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数学代写|运筹学作业代写operational research代考|MTH360

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如果你也在 怎样代写运筹学Operations Research 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。运筹学Operations Research为管理者、工程师和任何有更好解决方案的实践者提供更好的解决方案。这门科学诞生于第二次世界大战期间。虽然它最初用于军事行动，但它的应用以某种形式扩展到地球上的任何领域。

运筹学Operations Research是将科学方法应用于解决复杂问题，指导和管理工业、商业、政府和国防中由人、机器、材料和资金组成的大型系统。独特的方法是开发一个系统的科学模型，包括诸如变化和风险等因素的测量，以此来预测和比较不同决策、战略或控制的结果。其目的是帮助管理层科学地确定其政策和行动。

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。

数学代写|运筹学作业代写operational research代考|MTH360

数学代写|运筹学作业代写operational research代考|THE MINIMUM COST FLOW PROBLEM

The minimum cost flow problem holds a central position among network optimization models, both because it encompasses such a broad class of applications and because it can be solved extremely efficiently. Like the maximum flow problem, it considers flow through a network with limited arc capacities. Like the shortest-path problem, it considers a cost (or distance) for flow through an arc. Like the transportation problem or assignment problem of Chap. 8, it can consider multiple sources (supply nodes) and multiple destinations (demand nodes) for the flow, again with associated costs. In fact, all four of these previously studied problems are special cases of the minimum cost flow problem, as we will demonstrate shortly.
The reason that the minimum cost flow problem can be solved so efficiently is that it can be formulated as a linear programming problem so it can be solved by a streamlined version of the simplex method called the network simplex method. We describe this algorithm in the next section.
The minimum cost flow problem is described below.

  1. The network is a directed and connected network.
  2. At least one of the nodes is a suply node.
  3. At least one of the other nodes is a demand node.
  4. All the remaining nodes are transshipment nodes.
  5. 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. (If flow can occur in both directions, this would be represented by a pair of arcs pointing in opposite directions.)
  6. The network has enough arcs with sufficient capacity to enable all the flow generated at the supply nodes to reach all the demand nodes.
  7. The cost of the flow through each arc is proportional to the amount of that flow, where the cost per unit flow is known.
  8. The objective is to minimize the total cost of sending the available supply through the network to satisfy the given demand. (An alternative objective is to maximize the total profit from doing this.)

数学代写|运筹学作业代写operational research代考|Some Applications

Probably the most important kind of application of minimum cost flow problems is to the operation of a company’s distribution network. As summarized in the first row of Table 9.3, this kind of application always involves determining a plan for shipping goods from its sources (factories, etc.) to intermediate storage facilities (as needed) and then on to the customers.

For example, consider the distribution network for the International Paper Company (as described in the March-April 1988 issue of Interfaces). This company is the world’s largest manufacturer of pulp, paper, and paper products, as well as a major producer of lumber and plywood. It also either owns or has rights over about 20 million acres of woodlands. The supply nodes in its distribution network are these woodlands in their various locations. However, before the company’s goods can eventually reach the demand nodes (the customers), the wood must pass through a long sequence of transshipment nodes. A typical path through the distribution network is
Woodlands $\rightarrow$ woodyards $\rightarrow$ sawmills
$\rightarrow$ paper mills $\rightarrow$ converting plants
$\rightarrow$ warehouses $\rightarrow$ customers.
Another example of a complicated distribution network is the one for the Citgo Petroleum Corporation described in Sec. 3.5. Applying a minimum cost flow problem formulation to improve the operation of this distribution network saved Citgo at least \$16.5 million annually.

For some applications of minimum cost flow problems, all the transshipment nodes are processing facilities rather than intermediate storage facilities. This is the case for solid waste management, as indicated in the second row of Table 9.3. Here, the flow of materials through the network begins at the sources of the solid waste, then goes to the facilities for processing these waste materials into a form suitable for landfill, and then sends them on to the various landfill locations. However, the objective still is to determine the flow plan that minimizes the total cost, where the cost now is for both shipping and processing.

