2013建模美赛书信

合集下载

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年美国数学建模比赛提交论文-二等奖

Print This Page Close This WindowFor office use only T1________________T2________________T3________________T4________________Team Control Number22940Problem ChosenAFor office use only F1________________F2________________F3________________F4________________2013Mathematical 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 include the name of your school,advisor,or team members on this page.SummaryThis paper aims to design the optimal shape of Brownie pan on the given conditions.The influencing factors include the shape of pan,the ratio of width and the length of the oven,and the weight be conferred to the quantity of Brownie pans that can be put in a oven and evenness of heat distribution.We have a scientific study about the three typical shapes,namely rectangular circular and oval,and we get the following conclusion that the time required by circular pans is the shortest,while rectangular pan can maximize the use of space.To solve the problem,we simplify the influencing factors and assume that the area of being effective baking inside the oven can be exactly fulfilled by rectangular pans.So that we can easily solve the condition 1,and be aware of that the rectangular pans is the optimal shape;and the same to the condition 2,we believe that the temperature distribution on the outer edge of circular pans is completely uniform,so the optimal solution is circular.When it comes to the third condition,it’s a typical linear programming problem.We use the idea of normalizing to construct design objectivefunction:(1)s A nZ p p N A=+−.In the formula above,the calculation of area being effective heated adopt the view of discrete,which is very innovative (formula (k)).By using the above objective function,whose results is roughly the same with the results of MATLAB,This can be used to confirm each other.We obtain the desired optimal solution to a maximum of 1.215,which is larger than other shapes’maximum by 21.5%.Its advantages are obvious,so this further illustrates the correctness of the ideal optimal shape.So we can say the pan we design is the Ultimate Brownie Pan!ContentsI.Introduction (4)II.Assumptions (4)III.Analysis&Models (5)IV.Solutions (12)4.1rectangular Brownie Pans (12)4.2circular Brownie Pans (14)4.3oval Brownie Pans (15)V.Optimization of the model (12)5.1rectangular Brownie Pans (16)5.2circular Brownie Pans (16)5.3oval Brownie Pans (17)References (18)Advertising sheet (19)Appendix (20)List of Figures1.Region of discrete (8)2.Volumetric controlled by internal node (8)3.The simulation isotherms of rectangular pan (10)4.The simulation isotherms of circular pan (10)5.The simulation isotherms of oval pan (10)6.Temperature curve of the rectangular pan center (11)7.Temperature curve of the circular pan center (11)8.Temperature curve of the oval pan center (11)9.How the rectangular pans are placed in the oven (14)10.The placement of round Brownie Pan in the oven (14)11.The ideal shape of the Brownie Pan (15)12.The distribution of the pans in the oven (15)13.The placement of rectangular Brownie Pan in the oven (16)14.The zoning figure of the extended the oven (17)15.The arrangement of Brownie Pans in theoretical optimal shape in the oven (18)16.Distribution of temperature of rectangular pan (20)17.Distribution of temperature of circular pan (20)18.Distribution of temperature of oval pan (20)1Introduction:When baking in a rectangular pan heat is concentrated in the4corners 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.Since both have their own advantages and can complement each other,we consider that a combination of the two will be a better choice.The purpose of this article is to identify the pan in best shape to satisfy various needs.2Assumption:1.The heat inside the oven is Changeless,After a Brownie Pan have been put in the oven,the heat is conducted to the center from the edge.2.The initial temperature of the food is25C o.3.The coefficient of thermal conductivity of the food is0.1.4.To make full use of the area of a rectangular oven,We assume that the area of being effective baking inside the oven can just be fulfilled by rectangular Brownie Pans.Table 1NotationParameter MeaningFThe heat of per unit time through a given area λThermal conductivity TTemperaturetThe temperature field in the Cartesian coordinate system A Cross-sectional areaΔx,ΔyThe amount of change of the x-direction,y-direction τ∆Time variation amount ∆o F The number of grid FourierT ij The temperature of the arbitrary point (i,j)T oThe temperature of the center of the pant j i T ,The temperature of the arbitrary point (i,j)at time t 0div ∇The gradient value at a certain point3Analysis &ModelsWe solved this problem using numerical solution of heat conduction.The numerical solution is based on the basic knowledge of heat transfer,and a method of using partial differential equations solving a heat conduction problem.Heat conduction partial differential equations:⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧⎟⎟⎠⎞⎜⎜⎝⎛∂∂+∂∂+∂∂=∇=∂∂conditions Initial conditions dary B z T y T x T T t T oun 2222222ααThe basic idea:putting the problem of the temperature within the object continuously changing with time and space transformed into the problem of the temperature value in a finite number of discrete points within the area of the field of time and space.Further,using the temperature value of these discrete units to approximate the continuous temperature distribution.Models:1.The establishment of the physical modelAccording to the meaning of the questions,we construct three models,which are rectangular pan,circular pan and oval pan.2.The establishment of the mathematical modelOne of the main purpose of the study heat conduction problem is obtained under certain boundary conditions the state of objects within the temperature distribution with spatial location and time.