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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。
summaryOur solution paper mainly deals with the following problems:·How to measure the distribution of heat across the outer edge of pans in differentshapes and maximize even distribution of heat for the pan·How to design the shape of pans in order to make the best of space in an oven·How to optimize a combination of the former two conditions.When building the mathematic models, we make some assumptions to get themto be more reasonable. One of the major assumptions is that heat is evenly distributedwithin the oven. We also introduce some new variables to help describe the problem.To solve all of the problems, we design three models. Based on the equation ofheat conduction, we simulate the distribution of heat across the outer edge with thehelp of some mathematical softwares. In addition, taking the same area of all the pansinto consideration, we analyze the rate of space utilization ratio instead of thinkingabout maximal number of pans contained in the oven. What’s more, we optimize acombination of conditions (1) and (2) to find out the best shape and build a function toshow the relation between the weightiness of both conditions and the width to lengthratio, and to illustrate how the results vary with different values of W/L and p.To test our models, we compare the results obtained by stimulation and our models, tofind that our models fit the truth well. Yet, there are still small errors. For instance, inModel One, the error is within 1.2% .In our models, we introduce the rate of satisfaction to show how even thedistribution of heat across the outer edge of a pan is clearly. And with the help ofmathematical softwares such as Matlab, we add many pictures into our models,making them more intuitively clear. But our models are not perfect and there are someshortcomings such as lacking specific analysis of the distribution of heat across theouter edge of a pan of irregular shapes. In spite of these, our models can mainlypredict the actual conditions, within reasonable range of error.For office use onlyT1 ________________T2 ________________T3 ________________T4 ________________ Team Control Number18674 Problem Chosen AFor office use only F1 ________________ F2 ________________ F3 ________________ F4 ________________2013 Mathematical Contest in Modeling (MCM) Summary Sheet(Attach a copy of this page to your solution paper.)Type a summary of your results on this page. Do not includethe name of your school, advisor, or team members on this page.The Ultimate Brownie PanAbstractWe introduce three models in the paper in order to find out the best shape for the Brownie Pan, which is beneficial to both heat conduction and space utility.The major assumption is that heat is evenly distributed within the oven. On the basis of this, we introduce three models to solve the problem.The first model deals with heat distribution. After simulative experiments and data processing, we achieve the connection between the outer shape of pans and heat distribution.The second model is mainly on the maximal number of pans contained in an oven. During the course, we use utility rate of space to describe the number. Finally, we find out the functional relation.Having combined both of the conditions, we find an equation relation. Through mathematical operation, we attain the final conclusion.IntroductionHeat usage has always been one of the most challenging issues in modern world. Not only does it has physic significance, but also it can influence each bit of our daily life. Likewise,space utilization, beyond any doubt, also contains its own strategic importance. We build three mathematic models based on underlying theory of thermal conduction and tip thermal effects.The first model describes the process and consequence of heat conduction, thus representing the temperature distribution. Given the condition that regular polygons gets overcooked at the corners, we introduced the concept of tip thermal effects into our prediction scheme. Besides, simulation technique is applied to both models for error correction to predict the final heat distribution.Assumption• Heat is distributed evenly in the oven.Obviously, an oven has its normal operating temperature, which is gradually reached actually. We neglect the distinction of temperature in the oven and the heating process, only to focus on the heat distribution of pans on the basis of their construction.Furthermore, this assumption guarantees the equivalency of the two racks.• Thermal conductivity is temperature-invariant.Thermal conductivity is a physical quantity, symbolizing the capacity of materials. Always, the thermal conductivity of metal material usually varies with different temperatures, in spite of tiny change in value. Simply, we suppose the value to be a constant.• Heat flux of boundaries keeps steady.Heat flux is among the important indexes of heat dispersion. In this transference, we give it a constant value.• Heat conduction dom inates the variation of temperature, while the effects ofheat radiation and heat convection can be neglected.Actually, the course of heat conduction, heat radiation and heat convectiondecide the variation of temperature collectively. Due to the tiny influence of other twofactors, we pay closer attention to heat conduction.