数学代写|运筹学作业代写operational research代考|MTH360

运筹学代考

数学代写|运筹学作业代写operational research代考|THE MINIMUM COST FLOW PROBLEM

最小成本流问题在网络优化模型中占据中心位置，因为它包含了如此广泛的应用程序类别，并且因为它可以非常有效地解决。与最大流量问题一样，它考虑的是通过电弧容量有限的网络的流量。与最短路径问题一样，它考虑的是流过一条弧线的成本(或距离)。就像第8章的运输问题或分配问题一样，它可以考虑流的多个来源(供应节点)和多个目的地(需求节点)，同样也有相关的成本。事实上，前面研究的所有四个问题都是最小成本流问题的特殊情况，我们很快就会演示。
最小成本流问题能够如此有效地求解的原因是，它可以被表述为一个线性规划问题，因此它可以通过一种简化版的单纯形法来求解，称为网络单纯形法。我们将在下一节中描述这个算法。
最小成本流问题描述如下。

网络是一个有向连接的网络。

至少有一个节点是供应节点。

其他节点中至少有一个是需求节点。

其余节点均为转运节点。

只允许在箭头指示的方向流过弧，其中最大流量由该弧的容量给出。(如果流动可以在两个方向上发生，这将由一对指向相反方向的弧来表示。)

网络有足够的弧线和足够的容量，使所有在供应节点产生的流量到达所有需求节点。

通过每条弧线的流量成本与流量成正比，其中单位流量成本是已知的。

目标是最小化通过网络发送可用供应以满足给定需求的总成本。(另一个目标是使这样做的总利润最大化。)

数学代写|运筹学作业代写operational research代考|Some Applications

也许最小成本流问题最重要的一种应用是公司配电网的运行。正如表9.3第一行所总结的那样，这类应用程序总是涉及确定将货物从其来源(工厂等)运送到中间存储设施(根据需要)，然后再运送到客户的计划。

例如，考虑国际纸业公司的分销网络(如1988年3 – 4月刊《界面》所述)。该公司是世界上最大的纸浆、纸张和纸制品制造商，也是木材和胶合板的主要生产商。它还拥有或拥有大约2000万英亩林地的权利。其分销网络中的供应节点就是这些分布在不同位置的林地。然而，在公司的货物最终到达需求节点(客户)之前，木材必须经过一系列的转运节点。通过配电网的典型路径是
Woodlands $\right tarrow$ wooddyards $\right tarrow$锯木厂
$\右箭头$造纸厂$\右箭头$转换厂
$\右箭头$仓库$\右箭头$顾客。
另一个复杂分销网络的例子是第3.5节中描述的Citgo石油公司的分销网络。应用最小成本流问题公式来改善该分销网络的运作，每年为Citgo节省至少1650万美元。

对于最小成本流问题的一些应用，所有转运节点都是处理设施，而不是中间存储设施。如表9.3第二行所示，固体废物管理就是这种情况。在这里，通过网络的物料流从固体废物的源头开始，然后进入将这些废物处理成适合填埋的形式的设施，然后将它们送到各个填埋地点。然而，目标仍然是确定最小化总成本的流程计划，其中现在的成本是运输和加工的成本。

统计代写|运筹学作业代写operational research代考 请认准statistics-lab™

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语 广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

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。

随机分析代写


随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。 随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

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，工具箱允许您学习和应用专业技术。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。

R语言代写问卷设计与分析代写
PYTHON代写回归分析与线性模型代写
MATLAB代写方差分析与试验设计代写
STATA代写机器学习/统计学习代写
SPSS代写计量经济学代写
EVIEWS代写时间序列分析代写
EXCEL代写深度学习代写
SQL代写各种数据建模与可视化代写

数学代写|ECON2326 Operations Research

Posted on by statistics-lab

Statistics-lab™可以为您提供uconn.edu ECON2326 Operations Research运筹学的代写代考和辅导服务！

ECON2326 Operations Research课程简介

Operations Research utlizes mathematical modeling, techniques, and algorithms, to most appropriately allocate resources and meet goals. An alternative title for the course might be “Mathematical Theory of Management, Control and Decision Making” . In general the course is intended for students with interests in applied math, statistics, economics, engineering and science. A second semester of the course devoted to probabilistic techniques will be offered in the spring semester.