⑴.Equations of heat conductionThe equations of the temperature distribution in the description of thermal conductivity inside the object is called the partial differential equation of heat conduction,which also known as the heat diffusion w of conservation of energy and the Fourier law are its fundamental basis.Once solving the temperature distribution,the heat conduction rate of inside the object or surface at any point on can be obtained by Fourier law.That is the heat flux density.Unsteady heat conduction problem for the temperature field varies with time,by obtaining the temperature distribution inside the object at different times,can determine the thermal stress and thermal deformation of the various parts.The general form of the equation of heat conduction in the Cartesian coordinate system:v ztz y t y x t x t cφλλλτρ+∂∂∂∂+∂∂∂∂+∂∂∂∂=∂∂)(((................................(a)This formula reflects the relationship between the thermal conductivity of the object within the total energy conservation.In the calculation,the thermal conductivity is seen as a constant,so the above equation reduces to:c zty t x t t v ρφατ+∂∂+∂∂+∂∂=∂∂)(222222...............................................................(b)It is composed by three physical properties of a physical parameter,called thermal diffusivity,can also be referred to the coefficient of thermal conductive,whose unit iss m /2.It indicates the temperature tends to be uniform capacity of the object in the heating or cooling process.This comprehensive parameters in the non-steady-state heat transfer process is a very important parameter.⑵.Single -Valued Property conditionsAll pure heat conduction problem can be described with the equation of heat conduction in the corresponding coordinates,including one-dimensional and multidimensional,steady-state and non-steady-state,constant material properties and Variable physical properties,the internal heat source and no internal heat source heat conduction problems.Differential equations (The mathematical said general solution)must contain the pending integration constant.In addition to the differential equation,for these constants to be determined uniquely determined must be attached to a certain characteristics of solving this particular heat conduction problem and limited of external environment or supplementary explanation.These additional instructions and restriction condition is Single -Valued Property conditions,and mathematics known as boundary conditions.Mathematical model of any one specific full heat conduction problems,in addition to the equation of heat conduction in the appropriately selected coordinates,must be given the corresponding Single -Valued Property conditions.In terms of the general problem of thermal conductivity,the Single -Valued property conditions consist of the following two parts:①.Time conditionsFor unsteady-state heat conduction,the temperature field of the object at the start time must be given,so the time conditions are known as initial conditions.In the three-dimensional Cartesian coordinate system,the initial conditions can generally be expressed as the following form:),,(0z y x f t ==τ....................................................................(c)Under normal circumstances,the object having a completely uniform temperature at the beginning of the process,so we can think that the distribution function (),,(z y x f )is a constant in the above formula.②.Boundary conditionsThe boundary conditions refers to the contact and interaction in the heat exchange of hot objects between the boundary surface and the external environment.For unsteady heat conduction,it is often the external driving force making process take place and develop.Here we consider the first category.Stipulate:The surface temperature is a room temperature.),(y x f t w =,The boundary temperature.Notes:tcons t w tan =3.Discrete the solution domainDiscretization is divided the substantially continuous object into a series of tiny unit,while the center of the unit is called a node.Divided the spatial domain x into n subparagraph,the step is Δx,get 0,l,2,...,i,...,n x node of the n +1;divided the spatial domain y into n subparagraph,while the step is Δy,get 0,l,2,...,i,...,n y node of the n +1.Intersection point (i,j)of the separation line represents the spatial domain position.The size of space and time step depends on the specific circumstances of the problem,and sometimes cannot bearbitrarily selected,needing to consider the node temperature equation solving stability problems which are as shown in the following figure:Figure 1Region of discreteTime can also be divided into many inter-cell.The physical significance of the zone discretization is that we can think a node focus the heat capacity around its tiny area.Thus the node temperature is the average temperature of this tiny area.Thus,the temperature of all the nodes together represents the distribution of the temperature within this continuous region.4.Establish the difference equation of nodes’temperatureUsing difference quotient instead of the derivative,and then on the basis of the relationship of the heat balance we can a establish differential equation.The same as the steady heat conduction,using the method of heat balance,you can establish the temperature difference equations of objects internal nodes and boundary nodes at unsteady heat conduction.As shown in the following figure:Figure 2V olumetric controlled by internal nodeFor the two-dimensional unsteady heat conduction problems of constant material properties and no internal heat source,the internal nodes (i,j)represents the thermal eq uilibrium of the control volume is expressed as:In unit time,the heat flow rate jQ λand iQ λcoming from the adjacent control volume ),1()1j i j i +−，，（and)1,()1,(+−j i j i ，is equal to the control volume thermodynamic energy increase dU .dU Q Q j i =+λλ.Therefore,for each node Differential established as follows:tj,i 2222j,i y T x T t T ⎟⎟⎠⎞⎜⎜⎝⎛∂∂+∂∂α=⎟⎠⎞⎜⎝⎛∂∂................................................