• The area of ovens is a constant.I ntroduction of mathematic modelsModel 1: Heat conduction• Introduction of physical quantities:q: heat fluxλ: Thermal conductivityρ: densityc: specific heat capacityt: temperature τ: timeV q : inner heat sourceW q : thermal fluxn: the number of edges of the original polygonsM t : maximum temperaturem t : minimum temperatureΔt: change quantity of temperatureL: side length of regular polygon• Analysis:Firstly, we start with The Fourier Law:2(/)q gradt W m λ=- . (1) According to The Fourier Law, along the direction of heat conduction, positionsof a larger cross-sectional area are lower in temperature. Therefore, corners of panshave higher temperatures.Secondly, let’s analyze the course of heat conduction quantitatively.To achieve this, we need to figure out exact temperatures of each point across theouter edge of a pan and the variation law.Based on the two-dimension differential equation of heat conduction:()()V t t t c q x x y yρλλτ∂∂∂∂∂=++∂∂∂∂∂. (2) Under the assumption that heat distribution is time-independent, we get0t τ∂=∂. (3)And then the heat conduction equation (with no inner heat source)comes to:20t ∇=. (4)under the Neumann boundary condition: |W s q t n λ∂-=∂. (5)Then we get the heat conduction status of regular polygons and circles as follows:Fig 1In consideration of the actual circumstances that temperature is higher at cornersthan on edges, we simulate the temperature distribution in an oven and get resultsabove. Apparently, there is always higher temperature at corners than on edges.Comparatively speaking, temperature is quite more evenly distributed around circles.This can prove the validity of our model rudimentarily.From the figure above, we can get extreme values along edges, which we callM t and m t . Here, we introduce a new physical quantity k , describing the unevennessof heat distribution. For all the figures are the same in area, we suppose the area to be1. Obviously, we have22sin 2sin L n n n ππ= (6) Then we figure out the following results.n t M t m t ∆ L ksquare 4 214.6 203.3 11.3 1.0000 11.30pentagon 5 202.1 195.7 6.4 0.7624 8.395hexagon 6 195.7 191.3 4.4 0.6204 7.092heptagon 7 193.1 190.1 3.0 0.5246 5.719octagon 8 191.1 188.9 2.2 0.4551 4.834nonagon 9 188.9 187.1 1.8 0.4022 4.475decagon 10 189.0 187.4 1.6 0.3605 4.438Table 1It ’s obvious that there is negative correlation between the value of k and thenumber of edges of the original polygons. Therefore, we can use k to describe theunevenness of temperature distribution along the outer edge of a pan. That is to say, thesmaller k is, the more homogeneous the temperature distribution is.• Usability testing:We use regular hendecagon to test the availability of the model.Based on the existing figures, we get a fitting function to analyze the trend of thevalue of k. Again, we introduce a parameter to measure the value of k.Simply, we assume203v k =, (7) so that100v ≤. (8)n k v square 4 11.30 75.33pentagon 5 8.39 55.96hexagon 6 7.09 47.28heptagon 7 5.72 38.12octagon 8 4.83 32.23nonagon9 4.47 29.84 decagon 10 4.44 29.59Table 2Then, we get the functional image with two independent variables v and n.Fig 2According to the functional image above, we get the fitting function0.4631289.024.46n v e -=+.(9) When it comes to hendecagons, n=11. Then, v=26.85.As shown in the figure below, the heat conduction is within our easy access.Fig 3So, we can figure out the following result.vnActually,2026.523tvL∆==.n ∆t L k vhendecagons 11 187.1 185.8 1.3 0.3268 3.978 26.52Table 3Easily , the relative error is 1.24％.So, our model is quite well.• ConclusionHeat distribution varies with the shape of pans. To put it succinctly, heat is more evenly distributed along more edges of a single pan. That is to say, pans with more number of peripheries or more smooth peripheries are beneficial to even distribution of heat. And the difference in temperature contributes to overcooking. Through calculation, the value of k decreases with the increase of edges. With the help of the value of k, we can have a precise prediction of heat contribution.Model 2: The maximum number• Introduction of physical quantities:n: the number of edges of the original polygonsα: utility rate of space• Analysis:Due to the fact that the area of ovens and pans are constant, we can use the area occupied by pans to describe the number of pans. Further, the utility rate of space can be used to describe the number. In the following analysis, we will make use of the utility rate of space to pick out the best shape of pans. We begin with the best permutation devise of regular polygon. Having calculated each utility rate of space, we get the variation tendency.• Model Design:W e begin with the scheme which makes the best of space. Based on this knowledge, we get the following inlay scheme.Fig 4Fig 5According to the schemes, we get each utility rate of space which is showed below.n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 shape square pentagon hexagon heptagon octagon nonagon decagon hendecagon utility rate(％)100.00 85.41 100.00 84.22 82.84 80.11 84.25 86.21Table 4Using the ratio above, we get the variation tendency.Fig 6 nutility rate of space• I nstructions:·The interior angle degrees of triangles, squares, and regular hexagon can be divided by 360, so that they all can completely fill a plane. Here, we exclude them in the graph of function.·When n is no more than 9, there is obvious negative correlation between utility rate of space and the value of n. Otherwise, there is positive correlation.·The extremum value of utility rate of space is 90.69％,which is the value for circles.