PREREQUISITES 

Linear programming is a widely-used optimization technique that involves linear objective functions and linear constraints. The simplex algorithm is a widely-used algorithm for solving linear programming problems, and it involves moving from one feasible solution to another in an iterative fashion until an optimal solution is found. Sensitivity analysis involves examining the effects of changes in the objective function coefficients and constraint values on the optimal solution.

Dual problems are closely related to linear programming problems and involve the optimization of a dual objective function subject to dual constraints. The dual problem is used to provide information about the original problem, including bounds on the optimal objective value and information about the shadow prices of the constraints.

Integer programming is a type of linear programming that involves additional constraints that require the variables to take on integer values. Network models involve the modeling of complex systems using networks, such as transportation or communication systems. Dynamic programming is a powerful optimization technique that involves breaking down a problem into smaller subproblems and solving each subproblem in a recursive manner.

Finally, the KKT conditions are a set of necessary conditions for an optimization problem to have an optimal solution. These conditions involve the first-order conditions for optimality and the complementary slackness conditions, and they are used to analyze optimization problems and to derive insights about their solutions.

ECON2326 Operations Research HELP（EXAM HELP， ONLINE TUTOR）

问题 1.

A large bookshop sells 200 copies of the classic cookbook Grandma’s Secrets every month, with demand spread evenly over the month. The purchase cost is $€ 15$ per book. The bookshop’s total annual holding cost is $24 \%$ of the capital invested in inventory. The fixed ordering cost is estimated at $€ 35$.

(a) What is the optimal order quantity? Also calculate the minimum annual ordering and holding costs.

(b) What is the deviation from the theoretical minimum cost if the book can only be ordered by multiples of 50 ?

(c) When should an order be placed if the lead time is half a month?

(a) To determine the optimal order quantity, we can use the economic order quantity (EOQ) formula:

$EOQ = \sqrt{\frac{2DS}{H}}$

where:

  • D is the annual demand, which is 200 copies per month x 12 months = 2400 copies per year
  • S is the fixed ordering cost, which is €35
  • H is the holding cost per unit per year, which is 24% of the purchase cost per unit. Since the purchase cost is €15, the holding cost per unit per year is €15 x 24% = €3.60

Plugging in the values, we get:

$EOQ = \sqrt{\frac{2 \times 2400 \times 35}{3.60}} \approx 374.16$

Rounding up to the nearest whole number, the optimal order quantity is 375.

To calculate the minimum annual ordering and holding costs, we can use the following formulas:

$Annual ordering cost = \frac{DS}{EOQ} \times S$

$Annual holding cost = \frac{EOQ}{2} \times H$

Plugging in the values, we get:

$Annual ordering cost = \frac{2400 \times 35}{375} \approx 2240$

$Annual holding cost = \frac{375}{2} \times 3.60 \approx 675$

Therefore, the minimum annual ordering and holding costs are €2,240 and €675, respectively.

(b) If the book can only be ordered by multiples of 50, we need to find the order quantity that is closest to the EOQ and also a multiple of 50. The closest multiple of 50 to the EOQ of 374.16 is 350.

The annual ordering cost for an order quantity of 350 is:

$Annual ordering cost = \frac{2400 \times 35}{350} = 2400$

The annual holding cost can be calculated using the formula:

$Annual holding cost = \frac{350}{2} \times 3.60 = 630$

Therefore, the total annual ordering and holding costs for an order quantity of 350 are €2,400 + €630 = €3,030.

The deviation from the theoretical minimum cost is:

$\frac{3030 – 2915}{2915} \times 100% \approx 3.95%$

So the cost increase due to ordering in multiples of 50 is approximately 3.95%.