(d)for the left side:tT T t T (t j,i t t j ,i j ,i ∆−=∂∂∆+.................................................................(e)for the right side:2,1,,1,222)(x T T T x Tt j i t j i t j i j i ∆+−=∂∂−+........................................................(f)21,,1,,222(y T T T y Tt j i t j i t j i j i ∆+−=∂∂−+...................................................................(g)Making y x ∆=∆，2xF o ∆∆=∆τα，∆o F :called grid Fourier number.(e),(f)and (g)substituted into the formula (d).After finishing,the following equations can be obtained:t j i o t j i t j i t j i t j i o t t j i T F T T T T F T ,1,1,,1,1,)41()(∆+−+−∆∆+−++++=......(h)This equation can draw two conclusions:1).Any node temperature of an internal node at a particular moment can be obtained directly from the temperature of the node and its collar node at the former time.Do not have the simultaneous solution of the equations.This is the advantages of the explicit difference scheme.So we can start from the initial temperature sequentially obtainsτ∆,τ∆2,...,τ∆k each time nodes temperature.2).In front of the t j i T ,coefficient in the formula (h)can not be negative.So041≥−∆o F ,which is equivalent to 41≤∆o F ................................（i ）This is the stability conditions of two-dimensional non-steady-state thermal conductivity inside the object node temperature equation explicit difference scheme.Physical meaning:The temperature of any internal node at time t t ∆+all dependson the temperature of this node and its surrounding nodes at time t .When the temperature of surrounding nodes to are known,the higher the temperature of the node is,the higher the temperature of the node will be at time t t ∆+.Therefore,the coefficient in front of the formula can not be negative,which must meet the formula (i)that represents the stability condition.5.Solve temperature algebra equation and obtain the temperature of all nodes Assuming the area of each Brownie Pan is 2600cm A =and it will be heated for ing Matlab programming solve the problem of rectangular,round and oval pan heat conduct to the center from all sides into the incubator.As shown in the following figures:Figure 6The simulation isothermsFigure 7The simulation isothermsof rectangular panof circular panFigure 8The simulation isothermsof oval panyy102030405060708090100yWe can know from the figures above:whatever the shape of the pan is,the isothermal lines are arcuate curve,and they are nearly circular in its center position.Therefore,in the design of the shape of baking pan,we should try to design it in arc-shape in order to make the baking pan to be evenly heated.Temperature curve of the pan centerFigure 9Temperature curve of Figure 10Temperature curve ofthe rectangular pan centerthe circular pan centerFigure 11Temperature curve of oval pan center6.Conclusion⑴.The circular baking pan temperature rises very quickly.When the food reaches a certain temperature,the time required is the shortest.However,the space utilization of the pan in the rectangular oven is low,not efficient of using energy;⑵.Rectangular baking pan to maximize the use of space,effective use of energy.However,the temperature difference between the center and the periphery will cause the food being heated unevenly;⑶.The oval pan space utilization and degree of heat evenly is between the circular baking pan and rectangular baking pan.T e m p e r a t u r e /T e m p e r a t u r e /T e m p e r a t u r e /4Solutions Solutions::For the condition 1:Based on the assumption 4,it’s obvious that if we use a rectangular Brownie Pan,The Pans can exactly fulfill the oven.That’s to say,the number of pans in rectangular shape used is maximum.For the condition 2:①.How to determine the area where temperature is uniformly distributedWe use temperature variance of all the discrete points as a threshold value to determine whether a point is in the area where temperature is uniformly ly,whether its temperature can satisfy the following inequality:00ij T T div −≤∇..............................(j)If the above equation can be satisfied,we believe that this point is at the place where temperature is evenly distributed.②.The area of the region where temperature is uniformly distributedWhen we calculating the area of the region where temperature is uniformly distributed,We apply the previous concept of discretization.Firstly,we use MATLAB to calculate the number of points(N)in a given range,then,we can know that the area occupied by each discrete point isyx A d d d ×=...........................................(k)so,the total area occupied by all the discrete point isyx a d d N A ××=............................................(l)According to the temperature distribution figure of various Brownie Pan in geometric shapes in the first question,we measure the uniformity of the temperature distribution with the temperature variance22211((,))n navg i j T i j T n σ==−=∑∑..............................(m)and it’s obvious that:2rectangle22circle σσσ>>oval And we can know that only when Brownie Pan is round,can the heat on outer edge ofthe Brownie Pan be completely evenly distributed.For the condition 3:Based on the above two considerations,We think the combination of the above two optimization conditions we can use the following normalized method and get the Objective function ：(1)sA nZ p p N A=+−...................................(n)In addition,there are other restrictions:1.Different kinds of pan share the same area;2.The temperature reached after a specific time(6mins);3.The coefficient of thermal conductivity of the food is similar.(0.1)4.1Firstly,for rectangular Brownie Pans whose heating time is 6mins,we know in the objective function above,n=N,while the area where temperature is uniformly distributed ,A s ,need to be worked out.In this paper we can use the MATLAB program programmed in the Question 1,and we can get the area where temperature is uniformly distributed by directly substituting into the length and width of a rectangle,and the objective function turns into:(1)s AZ p p A=+−........................