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

## 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|AN EVALUATION OF PERT/CPM

PERT/CPM has stood the test of time. Despite being more than 40 years old, it continues to be one of the most widely used OR techniques. It is a standard tool of project managers.
The Value of PERT/CPM
Much of the value of PERT/CPM derives from the basic framework it provides for planning a project. Recall its planning steps: (1) Identify the activities that are needed to carry out the project. (2) Estimate how much time will be needed for each activity. (3) Determine the activities that must immediately precede each activity. (4) Develop the project network that visually displays the relationships between the activities. The discipline of going through these steps forces the needed planning to be done.

The scheduling information generated by PERT/CPM also is vital to the project manager. When can each activity begin if there are no delays? How much delay in an activity can be tolerated without delaying project completion? What is the critical path of activities where no delay can be tolerated? What is the effect of uncertainty in activity times? What is the probability of meeting the project deadline under the current plan? PERT/CPM provides the answers.

PERT/CPM also assists the project manager in other ways. Schedule and budget are key concerns. The CPM method of time-cost trade-offs enables investigating ways of reducing the duration of the project at an additional cost. PERT/Cost provides a systematic procedure for planning, scheduling, and controlling project costs.

In many ways, PERT/CPM exemplifies the application of OR at its finest. Its modeling approach focuses on the key features of the problem (activities, precedence relationships, time, and cost) without getting mired down in unimportant details. The resulting model (a project network and an optional linear programming formulation) are easy to understand and apply. It addresses the issues that are important to management (planning, scheduling, dealing with uncertainty, time-cost trade-offs, and controlling costs). It assists the project manager in dealing with these issues in useful ways and in a timely manner.

## 数学代写|运筹学代写Operations Research代考|Using the Computer

PERT/CPM continues to evolve to meet new needs. At its inception over 40 years ago, it was largely executed manually. The project network sometimes was spread out over the walls of the project manager. Recording changes in the plan became a major task. Communicating changes to crew supervisors and subcontractors was cumbersome. The computer has changed all of that.

For many years now, PERT/CPM has become highly computerized. There has been a remarkable growth in the number and power of software packages for PERT/CPM that run on personal computers or workstations. Project management software (for example, Microsoft Project) now is a standard tool for project managers. This has enabled applications to numerous projects that each involve many millions of dollars and perhaps even thousands of activities. Possible revisions in the project plan now can be investigated almost instantaneously. Actual changes and the resulting updates in the schedule, etc., are recorded virtually effortlessly. Communications to all parties involved through computer networks and telecommunication systems also have become quick and easy.

Nevertheless, PERT/CPM still is not a panacea. It has certain major deficiencies for some applications. We briefly describe each of these deficiencies below along with how it is being addressed through research on improvements or extensions to PERT/CPM.
Approximating the Means and Variances of Activity Durations
The PERT three-estimate approach described in Sec. 10.4 provides a straightforward procedure for approximating the mean and variance of the probability distribution of the duration of each activity. Recall that this approach involved obtaining a most likely estimate, an optimistic estimate, and a pessimistic estimate of the duration. Given these three estimates, simple formulas were given for approximating the mean and variance. The means and variances for the various activities then were used to estimate the probability of completing the project by a specified time.

Unfortunately, considerable subsequent research has shown that this approach tends to provide a pretty rough approximation of the mean and variance. Part of the difficulty lies in aiming the optimistic and pessimistic estimates at the endpoints of the probability distribution. These endpoints correspond to very rare events (the best and worst that could ever occur) that typically are outside the estimator’s realm of experience. The accuracy and reliability of such estimates are not as good as for points that are not at the extremes of the probability distribution. For example, research has demonstrated that much better estimates can be obtained by aiming them at the 10 and 90 percent points of the probability distribution. The optimistic and pessimistic estimates then would be described in terms of having 1 chance in 10 of doing better or 1 chance in 10 of doing worse. The middle estimate also can be improved by aiming it at the 50 percent point (the median value) of the probability distribution.

Revising the definitions of the three estimates along these lines leads to considerably more complicated formulas for the mean and variance of the duration of an activity. However, this is no problem since the analysis is computerized anyway. The important consideration is that much better approximations of the mean and variance are obtained in this way.

