## 数学代写|运筹学作业代写operational research代考|MATH3202

statistics-lab™ 为您的留学生涯保驾护航 在代写运筹学operational research方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写运筹学operational research代写方面经验极为丰富，各种代写运筹学operational research相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|运筹学作业代写operational research代考|Shortest Path in a Manhattan Network

An efficient algorithm for the shortest-path problem in Figure 5.2 is the recursive approach of dynamic programming. The basic principle of this approach is to divide the original problem into a series of related and easily solvable subproblems. The main observation of the recursive approach is that a shortest path from the starting point $A=(0,0)$ to the endpoint $B$ would be easy to calculate if a shortest path from each of the points $(1,0)$ and $(0,1)$ to $B$ were known. In general, one can observe that a shortest path from point $(x, y)$ to the endpoint $B$ could easily be calculated if the shortest path to $B$ from each of the points $(x+1, y)$ and $(x, y+1)$ were known. The original problem can therefore be divided into a series of nested subproblems of decreasing size. The smallest subproblem is the problem that calculates the shortest path to the endpoint $B$ from each of the points $(n-1, m)$ and $(n, m-1)$. The solution to this subproblem is trivial. To concretize the ideas, we define
$$f(x, y)=\text { minimum travel distance from }(x, y) \text { to the endpoint } B .$$
This function is called the value function and is crucial in dynamic programming. Note that this function is defined for every point $(x, y)$ even though the goal is to find $f(0,0)$. However, by defining $f(x, y)$ for every point $(x, y)$, it is possible to create a recursive algorithm for $f(x, y)$ that will eventually lead to the desired value $f(0,0)$ for the starting point $A=(0,0)$. The data of the problem are
\begin{aligned} & R(x, y)=\text { travel distance from point }(x, y) \text { to point }(x+1, y) \ & U(x, y)=\text { travel distance from point }(x, y) \text { to point }(x, y+1) . \end{aligned}
The algorithm is initiated with
$$f(n-1, m)=R(n-1, m) \quad \text { and } \quad f(n, m-1)=U(n, m-1)$$

The general recursion step of the dynamic programming algorithm is as follows:
$$f(x, y)=\min {R(x, y)+f(x+1, y), U(x, y)+f(x, y+1)} .$$
The argument behind this recursive relation is simple and generally applicable. Suppose that one knows the shortest path to the endpoint $B$ from each of the points $(x+1, y)$ and $(x, y+1)$. Then one finds the shortest path from point $(x, y)$ to $B$ by considering the following two paths:
(a) Go right to $(x+1, y)$ and then follow the shortest path from point $(x+1, y)$ to the endpoint $B$.
(b) Go up to $(x, y+1)$ and then follow the shortest path from point $(x, y+1)$ to the endpoint $B$.

## 数学代写|运筹学作业代写operational research代考|General Structure of Dynamic Programming Problems

Every dynamic programming problem consists of several key components. The problem can be divided into stages $n$, with a decision required at each stage. Stages are also called decision epochs: the moments at which a decision must be made. Each stage has a number of states associated with it. The state space $S_n$ is the set of possible states $i$ which can occur at stage $n$. The state contains all the information that is needed to make an optimal decision. Decisions are also called actions. The decision space $D_n(i)$ is the set of decisions $d$ which are feasible in state $i$ at stage $n$. As a consequence of a decision, two things happen: the decision maker receives an immediate reward, and there is a transition to another state in the next stage. We define $r_n(i, d)$ as the immediate reward during stage $n$ as a consequence of decision $d$ in state $i$. Naturally, these are rewards in a maximization setting and costs in a minimization setting. Next to the immediate reward, the decision $d$ in state $i$ at stage $n$ causes a transition to state $j$ in stage $n+1$. In deterministic dynamic programming problems, which we are considering right now, the decision chosen at any stage fully characterizes how the state at the current stage is transformed into the state at the next stage. The fact that a decision causes an immediate reward as well as a transition to another state is at the heart of optimization in dynamic programming problems: a decision is optimal if it achieves the maximum value of the sum of the immediate reward and the rewards that can be earned from the next stage onward.