• Usability testing:We pick regular dodecagon for usability testing. Below is the inlay scheme.Fig 7The space utility for dodecagon is 89.88％, which is around the predicted value. So, we’ve got a rather ideal model.• Conclusion:n≥), the When the number of edges of the original polygons is more than 9(9 space utility is gradually increasing. Circles have the extreme value of the space utility. In other words, circles waste the least area. Besides, the rate of increase is in decrease. The situation of regular polygon with many sides tends to be that of circles. In a word, circles have the highest space utility.Model 3: Rounded rectangle• Introduction of physical quantities:A: the area of the rounded rectanglel: the length of the rounded rectangleα: space utilityβ: the width to length ratio• Analysis:Based on the combination of consideration on the highest space utility of quadrangle and the even heat distribution of circles, we invent a model using rounded rectangle device for pans. It can both optimize the cooking effect and minimize the waste of space.However, rounded rectangles are exactly not the same. Firstly, we give our rounded rectangle the same width to length ratio (W/L) as that of the oven, so that least area will be wasted. Secondly, the corner radius can not be neglected as well. It’ll give the distribution of heat across the outer edge a vital influence. In order to get the best pan in shape, we must balance how much the two of the conditions weigh in the scheme.• Model Design:To begin with, we investigate regular rounded rectangle.The area224r ar a A π++= (10) S imilarly , we suppose the value of A to be 1. Then we have a function between a and r :21(4)2a r r π=+--(11) Then, the space utility is()212a r α=+ (12) And, we obtain()2114rαπ=+- (13)N ext, we investigate the relation between k and r, referring to the method in the first model. Such are the simulative result.Fig 8Specific experimental results arer a ∆t L k 0.05 0.90 209.2 199.9 9.3 0.98 9.49 0.10 0.80 203.8 196.4 7.4 0.96 7.70 0.15 0.71 199.6 193.4 6.2 0.95 6.56 0.20 0.62 195.8 190.5 5.3 0.93 5.69 0.25 0.53 193.2 189.1 4.1 0.92 4.46Table 5According to the table above, we get the relation between k and r.Fig 9So, we get the function relation3.66511.190.1013r k e -=+. (14) After this, we continue with the connection between the width to length ratioW Lβ=and heat distribution. We get the following results.krFig 10From the condition of heat distribution, we get the relation between k and βFig 11And the function relation is4.248 2.463k β=+ (15)Now we have to combine the two patterns together:3.6654.248 2.463(11.190.1013)4.248 2.463r k e β-+=++ (16)Finally, we need to take the weightiness (p) into account,(,,)()(,)(1)f r p r p k r p βαβ=⋅+⋅- (17)To standard the assessment level, we take squares as criterion.()(,)(1)(,,)111.30r p k r p f r p αββ⋅⋅-=+ (18) Then, we get the final function3.6652(,,)(1)(0.37590.2180)(1.6670.0151)1(4)r p f r p p e rββπ-=+-⋅+⋅++- (19) So we get()()3.6652224(p 1)(2.259β 1.310)14r p f e r r ππ--∂=-+-+∂⎡⎤+-⎣⎦ (20) Let 0f r∂=∂,we can get the function (,)r p β. Easily,0r p∂<∂ and 0r β∂>∂ (21) So we can come to the conclusion that the value of r decreases with the increase of p. Similarly, the value of r increases with the increase of β.• Conclusion:Model 3 combines all of our former analysis, and gives the final result. According to the weightiness of either of the two conditions, we can confirm the final best shape for a pan.• References:[1] Xingming Qi. Matlab 7.0. Beijing: Posts & Telecom Press, 2009: 27-32[2] Jiancheng Chen, Xinsheng Pang. Statistical data analysis theory and method. Beijing: China's Forestry Press, 2006: 34-67[3] Zhengshen Fan. Mathematical modeling technology. Beijing: China Water Conservancy Press, 2003: 44-54Own It NowYahoo! Ladies and gentlemen, please just have a look at what a pan we have created-the Ultimate Brownie Pan.Can you imagine that just by means of this small invention, you can get away of annoying overcookedchocolate Brownie Cake? Pardon me, I don’t want to surprise you, but I must tell you , our potential customers, that we’ve made it! Believing that it’s nothing more than a common pan, some people may think that it’s not so difficult to create such a pan. To be honest, it’s not just a simple pan as usual, and it takes a lot of work. Now let me show you how great it is. Here we go!Believing that it’s nothing more than a common pan, some people may think that it’s not so difficult to create such a pan. To be honest, it’s not just a simple pan as usual, and it takes a lot of work. Now let me show you how great it is. Here we go!Maybe nobody will deny this: when baked in arectangular pan, cakes get easily overcooked at thecorners (and to a lesser extent at the edges).But neverwill this happen in a round pan. However, round pansare not the best in respects of saving finite space in anoven. How to solve this problem? This is the key pointthat our work focuses on.Up to now, as you know, there have been two factors determining the quality of apan -- the distribution of heat across the outer edge of and thespace occupied in an oven. Unfortunately, they cannot beachieved at the same time. Time calls for a perfect pan, andthen our Ultimate Brownie Pan comes into existence. TheUltimate Brownie Pan has an outstandingadvantage--optimizing a combination of the two conditions. As you can see, it’s so cute. And when you really begin to use it, you’ll find yourself really enjoy being with it. By using this kind of pan, you can use four pans in the meanwhile. That is to say you can bake more cakes at one time.So you can see that our Ultimate Brownie Pan will certainly be able to solve the two big problems disturbing so many people. And so it will! Feel good? So what are you waiting for? Own it now!。