## 数学代写|运筹学代写Operations Research代考|AN EVALUATION OF PERT/CPM

PERT/CPM经受住了时间的考验。尽管有40多年的历史，它仍然是最广泛使用的手术室技术之一。它是项目经理的标准工具。
PERT/CPM的值
PERT/CPM的大部分价值来源于它为规划项目提供的基本框架。回顾其计划步骤:(1)确定执行项目所需的活动。(2)估计每项活动需要多少时间。(3)确定每项活动之前必须立即进行的活动。(4)建立项目网络，直观地展示活动之间的关系。通过这些步骤的纪律迫使完成所需的计划。

PERT/CPM生成的进度信息对项目经理也是至关重要的。如果没有延误，每个活动什么时候开始?在不延迟项目完成的情况下，活动的延迟可以容忍到什么程度?不能容忍延迟的活动的关键路径是什么?不确定性对活动时间的影响是什么?在目前的计划下，在项目截止日期前完成的概率是多少?PERT/CPM提供了答案。

PERT/CPM还以其他方式协助项目经理。进度和预算是关键问题。时间成本权衡的CPM方法允许研究以额外成本减少项目持续时间的方法。PERT/Cost为计划、调度和控制项目成本提供了一个系统的程序。

## 数学代写|运筹学代写Operations Research代考|Using the Computer

PERT/CPM继续发展以满足新的需求。在40多年前开始的时候，它主要是手工执行的。项目网络有时会在项目经理的办公室里展开。记录计划的变化成了一项主要任务。与机组主管和分包商沟通变化是很麻烦的。电脑改变了这一切。

## 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。

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

## 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|SCHEDULING A PROJECT WITH PERT/CPM

At the end of Sec. 10.1, we mentioned that Mr. Perty, the project manager for the Reliable Construction Co. project, wants to use PERT/CPM to develop answers to a series of questions. His first question has been answered in the preceding section. Here are the five questions that will be answered in this section.

Question 2: What is the total time required to complete the project if no delays occur?
Question 3: When do the individual activities need to start and finish (at the latest) to meet this project completion time?
Question 4: When can the individual activities start and finish (at the earliest) if no delays occur?
Question 5: Which are the critical bottleneck activities where any delays must be avoided to prevent delaying project completion?
Question 6: For the other activities, how much delay can be tolerated without delaying project completion?

The project network in Fig. 10.1 enables answering all these questions by providing two crucial pieces of information, namely, the order in which certain activities must be performed and the (estimated) duration of each activity. We begin by focusing on Questions 2 and 5.
The Critical Path
How long should the project take? We noted earlier that summing the durations of all the activities gives a grand total of 79 weeks. However, this isn’t the answer to the question because some of the activities can be performed (roughly) simultaneously.
What is relevant instead is the length of each path through the network.
A path through a project network is one of the routes following the arcs from the START node to the FINISH node. The length of a path is the sum of the (estimated) durations of the activities on the path.
The six paths through the project network in Fig. 10.1 are given in Table 10.2, along with the calculations of the lengths of these paths. The path lengths range from 31 weeks up to 44 weeks for the longest path (the fourth one in the table).

So given these path lengths, what should be the (estimated) project duration (the total time required for the project)? Let us reason it out.