(c) If the lead time is half a month, we need to adjust the EOQ formula to include the additional demand during the lead time. The modified EOQ formula is:

$EOQ = \sqrt{\frac{2DS}{H} \times \frac{P}{P+L}}$

where:

  • P is the number of periods in a year, which is 12
  • L is the lead time in months, which is 0.5

Plugging in the values, we get:

$EOQ = \sqrt{\frac{2 \times 2400 \times 35}{3.60} \times \frac{12}{12+0.5}} \approx 382.99$

Rounding up to the nearest multiple of 50, the optimal order quantity is 400.

To determine when to place an order, we need to calculate the reorder point. The reorder point is the level of inventory at which a new order should be placed. It can be calculated using the following formula:

$Reorder point = D \times \frac{L+P}{P} = 2400 \times \frac{0.5+12}{12}

问题 2.

A specialist store sells 3500 Golden Tip ballpoint pens a year, with demand spread evenly over the year. The store estimates the annual holding cost at $18 \%$ of the capital invested in inventory. The purchase cost of the ballpoint pen is $€ 7.00$ if at least 350 pens are ordered; otherwise, the cost is $€ 7.50$. Every time the shop places an order with the supplier, it costs $€ 25$. What is the optimal order quantity?

To determine the optimal order quantity, we can use the economic order quantity (EOQ) formula:

$EOQ = \sqrt{\frac{2DS}{H}}$

where:

  • D is the annual demand, which is 3500 pens per year
  • S is the fixed ordering cost, which is €25 per order
  • H is the holding cost per unit per year, which is 18% of the purchase cost per unit.

We need to consider two different purchase costs: €7.00 if at least 350 pens are ordered, and €7.50 otherwise. Let’s first calculate the EOQ for the lower purchase cost of €7.00:

$EOQ_1 = \sqrt{\frac{2 \times 3500 \times 25}{0.18 \times 350}} \approx 279.72$

Rounding up to the nearest whole number, the optimal order quantity for the lower purchase cost is 280.

To check whether we should order at the lower purchase cost, we need to calculate the total cost of ordering 280 pens at €7.00:

$Total cost_1 = \frac{DS}{EOQ_1} \times S + \frac{EOQ_1}{2} \times H_1 + DP_1$

where:

  • D is the annual demand, which is 3500 pens per year
  • S is the fixed ordering cost, which is €25 per order
  • EOQ_1 is the optimal order quantity for the lower purchase cost of €7.00, which is 280 pens
  • H_1 is the holding cost per unit per year for the lower purchase cost, which is 18% of €7.00 = €1.26 per pen per year
  • P_1 is the purchase cost per pen, which is €7.00
  • D is the annual demand, which is 3500 pens per year

Plugging in the values, we get:

$Total cost_1 = \frac{3500 \times 25}{280} \times 25 + \frac{280}{2} \times 1.26 + 3500 \times 7 = 6325$

Now let’s calculate the EOQ for the higher purchase cost of €7.50:

$EOQ_2 = \sqrt{\frac{2 \times 3500 \times 25}{0.18 \times 350}} \approx 264.58$

Rounding up to the nearest whole number, the optimal order quantity for the higher purchase cost is 265.

To check whether we should order at the higher purchase cost, we need to calculate the total cost of ordering 265 pens at €7.50:

$Total cost_2 = \frac{DS}{EOQ_2} \times S + \frac{EOQ_2}{2} \times H_2 + DP_2$

where:

  • D is the annual demand, which is 3500 pens per year
  • S is the fixed ordering cost, which is €25 per order
  • EOQ_2 is the optimal order quantity for the higher purchase cost of €7.50, which is 265 pens
  • H_2 is the holding cost per unit per year for the higher purchase cost, which is 18% of €7.50 = €1.35 per pen per year
  • P_2 is the purchase cost per pen, which is €7.50
  • D is the annual demand, which is 3500 pens per year

Plugging in the values, we get:

$Total cost_2 = \frac{3500 \times 25}{265} \times

Textbooks


• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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数学代写|ECON2326 Operations Research

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