美赛数学建模A题翻译版论文The document was finally revised on 2021数学建模竞赛(MCM / ICM)汇总表基于细胞的高速公路交通模型自动机和蒙特卡罗方法总结基于元胞自动机和蒙特卡罗方法,我们建立一个模型来讨论“靠右行”规则的影响。
首先,我们打破汽车的运动过程和建立相应的子模型car-generation的流入模型,对于匀速行驶车辆，我们建立一个跟随模型,和超车模型。
然后我们设计规则来模拟车辆的运动模型。
我们进一步讨论我们的模型规则适应靠右的情况和,不受限制的情况, 和交通情况由智能控制系统的情况。
我们也设计一个道路的危险指数评价公式。
我们模拟双车道高速公路上交通(每个方向两个车道,一共四条车道),高速公路双向三车道（总共6车道)。
通过计算机和分析数据。
我们记录的平均速度,超车取代率、道路密度和危险指数和通过与不受规则限制的比较评估靠右行的性能。
我们利用不同的速度限制分析模型的敏感性和看到不同的限速的影响。
左手交通也进行了讨论。
根据我们的分析,我们提出一个新规则结合两个现有的规则(靠右的规则和无限制的规则)的智能系统来实现更好的的性能。
1介绍术语假设2模型设计的元胞自动机流入模型跟随模型超车模型超车概率超车条件危险指数两套规则CA模型靠右行无限制行驶规则3补充分析模型加速和减速概率分布的设计设计来避免碰撞4模型实现与计算机5数据分析和模型验证平均速度快车的平均速度密度超车几率危险指数6在不同速度限制下敏感性评价模型7驾驶在左边8交通智能系统智能系统的新规则模型的适应度智能系统结果9结论10优点和缺点优势弱点引用附录。
1 Introduction今天,大约65%的世界人口生活在右手交通的国家和35%在左手交通的国家交通流量。
[worldstandards。
欧盟,2013] 右手交通的国家,比如美国和中国,法规要求驾驶在靠路的右边行走。
多车道高速公路在这些国家经常使用一个规则,要求司机在最右边开车除非他们超过另一辆车,在这种情况下,他们移动到左边的车道、通过，返回到原来的车道。
终极布朗尼烤烤盘一、摘要根据题意，我们把把要解决的分成三个问题；第一个就是建立一个模型来表示整个烤盘的外边缘热量的分布。
第二个就是优化组合题目中条件1和条件2，使得权重p和(1- p)能够描述随着W/L和p值的改变，最佳的烤烤盘形状和热量分布情况是如何改变的第三个问题就是为布朗尼美食家杂志准备一到两页的宣传广告，需要突出设计和结果。
对于第一个问题，我们结合傅里叶定律构建了二维热传导模型；然后通过模型中的S来限定范围得到六种不同形状烤盘对应的热传导偏微分方程。
然后对模型赋值和第二类边界条件（Neumann边界条件）下，应用comsol得出六种烤盘稳定热量分布图像和烤盘外边缘热量分布图像。
通过输出的图像，我们得出结论：矩形四角处温度较高，圆形外边缘热量分布比较均匀；随着烤盘边数的增加，烤盘外边缘热量分布愈加均匀，但在角处温度仍然会高一些对于问题二对于问题三关键词：二、问题重述当用一个长方形的平底烤盘（盘）烘烤时，热量被集中在4个角，在角落处，食物可能被烤焦了，而边缘处烤的不够熟。
在一个圆形的平底烤盘（盘）热量被均匀地分布在整个外边缘，在边缘处食物不会被烤焦。
但是，大多数的烤箱的形状是矩形的，采用了圆形的烤盘（盘）相对于烤箱的使用空间而言效率不高。
为所有形状的烤盘（盘）----包括从矩形到圆形以及中间的形状，建立一个模型来表示整个烤盘（盘）的外边缘热量的分布。
假设：1. 形状是矩形的烤箱宽长比为W/L；2. 每个烤烤盘（盘）的面积为A；3. 每个烤箱最初只有两个均匀放置的烤架。
根据以下条件，建立一个能使用的最佳类型或形状的烤烤盘（盘）：1.放入烤箱里的烤烤盘（盘）数量的最大值为(N)；2.烤烤盘（盘）的平均分布热量最大值为(H)；3.优化组合条件1和条件2，使得权重p和(1- p)能够描述随着W/L和p值的改变，最佳的烤烤盘形状和热量分布情况是如何改变的。
除了完成规定的解决方案，为布朗尼美食家杂志准备一到两页的宣传广告，需要突出你的设计和结果。
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。
中国水资源战略摘要Summary为了确定中国最佳的水资源战略，将中国分为九大流域，首先借助MATLAB建立多项式拟合模型来预测出中国2013年到2025年每年各流域的供水量和需水量，接着在可持续发展的原则指导下建立区域水资源合理配置模型，对每一个流域，采用水资源综合短缺度最小为目标函数, 对地表水、地下水等多种水源统筹考虑, 用权重区别对待工业、农业、生活、生态环境等不同领域的用水需求, ,从而求出各个流域最小的缺水量。
再根据前面的两个模型所预测出来的各流域的缺水量，建立最佳的补水模型解决缺水问题：通过对实际问题的分析，可能的补水方案有两个：方案一是直接从珠江流域调水到缺水的流域，方案二是沿海流域采取海水淡化补水，内陆流域采取直接从珠江流域调水过去，经过分析、计算发现方案二是最佳的。
最后，我们统筹考虑我们所制定的水策略，发现其无论是对经济、社会还是生态环境都将产生重大影响。
In order to determine the best water resources strategy, we divided China into nine basins. Firstly, we established polynomial fitting model with the use of MATLAB to predict the water supply and the water demand of every basin from 2013 to 2025. Secondly, we established the regional water resources rational allocation model under the guidance of the principle of sustainable development. In this model, through taking the minimum comprehensive water shortage degree as objective , surface water , groundwater and other water are considered, and different weightings are used for industrial, agricultural, domestic and ecological water users in order to realize regional water resources rational allocation .In this way can we obtained the minimum amount of water scarcity in every basin. Thirdly, according to the data predicted based on the previous two models, we can establish the optimal replenishment model to solve the problem of water shortage. We identified two possible replenishment program based on the analysis of the actual problems. One is to transfer the water of the Pearl River to basins where lack of water resources, another is to transfer the water of the Pearl River to inland basins directly while we meet the water shortage of coastal basins by desalination. After analysis and calculation, we find second program is the best. Finally, we find the water strategy we developed has a significant impact on the economic, social and ecological environment after we considered the models we established.关键字：水策略多项式拟合模型区域水资源合理配置模型补水模型Keywords：Water strategythe Polynomial fitting modelThe Regional water resources rational allocation modelthe Replenishment model§1.问题重述Problem restatement水是生命之源, 是人类生存和发展不可替代的资源, 是经济、社会可持续发展的基础。
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。
PROBLEM A: The Ultimate Brownie Pan问题A：终极布朗尼蛋糕盘When 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).当蛋糕在矩形盘里加热时，热量会集中在蛋糕的四个角导致加热过度（在较小的边缘处也有这种情况）。
In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges.在圆形的盘子里，热量均匀的分布在蛋糕表面，所以不会加热过度。
However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven.然而，由于大多数的烤箱是矩形的，使用的盘子是圆形的，不能有效的利用烤箱里的空间。
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.建立模型来显示在使用不同形状的盘子时（方形，圆形和两者之间的其他形状），热量在其表面的分布情况。
Assume假设1. A width to length ratio of W/L for the oven which is rectangular in shape.1，烤箱是矩形的，长为l，宽为W。
PROBLEM 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 A: 终极布朗尼锅当在一个矩形的锅里烹煮食物时，受热集中在锅的4个角落里，因此食品在这4个拐角处被过度烹饪（在边缘程度会稍微轻点）。