## 数学代写|运筹学代写Operations Research代考|Scheduling Individual Activities

The PERT/CPM scheduling procedure begins by addressing Question 4: When can the individual activities start and finish (at the earliest) if no delays occur? Having no delays means that (1) the actual duration of each activity turns out to be the same as its estimated duration and (2) each activity begins as soon as all its immediate predecessors are finished. The starting and finishing times of each activity if no delays occur anywhere in the project are called the earliest start time and the earliest finish time of the activity. These times are represented by the symbols
\begin{aligned} & \mathrm{ES}=\text { earliest start time for a particular activity, } \ & \mathrm{EF}=\text { earliest finish time for a particular activity, } \end{aligned}
where
$$\mathrm{EF}=\mathrm{ES}+(\text { estimated }) \text { duration of the activity. }$$
Rather than assigning calendar dates to these times, it is conventional instead to count the number of time periods (weeks for Reliable’s project) from when the project started. Thus,
Starting time for project $=0$.
Since activity $A$ starts Reliable’s project, we have
\text { Activity A: } \quad \begin{aligned} \mathrm{ES} & =0, \ \mathrm{EF} & =0+\text { duration (2 weeks) } \ & =2, \end{aligned}
where the duration (in weeks) of activity $A$ is given in Fig. 10.1 as the boldfaced number next to this activity. Activity $B$ can start as soon as activity $A$ finishes, so
Activity $B$ :
\begin{aligned} \mathrm{ES} & =\mathrm{EF} \text { for activity } A \ & =2, \ \mathrm{EF} & =2+\text { duration (4 weeks) } \ & =6 . \end{aligned}
This calculation of ES for activity $B$ illustrates our first rule for obtaining ES.
If an activity has only a single immediate predecessor, then
$\mathrm{ES}$ for the activity $=\mathrm{EF}$ for the immediate predecessor.

## 数学代写|运筹学代写Operations Research代考|Scheduling Individual Activities

PERT/CPM调度程序从解决问题4开始:如果没有延迟发生，单个活动何时开始和结束(最早)?无延迟意味着(1)每个活动的实际持续时间与其估计持续时间相同，(2)每个活动在其所有前一个活动完成后立即开始。如果在项目中没有任何延迟，则每个活动的开始和结束时间称为该活动的最早开始时间和最早结束时间。这些时间由符号表示
\begin{aligned} & \mathrm{ES}=\text { earliest start time for a particular activity, } \ & \mathrm{EF}=\text { earliest finish time for a particular activity, } \end{aligned}

$$\mathrm{EF}=\mathrm{ES}+(\text { estimated }) \text { duration of the activity. }$$

\text { Activity A: } \quad \begin{aligned} \mathrm{ES} & =0, \ \mathrm{EF} & =0+\text { duration (2 weeks) } \ & =2, \end{aligned}

\begin{aligned} \mathrm{ES} & =\mathrm{EF} \text { for activity } A \ & =2, \ \mathrm{EF} & =2+\text { duration (4 weeks) } \ & =6 . \end{aligned}

$\mathrm{ES}$表示活动$=\mathrm{EF}$表示直接前身。

## 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。

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

## 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 MAXIMUM FLOW PROBLEM

Now recall that the third problem facing the Seervada Park management (see Sec. 9.1) during the peak season is to determine how to route the various tram trips from the park entrance (station $O$ in Fig. 9.1) to the scenic wonder (station $T$ ) to maximize the number of trips per day. (Each tram will return by the same route it took on the outgoing trip, so the analysis focuses on outgoing trips only.) To avoid unduly disturbing the ecology and wildlife of the region, strict upper limits have been imposed on the number of outgoing trips allowed per day in the outbound direction on each individual road. For each road, the direction of travel for outgoing trips is indicated by an arrow in Fig. 9.6. The number at the base of the arrow gives the upper limit on the number of outgoing trips allowed per day. Given the limits, one feasible solution is to send 7 trams per day, with 5 using the route $O \rightarrow B \rightarrow E \rightarrow T, 1$ using $O \rightarrow B \rightarrow C \rightarrow E \rightarrow T$, and 1 using $O \rightarrow B \rightarrow C \rightarrow$ $E \rightarrow D \rightarrow T$. However, because this solution blocks the use of any routes starting with $O \rightarrow C$ (because the $E \rightarrow T$ and $E \rightarrow D$ capacities are fully used), it is easy to find better feasible solutions. Many combinations of routes (and the number of trips to assign to each one) need to be considered to find the one(s) maximizing the number of trips made per day. This kind of problem is called a maximum flow problem.
In general terms, the maximum flow problem can be described as follows.

