2013建模美赛

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2013年美赛MCM题目A评委点评中文翻译

介绍今年的焦点问题是如何实现质量和数量的平衡。

在质量方面，尽可能使热量均匀地分布。

目标是降低或避免矩形烤盘四个边角发生热量聚集的情况。

所以解决热量均匀分布这方面的问题，使用圆形烤盘是最佳的选择。

在数量方面，应该使烤盘充分的占据烤箱的空间。

所以我们的目的是使用尽可能多的烤盘来充分占据烤箱的空间，此时矩形烤盘是最佳选择。

对于这方面的问题的解决，就要考虑烤盘在烤箱水平截面上所占的比率。

在这个评论中，我们首先描述判断步骤，然后再讨论队伍对于三个问题的求解。

下一个话题就是论文的灵敏度和假设，紧随其后讨论确定一个给定方法的优势和劣势。

最后，我们简短的讨论一下参考和引用之间的区别。

过程第一轮的判别被称为“分流轮”。

这些初始轮的主要思想是确定论文应被给予更详细的考虑。

每篇论文应该至少阅读两次。

在阅读一篇论文的时候，评审的主要问题是论文是否包含所有必要的成分,使它成为一个候选人最详细的阅读。

在这些初始轮中，评审的时间是有限制的，所以我们要尽量让每一篇论文得到一个好的评判。

如果一篇论文解决了所有的问题，就会让评审觉得你的模型建立是合理的。

然后评审可能会认为你的论文是值得注意的。

有些论文在初轮评审中可能会得到不太理想的评论。

特别值得注意的是，一篇好的摘要应该要对问题进行简要概述，另外，论文的概述和方法，队员之间应该互相讨论，并且具体的结果应该在某种程度上被阐述或者表达出来。

在早期的几轮中，一些小细节能够有突出的表现，包括目录，它更便于评委看论文，同时在看论文的时候可能会有更高的期待。

问题求解也很重要。

最后，方法和结果要清晰简明的表达是至关重要的。

另外，在每个部分的开始，应该对那个部分进行一个概述。

在竞赛中，建模的过程是很重要的，同时也包括结论的表达。

如果结果没有确切和充分的表达，那么再好的模型和再大努力也是没有用的。

最后的回合最后一轮阅读的第一轮开始于评委会会议。

在这个会议中，评委将进行讨论，他们会分享他们各自认为的问题的关键方面。

2013年数学建模美赛题目中文翻译_共4页

Problem c:背景：社会正致力于运用和开发模型来预测地球的生物和环境情况。

很多科学研究总结了逐渐增长的地球环境和生物系统压力，但很少有人用全球范围的模型来检测这些观点。

联合国发表的千年生态系统评估综合报告发现：近三分之二的地球生命支持生态系统——包括净水，洁净的空气，稳定的气候——正在因非可持续性使用而逐渐衰减。

其中大部分破坏归咎于人类行为。

暴增的对于食物，淡水，燃料，木材的需求导致了剧烈的环境变化；从森林砍伐到空气，土壤和水污染。

尽管已存在大量关于局部习惯和地区因素的研究，目前的模型还不能告知决定人他们的局部策略是如何影响整个地球的健康的。

许多模型忽略了复杂的全球因素，这些模型无法判断重大政策的长期影响。

尽管科学家们意识到巨大环境和生物系统中存在的复杂关系和交叉作用，当前的模型通常忽略这些管理或限定了系统间的影响。

系统的复杂性体现在多元交互（多个元素的相关性），反馈，突发行为，即将发生的状态变化或触发点。

最近的自然杂志中一篇由22位国际知名科学家撰写的题为“迫近地球生物圈的状态变化”的文章讨论了许多有关科学模型对于预测行星健康系统潜在状态变化的重要性与必需性。

文章提供了两种具体定性的模型，并寻求更好的预测模型：1）通过在全球模型中加入相关系统的复杂性（包括局部情况对全球系统的影响，反之亦然）来优化生物状态预测。

2）辨别不同因素在产生非健康全球状态变化中的作用并展示如何运用有效的生态系统管理来预防或限制这些即将发生的状态变化。

研究最终归结于问题：我们是否能利用全球健康的局部或地区性组成部分预测潜在状态变化来帮助决策者制定基于对全球健康状况潜在影响的，有效的策略。

尽管有越来越多的警示信号出现，没人知道地球是否确实在接近全球性的转折点（极端状态），这种极端的状态是否是不可避免的。

自然杂志等研究指出了地球生态系统中的一些重要工作元素。

（例如：局部因素，全球变化，多维元素与关系，变化的时间与空间范围）。

2013年美国数学建模竞赛B题

水资源计划摘要本文是要设计一个有效的，可行的，低成本的用水计划，来满足某国2025年的用水需求。

我们选择中国为研究对象，根据中国各地区历年的水资源总量并求出其均值，参考各地区历年用水总量来预测2025年的用水总量，将两者相减得出差值，并以此为依据将中国各地区分为缺水地区，不缺水地区，水资源丰富地区三类。

经研究分析有两种可行性高的方案。

第一种，由水资源丰富地区向缺水地区提供水。

第二种，是由沿海缺水城市进行海水淡化并运往其他缺水城市。

我们主要考虑经济因素对两种方案进行分析研究，最终得出结论由水资源丰富地区铺设管道向缺水地区提供水为最优方案。

并以各省的省会作为核心城市，说明全省的需水和调水情况，并以省会城市或直辖市为顶点构成一个赋权图，即把问题转换为求水资源丰富地区到缺水地区的最短路问题，并用图论的知识来解决问题。

在此基础上考虑到此方案会改变就业，生产力，水资源利用等因素，从而对经济，物理，环境产生不同程度的影响，并用层次分析加以研究，最终以报告的方式向政府反映。

关键词：回归分析最小生成树层次分析法一、问题重述淡水是世界大部分地区的发展限制。

试建立一个数学模型，用来确定一个有效的、可行的和低成本的水资源战略，以满足2025年预计的用水需求，特别是，您的数学模型必须解决存储和输送，去盐碱化和环境保护等问题。