Modeling and Simulation of the heat conduction of pans and food in oven by using finite element methodAbstractWe build a mathematical model to show the heat distribution in the pan and select the best type of Brownie Pan for the oven.For the First Model, we use Three-dimensional modeling software to create a virtual oven and discuss distribution of heat in the oven. Then we use finite element to simulate the heat distribution of the pan in three-dimensional space. Through the simulation we find the amount of heat pass through a layer of pan is similar. Based on the Fourier’s Law and Neumann boundary condition we develop a model to show the distribution of heat across the outer edge of pans. Last, we use 6 different types of pans to solve and test our model. We also extract the data from the testing result and analyze the maximum temperature, average temperature and dispersion degree of different kinds of pans to discuss the difference between them.For the Second Model, according to the demand of the Question, we build a optimization objective function about the shape of the pan, the ratio of W L of oven and the weight value p to select the best type of the pan. To solve /the model , first , we simulate how the distribution of heat vary with the quantities of the pan,which depend on the ratio of /W L with the method of numercial modeling and use MATLAB PDE as a auxiliary tool. Then take p into account, we analyze the relationship of these three variables and statistic how N and H vary with p and calculate the best type of the pan. Finally we select a best type of the pan, its shape is resemble to a track.At last , we combine our model and reality to show why our conclusion is best . Then we design a scheme to initiate our conclusion.Keywords: Heat Conduction; Pan; Oven; Finite Element Method; PDE1. IntroductionWe tackle two main problems:∙Develop a model to show the distribution of heat across the outer edge of a pan of different shapes from rectangular to circular under the basic assumption.∙Develop a model that can be used to select the best shape of pan under certain conditions.For the First Task, we focus primarily on the most conspicuous characteristic of pan, namely, shape. We use the knowledge of heat transfer to solve the problem. Because the temperature of pan and food are almost equal at the outer edge, thus we can discuss the heat distribution of food, such as bread, to show the heat distribution of pan. First,we use finite element theory to simulate the distribution of heat in the oven and discuss the heat distribution of different types of pans. With the help of Fourier’s Law and Partial Differential Equation we can build a two-dimensional heat conduction model. Then we set values to the parameter of food and use MATLAB PDE tool to simulate the temperature field. Next, we use 6 different types of pans with the same area to solve and test the model. Finally we statistic all the data and do some analysis. Such as comparing the maximum temperature, average temperature and dispersion degree of each pan.For the Second Task, to build a optimization function about three variables (the shape of the pan, the ratio of /W L, the weight value p), we use the thoughts of the method of Successive approximation and Stepwise. First, we discuss the influence of the shape of oven (namely the ratio of /W L) to the maximum numbers of pans can fit in the oven. Next, we analyze the effect of the numbers of pan N on the even distribution of heat H. Finally, we introduce a weight value p, and consider the relatioship of these three variables to build the objective function . When solving this model, we simplify it through the solution that regarding a variable is constant firstly , discuss the other two variables,then introduce another variable to analyze the effects on each other of three variables.To solve the influence of the shape of pan and the ratio of /W L to the maximum numbers of pans can fit in the oven, we build a simple integer programming formulation, then we use LINGO to solve it. In the model of the even distribution of heat ,we make use of the idea of the rule of radiation during the space and help us to analyze this model. For the optimization objective function, we simplify it to p is a constant, discuss the distributionof heat and the quantities of the pan vary with the ratio of /W L. In next step, we statistics how N and H vary with p and calculate the best type of the pan.2. Modeling and results2.1 Effects of the heating source of oven on the heat distribution2.1.1 Struction of ovenBased on the reality, we use CATIA (a Three-dimensional modeling software) to create a virtual oven. This oven has two radiation sources on the upper side and bottom in the oven. Other properties as the figure1showIn reality, the oven is heated by infrared radiation in the up and down sides. Here we give a broad outline of the oven.Figure 1 Structure diagram of oven2.1.2 Simulation of heat distribution of ovenThe heat transfer of oven mainly been realized by heat conduction of the air in oven and heat radiation of the heating sources. The heat convection in oven can been ignored. Generally, the heat conduction of the air has less influence than the heat radiation. In order to study the effect of heat conduction on the temperature of air in oven, we build a heat conduction model and solved the model by finite element method, which are completed by the Comsol software.For the heat conduction model, we only considerate two heating sources at the center of bottom and top plane of the oven which only fill air. The shape of the heating sources is specified to be a round plane and radius is 15cm. The shape of oven is specified to be a rectangular and the size of width, depth and height is 60cm, 50cm and 50cm, respectively. Thermal conductivety, density and heat capacity of air are set as 0.023W/(m.K), 1.49Kg/m^3 and 1,005 J/(kg.K), which is showed in table 1.The initial temperature of air is 293K. The six wall of oven is thermal insulation. The two heating sources are represented by two inflow heat flex of 40000W/m2, which means a Neumann boundary condition.Table 1 The properties of airThermal conductivety Density Heat capacity0.023W/(m.K) 