1. All flow through a directed and connected network originates at one node, called the source, and terminates at one other node, called the sink. (The source and sink in the Seervada Park problem are the park entrance at node $O$ and the scenic wonder at node $T$, respectively.)
2. All the remaining nodes are transshipment nodes. (These are nodes $A, B, C, D$, and $E$ in the Seervada Park problem.)
3. Flow through an arc is allowed only in the direction indicated by the arrowhead, where the maximum amount of flow is given by the capacity of that arc. At the source, all arcs point away from the node. At the sink, all arcs point into the node.
4. The objective is to maximize the total amount of flow from the source to the sink. This amount is measured in either of two equivalent ways, namely, either the amount leaving the source or the amount entering the sink.

## 数学代写|运筹学代写Operations Research代考|Some Applications

Here are some typical kinds of applications of the maximum flow problem.

1. Maximize the flow through a company’s distribution network from its factories to its customers.
2. Maximize the flow through a company’s supply network from its vendors to its factories.
1. Maximize the flow of oil through a system of pipelines.
2. Maximize the flow of water through a system of aqueducts.
3. Maximize the flow of vehicles through a transportation network.

For some of these applications, the flow through the network may originate at more than one node and may also terminate at more than one node, even though a maximum flow problem is allowed to have only a single source and a single sink. For example, a company’s distribution network commonly has multiple factories and multiple customers. A clever reformulation is used to make such a situation fit the maximum flow problem. This reformulation involves expanding the original network to include a dummy source, a dummy $\operatorname{sink}$, and some new arcs. The dummy source is treated as the node that originates all the flow that, in reality, originates from some of the other nodes. For each of these other nodes, a new arc is inserted that leads from the dummy source to this node, where the capacity of this arc equals the maximum flow that, in reality, can originate from this node. Similarly, the dummy sink is treated as the node that absorbs all the flow that, in reality, terminates at some of the other nodes. Therefore, a new arc is inserted from each of these other nodes to the dummy sink, where the capacity of this arc equals the maximum flow that, in reality, can terminate at this node. Because of all these changes, all the nodes in the original network now are transshipment nodes, so the expanded network has the required single source (the dummy source) and single sink (the dummy sink) to fit the maximum flow problem.

## 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。

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

## 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 SHORTEST-PATH PROBLEM

Although several other versions of the shortest-path problem (including some for directed networks) are mentioned at the end of the section, we shall focus on the following simple version. Consider an undirected and connected network with two special nodes called the origin and the destination. Associated with each of the links (undirected arcs) is a nonnegative distance. The objective is to find the shortest path (the path with the minimum total distance) from the origin to the destination.

A relatively straightforward algorithm is available for this problem. The essence of this procedure is that it fans out from the origin, successively identifying the shortest path to each of the nodes of the network in the ascending order of their (shortest) distances from the origin, thereby solving the problem when the destination node is reached. We shall first outline the method and then illustrate it by solving the shortest-path problem encountered by the Seervada Park management in Sec. 9.1.
Algorithm for the Shortest-Path Problem.
Objective of $\mathrm{n}$ th iteration: Find the $n$th nearest node to the origin (to be repeated for $n=1,2, \ldots$ until the $n$th nearest node is the destination.
Input for $\mathrm{n}$ th iteration: $n-1$ nearest nodes to the origin (solved for at the previous iterations), including their shortest path and distance from the origin. (These nodes, plus the origin, will be called solved nodes; the others are unsolved nodes.)

Candidates for $\mathrm{n}$ th nearest node: Each solved node that is directly connected by a link to one or more unsolved nodes provides one candidate – the unsolved node with the shortest connecting link. (Ties provide additional candidates.)
Calculation of $\mathrm{n}$ th nearest node: For each such solved node and its candidate, add the distance between them and the distance of the shortest path from the origin to this solved node. The candidate with the smallest such total distance is the $n$th nearest node (ties provide additional solved nodes), and its shortest path is the one generating this distance.