More formally, when optimizing in a dynamic programming problem, the objective is to maximize the total reward over all stages:
$$\max \left{\sum_{n=0}^N r_n(i, d)\right} .$$
This is done recursively, using the optimal value function $f_n(i)$, which is the maximum total reward that can be obtained from stages $n$ through $N$ if the system is in state $i$ at stage $n$. The optimal value function is characterized by a recursion relation which has the following general structure:
$$f_n(i)=\max {d \in D_n(i)}\left{r_n(i, d)+f{n+1}(d)\right}$$
At the final stage, $N$, there is no transition to another state in the next stage, so only the immediate reward plays a role. Hence, $f_N(i)$ can be found easily for all states $i$ and is therefore a natural starting point for the recursive calculations. $f_N(i)$ is called the salvage value. An important principle in dynamic programming is the principle of optimality: given the current state, the optimal decision for each of the remaining stages must not depend on previously reached states or previously chosen decisions.

In summary, a dynamic programming problem consists of stages, states, decisions, and immediate rewards, which come together in an optimal value function.

# 运筹学代考

## 数学代写|运筹学作业代写operational research代考|Shortest Path in a Manhattan Network

$f(x, y)=$ minimum travel distance from $(x, y)$ to the endpoint $B$.

$R(x, y)=$ travel distance from point $(x, y)$ to point $(x+1, y) \quad U(x, y)=$ travel di

$$f(n-1, m)=R(n-1, m) \quad \text { and } \quad f(n, m-1)=U(n, m-1)$$

$$f(x, y)=\min R(x, y)+f(x+1, y), U(x, y)+f(x, y+1) .$$

(a) 右转 $(x+1, y)$ 然后沿着点的最短路径 $(x+1, y)$ 到终点 $B$.
(b) 上升到 $(x, y+1)$ 然后沿着点的最短路径 $(x, y+1)$ 到终点 $B$.

## 数学代写|运筹学作业代写operational research代考|General Structure of Dynamic Programming Problems

$\backslash \max \backslash$ \eft $\backslash$ \sum_{ $\left.{\mathrm{n}=0}^{\wedge} \mathrm{N} r_{-} \mathrm{n}(\mathrm{i}, \mathrm{d}) \backslash r i g h t\right}$ 。

$f_{_} n(i)=\backslash \max \left{d \backslash i n _D _n(i)\right} \backslash l e f t\left{r_{-} n(i, d)+f{n+1}(d) \backslash r i g h t\right}$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 数学代写|运筹学作业代写operational research代考|MAT2200

statistics-lab™ 为您的留学生涯保驾护航 在代写运筹学operational research方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写运筹学operational research代写方面经验极为丰富，各种代写运筹学operational research相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|运筹学作业代写operational research代考|An Oil Drilling Problem

The Wildcat Company of the business duo Jacobse \& Van Es has acquired an option in Texas to search for oil on a specific piece of land. The option expires if drilling does not start within the next two weeks. The company must therefore decide soon; the business partners have three possible choices: 1 . drill immediately, 2. get a seismic test before deciding whether or not to drill, and 3. let the option expire. At the moment, the company’s working capital is 1275000 euros. Moreover, the following data are relevant to the decision-making process. Based on the currently available information on the condition of the site, the estimated probability that the relevant piece of land contains oil is equal to 0.50 . In order to obtain more information, it is also possible to have a seismic test carried out within a few days by an experienced geologist. In cases where oil drilling took place and the geologist had given prior advice, the geologist had advised to drill in $85 \%$ of the cases where oil was found and in $25 \%$ of the cases where no oil was found. The seismic test costs 100000 euros. If, after the test, the company decides to drill for oil, sufficient funds will still be available to finance the drilling cost of 1000000 euros. A large oil company has agreed with the duo Jacobse \& Van Es to take over the rights to the area for 3 million euros if oil is found. What should the business partners do?
This decision problem under uncertainty can be structured clearly in the form of a decision tree; see Figure 4.6. As usual, the tree is built up chronologically over time from left to right, where a decision node is indicated by a square and a chance node by a circle. The numbers in parentheses along the decision branches give the direct revenue of the decisions. At the end of each path in the tree, the end capital obtained by following the path is given. The end capital for a path is obtained by taking the sum of the starting capital of 1275000 euros and the amounts along the decision branches of the path. The numbers in parentheses along the chance branches give the probabilities of the different events that can occur. These probabilities require some explanation. The probability of oil being present changes when information from the seismic test becomes available. The numerical values $\frac{17}{22}$ and $\frac{5}{22}$ along the chance branches from chance node $4 a$ give the probability of there being oil given the advice to drill and the probability of there not being oil given the advice to drill. The numerical values $\frac{1}{6}$ and $\frac{5}{6}$ along the chance branches from chance node $4 b$ give the probability of there being oil given the advice not to drill and the probability of there not being oil given the advice not to drill. These probabilities can easily be deduced from Bayes’ rule in odds form. To calculate the posterior probability of finding oil given the advice to drill, apply this rule with $H$ the hypothesis that oil is present and $E$ the evidence that the geologist has advised to drill.