(o)the equation shows objective function becomes a single-valued function of p,and ithas no relationship with WL,while 'L NL =By using the MATLAB program written in the Analysis &Models,we can get we can obtain the output of 12sets of data(in the table 2),Fitting the relationship between a and b,and we can get the following function:2234.49()469.05599.16s W WA L L=−+Notes:to obtain their solutions,we assume that the area of a rectangle is 600cm 2,And the objective function turns into600/)16.59905.469(49.234)(1(2+−−+=LWL W p p Z With the constrains:..(0,1)[0,1]s t p W L∈∈And the optimal solution is Z=1,if and only if 1&Wp L=it can be satisfied Table 2Scatter values calculated by the MATLABL W 0.800.760.630.600.580.55A s /cm 2374.78378.69397.52402.94406.01412.92LW 0.490.450.420.380.350.32A s /cm 2426.43436.38444.33455.59464.53473.89Figure 12How the rectangular pans are placed in the oven4.2Secondly,for circular Brownie Pan which have been heated in the oven for 6mins,We already know from the question1that heat is evenly distributed on the outer edge of the round pan.Therefore,we can think in round Brownie Pans A A=and the objective function turns into:1nZ p p N=+−........................................(p)Figure 13The placement of round Brownie Pan in the ovenTo simplify the calculation,We make'W L for the fixed value k,namely 'Wk L =,And we know that by calculating the horizontal distance between two round pan is(n −............................(q)And we can obtain constraint condition of circular as following:(1n L −+≤.....................(r)4.3thirdly,according to the above calculation results,we think it will be a good ideato combine the advantages of round Brownie Pan and rectangle Brownie Pan.And we can draw the following diagram of ideal shape of Brownie Pan:We will have a study on the ideal Brownie Pan.In order to ensure the area of the pan unchanged,we can see how Brownie Pan's width changes:224'''2r r L L W rπ−=+−.............(s)To ensure the pan will not exceed the scope of the oven,which is impossible,we can get a restriction on the oven:224''2r r n L NL W r π⎛⎞−+≤⎜⎟−⎝⎠..........(t)Figure 14The ideal shape of the Brownie Pan And the distribution of the pans in the oven as follows.Figure 15The distribution of the pans in ideal shape in ovenAnd the objective function turns into:(1)s A nZ p p N A=+−........................................................(u)By using the MATLAB as well,we know that the optimal solution isWhen p=0.75and 0.8WL=,Z can get the optimal solution,andZ=1.215Taking that the oven has two racks into account ,all of the above data required to be multiplied by a expansion of the coefficient α=2,which has no influence on the relative difference among different shape.5Optimization of the modelIn the optimized design above,we consider that the Brownie Pans can only be placed in a row in the oven.Next,we intend to expand the previously conclusion.And we consider that the Brownie Pans can be placed in N rows in the oven.5.1Firstly,for rectangular Brownie PansFigure 16The placement of rectangular Brownie Pan in the ovenAt this point,we know that:the specifications of Brownie pan is:L W N M×It’s obvious that the Brownie Pan optimization function does not change,which still is:**(1)(1)**ss A A M N Z p p p p M N A A =+−=+− (v)Namely max 1Z =.5.2Secondly,for round Brownie Pans,we divide the extended the oven into threeparts,which is as follows:Figure 17The zoning figure of the extended ovenThe reasons why we adopt the above area dividing method are based on the following two considerations:1.To Simplify the calculation;2.To take advantage of the inequality (r)above.And we can obtain constraint condition of circular are as follows:1.Region 1:the number of the circulars is1([1)([]1)W Ln d d=−×−...........................................(w)Notes:[]tan W Ws ds for the Gauss number ofd d2.Region 2:We assume the number of the circulars is n 2and the constraint condition is(21n L −≤..............................(x)Notes:L L =',d dWW W )1(['−−=3.Region 2:We assume the number of the circulars is n 3and the constraint condition is(31'n L −≤....................(y)Notes:d n L L 1''−=,d n W 1''=and it’s obvious that max 1Z <5.3Thirdly,for Brownie Pans in theoretical optimal shapeFigure18The arrangement of Brownie Pans in theoretical optimal shape in the oven It’s obvious that in the oven that can arrange multi-row Brownie Pans the optimal solution keeps unchanged.Namely,Z f=Z=1.215From all above,we can know that the Brownie Pan in the theoretical optimal shape can get greater optimal solution than either rectangular pans or round pans.So it’s a smart idea to choose a Brownie Pan in the theoretical optimal shape to cook your Brownie cakes!6References[1]Lars Mönch Robert Unbehaun You In Choung,Minimizing Earliness–Tardiness on a Single Burn-in Oven with a Common due Date and Maximum Allowable Tardiness Constraint, /static/pdf/374/art%253A10.1007%252Fs00291-005-0013-4.pdf?aut h66=1360982955_54758b58d2754d71008329400df3c11e&ext=.pdf,10Dec.2005[2]K.Venkateshmurthy&K.S.M.S.Raghavarao,Analysis of Modes of Heat Transfer in Baking Indian Rice Pan Cake(Dosa,)a Breakfast Food, /static/pdf/517/art%253A10.1007%252Fs13197-010-0204-0.pdf?aut h66=1360983288_e6bf23e753296afd9fcd43eea80dad4a&ext=.pdf,06Dec.2010[3]Jiang Qiyuan and Xie Jinxing,Mathematical model Mathematical programming model,3rd edn, Mathematical Modeling,Accessed Feb.2003[4]D.Pitts et al,Schaum's Outline of Theory and Problems of Heat Transfer,2nd edn,Science Press,2002[5]M.N.Aoqi Sigg,Heat Conduction,Higher Education Press,1984A dvertising sheet:AppendixFigure3Distribution of temperature of rectangular pan199.5198.5Figure4Distribution of temperature of circular panFigure5Distribution of temperature of oval pan。