## 数学代写|运筹学代写Operations Research代考|Applying This Algorithm to the Seervada Park Shortest-Path Problem

The Seervada Park management needs to find the shortest path from the park entrance (node $O$ ) to the scenic wonder (node $T$ ) through the road system shown in Fig. 9.1. Applying the above algorithm to this problem yields the results shown in Table 9.2 (where the tie for the second nearest node allows skipping directly to seeking the fourth nearest node next). The first column $(n)$ indicates the iteration count. The second column simply lists the solved nodes for beginning the current iteration after deleting the irrelevant ones (those not connected directly to any unsolved node). The third column then gives the candidates for the $n$th nearest node (the unsolved nodes with the shortest connecting link to a solved node). The fourth column calculates the distance of the shortest path from the origin to each of these candidates (namely, the distance to the solved node plus the link distance to the candidate). The candidate with the smallest such distance is the $n$th nearest node to the origin, as listed in the fifth column. The last two columns summarize the information for this newest solved node that is needed to proceed to subsequent iterations (namely, the distance of the shortest path from the origin to this node and the last link on this shortest path).

Now let us relate these columns directly to the outline given for the algorithm. The input for $\mathrm{n}$ th iteration is provided by the fifth and sixth columns for the preceding iterations, where the solved nodes in the fifth column are then listed in the second column for the current iteration after deleting those that are no longer directly connected to unsolved nodes. The candidates for $\mathrm{n}$ th nearest node next are listed in the third column for the current iteration. The calculation of $\mathrm{n}$ th nearest node is performed in the fourth column, and the results are recorded in the last three columns for the current iteration.

After the work shown in Table 9.2 is completed, the shortest path from the destination to the origin can be traced back through the last column of Table 9.2 as either $T \rightarrow D \rightarrow E \rightarrow B \rightarrow A \rightarrow O$ or $T \rightarrow D \rightarrow B \rightarrow A \rightarrow O$. Therefore, the two alternates for the shortest path from the origin to the destination have been identified as $O \rightarrow A \rightarrow B \rightarrow$ $E \rightarrow D \rightarrow T$ and $O \rightarrow A \rightarrow B \rightarrow D \rightarrow T$, with a total distance of 13 miles on either path.

## 数学代写|运筹学代写Operations Research代考|THE SHORTEST-PATH PROBLEM

$\ maththrm {n}$第n次迭代的目标:找到离原点最近的$n$节点(对于$n=1,2， \ldots$重复，直到$n$最近的节点是目的地)。
$\mathrm{n}$ th迭代的输入:$n-1$离原点最近的节点(在前一次迭代中求解)，包括它们的最短路径和到原点的距离。(这些节点加上原点，称为已解节点;其他是未解决的节点。)

$\ mathm {n}$第th最近节点的候选节点:每个通过链路直接连接到一个或多个未解决节点的已解决节点提供一个候选节点-具有最短连接链路的未解决节点。(领带提供了更多的候选人。)
$\mathrm{n}$ th最近节点的计算:对于每个已解节点及其候选节点，将它们之间的距离和从原点到该已解节点的最短路径的距离相加。总距离最小的候选节点是第n个最近的节点(领带提供了额外的已解节点)，它的最短路径是产生这个距离的节点。

## 数学代写|运筹学代写Operations Research代考|Applying This Algorithm to the Seervada Park Shortest-Path Problem

Seervada公园管理需要通过图9.1所示的道路系统，找到从公园入口(节点$O$)到景区奇观(节点$T$)的最短路径。将上述算法应用于此问题会产生如表9.2所示的结果(其中第二个最近节点的平局允许直接跳转到接下来寻找第四个最近节点)。第一列$(n)$表示迭代计数。第二列只是列出了在删除不相关节点(没有直接连接到任何未解决节点的节点)之后开始当前迭代的已解决节点。然后，第三列给出了第n个最近节点的候选节点(到已解节点的连接链路最短的未解节点)。第四列计算从原点到每个候选节点的最短路径的距离(即到已解节点的距离加上到候选节点的链接距离)。距离最小的候选节点是距离原点第n个最近的节点，如第五列所示。最后两列总结了这个最新解决的节点的信息，这些信息用于后续迭代(即，从原点到该节点的最短路径的距离以及该最短路径上的最后一个链接)。

## 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。

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

## 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 Transportation Problem Model

To describe the general model for the transportation problem, we need to use terms that are considerably less specific than those for the components of the prototype example. In particular, the general transportation problem is concerned (literally or figuratively) with distributing any commodity from any group of supply centers, called sources, to any group of receiving centers, called destinations, in such a way as to minimize the total distribution cost. The correspondence in terminology between the prototype example and the general problem is summarized in Table 8.4.

