Posted on Categories:Operations Research, 数学代写, 运筹学

# 数学代写|运筹学代写Operations Research代考|STAT721 The Chinese Postman Problem

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|运筹学代写Operations Research代考|The Chinese Postman Problem

Consider, again, a connected, undirected graph. Suppose that no Euler circuit exists. Then a natural question is how to construct the smallest cycle that passes through every edge of the graph at least once. More generally, suppose that positive costs of $c(a)$ are made every time edge $a$ is traversed. The question is then how to construct, at the lowest possible cost, a circuit such that every edge of the graph is passed through at least once. This problem is known as the Chinese postman problem, named after the Chinese scholar M.K. Kwan. The Chinese postman problem has many real-world applications:

1. Trash pickup. Suppose that a garbage truck has to pick up the trash in a specific district. The edges of the graph represent the streets, and the nodes correspond to street intersections. The cost assigned to an edge could be the street length.
2. Mail or newspaper delivery. This is also typically an example of a problem where a route is sought such that every street is visited at least once and the total distance traveled is as small as possible.

If the graph contains an Euler circuit, then the Euler circuit provides an optimal solution to the Chinese postman problem. Suppose that the graph does not contain an Euler circuit because at least one of the vertices has odd degree. To prepare for how we then proceed, we define
$X^{-}=$set of vertices in the graph with odd degree.

## 数学代写|运筹学代写Operations Research代考|The Monty Hall Problem

Rarely has a probability problem aroused emotions as much as the Monty Hall problem. This problem, named after the quizmaster of a then-famous TV show, became known worldwide when the American columnist Marilyn vos Savant raised the problem in Parade magazine in 1990. The finalist of a television quiz has to choose one of three doors. Behind one door is an expensive car, and behind the other two are goats. The finalist randomly chooses one of the doors. As promised beforehand, the quizmaster then opens one of the two remaining doors hiding a goat. He then asks the candidate whether she wants to switch doors. The candidate faces a dilemma. What should she do? Vos Savant’s advice was to switch and take the only remaining door. In this way, the candidate would increase the probability of winning the car to $\frac{2}{3}$. Vos Savant was inundated by thousands of letters from readers who for the most part did not agree with her solution. The challenges to her solution were sometimes worded quite fiercely. Ninety percent of the letter writers, including mathematicians, said that there was no point in switching doors.

They argued that the two unopened doors that remained were each hiding the car with probability $\frac{1}{2}$. In other countries, too, the problem received the necessary attention and provoked many emotional reactions. In the Netherlands, a reader wrote to his newspaper: “I am aware that it is of great brutality, as someone who has always failed mathematics, to dispute the conclusion that the probability of winning becomes $\frac{2}{3}$ when switching doors. Let me use an analogous example to show that the columnist is wrong. Suppose that there are 100 doors and that the candidate stands in front of door 1. She then has probability $1 \%$ of standing in front of the correct door, while the probability that the car is behind one of the other doors is $99 \%$. The quizmaster then opens doors 2 through 99 . The car is behind none of them. Then the car must be behind door 1 or door 100 . If I were to follow the columnist’s reasoning, the entire $99 \%$ probability would transfer to door 100. Needless to say, this is complete nonsense. Here, too, a new situation has arisen with only two possibilities, each with an equal probability.” This reader is not only completely wrong but inadvertently contributes a strong argument for switching doors. Another reader wrote: “As a professional mathematician, I am increasingly concerned about the lack of mathematical understanding among the general public. Just realize that the probability must be $\frac{1}{2}$ and, in the future, refrain from commenting on matters you do not understand.” Martin Gardner, the spiritual father of the Monty Hall problem, wrote a long time ago: “In no other branch of mathematics is it so easy for experts to blunder as in probability theory.”

## 数学代写|运筹学代写Operations Research代考|The Chinese Postman Problem

1. 捡垃圾。假设一辆垃圾车必须在特定地区捡垃圾。图的边代表街道，节点对应街道交叉口。分配给边缘 的成本可以是街道长度。
2. 邮件或报纸递送。这也是一个典型的问题示例，其中寻找一条路线，使得每条街道至少被访问一次并且 行进的总距离尽可能小。
如果图中包含欧拉回路，则欧拉回路提供了中国邮递员问题的最优解。假设该图不包含欧拉回路，因为至少有 一个顶点的度数为奇数。为了准备我们接下来如何进行，我们定义
$X^{-}=$图中具有奇数度的一组顶点。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。