2013美国大学生数学建模竞赛论文

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!。

写给要参加美国数学建模竞赛的同学

写给要参加美国数学建模竞赛的同学美国赛分为两种：MCM(The Mathematical Contest in Modeling数学建模竞赛)和ICM(Interdisciplinary Contest in Modeling跨学科建模竞赛)。

每年的美国赛共有A,B,C三个题，如果选做MCM竞赛，那么就在A,B题中选一题做，如果做ICM竞赛，那么就只能做C题。

美国赛一般是在每年的二月中旬举行，2011年的美国赛将在在2011年2月10号到14号举行。

MCM规定，一名指导老师最多只能指导两个队，每个队的报名费为100美元。

美国赛面向全球，通过网络报名，参赛队交费通过via Mastercard or Visa卡通过网络支付，交费报名的同时取得唯一的报名号。

竞赛期间的规则和全国赛差不多，这里不再重复。

美国赛参赛队需要将paper和summary准备三份在deadline之前寄到组委会。

三份论文会分别交给：The Institute for Operations Research and the Management Sciences (美国运筹与管理科学协会)，The Society for Industrial and Applied Mathematics（美国工业与应用数学协会），The Mathematical Association of America（美国数学会）进行评选，summary及其重要， summary写的出色，judge通过你的summary才能产生进一步看你paper欲望，因为参加竞赛的team是很多的，summary不能打动judge,那么直接就被fire掉了。

每个协会的judge会给每个题中评出的一个outstanding奖得主（有时一个题最后评出来的outstanding 奖有5，6支队，但是这三个协会都只会提名一只队伍作为该协会的outstanding奖得主，有时还会出现一只队获得了两个协会的同时提名的情况:06年做A题的科罗拉多大学的一个队就同时获得了美国工业与应用数学协会和美国数学会的提名。