如果可能的话，用你的模型探讨此战略在经济，物理和环境等方面的影响。

试提供一个非技术性的文件，向政府相关部门介绍你的方法以及其可行性和成本，并说明为什么它是“最好的水战略”。

二、符号说明ˆy：预测得出的2025年用水量；S：输水的造价；1S：海水淡化的造价；2d1: 输水工程的单位造价；d2：海水淡化的单位造价；2R：拟合度.三、模型假设1.从2013年到2025年各外部因素对水资源总量无影响，例如：雪灾、地震、洪水、战争等对环境的影响；2.各地区海水淡化单位费用相同；3.不同地区淡水转移的单位费用相同；4.人们的消费水平及劳动力费用不会随意外事故发生明显改变。

2013年美国大学生数学建模竞赛(MCMICM)参赛规则中英文对照

2013年美国大学生数学建模竞赛（MCMICM）参赛规则中英文对照2 ICM:The InterdisciplinaryContest in ModelingICM：交叉学科建模竞赛ContestRules, Registration and Instructions比赛规则，报名注册和指导(All rules and instructions apply to both ICM and MCM contests, except where otherwisenoted.)（所有MCM的说明和规则除特别说明以外都适用于ICM）To participate in a contest, each team must be sponsored by a faculty advisor fromits institution.参加MCM的每个队伍需有一名在职的高校老师负责指导。

TeamAdvisors: Please read these instructions carefully. It isyour responsibility to make sure that teams are correctly registered and thatall of the following steps required for participation in the contest arecompleted:Pleaseprint a copy of these contest instructions for reference before, during, andafter the contest. Click here for the printer friendly version.指导老师：请认真阅读这些说明事项，确保完成了所有相关的项。

每位指导教师的责任包括确保每个参赛队正确注册并正确完成参加MCM/ICM所要求的相关步骤。

2009-2013美赛建模题目

2009 Contest ProblemsMCM PROBLEMSPROBLEM A: Designing a Traffic CircleMany cities and communities have traffic circles—from large ones with many lanes in the circle (such as at the Arc de Triomphe in Paris and the Victory Monument in Bangkok) to small ones with one or two lanes in the circle. Some of these traffic circles position a stop sign or a yield sign on every incoming road that gives priority to traffic already in the circle; some position a yield sign in the circle at each incoming road to give priority to incoming traffic; and some position a traffic light on each incoming road (with no right turn allowed on a red light). Other designs may also be possible.The goal of this problem is to use a model to determine how best to control traffic flow in, around, and out of a circle. State clearly the objective(s) you use in your model for making the optimal choice as well as the factors that affect this choice. Include a Technical Summary of not more than two double-spaced pages that explains to a Traffic Engineer how to use your model to help choose the appropriate flow-control method for any specific traffic circle. That is, summarize the conditions under which each type of traffic-control method should be used. When traffic lights are recommended, explain a method for determining how many seconds each light should remain green (which may vary according to the time of day and other factors). Illustrate how your model works with specific examples.MCM2009问题A : 设计一个交通环岛在许多城市和社区都建立有交通环岛，既有多条行车道的大型环岛（例如巴黎的凯旋门和曼谷的胜利纪念碑路口），又有一至两条行车道的小型环岛。

2013年美国大学生数学建模大赛A题一等奖

最终的布朗尼蛋糕盘Team #23686 February 5, 2013摘要Summary/Abstract为了解决布朗尼蛋糕最佳烤盘形状的选择问题，本文首先建立了烤盘热量分布模型，解决了烤盘形态转变过程中所有烤盘形状热量分布的问题。

又建立了数量最优模型，解决了烤箱所能容纳最大烤盘数的问题。

然后建立了热量分布最优模型，解决了烤盘平均热量分布最大问题。

最后，我们建立了数量与热量最优模型，解决了选择最佳烤盘形状的问题。

模型一：为了解决烤盘形态转变过程中所有烤盘形状热量分布的问题，我们假设烤盘的任意一条边为半无限大平板，结合第三边界条件下非稳态导热公式，建立了不同形状烤盘的热量分布模型，模拟出不同形状烤盘热量分布图。

最后得到结论：在烤盘由多边形趋于圆的过程中，烤焦的程度会越来越小。

模型二：为了解决烤箱所能容纳最大烤盘数的问题，本文建立了随烤箱长宽比变化下的数量最优模型。

求解得到烤盘数目N 随着烤箱长宽比和烤盘边数n 变化的函数如下：AL W L W cont cont cont N 4n2nsin 1222⎪⎭⎫ ⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+⋅--=π模型三：本文定义平均热量分布H 为未超过某一温度时的非烤焦区域占烤盘边缘总区域的百分比。

为了解决烤盘平均热量分布最大问题，本文建立了热量分布最优模型，求解得到平均热量分布随着烤箱长宽比和形状变化的函数如下：n sin n cos -n 2nsin 22ntan1H ππδπδπ⎪⎪⎪⎪⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛⋅-=A结论是：当烤箱长宽比为定值时，正方形烤盘在烤箱中被容纳的最多，圆形烤盘的平均热量分布最大。

当烤盘边数为定值时，在长宽比为1:1的烤箱中被容纳的烤盘数量最多，平均热量分布H 最大。

模型四：通过对函数⎪⎭⎫ ⎝⎛n ,L W N 和函数⎪⎭⎫⎝⎛n ,L W H 作无量纲化处理，结合各自的权重p 和()p -1，本文建立了数量和热量混合最优模型，得到烤盘边数n 随p值和LW的函数。

2013数学建模美赛题目(中文版)[1]