1.49Kg/m^3 1,005 J/(kg.K)The simulation result of temperature distribution in the oven at 180s is showed in figure 2. From the figure 2, we see the high temperature mainly focus on the heating source area. The air temperature is low in the oven, which indicates the heat conduction by air is small.Figure 2 Temperature distribution of empty oven by air heat conductionThe results means the heat conduction of oven is not the main reason for raising the temperature of the oven. So, we need to discuss the heat radiation.2.1.3 Cosine radiator for a heat sourceAccording to the section 2.1.2, we know the heat radiation of two heating sources. In this section we will discuss how the heating sources influence the heat distribution.The heating sources can be considered as a consine radiator, which is showed in figure 1. To deduce the formula, several terminologies are introduced firstly. Radiation flux is a measurement of radiation power of emission transmission or receiving radiation. Radiation intensity is the size of the radiation flux in a particular direction referring to source. Irradiance represents per unit area to receive radiation through put density. As the radiation source in the up and down sides is identical. To simply the problem, we just consider the infrared radiation come from the upside and simplify the radiation source to a small sphere. Its radiation intensity is:d d I ωΦ=(1)Figure 3 Diagram of cosine radiatorIn the spherical coordinates, the direction can be remarked by ϕ and i .M eanwhile, radiation intensity I can be remarked by ϕ and i . So we can get radiation flux Φ .220000=(,)d (,)d (,)sin d d I i I i I i i i ππππϕωϕωϕϕΦ==⎰⎰⎰⎰⎰ (2)When the radiation intensity (,)I i ϕ is axial symmetry, namely(,)=()I i I i ϕwe can calculate radiation flux Φ by using equation (3).0=2()sin d I i i i ππΦ⎰ (3) By making definition, Irradiance is:d d E SΦ= (4)According to aboved assume, we simplify the radiation source to a small sphere, its radius is R and its radiation intensity 0I . The flux is 04I π, Spherical area is 24R π, we can receive the irradiance at the distance ofR002244I I E R R ππ==d S OFigure 4 The law of o blique illuminationIf the plane of illumination is skew to the radiation direction,the radiation intensity of O is 0I , the area of the plane of illumination is dS ,the distance between O and the plane of illumination is r , 2cos dw dS r θ=⋅, t he radiation through dw can be illustrated by :200cos dS d I I r dw θ⋅Φ==Therefore, the irradiance of the dS is02cos E I d dS rθ=Φ= This formula can be used to calculate the heat radiation energy of pan in an uneven distribution form.2.3 Heat distribution of pan with different shapeIn order to calculate the pan of heat distribution, the heat radiation of heating sources and the heat conduction of pan need to consideration.The material of pan chooses to be stainless steel. The steel properties are shown in the table 2. And the pan is putted on the upper racks. The types of pan include rectangular (W*D*H= 60*40*5 cm 3), circular (R= 27.6cm, H= 5cm) and elliptical (a= 30cm, b= 25.5cm, H= 5cm).Table 2 Material properties of the stainless steel panThermal conductivety Density Heat capacity Area 40.5W/(m.K) 7470Kg/m^3 500J/(kg.K) 0.24 2mThe heat conduction of steel pan has following equation:2222222()u u u u c t x y z ∂∂∂∂=++∂∂∂∂where c is a material parameter. The initial temperature of pan is 293K. The surface of the pan is heated by the heat radiation, which satisfy the equation of irradiance.The simulation results of different type pans are showed in figure 5. From the figure 5, we see the corners of pans are in the highest temperature points, and edges are lower than the corner point for rectangular. The edges of circle and ellipse pan shown higher heat than the bottom plane.From the temperature of the side plane of the six pans, we find the temperature is equal for the same high point on the planes in a same pan. This means the slice food in the pan get the same heat energy from the side plane. Rectangular: W*D*H=60*40*5cm 3The time of heating 30sThe time of heating 60sCircle: R=27.6cm, H=5cmThe time of heating 30sThe time of heating 60sEllipse: a=30,b=25.5The time of heating 30s The time of heating 60sFigure 5 Temperature distribution of pan at different heating timesIn order to get a quantitative analysis of the temperature distribution of different type pan, we analyze the maximum, minimum and range temperature of each pan. The result is showed in the figure 6.We use both range and dispersion degree to be the standard which shows theheat fluctuation and make comparison between them. As table shows belowFrom the table above we come to a conclusion that the pan can make the heat distribution evenly in each layer of the pan. Thus we can discuss the distribution of food in the pan with the help of Fourier’s Law and Neumann boundary condition.2.3 Heat distribution of food in panTask1Build a mathematical model to show the distribution of heat across the outer edge of different shapes of pan when heating the bread.Based on the knowledge of heat transfer we build a mathematical model to test different kinds of pans and show the distribution of heat across the outer edge. Basic ModelAccording to the section 2 we make these assumptions∙The metal material is distributed equally in the pan.∙The temperature of food and pan is similar at the outer edge.∙We use bread to be our foot material and regard the bread can take place of all food.∙The amount of heat which passes through each point on unit area in unit time is equal.