2013mcmContest Rules美赛规则

Contest Rules, Registration and Instructions(All rules and instructions apply to both ICM and MCM contests, except whereotherwise noted.)To participate in a contest, each team must be sponsored by a faculty advisor from its institution.Team Advisors: Please read these instructions carefully. It is your responsibility to make sure that teams are correctly registered and that all of the following steps required for participation in the contest are completed:Please print a copy of these contest instructions for reference before, during, and after the contest. Click here for the printer friendly version.COMAP is pleased to announce a new supplement to the MCM/ICM contest. Click here to read more details.I. BEFORE THE CONTEST BEGINS:A. RegistrationB. Choose your team membersII. AFTER THE CONTEST BEGINS:A. View the contest problems via the contest web siteB. Choose a problemC. Teams prepare solutionsD. Print Summary Sheet and Control SheetIII. BEFORE THE CONTEST ENDS:A. Send electronic copy of Solution Paper by emailIV. WHEN THE CONTEST ENDS:A. Prepare Solution PacketB. Mail Solution PacketV. AFTER THE CONTEST IS OVER:A. Confirm that your team’s solution was receivedB. Check contest resultsC. CertificatesD. PrizesIMPORTANT NOTES:∙COMAP is the final arbiter of all rules and policies, and may disqualify or refuse to register any team that, in its sole discretion, does not follow these contest regulations and procedures.∙If a team is caught violating the rules, the faculty advisor will not be permitted to advise another team for one year, and the advisor’s in stitution will be put on probation for one year.∙If a team from the same institution is caught violating the rules a second time, then that school will not be allowed to compete for a period of at least one year.∙All times given in these instructions are in terms of Eastern Standard Time (EST).(COMAP is located in the U.S. Eastern Time zone.)I. BEFORE THE CONTEST BEGINS:A. RegistrationAll teams must be registered before 2PM EST on Thursday, January 31, 2013. We recommend that all teams complete the registration process well in advance, since the registration system will not accept any new team registrations after the deadline. COMAP will not accept late registrations for MCM/ICM 2013 under anycircumstances. NO EXCEPTIONS WILL BE MADE.1.Register your team online via the contest web site: Go to/undergraduate/contests/mcm.a. If you are registering your first team for this year’scontest, click on Register for 2013 Contest on the left-handside of the screen.Enter all the required information, including your emailaddress and contact information.IMPORTANT: Be sure to use a valid and current email addressso that we can use it to contact you at any point before, during, and after the contest, if necessary.b. If you have already registered a team for this year’s contestand want to register a second team, click on Advisor Login, then log in with the same email address and password that you used when you registere d your first team. Once you’re logged in, click on Register Another Team near the upper right corner of the page, then follow the instructions there.Note:An advisor may register no more than two (2) teams. If you already registered two teams, the Register Another Teamlink will not appear. The system will not allow you toregister more than two teams.c. Although each advisor can register only two teams, there isno restriction on the number of advisors or teams that can register from any particular institution or department.2.Registration FeeA $100 registration fee per team is required.For an additional $100 fee per team, you can receive a Judges Commentary written specifically about your team’s paper.We accept payment with Mastercard or Visa only via our secure web site. We cannot accept other forms of payment. Our secure site will process your credit card payment, so your credit card number is protected. Our system will not store your credit card number after it processes your payment.3.After we receive approval from your financial institution (thistakes only a few seconds), the system will issue a control number for your team. Your team is not officially registered until you have received a team control number. Print the page thatdisplays your team control number: It is your only confirmation that your team has been registered. This page also lists the email address and password that you entered when registering;you will need this information to complete the contestprocedures.You will NOT receive an email confirmation of your registration.4.If you need to change any of the information (name, address,contact information, etc.) that you specified when youregistered, you can do so at any point before or during thecontest by logging in to the contest web site with the same emailaddress and password that you used when registering (click onthe Advisor Login link on the left side of the screen). Oncelogged in, click on the Edit Advisor or Institution Data linknear the upper right corner of the page.5.Check the contest web site regularly for any updatedinstructions or announcements about the contest. Except inextreme circumstances, COMAP will not send any confirmation,reminders, or announcements by email. All communicationregarding the contest will be via the contest web site.6.You will return to the contest web site during the contest toenter and confirm information about your team, and to print outyour team’s Control Sheet and Summary Sheets, which you wil luse when preparing your team’s solution packet. Details onthese steps follow in the instructions below.B. Choose your team members:1.You must choose your team members before the contest beginsat 8PM EST on Thursday January 31, 2013. Once the contest beginsyou may not add or change any team members (you may, however,remove a team member, if he or she decides not to participate).2.Each team may consist of a maximum of three students.3.Each student may participate on only one team.4.Team members must be enrolled in school at the time of thecontest, but they need not be full-time students. Team membersmust be enrolled at the same school as the advisor and other teammembers.II. AFTER THE CONTEST BEGINS:A. View the contest problems via the contest web site:Teams can view the contest problems via the contest web site when the contest begins at 8PM EST on Thursday January 31, 2013:1.The contest problems will become available precisely at 8PM ESTon Thursday January 31, 2013; team members can view them byvisiting /undergraduate/contests/mcm. Nopassword will be needed to view the problems; simply go to thecontest web site at or after 8PM EST on Thursday, January 31,2013 and you will see a link to view the problems.2. The contest problems will become available precisely at 7:50PMEST on Thursday January 31, 2013 on the following mirror sites:/mcm/index.html/mcm/index.html/mcm/index.htmlIf you cannot access any of the sites, there may be a problemwith your local Internet connection. Contact your local Internetservice provider to resolve the issue.B. Choose a problem:Each team must choose one of the three problem choices according to the following rules:•MCM teams must choose either Problem A or Problem B; an MCM team may submit a solution to only one of the problems. (MCMteams should NOT choose Problem C.)•ICM teams must choose Problem C. There is no choice for ICM teams. (ICM teams should NOT choose Problem A or Problem B.)C. Teams prepare solutions:1.Teams may use any inanimate source of data or materials:computers, software, references, web sites, books, etc. ALLSOURCES USED MUST BE CREDITED. Failure to credit a source willresult in a team being disqualified from the competition.2.Team members may not seek help from or discuss the problem withtheir advisor or anyone else, except other members of the sameteam. Input in any form from anyone other than student teammembers is strictly forbidden. This includes email, telephonecontact, and personal conversation, communication via web chator other question-answer systems, or any other form ofcommunication.3.Partial solutions are acceptable. There is no passing orfailing cut-off score, and numerical scores will not be assigned.The MCM/ICM contest judges are primarily interested in the team’s approach and methods.4.Summary Sheet:The summary is an essential part of your MCM/ICMpaper. The judges place considerable weight on the summary, and winning papers are often distinguished from other papers based on the quality of the summary.To write a good summary, imagine that a reader will choose whether to read the body of the paper based on your summary: Your concise presentation in the summary should inspire a reader to learn about the details of your work. Thus, a summary should clearly describe your approach to the problem and, mostprominently, your most important conclusions. Summaries that are mere restatements of the contest problem, or are acut-and-paste boilerplate from the Introduction are generally considered to be weak.Besides the summary sheet as described each paper should contain the following sections:∙Restatement and clarification of the problem: State in your own words what you are going to do.