As indicated by the fourth and fifth rows of the table, each source has a certain supply of units to distribute to the destinations, and each destination has a certain demand for units to be received from the sources. The model for a transportation problem makes the following assumption about these supplies and demands.
The requirements assumption: Each source has a fixed supply of units, where this entire supply must be distributed to the destinations. (We let $s_i$ denote the number of units being supplied by source $i$, for $i=1,2, \ldots, m$.) Similarly, each destination has a fixed demand for units, where this entire demand must be received from the sources. (We let $d_j$ denote the number of units being received by destination $j$, for $j=1,2, \ldots, n$.)
This assumption that there is no leeway in the amounts to be sent or received means that there needs to be a balance between the total supply from all sources and the total demand at all destinations.
The feasible solutions property: A transportation problem will have feasible solutions if and only if
$$\sum_{i=1}^m s_i=\sum_{j=1}^n d_j$$

## 数学代写|运筹学代写Operations Research代考|Using Excel to Formulate and Solve Transportation Problems

To formulate and solve a transportation problem using Excel, two separate tables need to be entered on a spreadsheet. The first one is the parameter table. The second is a solution table, containing the quantities to distribute from each source to each destination. Figure 8.4 shows these two tables in rows $3-9$ and 12-18 for the P\&T Co. problem.

The two types of functional constraints need to be included in the spreadsheet. For the supply constraints, the total amount shipped from each source is calculated in column $\mathrm{H}$ of the solution table in Fig. 8.4. It is the sum of all the decision variable cells in the corresponding row. For example, the equation in cell H15 is ” $=\mathrm{D} 15+\mathrm{E} 15+\mathrm{F} 15+\mathrm{G} 15$ ” or “=SUM(D15:G15).” The supply at each source is included in column J. Hence, the cells in column H must equal the corresponding cells in column J.

For the demand constraints, the total amount shipped to each destination is calculated in row 18 of the spreadsheet. For example, the equation in cell D18 is “=SUM(D15:D17).” The demand at each destination is then included in row 20 .

The total cost is calculated in cell H18. This cost is the sum of the products of the corresponding cells in the main bodies of the parameter table and the solution table. Hence, the equation contained in cell H18 is “=SUMPRODUCT(D6:G8,D15:G17).”

Now let us look at the entries in the Solver dialogue box shown at the bottom of Fig. 8.4. These entries indicate that we are minimizing the total cost (calculated in cell H18) by changing the shipment quantities (in cells D15 through G17), subject to the constraints that the total amount shipped to each destination equals its demand (D18:G18=D20:G20) and that the total amount shipped from each source equals its supply (H15:H17=J15:J17). One of the selected Solver options (Assume Non-Negative) specifies that all shipment quantities must be nonnegative. The other one (Assume Linear Model) indicates that this transportation problem is a linear programming problem.

The values of the $x_{i j}$ decision variables (the shipment quantities) are contained in the changing cells (D15:G17). To begin, any value (such as 0) can be entered in each of these cells. After clicking on the Solve button, the Solver will use the simplex method to solve the problem. The optimal solution obtained in this way is shown in the changing cells in Fig. 8.4, along with the resulting total cost in cell H18.

Note that the Solver simply uses the general simplex method to solve a transportation problem rather than a streamlined version that is specially designed for solving transportation problems very efficiently, such as the transportation simplex method presented in the next section. Therefore, a software package that includes such a streamlined version should solve a large transportation problem much faster than the Excel Solver.
We mentioned earlier that some problems do not quite fit the model for a transportation problem because they violate the requirements assumption, but that it is possible to reformulate such a problem to fit this model by introducing a dummy destination or a dummy source. When using the Excel Solver, it is not necessary to do this reformulation since the simplex method can solve the original model where the supply constraints are in $\leq$ form or the demand constraints are in $\geq$ form. However, the larger the problem, the more worthwhile it becomes to do the reformulation and use the transportation simplex method (or equivalent) instead with another software package.
The next two examples illustrate how to do this kind of reformulation.