∙We neglect the influence of pan to the heat field.∙The thermal conductivity of food is not change with the temperature of oven. ∙Initially bread is get from the refrigerator and the temperature is0℃.Step1: Application of Fourier’s Law and the Heat Conduction Equation We regard a piece of bread as a point D in the three dimensional space and use ()T f x y z t to describe the temperature of D, using ()=,,,x y z describe the,,,location of D. As figure1 showsFigure 1According to Fourier ’s Law [1], we can derive this heat conduction equation:()=-k ,,T dQ x y z dSdt n ∂∂ (1)WheredQ is rate of heat transfer in all direction by conduction, W ℃k is thermal conductivity, W/m ℃dS is area through which heat flows, m 2T is temperature, ℃n is length, variable, mWhen conduction occurs in all three coordinate directions, the balance equationcontains y - and z -derivatives analogous to the x -derivative. The balanceequation then becomes=++T k T k T k T vt x x y y z z ρ⎛⎫∂∂∂∂∂∂∂⎛⎫⎛⎫ ⎪ ⎪ ⎪∂∂∂∂∂∂∂⎝⎭⎝⎭⎝⎭Whereρis the density of bread . v is the special heat capacity of food.When k is constant, it can be taken outside the derivatives and when=k c v ρThe equation can be represented as 2222222=++T T T T c t x y z ⎛⎫∂∂∂∂ ⎪∂∂∂∂⎝⎭Thus we obtain the three-dimensional unsteady heat conduction.Step2: Cutting the bread into different layers and transfer the problem totwo-dimensional heat conduction question .Since it ’s too complicated to get ()=,,,T f x y z t in the oven, we assume that theamount of heat which passes through each point on unit area in unit time isequal. Then we can apply Neumann boundary condition and Fourier ’s Law tocalculate the distribution of heat in each layer.We can simplify this three-dimensional heat transfer problem to atwo-dimensional question by cutting the whole bread into different layers. Asfigure2 shows(a) (b)Figure2Then we analyze the distribution of heat on a single layer. Since we only focuson two-dimensional heat conduction, the new heat equation is:22222=+T T T c t x y ⎛⎫∂∂∂ ⎪∂∂∂⎝⎭ Thus we get the distribution of heat in each layer and we can derive thedistribution of heat in the bread. Let the bread vary in different shapes, thus wecan find the distribution of heat across the outer edge of different types of pans.Solving the modelStep1: Determine the parameter and initial conditionIn order to solve the Partial Differential Equation we should determine theNeumann boundary condition, as the equation showsXHeat Bread(),=L x y Tn ϕ∈∂∂While L is determined by the properties of bread, such as the density of breadρthe special heat capacity of bread v and the thermal conductivity k. We canfind these parameter in [], so the Neumann boundary condition can be settled.The initial condition can be represented as()0,,=FT x y Based the above assumption that the initial temperature of bread is 0℃, we canget the initial condition is()0,,=0T x y ℃ Then we use Finite element method to solve the Partial Differential Equation.It ’s too difficult for us to solve the equation, however with the help ofMATLAB we can do it.Step2: Use MATLAB PDE tool to simulate heat field and test differenttypes of pans.First, we select 6 types of pans which have the same area and determine theparameter of the pan. As table1 showsTable1 different types of pansA(area ) 20.04m 20.04m 20.04m 20.04m 20.04m 20.04m H(height )0.05m 0.05m0.05m0.05m0.05m 0.05mThen we use MATLAB PDE tool to simulate the distribution of heat in thebread, thus get the distribution of heat across the outer edge of the pan.In daily life the bread will be cooked between 5to 8 minutes, ideally we set thetime of simulation is 5 minutes. The result is as table 2showsTable2 (The distribution of heat across the outer edge of different pans)A B BreadRectangular pan Square pan℃D CStep4: Analysis of the resultsWe can export all the data from the PDE tool and use Excel to do statisticalanalysis. We use data of t=256s to analyze the result.This is the result we get. As the figure1 and figure2 showE F Circular panElliptical pan ℃ m℃m BreadFigure1Figure2Wheremax is maximum temperature of the panmin is the minimum temperature of the panaverage is the average temperature of the panrange is the maximum value minute the minimum valueFrom the graph above, we can find that A and B types of pan have the samemaximum, minimum and range value, however the C D E and F are not. Sincethe dispersion degree indicates the temperature fluctuation, we can classifythese 6 pans into two classes. Depend on the dispersion degree we can discoverthat if a pan have a transition fillet it will have a small chance the bread beovercooked at the edge of the pan.A CB D E F A BCD EF2.4 Optimization of oven shape and arranging pan on racksQuestion 2Assumption1 the area of different shapes of pans is idetical, A =0.042m2 difference of the shape of the oven is caused by /W L ,namely L is known, inour model, we assume it is 600mm .3 the radiation source can be insteaded by a small sphere, and the radiationsource in the up and down sides is identical.In first question, we have discussed the temperature distribution of a browniepan in the oven. But in reality, we often put several pans into the oven. In suchconditions, how the temperature of pans distribute, how many pans can be putinto, and how we select the shape of the pan to get the maximum of both ofthem.Imagine if you are a boss of the bakery, the more numbers of bread you make,the more profit you can get. Besides, the higher quality of you bread, the betterprestige of