2013 ICM问题problem A:当用矩形平底锅高温加热物品时，热量一般集中于4个角落，因而在角落的物品会被焙烧过度（较小程度在角落的物品一部分会被焙烧过度）。

当用一个圆锅加热物品时，热量是均匀分布在整个外缘，因而物品不会在边缘被焙烧过度。

然而，大多数烤箱是长方形的，而圆型的锅被认为效率低的。

建立一个模型以显示不同形状如矩形圆形或者其他介于两者之间的形状的锅在整个外缘的热量分布。

假设1长方形烤箱的宽/长=W/L ；2 每个锅的面积是确定的常熟A；3 最初，烤箱里的烤架两两之间间隔均匀。

建立一个模型，该模型可用于在下列条件之下选择最佳形状的锅：1烤箱中，锅数量（N）最大；2均匀分布的热量（H）最大的锅；3 优化组合条件1和条件2，以比重p和（1-p）的不同分配来说明结果与W/L 和p的不同值的关系。

problem B :对世界来说，新鲜的水资源是限制发展的制约因素。

对2013年建立一个确实有效的，可行的和具有成本效益的水资源战略数学模型，以满足2025年[从下面的列表选择一个国家]预计的用水需求,并确定最佳水资源战略。

尤其是，你的数学模型必须解决水的存储，运动，盐碱化和保护等问题。

如果可能的话，用你的模型，探讨经济，物理和环境对于你的战略的影响。

提供一个非技术性的文件，向政府领导介绍你的方法，介绍其可行性和成本，以及为什么它是“最好的的水战略选择。

”国家有：美国，中国，俄罗斯，埃及，沙特阿拉伯3.网络建模的地球的健康背景:社会是感兴趣的发展和使用模型来预测生物和环境卫生条件我们的星球。

许多科学研究认为越来越多的压力在地球的环境和生物吗系统,但是有很少的全球模型来测试这些索赔。

由联合国支持的年生态系统评估综合报告》显示,近三分之二的地球的维持生命的生态系统——包括干净的水,纯净的空气,和稳定的气候-正在退化,被不可持续的使用。

人类是归咎于很多这次的损坏。

不断飙升的要求食品、新鲜水、燃料和木材有贡献到戏剧性的环境变化,从森林砍伐,空气,土地和水的污染。

2013年数学建模美赛B题论文

2013建模美赛B题思路摘要水资源是极为重要生活资料，同时与政治经济文化的发展密切相关，北京市是世界上水资源严重缺乏的大都市之一。

本文以北京为例，针对影响水资源短缺的因素，通过查找权威数据建立数学模型揭示相关因素与水资源短缺的关系，评价水资源短缺风险并运用模型对水资源短缺问题进行有效调控。

首先，分析水资源量的组成得出影响因素。

主要从水资源总量（供水量）和总用水量（需水量）两方面进行讨论。

影响水资源总量的因素从地表水量，地下水量和污水处理量入手。

影响总用水量的因素从农业用水，工业用水，第三产业及生活用水量入手进行具体分析。

其次，利用查得得北京市2001-2008年水量数据，采用多元线性回归，建立水资源总量与地表水量，地下水量和污水处理量的线性回归方程yˆ=-4.732+2.138x1+0.498x2+0.274x3根据各个因数前的系数的大小，得到风险因子的显著性为r x1>r x2>r x3(x1, x2，x3分别为地表水、地下水、污水处理量)。

再次，利用灰色关联确定农业用水、工业用水、第三产业及生活用水量与总用水量的关联程度r a=0.369852，r b= 0.369167，r c=0.260981。

从而确定其风险显著性为r a>r b>r c。

再再次，由数据利用曲线拟合得到农业、工业及第三产业及生活用水量与年份之间的函数关系，a=0.0019（t-1994）3-0.0383（t-1994）2-0.4332（t-1994）+20.2598;b=0.014（t-1994）2-0.8261t+14.1337;c=0.0383（t-1994）2-0.097（t-1994）+11.2116;D=a+b+c；预测出2009-2012年用水总量。

最后，通过定义缺水程度S=（D-y）/D=1-y/D，计算出1994-2008的缺水程度，绘制出柱状图，划分风险等级。

我们取多年数据进行比较，推测未来四年地表水量和地下水量维持在前八年的平均水平，污水处理量为近三年的平均水平，得出2009-2012年的预测值，并利用回归方程yˆ=-4.732+2.138x1+0.4982x2+0.274x3计算出对应的水资源总量。

2013美国数学建模题目以及部分翻译

2013 Contest ProblemsMCM PROBLEMSPROBLEM A: The Ultimate Brownie PanWhen baking in a rectangular pan, heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges).矩形的平锅In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. 圆形锅，热量均匀地分布在所有的边缘However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven.大多数的灶是矩形的，用圆锅没有效率Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between.建立模型，展示不同形状的锅子外部边缘上热量的分布，Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.炉灶的长宽比是矩形2. Each pan must have an area of A.每个平锅面积都是A3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape)under the following conditions:1. Maximize number of pans that can fit in the oven (N)锅子最多2. Maximize even distribution of heat (H) for the pan最大化炉灶的热量分布3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different valuesof W/L and p.完善一二条件的组合，权重为p和1-p，阐明不同情况下的结果In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.PROBLEM B: Water, Water, EverywhereFresh water is the limiting constraint for development in much of the world. Build a mathematical model for determining an effective, feasible, andcost-efficient water strategy for 2013 to meet the projected water needs of [pick one country from the list below] in 2025, and identify the best water strategy.建立数学模型来决定一个有效的、可行的、成本效率的2013水战略，以满足2025年以下某一个国家计划的用水需求。