## 数学代写|运筹学代写Operations Research代考|The Transportation Problem Model

$$\sum_{i=1}^m s_i=\sum_{j=1}^n d_j$$

## 数学代写|运筹学代写Operations Research代考|Using Excel to Formulate and Solve Transportation Problems

$x_{i j}$决策变量(装运数量)的值包含在变化单元格中(D15:G17)。首先，可以在每个单元格中输入任何值(例如0)。单击Solve按钮后，求解器将使用单纯形法求解问题。通过这种方法得到的最优解如图8.4变化的单元所示，以及得到的总成本在H18单元中。

## 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。

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

## 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 UPPER BOUND TECHNIQUE

It is fairly common in linear programming problems for some of or all the individual $x_j$ variables to have upper bound constraints
$$x_j \leq u_j$$
where $u_j$ is a positive constant representing the maximum feasible value of $x_j$. We pointed out in Sec. 4.8 that the most important determinant of computation time for the simplex method is the number of functional constraints, whereas the number of nonnegativity constraints is relatively unimportant. Therefore, having a large number of upper bound constraints among the functional constraints greatly increases the computational effort required.

The upper bound technique avoids this increased effort by removing the upper bound constraints from the functional constraints and treating them separately, essentially like nonnegativity constraints. Removing the upper bound constraints in this way causes no problems as long as none of the variables gets increased over its upper bound. The only time the simplex method increases some of the variables is when the entering basic variable is increased to obtain a new BF solution. Therefore, the upper bound technique simply applies the simplex method in the usual way to the remainder of the problem (i.e., without the upper bound constraints) but with the one additional restriction that each new BF solution must satisfy the upper bound constraints in addition to the usual lower bound (nonnegativity) constraints.

To implement this idea, note that a decision variable $x_j$ with an upper bound constraint $x_j \leq u_j$ can always be replaced by
$$x_j=u_j-y_j$$
where $y_j$ would then be the decision variable. In other words, you have a choice between letting the decision variable be the amount above zero $\left(x_j\right)$ or the amount below $u_j\left(y_j=u_j-x_j\right)$. (We shall refer to $x_j$ and $y_j$ as complementary decision variables.) Because
$$0 \leq x_j \leq u_j$$
it also follows that
$$0 \leq y_j \leq u_j .$$
Thus, at any point during the simplex method, you can either
Use $x_j$, where $0 \leq x_j \leq u_j$,
or 2. Replace $x_j$ by $u_j-y_j$, where $0 \leq y_j \leq u_j$.

## 数学代写|运筹学代写Operations Research代考|AN INTERIOR-POINT ALGORITHM

In Sec. 4.9 we discussed a dramatic development in linear programming that occurred in 1984, the invention by Narendra Karmarkar of AT\&T Bell Laboratories of a powerful algorithm for solving huge linear programming problems with an approach very different from the simplex method. We now introduce the nature of Karmarkar’s approach by describing a relatively elementary variant (the “affine” or “affine-scaling” variant) of his algorithm. ${ }^1$ (Your OR Courseware also includes this variant under the title, Solve Automatically by the Interior-Point Algorithm.)

Throughout this section we shall focus on Karmarkar’s main ideas on an intuitive level while avoiding mathematical details. In particular, we shall bypass certain details that are needed for the full implementation of the algorithm (e.g., how to find an initial feasible trial solution) but are not central to a basic conceptual understanding. The ideas to be described can be summarized as follows:

Concept 1: Shoot through the interior of the feasible region toward an optimal solution. Concept 2: Move in a direction that improves the objective function value at the fastest possible rate.
Concept 3: Transform the feasible region to place the current trial solution near its center, thereby enabling a large improvement when concept 2 is implemented.
To illustrate these ideas throughout the section, we shall use the following example:
Maximize $Z=x_1+2 x_2$,
subject to
$$x_1+x_2 \leq 8$$
and
$$x_1 \geq 0, \quad x_2 \geq 0 .$$
This problem is depicted graphically in Fig. 7.3, where the optimal solution is seen to be $\left(x_1, x_2\right)=(0,8)$ with $Z=16$.

## 数学代写|运筹学代写Operations Research代考|THE UPPER BOUND TECHNIQUE

$$x_j \leq u_j$$

## 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。