∙Explain assumpt ions and rationale/justificat ion: Emphasize the assumptions that bear on the problem. Clearly list all variables used in yourmodel.∙Include your model design and just ificat ion for type model used or developed.∙Describe model testing and sensitivity analysis, including error analysis, etc.∙Discuss the strengths and weaknesses of your model or approach.5.The judges will evaluate the quality of your writing in theSolution Paper:∙Conciseness and organization are extremely important.∙Key statements should present major ideas and results.∙Present a clarification or restatement of the problem, as appropriate.∙Present a clear exposition of all variables, assumptions, and hypotheses.∙Present an analysis of the problem, including the motivation or justification for the model that is used.∙Include a design of the model.∙Discuss how the model could be tested, including error analysis and stability (conditioning, sensitivity, etc.).∙Discuss any apparent strengths or weaknesses in your model or approach.6.Papers must be typed and in English.7.The solution must consist entirely of written text, andpossibly figures, charts, or other written material, on paper only. No non-paper support materials such as computer files or disks will be accepted.8.The Solution Paper must display the team control number andthe page number at the top of every page; for example, use the following page header on each page:Team # 321 Page 6 of 139.The names of the students, advisor, or institution should NOTappear on any page of the print solution or electronic solution.The solution should not contain any identifying informationother than the team control number.10.Failure to adhere to any preparation rule is grounds for teamdisqualification.D. Print Summary Sheet and Control Sheet:After the contest begins at 8PM EST on Thursday January 31, 2013, and while the teams are preparing their solutions, the advisor should:1.Login to the contest web site (go to/undergraduate/contests/mcm. Click onAdvisor Login, then enter your email address and password).2.Enter the team member names and confirm that each name isspelled correctly. This determines how the names will appear on the contest certificates. COMAP will not make any changes or reprint certificates for any reason.3.Specify the problem that your team has chosen to solve.4.Print one copy of the Control Sheet.5.Print one copy of the team Summary Sheet.III. BEFORE THE CONTEST ENDS:A. Send electronic copy of Solution Paper by email:1.Each team is required to submit an electronic copy of itssolution paper by email to solutions@. Any team memberor the advisor may submit this email.a. Your email MUST be received at COMAP by the submissiondeadline of 8PM EST on February 4, 2013.b. Failure by a team to submit a solution via email by 8PM ESTon February 4, 2013 constitutes a violation of the contestrules and will result in that team’s disqualification.c. No further modifications, enhancements, additions, orimprovements may be made to the team’s solution paper afterthis email submission. Each team’s electronic submissionwill be cross-checked for consistency with their papersubmission. Any changes to the paper version will constitutea violation of the contest rules and may result indisqualification.2.In the subject line of your email write: COMAP and your team’scontrol number. For example:Subject: COMAP 2222e your team’s control number as the name of your fileattachments.AP will accept only an Adobe PDF or Microsoft Word file ofyour solution. DO NOT include programs or software with youremail as they will not be used in the judging process. Limit onesolution per email. The names of the students, advisor, orinstitution should NOT appear on any page of the electronicsolution. Your team's summary should be included as the firstpage of your file.5.See the rules for submitting the paper copy and CD via mailbelow.IV. WHEN THE CONTEST ENDS:A. Prepare Solution Packet:When the contest ends at 8PM EST on February 4, 2013:1.Each team member must sign the Control Sheet to pledge that heor she abided by the contest rules and instructions.2. Make one copy of y our team’s Solution Paper.3. Place the Summary Sheet on top of the Solution Paper.4. Staple the Control Sheet on top the Summary Sheet and SolutionPaper.*So the paper order should be: Control Sheet on top followed by SummarySheet followed by the Solution Paper.5.Enclose an electronic copy (PDF or Word file) of your team’sSolution Paper on a CD-ROM. Programs and software will not beused in the judging process. DO NOT include them on the CD.If you advise more than one team, please include both teams’files on a single CD-ROM and label it with contest, year, andboth teams’ control numbers.Example:Year Contest Control Numbers2013 MCM/ICM 10004, 10005B. Mail Solution Packet:1.Mail your team’s complete Solution Packet to:MCM/ICM CoordinatorCOMAP, Inc.175 Middlesex Turnpike., Suite 3BBedford, MA 01730USAAP must receive your Solution Packet via mail on or beforeWednesday February 13 , 2013. It is your responsibility to makesure that your team’s Solution Packe t arrives at COMAP by thisdeadline. Use registered or express mail, if necessary, toinsure that it arrives at COMAP by Wednesday February 13, 2013.AP will not accept late solutions under any circumstances.4.If you require confirmation that your paper was received byCOMAP, send the packet via a carrier that provides packagetracking. Due to the number of papers received, COMAP can notanswer receipt inquiries or emails.V. AFTER THE CONTEST IS OVER:A. Confirm that your team’s solutio n was received:1.You may login to the contest web site using the Advisor Loginlink to verify that your team’s Solution Packet was receivedat COMAP. After mailing your Solution Packet, please allowseveral days for us to process your packet before expecting tosee this confirmation.B. Check contest results:1.Judging: Judging will be completed in March and the results willbe posted on April 29, 2013. The Solution Papers will berecognized as Unsuccessful, Successful Participant, HonorableMention, Meritorious, Finalist, or Outstanding Winner.2.We will post the contest results on the web site as soon as theyare available, so visit the contest web site regularly to checkfor updates. It will take several weeks for the judges toevaluate the solutions and for COMAP to process the results.Please do NOT call or email COMAP regarding contest results.C. Certificates:After the results are issued, each successfully participating team will receive a certificate of participation. The certificate will be mailed or emailed to the advisor at the address used during the registration process. All international teams will receive ONLY an electronic (PDF) certificate. Please allow several weeks after theresults are posted to the contest web site to receive yourcertificate.D. Prizes:∙The Institute for Operations Research and the Management Sciences (INFORMS) will designate an Outstanding team from eachof the three problems as an INFORMS winner.∙The Society for Industrial and Applied Mathematics (SIAM) will designate one Outstanding team from each problem as a SIAMwinner.∙The Mathematical Association of America (MAA) will designate one Outstanding team from each problem for the MCM as a MAAwinner.∙The Ben Fusaro Award, Typically, among the final MCM papers from which the Outstanding ones are selected is a paper that isespecially creative but contains a flaw that prevents it fromattaining the Outstanding designation. In accord with Ben’swishes, the award will recognize such teams.∙The Mathematical Contest in Modeling’s Frank R. Giordano Award began in 2012. It honors Brig. Gen. (ret) Frank Giordano whodirected the MCM for 20 years. This award goes to a paper thatdemonstrates true excellence in the execution of the modelingprocess.If you agree to abide by these terms and instructions, indicate so by entering your email address and clicking on the I A GREE button. If, for any reason, you cannot agree to these terms, click on the I DISA GREE button. If you click on I DISA GREE, the information you have entered will be lost and you will have to start the registration process over again if you decide to register later.。