## 数学代写|运筹学作业代写operational research代考|Complexity Analysis

For larger networks, it is practically impossible to determine all possible paths and their lengths. To show the fundamental difficulty of the method of complete enumeration, we consider the problem in a more general form. Consider an $n \times m$ Manhattan network as shown in Figure 5.2. It is easy to order the intersections $(x, y)$ in a naturally way, where $x=0,1, \ldots, n$ and $y=0,1, \ldots, m$. We call the point $(0,0)$ at the bottom left $A$ and the point $(n, m)$ at the top right $B$. Only one-way traffic is allowed in the network. The intersection $(x, y)$ only has direct connections to the intersections $(x+1, y)$ and $(x, y+1)$, via a line segment to the right and a line segment going up. Suppose that each segment has a given travel distance. The problem is finding the shortest path from the starting point $A$ to the endpoint $B$. To show the numerical complexity of the method in which all possible paths are enumerated, we determine the total number of possible paths from $A$ to $B$. Combinatorics tells us that
total number of paths from $A$ to $B=\left(\begin{array}{c}n+m \ n\end{array}\right)=\frac{(n+m) !}{n ! m !}$.
The argument is simple. To go from $A$ to $B$, one needs to go right $n$ steps and up $m$ steps. The total number of paths from $A$ to $B$ is then the same as the number of different ways of placing $n$ elements of one kind and $m$ elements of another kind in a sequence. In particular, for an $n \times m$ network with $m=n$, we have
$$\text { total number of paths from } A \text { to } B=\frac{(2 n) !}{n ! n !} \text {. }$$
To estimate the order of magnitude of this number, we use Stirling’s approximation $n ! \approx \sqrt{2 \pi} n^{n+\frac{1}{2}} e^{-n}$ for large $n$.
In practice, this formula can already be used for $n \geq 10$. The relative error in the approximation is about $(100 / 12 n) \%$. By applying Stirling’s approximation, we find for the $n \times n$ network that
total number of paths from $A$ to $B \approx \frac{1}{\sqrt{\pi n}} 2^{2 n}$.
Every path requires $2 n-1$ additions. So for the method of total enumeration, we have
$$\text { total number of operations } \approx \sqrt{(n / \pi)} 2^{2 n+1} \text {. }$$

# 运筹学代考

## 数学代写|运筹学作业代写operational research代考|Complexity Analysis

total number of paths from $A$ to $B=\frac{(2 n) !}{n ! n !}$.

total number of operations $\approx \sqrt{(n / \pi)} 2^{2 n+1}$.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 数学代写|运筹学作业代写operational research代考|MAT2200

statistics-lab™ 为您的留学生涯保驾护航 在代写运筹学operational research方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写运筹学operational research代写方面经验极为丰富，各种代写运筹学operational research相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|运筹学作业代写operational research代考|General Modeling Tricks

In this section, we discuss a number of generally applicable modeling tricks to model “almost-linear” programming problems as (mixed) ILP problems. Some of these tricks have already been used in the examples in Section 2.2. Other tricks will come back in the exercises.
Almost-Linear Objective Function
Suppose that a production-stock problem is given, with aim to minimize the costs, and the objective function contains an almost-linear term $P(x)$, where $P(x)$ represents the production costs of $x$ units. The following three cases are of interest for applications:
(a) variable production costs plus fixed setup cost:
$$P(x)=\left{\begin{array}{cc} K+c x & \text { for } x>0 \ 0 & \text { for } x=0 \end{array}\right.$$
where $K$ and $c$ are constants with $K>0$ and $x$ is a variable;
(b) piecewise-linear production costs:
$$P(x)=\left{\begin{array}{cl} c_1 x & \text { for } 0 \leq x \leq a_1, \ c_1 a_1+c_2\left(x-a_1\right) & \text { for } x>a_1, \end{array}\right.$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。