you bakery. To get a balance of numbers and quality, we introduce aweight p . It dely on what you like.Question 2 require us to develop a model to select the best type of the pan(shape) under the following conditions:1 Maximize the quantities of brownie pans N that can fit in the oven2 Maximize the distribution of heat H for the pan3 Optimize a contributions 1 and 2 where weights p and (1-p ) areassigned to how the results vary with different values of /W L and p .To maximize the quantities of brownie pans N, the distribution of heat H forthe pan, and to optimize a combination of each other. We have a few steps toanalysis this question.ModelStep 1:compute the number of pans NFor pans of every shape, the quantities of pans can fit in the oven vary with/W L of the oven. So we build a objective function to discuss this phenomenon.Suppose the area of pans of every shape A is 0.042m .()()2,N xL x y A y =()(),N x y a y L⋅≤.St()(),N x y b y W⋅≤()()2,N x y A y xL⋅≤(),N x y N+∈Where N is the quantities of pans.x is the ratio of /W L, according to the reference data, we assume range of variation is 0.5-1.5.y is the mark of the shape of every pan, we select 3 kinds of pans from Question 1.L, W is the length and wideth of the oven.()A y is the area of pans, ()A y=A=0.042m.()a y,()b y is the the length and wideth ofevery kind of pan.We use LINGO to solve this model, then get 3⨯10 matrix table(the step of x is 0.1), it shows how the quantities of pans N vary with the ratio of /W L of the oven. Step 2:calculation of the distribution of heat HThe oven is heated by infrared radiation in the up and down sides. Here we give a broad outline of the oven.As the radiation source in the up and down sides is identical. To simply the problem, we just consider the infrared radiation comes from the upside and simply the radiation source to a small sphere.d S OFigureIn reality, the plane of illumination is skew to the radiation direction,the radiation intensity of O is 0I , the area of the plane of illumination is dS ,the distance between O and the plane of illumination is r , 2cos dw dS r θ=⋅, t he radiation through dw can be illustrated by :200cos dS d I I r dw θ⋅Φ==Therefore, the irradiance of the dS is 02cos E I d dS rθ=Φ= Through Step 1, we can get the classification of pans under every value of /W L . Every (),N x y is corresponding to a H (the even distribution of heat of a kind of pan), we can receive a formula of (),N x y and (),H x y vary with different /W L .Step 3:optimization of N and HIn this step, we take p into account, and combine the influence to the (),N x y ,(),H x y of /W L with p . We difine a function to anlysis the rule of(),N x y and (),H x y vary with different /W L and p .()(),(1),S p N p H x y x y =⋅+-⋅This is a function with three variables(p ,x ,y ). Objective is to get themaximum value of S .To solve this problem, it needs us to do three partial derivatives, then to destermine the extreme values. But in this model, there aren’t enough constraints, it is too difficult for us to solve it in limited time. Here we just do a partial derivative to x .()(),,N x y H x y S x x x∂∂∂=+∂∂∂ Through it, we can get a extreme value.whereas it may not be the best objective.To make up this shortage, we have an another method to solve it.Weassume a variables is constant first, study the relationship of another two variables. Next, take this value into account and consider the relatioships of these three variables.In the step 1 and 2, we haven’t considered p ,which is equal to p is a constant and get a formula of (),N x y and (),H x y vary with different /W L .Now we regard p as a variable, make its length of one step is 0.1.( ()0,1p ∈)in each value of p , we discuss the rule of (),N x y and (),H x y vary with different /W L , in this way, we can get 10⨯10 matrix table and calculate S right now.From these data, we select the best shape of pan when S has a maximum under the conditions of p ,(),N x y ,(),H x y are a reasonable value.Solving the ModelTo simply the model, here we only anlysis x is 0.5,0.95,1.5, three kinds of shapes of pans, and discuss how p affect the objective function. Whenx =0.95 (W L ⨯=570⨯600),p =0.1, under the conditions of the shape ofthe oven and the size of pans with different kinds of shape is constant. We can easily get the maximum quantities of pans with different shapes through the algorithm of step 1.(0.95,1)N =5 (0.95,2N =6 (0.95,4)N =7For every kinds of pan, we do a MATLAB PDE simulation.From the result, we can get the standard deviation of the even distribution.()H= 2.3577 ()H=2.84670.95,10.95,1H=0.5428 ()0.95,2The following is the distribution of heat with different shapes:Figure the distribution of heat of rectangleFigure the distribution of heat ofellipseFigure the distribution of heat of circleHere p =0.1, we can compute()()(),,,(1),S p N p H p x y x y x y =⋅+-⋅()0.1,0.95,1S =0.98852 ()0.1,0.95,2S = 2.82193()0.1,0.95,3S =3.16203In the same method, we can receive other data of (),N x y ,(),H x y ,(),,S p x y . These data can form three 3⨯3 matrix table.We can draw up a scatter diagram between S and /W L.Then we use similar method to do it when p is 0.2,0.3 1.We can get the ellipse is the best shape in our model.Actually we need to consider every shape between rectangle and circle.So we compare ellipse with other shapes of pans and repeat aboved procedure. According to our simulation and comparison, we get the best type of shape as follows:it is resemble to a track.Strengths and WeaknessesSummary and Recommendation Reference。
