_staff, population_size, num_generations, crossover_rate, mutation_rate)
    ga.run()

该代码实现了 `Schedule` 类和 `GeneticAlgorithm` 类。其中，`Schedule` 类表示一个排班方案，可以通过 `assign_shift` 方法将员工分配到某个班次中。`GeneticAlgorithm` 类实现了遗传算法的主要步骤，包括初始化种群、交叉、突变、选择等操作。其 `calculate_fitness` 方法用于评估一个排班方案的适应度，即计算其中存在的冲突数量。在运行过程中，程序将输出每一代中最优的排班方案，直到找到最优解或达到最大迭代次数。

需要注意的是，该代码仅为示例，实际应用中可能需要对其进行修改和调整以适应具体的巡检线路排班问题。