2013年数学建模美赛A题二等奖作品

The perfect pan for ovenThe heat transfer in the oven includes heat conduction, heat radiation and heatconvection. We use two-dimensional Fourier heat conduction equation ∂u∂t −α(ð2u∂x2+ð2u∂y2)=f(x, y, t) to make a research on distribution of heat for the pan. Heat source heats the pan by heat radiation. The pan interacts with air in the oven in the way of natural convection, so the pan realizes heat dissipation.We calculate heat radiation based on radiation ability of heat source and heating tube area. We use heat dissipation function to show the pan's different parts' loss of heat caused by natural convection. Both of them consist in heat source function f.The area of the pan is fixed at 0.085m2in this paper. When comparing temperatures at the edges of rectangular pans with different length to width ratios ξ, we can get that the smaller ξ is, the lower the temperature of the edges is. But as long as it is still a rectangle, the amplitude of its drop won't be very big. When we make the pans with fixed area vary from square to round square to round, we find that the bigger the fillet radius is, the lower the temperature of its corners is and the extent of temperature's reducing is large.We fix the bottom area of the oven and area of the pan. Through study, we find that round square's capacity for uniform distribution of heat is far higher than other shape's (except round). The larger the fillet radius of the round square is, the larger the pan’s waste of space is. But heat distribution is more uniform. We work out the optimal solution of pan’s size under different weights p through optimizing the relationship between two conditions. Then we get several oven's width to length ratios of W/L by arranging the pans with the optimal size.I. IntroductionThe temperature of each point in the pan is different. For a rectangular pan, the corners have the highest temperature, so the food is easily overcooked. While the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges in the round pan.To illustrate the model further, the following information is worth mentioning1.1 Floor space of the panThe floor space of each pan is not the square itself necessarily. In this paper, there are 3 kinds of pans with different shapes, as rectangular pans, round pans and round rectangle pans.For rectangular pan, the floor space is the square itself, and the pans can connect closely without space.For round pans, the diagrammatic sketch of the floor space is as follows:shade stands for the round pan; square stands for the floor spaceFigure 1Round pans have the largest floor space for a certain area. The space between each pan is larger than other two kinds of pans. The coefficient of utilization for the round pans is the lowest.For round rectangle pans, the diagrammatic sketch of the floor space is as follows:shade stands for the round rectangle pan; square stands for the floor spaceFigure 2If the area of the round rectangle is the same as the other two, its floor space is between them. The coefficient of utilization of oven decreases with the radius of expansion.1.2 Introduction of ovenThe oven is usually a cube, no matter it is used in home or for business. A width to length ratio for the oven is not a certain number. There are always two racks in the oven, evenly spaced. There are one or more pans on each rack. To preserve heat for the oven, food is heated by radiation. Heating tube can be made of quartz or metal. The temperature of the tube can reach 800℃ high when the material is quartz. The heating tube is often in the top and bottom of the oven. Heating mode can be heating from top or heating from bottom, and maybe both[1].1.3 Two dimensional equation of conductionTo research the heat distribution of pan, we draw into two dimensional equation[2]of conduction:∂u ∂t −α(ð2u∂x2+ð2u∂y2)=f(x, y, t)In this equation:u- temperature of the pant- time from starting to heatx- the abscissay-ordinateα- thermal diffusivityf- heat source functionThe heat equation is a parabolic partial differential equation which describes the distribution of heat (or variation in temperature) in a given region over time. The heat equation is of fundamental importance in diverse scientific fields. In mathematics, it is the prototypical parabolic partial differential equation. In probability theory, the heat equation is connected with the study of Brownian motion via the Fokker–Planck equation. The diffusion equation, a more general version of the heat equation, arises in connection with the study of chemical diffusion and other related processes.II. The Description of the Problem2.1 The original problemWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient withrespect to using the space in an oven.Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes -rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.2.2 Problem analysisWe analyze this problem from 3 aspects, showing as follows:2.2.1 Why the edge of the pan has the highest temperature?The form of heat transfer includes heat radiation, heat conduction and heat convection. Energy of heat radiation comes from heat resource. The further from heat resource, the less energy it gets. Heat conduction happens in the interior of the pan, and heat transfers from part of high temperature to the part of low with the temperature contrast as its motivation. With the two forms of heat transfer above, we find the result is that pan center has the high temperature and the boundary has the low. D epending on that, we can’t explain why the product gets overcooked at the corners while at the edges not. We think that there is natural convection between pan and gas, because the temperature of pan is much higher than that of gas. The convection is connected with the contact area. The point in pan center has a larger contact range with air, so the energy loss from convection is more. While the point in the corners of a rectangular pan has a narrow contact area, the energy loss is less than that in the inner part. Because that above, the energy in the pan center is more than that in corner, and the corners have higher temperature.2.2.2 Analysis of heat distribution in pans in different shapesThe shape of pan includes rectangle, round rectangle and round. When these pans' area is fixed, the rectangles with different shapes can be shown with different length to width ratios. Firstly, we study temperature (maximum temperature) in the corners of rectangles with different length to width ratios. Then we study how temperature in the corners changes when the pans vary from square to round square to round. After that, we select some rectangular panand make it change from rectangle to round rectangle to study changes of temperature in the corners. To calculate distribution of heat for the pan, we would use three main equations. The first one is the Fourier equation, namely heat conduction equation, the second one is the radiation transfer equation of heat source and the third one is the equation of heat dissipation through convection. In the radiation transfer equation of heat source, we take heat source as a point. We get its radiating capacity through its absolute temperature, blackening and Stefan-Boltzmann law. We combine radiating capacity with surface area of quartz heating tube to get quantity of heat emitted by heat source per second, then we can get heat flux at each point of the pan. In the equation of heat loss through convection, Heat dissipating capacity is proportional to area of heat dissipation, its proportional coefficient can be found from the related material. Through the establishment of above three main equations, we can use pdetool in matlab to draw the figure about distribution of heat for the pan.2.2.3 How to determine the shape of the pan?To make heat distribution of the pan uniform, we must make it approach round. But under the circumstances of the pan's fixed area, the closer the pan approaches round, the larger its floor space is. In other words, the closer the pan approaches round, the lower the utilization rate of the oven is.More uniform distribution of heat for the pan is, the lower temperature in the corners of the pan is. Assuming the bottom area of the oven is fixed, the number of most pans which the oven can bear is equal to the quotient of the bottom area of the oven divided by floor space per pan. So in a certain weight P, we can get the best type of pan (shape) by optimizing the relationship between temperature in the corners and the number of most pans which the oven can bear. Then we get the oven's width to length ratio of W/L by arranging the pans with the optimal size.2.3 Practical problem parameterizationu: temperature of each point in the oven;t: heating time;x: the abscissa values;y: the ordinate value;α:thermal diffusivity;ε: degree of blackness of heat resource;E: radiating capacity of heat resource;T: absolute temperature of heat resource;T max: Highest temperature of pans’ edge;ξ: the length to width ratio for a rectangular pan;R: radius for a round pan;L: length for the oven;W: width for the oven;q: heat flux;k: coefficient of the convective heat transfer;Q: heat transfer rate;P: minimum distance from pan to the heat resource;Other definitions will be given in the specific models below2.4 Assumption of all models1. We assume the heat resource as a mass point, and it has the same radiation energy in all directions.2. The absorbtivity of pan on the radiation energy is 100%.3. The rate of heat dissipation is proportional to area of heat dissipation.4. The area of heat dissipation changes in a linear fashion from the centre of the pan to its border.5. Each pan is just a two-dimensional surface and we do not care about its thickness.6. Room temperature is 25 degrees Celsius.7.The area of the pan is a certain number.III. ModelsConsider the pan center as origin, establishing a coordinate system for pan as follows:Figure 33.1 Basic ModelIn order to explain main model better, the process of building following branch models needs to be explained specially, the explanation is as follows:3.1.1 Heat radiation modelFigure 4The proportional of energy received by B accounting for energy from A is4π(x2+y2+P2) The absolute temperature of heat resource A (the heating tube made of quartz) T= 773K when it works. The degree of blackness for quartz ε=0.94.The area of quartz heating tube is 0.0088m2.Depending on Stefan-Boltzmann law[3]E=σεT4 (σ=5.67*10-8)we can get E=18827w/m2.The radiation energy per second is 0.0088E.At last, we can get the heat flux of any point of pan. =0.0088E4π(x2+y2+P2)Synthesizing the formulas above, we can get:=13.24π(x2+y2+P2)(1) 3.1.2 Heat convection model3.1.2.1 Round panheat dissipation area of round panFigure 4When the pan is round, the coordinate of any point in the pan is (x,y). When the point is in the centre of a circle, its area of heat dissipation is dxdy. When the point is in theboundaries of round, its area of heat dissipation is 12dxdy. According to equation of heatPdissipation through convection dQ=kdxdy and assumption that the area of heat dissipation changes in a linear fashion along the radius[4], we can get the pan's equation of heat dissipation:q =k[1-0.5(x2+y2)0.5/R] (2) 3.1.2.2 Rectangular panHeat dissipation area of rectangular pan(length: M width: N)Figure 5When the pan is rectangular, the coordinate of any point in the pan is (x,y). When the point is in the centre of a rectangle, its area of heat dissipation is largest, namely dxdy. Whenthe point is in the center of the rectangular edges, its area of heat dissipation is 1dxdy. When2the point is in rectangular vertices, its area of heat dissipation is minimum, namely According to equation of heat dissipation dQ=kdxdy and assumption that the area of heat dissipation changes in a linear fashion from the centre of a rectangle to the center of the rectangular edges.The area of any other point in the pan can be regarded as the result of superposing two corresponding points' area in two lines.Heat dissipating capacity of any point is:=1−y/N(3)1−x/M3.2 Pan heat distribution Model3.2.1 Heat distribution of rectangular pansAssume that the rectangle's length is M and its width is N,the material of the pan is iron.For rectangular pans, we change its length to width ratio, establishing a model to get thetemperature of the corners (namely the highest temperature of the pan).We assume the area of pan is a certain number 0.085m2,The distance from the pan above to the top of the oven P=0.23m,By checking the data, we can know that the coefficient of the convective heat transfer k is approximately 25 if the temperature contrast between pan and oven is 100~200℃.then get a several kinds rectangular pans following:In the two dimensional equation of conduction,we get the thermal diffusivity of iron is 0.000013m2/s through checking data.Heat source function equals received thermal radiation minus loss of heat caused by heat dissipation through ly equation (1) minus equation (3).In this way, we get a more complicated partial differential equation. For example, through analyzing the NO.1 pan, we can get the following partial differential equation.∂u ∂t −α(ð2u∂x2+ð2u∂y2)=13.24π(x2+y2+0.529)−1−y/0.291551−x/0.29155It is hardly to get the analytic solutions of the partial differential equation. We utilize the method of finite element partition to analyze its numerical solution, and show it in the form of figures directly.Pdetool in matlab can solve the numerical solution to differential equation in the regular form quickly and show distribution of heat by the three-dimensional image[5]. We enter partial differential equation of two-dimensional heat conduction into it, then we get the figures about distribution of heat in different pans.When we use pdetool, we take Neumann condition as boundary condition, and we suppose that the boundary is insulated, In fact, it is not insulated, and the heat dissipation will show in the heat source function.We heat the pan for 8 minutes no matter what kind of shape the pan is. By Pdetool, the heat distribution of each pan shows as follows:Figure 6(heat distribution for NO.1 pan) Figure 7(heat distribution for NO.2 pan) Figure 8(heat distribution for NO.3 pan)This pan is just a square pan, with the lowest length to width ratio. From the figure, we can know that corners have the highest temperature which is 297.7℃.The corners have the highest temperature for the pan, which is 296℃The corners have the highest temperature for the pan, which is 294.8℃The temperature ofcorners for the pan isslightly lower than thehighest temperature, andthe highest temperature is293.7 ℃.Figure 9(heat distribution for NO.4 pan)To get a more accurate relationship between the length to width ratio(ξ) and highest temperature(T max), we make several more figures of heat distribution based on the different length to width ratio. At last, we can get its highest temperature. The specific result is as follows:ξ is the argument and T max is the dependent variable. The points in the chart are scaled out in the coordinate system by mathematical software Origin. Connecting the points by smooth curve, we can get the figure following:Figure 10(the relationship between ξ and T max )From the figure we can find that, with the length to width ratio increases, the temperature of corners will sharply fall at first, and it is namely that the heat distributes evenly . When the length to width ratio increases further, the temperature of corners drops obscurely . When the ratio reaches about 2.125, the length to width ratio of rectangular pans has little effect on the heat distribution. In contrast, the ratio is too big, it is difficult for practical application.In addition, we can also find that as long as the shape of the pan is a rectangle. The highest temperature in the corners of the pan won't change a lot whether its length to width ratio changes, From the figure 10, we can find that maximum range is about 6 degrees Celsius. So in general, it is very difficult to change high temperature in the corners of the rectangular pan.3.2.2 Heat distribution of square pans to round pans 3.2.2.1 Size definition of round squareWe need some sizes to define round square, our definition isas follows:T maxThe size of round square isdecided by l and r (l stands forthe length of the straight flange;r stands for the radius of thefillet).Figure 11We assume that the area of the pan is 0.085m2. The r of round square has its range, which is 【0，0.164488】，When the r reaches the two extremums, round square becomes square and circle.3.2.2.2 ModelWhen we make this kind of pans’ heat distribution figures, we take the heat source function as (1) and (3). When the round square becomes circle, the heat source function is (1) and (2).We get a several round squares with different r and l, and take a circle as an example. The specific examples show in the table below:The pan is still made of iron. Here we also heat the pan for 8 minutes.By Pdetool, we can obtain their heat distribution figures as follows:Figure 11(heat distribution for NO.1 pan) Figure 12(heat distribution for NO.2 pan) Figure 13(heat distribution for NO.3pan)The temperature of the corners about the pan is relatively low, and the highest temperature is 264℃. On the contrary, the heat distributes quite evenly.The highest temperature of the pan is 271.1℃The highest temperature of the pan is 274℃The highest temperatureof the pan is 279.8℃Figure 14(heat distribution for NO.4pan)The highest temperatureof the pan is 283.7℃Figure 15(heat distribution for NO.5pan)To get the different relationship between l/r and T max, we take l/r as argument, and T max as dependent variable. We change l and r of round square, and make several heat distribution figures. In this way, we can get the highest temperature T max, and the results show in the table below:Here, we alsouse the mathematical drawing software origin. We use l/r and T max to express coordinates. We use smooth curve to connect points, then we can get the trend line.Figure 16(the relationship between l/r and T max )From the figures we can know that, the value of l/r is smaller, the temperature of the corners about pan is higher. It is namely that the pan is more closely to circle, and heat distribution is more evenly. With the value of l/r increases, the temperature of corners rise very quickly at first, then the amplitude is getting smaller. When the value of l/r is infinitely great, the highest temperature of the pan go to a certain number.Analyzing in a theoretical way , the shape of pan goes to square when the value of l/r is infinitely great. At the same time, the temperature of the round square’s corners approach to that of square’s. From the figures, we can know that this function has a upper boundary ,T maxwhose value is close to the corner temperature of square. Through this, we can verify the correctness of our models.3.2.3 Heat distribution of round rectangle (except round square)From the model about heat distribution of rectangular pan, we can learn that drop of temperature in the corners of rectangular pans will be very little when its length to width ratio is bigger than 2.125. So we select the rectangle with length to width ratio of 2.125. We let the pan vary from the rectangle to round rectangle. So we can study changes of temperature in the corners of the pan.3.2.3.1 Size definition of round rectangleThe specification of the roundsquare is decided by l and r. Weset its width the same as therectangular pan before, namely0.2m.(l stands for the length ofstraight long side, and r stands forthe radius of the fillet.)Figure 17We assume that the area of the pan is 0.085m2, For round rectangle pans, with r decreases constantly, the pan finally approaches the rectangular pan before .If the r increases constantly, its shape will become that of playground. The range of r is 【0,0.1】3.2.3.2 ModelWhen we draw the figure about heat distribution of this kind of pan, heat source function equals equation (1) minus equation (2).We can get some different round rectangles by changing the value of r and l. Their detailed specifications are shown in the following table:The pan is still made of iron. Here we also heat the pan for 8 minutes.By Pdetool, we can obtain their heat distribution figures as follows:The corner temperaturewhich is 291.7℃andslightly lower than thehighest temperature ofthe pan.Figure 18(heat distribution for NO.1pan)The corner temperaturewhich is 292.54℃andslightly lower than thehighest temperature ofthe pan.Figure 19(heat distribution for NO.2pan)The corner temperaturewhich is 293.5℃andslightly lower than thehighest temperature ofthe pan.Figure 20(heat distribution for NO.3pan)To get the different relationship between l/r and T max, we take l/r as argument, and T max as dependent variable. We change l and r of round square, and make several heat distribution figures. In this way, we can get the highest temperature T max, and the results show in the table below:Here, we also use the mathematical drawing software origin. We use l/r and T max to express coordinates. We use smooth curve to connect points, then we can get the trend line.Figure 21(T max)From the figures we can get that, with l/r increases, the highest temperature of pan goes down. That is to say, the bigger radius of the fillet is, the more evenly heat distributes. In addition, with l/r increases, T max rises quickly at first, then the extent is smaller. When l/r is infinitely great, the round rectangle pans become rectangle pans, and the temperature approaches to that of the rectangle pan before.Secondly, from the figure, we can learn that change of the largest temperature is very little and its largest temperature is all very high when the pan varies from rectangle to round rectangle. compared with round square pan, round rectangle pan has much worse capacity of distributing heat.3.3 Best type of pan selection ModelFrom the model of heat distribution, we can know that the extent of heat distribution for round square pan is more than the extent of other pans with other kinds of shape( except T maxcircle). It is difficult to accept it for people because the food is easily overcook, no matter what the number of pan is. Depending on that, people will choose the round square pan.In the following models, we mainly discuss the advantages and disadvantages of round square pans with different specifications.3.3.1 Local parametersFor study's convenience, we take commercial oven of bottom area 1.21m2as an example. The distance between heating tube on the top and the nearest rack is P=0.23m.The number of pans which the oven can contain is n;The fillet radius of round square is r;The floor space of pan is S;The weight of the number of pans in the oven is P;The area of pan is still 0.085m2, the material is still iron.3.3.2 The relationship between T max and rThrough the models before, we know the relationship between l/r and T max. We transform it as the function of r and T max, and shows in the form of table below:We take r as the argument, and T max is the dependent variable. Making the dots in the coordinate system by the mathematical software Origin[6].The figure is as follows:Figurebetween r and T max )To our surprise, we can find that there is nearlylinear relation between r and T max fromthe figure. We may fit the relation with a linear function, so we can get the function of r and T max .FigureThe function we are getting is: T max =-174.5r+291.5 (4)3.3.3 The relationship between n and rWe have introduced that the floor space is not the area of the pan itself in theT maxT maxIntroduction. So we can get the formula below of the area of round square pan and r:S=r2(4−π)+0.085We have already known the floor space of the oven. So we can know the maximum number of pans that the oven can hold , in which condition the shape of pan is sure. Above all, we can get the function of n and r.n= 1.21r2(4−π)+0.085(5) 3.3.3 The optimum solutionThe weight of the number of pans which the oven can hold is P, while the weight of heat distribution is (1-P). The dimensions of the two are different. The effects are also different with the unit change of r. Depending on the message above, in order to induce the weight P. we need to eliminate their dimensions[7].Through observing the figure 23, we can know that the range of T max is 27℃with the domain of r.Then we change the value of r in its domain of definition, then we can get n's approximate range: 3.05mThen we eliminate their dimension, so they are transformed into value which could be compared.They are T max/27 and n/3.5 respectively.We hope that we can get a smaller T max and a larger n. We let T max/27 multiply by -1, then add them (T max/27 and n/3.5) together, the final result is K. K has no practical significance, and we just want to know its relative value.K=-(1-P)T max/27+P n/3.5After simplification, we can obtain:K=-(1-P)(-6.463r+10.796)+0.345Pr2(4−π)+0.085(6)How to get the pan we want with the idealized shape and its corresponding width to length ratio of the oven by using this formula?We explain it by an exampleIf some one’s ideal weight P is 0.6, the function(6) of K becomes:K=-0.4(-6.463r+10.796)+0.207r2(4−π)+0.085We can get the figure of K within the domain of r, by using the matlab. The figure is presented as follow:Figure 24(the relationship between K and r)We plugged the value of r into the equation (5), then we can get n=13.76Because the number of the pan should be an integer, we round up n, namely n=13.The largest number of pans which the oven can contain is prime number, the oven has only a width to length ratio of 1/13.So at last, we get the following conclusion:When P=0.6, n=13, W/L=1/13, r=0.058m is the best solution.To make the coefficient of oven reaches the top (namely without space), we take the radius of round square into equation (5). We should notice that n must be an integer. We adopt the method of exhaustion, and the result shows in the table below:This table will be used in the following advertizing.3.3.4 Model verificationWe can see the equation (6) , when the P tends to 0, which means the largest number of pans the oven can contain make no sense, the equation becomes: K=- (-6.463r+10.796) ; the optimum solution is r=0.164488m. which means maximize even distribution of heat for the pan is most important.On the contrary, when the P tends to 1, which means the maximize even distribution of heat for the pan make no sense, the equation becomes: K =0.345r 2(4−π)+0.085; the optimum solution is r=0m. which means the largest number of pans the oven can contain is most The value of r for thecorresponding peak valuein the figure is 0.058m。

2013年美赛数学建模b题的分析

Байду номын сангаас
美国2025年可提供的实际淡水量怎样预测？
（可以先预测不同地区不同领域供水量，再相加）
如果实际淡水量小于需求量，可通过那些方法提供不足的淡水？（例如：海水淡化、不同地区淡水转移）
这些方法那个更好？（比较的依据是什么？例如费用，环境）需要做出那些假设？
1. 从现在到2025年之前不会有战争，自然灾害等影响因素 2.同一个地区的居民年用水量相同 3.同一个地区的居民量的年增长率为常数 4.不同地区海水淡化的单位费用相同 5.不同地区淡水转移的单位费用相同
注：模型假设一般是在问题分析中根据需要提出的，所以可以先提出基本假设，以后再补充
Vx pre S
1.预测2025年供水量的模型（以一个州的降水量为例）
其中，表示州i 的可用降水量，表示该州的单位面积平均年降水量， Si 表示该州的淡水地表面积
Vx pre S
2.预测2025年需水量的模型（以一个州的居民用水量为例）
若 V大于0，则说明2025年美国的淡水量是足够的，然后再看每个州需要补充的淡水量，即 V V ，
si ri
若每个州需要补充的淡水量均大于0，则不需要采取任何措施，
否则，需要在不通州之间进行淡水调度。
若 V小于0，则说明2025年美国的淡水量是不足的，需要采取海水淡化等措施
注：为使表达更清晰，可以在论文中采用图、表等形式
其中，表示州i 在2025年人口数量，表示该州在2010年人口数量，表示该州的人口平均年增长率则该州在2025年的居民用水量为
i Vri w N 2 0 0 5
其中，w 表示该州居民的平均年用水量
3. 2025年美国需要补充的淡水量

2013美赛结果分析

2013年美国大学生数学建模竞赛结果发布COMAP非常高兴地宣布第29届大学生数学建模竞赛（MCM）结果。

今年共有5636支队伍参加了比赛，分别代表14个国家和地区。

以下11支队伍提交的论文被评定为优胜论文（OUTSTANDING WINNERS）：Beijing Univ. of Posts and Telecomm, China（北京邮电大学：郭众鑫、吴帆、王蓓丹；指导教师：贺祖国）Bethel University, Arden Hills, MNColorado College, Colorado Springs, COFudan University, China（复旦大学：王坤睿、许晶、曾溦；指导教师：杨翎）Nanjing University, China（南京大学：陈炜、刘威志、杨岑莹；指导教师：瞿慧）Peking University, China（北京大学：金冲、刘博闻、吴蒙; 指导教师：刘旭峰）Shandong University, China（山东大学：宋炎侃、徐珂、伊凡；指导教师：Hengxu Zhang）Shanghai Jiaotong University, China（上海交通大学:文理斌、吴婧元、王聪; 指导教师：Yuehui Zhang）Tsinghua University, China（清华大学: 高鹏飞、何博硕、邹天忻；指导教师：吴昊）University of Colorado Boulder, Boulder, CO (2)今年的竞赛时间是从2013年1月31日（星期四）到2013年2月4日（星期一）。

在这段时间里，由三名学生组成的本科生或高中生队伍从两个竞赛问题中选择一个，认真研究并建模，最终提交一份解决方案。

今年MCM的主要形式通过网络展开。

参赛队伍需要在规定的时间内通过COMAP的MCM网站注册、获得竞赛材料并下载题目和数据。

今年MCM的两个问题被公认为具有很大的挑战性。

美国大学生数学建模竞赛2013 获奖论文

Team #111111
Page 3 of 22
General Assumptions
The heat can only transfer to the pan from its outer edge through the air. Since the food placed on it prevents the heat from conducting to it, this is a reasonable. The temperature in oven is even since the air is flowing. That there is only one kind of pans in the ovens. Initially there are two racks in the oven, evenly spaced. We suppose that the temperature and heat are equivalent and constant, so we just considerate one rack and the other one is the same with it The ratio of the oven plane’s width and length is W/L. Every pan shares the same area of A. The data we cited in the models are true. The area of the oven is S 750cm 2 , and the ratio between width and length is W / L 22 : 34 . Moreover, the area of pan is A 100cm2 .[2]

【数学建模2013美赛特等奖】14297

SummaryLeaf shape plasticity has long been a heated topic in the field of morphology and taxonomy. Although numerous researchers have been involved in the study of how the shape formation is controlled by genes and environmental stress, little has been done to address the interactions between different plant organs. In this paper, we approach this issue in a creative way: starting from isolated parts (leaves), building connections (leaf-leaf, leaf-branch) and drawing the big picture of the entire system (tree).The mathematical model introduced in this paper describes morphological, physiological and partially ecological properties of the vegetations. First and foremost, relative errors of the numerical outputs are limited within 7.5% and all the values predicted are within the right order of magnitude. Moreover, biological intuition is presented in the consistent manner. For instance, the dynamic positive feedback is applied to describe leaf growth pattern and transportation process. It is also worth noticing that a much simple and flexible algorithm is achieved by executing ingenious geometric strategy.Interestingly findings were generated in the procedures of solving the problems. The seemly unrelated bio-phenomena could be explained when morphological, physiological and partially ecological interactions were taken into account. To illustrate, as we conducted the leaf mass estimation, it could give us a rough picture of leaf-leaf, leaf-branch, and leaf-tree relations.Dear editor,The topic of this paper is about mathematical modeling of tree leaves as well as how leaf traits impose influence on the tree system.In this paper, we addressed the morphological, physiological and partially ecological properties of the vegetations by applying a series of modeling process to related issues. The performance of our model in the sensitivity test satisfied our expectation. Different from the previous research, we approached this issue in a creative way: starting from isolated parts (leaves), building connections (leaf-leaf, leaf-branch) and drawing the big picture of the entire system (tree).Interestingly, the seemly unrelated bio-phenomena could be explained when morphological, physiological and partially ecological interactions were taken into account. Properties of the individual leaves are closely related to the characteristics of the entire tree system.We are looking forward to your comments and revision. Hopefully, our paper could be published in this Journal.Your sincerelyIntroductionThe mechanism of plant shape plasticity has long been a topic of numerous researches in the field of morphometrics. And leaves have been always favored as the subject because of its two-dimensionality and relative simplicity. The change of leaf shapes are generally regarded as both a genetically determination (Howell, 1998) and results of structural optimization to environmental stress (Hemsley and Poole, 2004). However, little attention is drawn to the interactions between different plant organs (e.g. leaf, branch) and the influence from the system as an entity. To fill this gap, we construct a mathematical model to interpret such in-leaf communications, and perform our analysis on simple leaves.As is shown in some literatures (Stern et al., 2008), simple leaves are classified into different shape catalogs (Figure 1), such as flabellate leaves (e.g. ginkgo), palmate leaves (e.g. maple), and cordate leaves (e.g. Hibiscus tiliaceus). The major criteria of classification are the shape apex, base, and margin of leaves.Figure 1 Shape of LeavesLeaf venation is also a critical component of leaf types. There are three major patterns of major vein organization: pinnate, palmate, and parallel, as are shown in Figure 2.Figure 2 Leaf VenationBesides the diversification in leaf shapes and venation, arrangement of leaves on a stem (phyllotaxy) are generally in three distinct patterns (Figure 3), which are defined by the number of leaves attached to a single node. In most species, leaves are attached alternately or in a spiral along a stem, with one leaf per node. This is called an alternate arrangement. If two leaves are attached at each node, they provide an opposite arrangement. Leaves are whorled when three or more occur at a node. A special case in opposite arrangement is called decussate, if successive pairs are orientated at 90o to each other.Figure 3 Leaf ArrangementReview of previous researches shows that most models developed are based on statistical results (Greig-Smith, 1983). And so far, no model is built to indicate the effect of plant organ interactions (leaf-leaf, leaf-branch) on leaf shapes. The mathematical model presented by us demonstrates the logic of plant-environment and in-plant communications at physiology, inter-organ, and tree system levels. We first。

2013美国大学生数学建模竞赛

1.热力学模型：
1.1热力学内容分析：主要考虑热传导问题，热辐射。

更深入还需考虑：蛋糕和盘子之间的热量传递，烤盘之间热量传递的量化分析，参数计算等等。

1.2过度形状的如何优化选择
1.3偏微分方程的解法（这里要考虑到过度形状的选择），主要软件有annsy，pde等等。

对偏微分方程数值解的误差分析，收敛性分析证明等等。

1.4均匀化程度的量化指标，以及如何改进软件实现各种过度形状的并行计算。

2.平面装箱问题（这里的处理需要联系到热力学模型的烤盘间热量传递分析）：
2.1.矩形的装箱处理（这里估计需要简化，如何实现一般的横竖放置处理是一个难点）
2.2.椭圆形的装箱处理
2.3,过度形状处理
3.组合权重模型主要是解决两个核心问题：
3.1考虑权重后的一般评价函数的确定（考虑非线性处理）
3.2不同烤箱形状对最终结果影响的一般分析。

4.其他
4.1关于2和3的灵敏度分析（注意到关于烤箱形状，N未必一定是稳定的。

因此如何对其作出一般的灵敏度考量是一个关键问题）
4.2广告的处理（主要问题是广告的一般性主旨是宣传，但是又要加上模型的结果但是不能直接标出参数，这两个之间的权衡也十分重要）。

2013全美数学建模题目

2013 Contest ProblemsMCM PROBLEMSPROBLEM A: The Ultimate Brownie PanWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven.Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.PROBLEM B: Water, Water, EverywhereFresh water is the limiting constraint for development in much of the world. Build a mathematical model for determining an effective, feasible, and cost-efficient water strategy for 2013 to meet the projected water needs of [pick one country from the list below] in 2025, and identify the best water strategy. In particular, your mathematical model must address storage and movement; de-salinization; and conservation. If possible, useyour model to discuss the economic, physical, and environmental implications of your strategy. Provide a non-technical position paper to governmental leadership outlining your approach, its feasibility and costs, and why it is the “best water strategy choice.”Countries: United States, China, Russia, Egypt, or Saudi ArabiaICM PROBLEMPROBLEM C: Network Modeling of Earth's HealthClick the title below to download a PDF of the 2013 ICM Problem. Your ICM submission should consist of a 1 page Summary Sheet and your solution cannot exceed 20 pages for a maximum of 21 pages.Network Modeling of Earth's Health© 2013 COMAP, The Consortium for Mathematics and Its Applications May be reproduced for academic/research purposesFor More information on COMAP and this projectvisit 。