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年美国数学建模竞赛

2013 ICM ProblemNetwork Modeling of Earth's HealthBackground: Society is interested in developing and using models to forecast the biological and environmental health conditions of our planet. Many scientific studies have concluded that there is growing stress on Earth's environmental and biological systems, but there are very few global models to test those claims. The UN-backed Millennium Ecosystem Assessment Synthesis Report found that nearly two-thirds of Earth's life-supporting ecosystems— including clean water, pure air, and stable climate— are being degraded by unsustainable use. Humans are blamed for much of this damage. Soaring demands for food, fresh water, fuel, and timber have contributed to dramatic environmental changes; from deforestation to air, land, and water pollution. Despite the considerable research being conducted on local habitats and regional factors, current models do not adequately inform decision makers how their provincial polices may impact the overall health of the planet. Many models ignore complex global factors and are unable to determine the long-range impacts of potential policies. While scientists realize that the complex relationships and cross-effects in myriad environmental and biological systems impact Earth's biosphere, current models often ignore these relationships or limit the systems' connections. The system complexities manifest in multiple interactions, feedback loops, emergent behaviors, and impending state changes or tipping points. The recent Nature article written by 22 internationally known scientists entitled "Approaching a state shift in Earth's biosphere" outlines many of the issues associated with the need for scientific models and the importance of predicting potential state changes of the planetary health systems. The article provides two specific quantitative modeling challenges in their call for better predictive models:1) To improve bio-forecasting through global models that embrace the complexityof Earth's interrelated systems and include the effects of local conditions on theglobal system and vice versa.2) To identify factors that could produce unhealthy global state-shifts and to showhow to use effective ecosystem management to prevent or limit these impending state changes.The resulting research question is whether we can build global models using local or regional components of the Earth's health that predict potential state changes and help decision makers design effective policies based on their potential impact on Earth's health. Although many warning signs are appearing, no one knows if Planet Earth is truly nearing a global tipping point or if such an extreme state is inevitable.The Nature article and many others point out that there are several important elements at work in the Earth's ecosystem (e.g., local factors, global impacts, multi-dimensional factors and relationships, varying time and spatial scales). There are also many other factors that can be included in a predictive model — human population, resource and habitat stress, habitat transformation, energy consumption, climate change, land use patterns, pollution, atmospheric chemistry, ocean chemistry, bio diversity, and political patterns such as social unrest and economic instability. Paleontologists have studied and modeled ecosystem behavior and response during previous cataclysmic state shifts and thus historic-based qualitative and quantitative information can provide background for future predictive models. However, it should be noted that human effects have increased significantly in our current biosphere situation.Requirements:You are members of the International Coalition of Modelers (ICM) which will soon be hosting a workshop entitled "Networks and Health of Planet Earth" and your research leader has asked you to perform modeling and analysis in advance of the workshop. He requires your team to do the following:Requirement 1: Build a dynamic global network model of some aspect of Earth's health (you develop the measure) by identifying local elements of this condition (network nodes) and appropriately connecting them (network links) to track relationship and attribute effects. Since the dynamic nature of these effects is important, this network model must include a dynamic time element that allows the model to predict future states of this health measure. For example, your nodes could be nations, continents, oceans, habitats, or any combination of these or other elements which together constitute a global model. Your links could represent nodal or environmental influences, or the flow or propagation of physical elements (such as pollution) over time. Your health measure could be any element of Earth's condition to include demographic, biological, environmental, social, political, physical, and/or chemical conditions. Be sure to define all the elements of your model and explain the scientific bases for your modeling decisions about network measures, nodal entities, and link properties. Determine a methodology to set any parameters and explain how you could test your model if sufficient data were available. What kinds of data could be used to validate or verify the efficacy of your model? (Note: If you do not have the necessary data to determine parameters or perform verification, do not throw out the model. Your supervisor realizes that, at this stage, good creative ideas and theories are as important as verified data-based models.) Make sure you include the human element in your model and explain where human behavior and government policies could affect the results of your model.Requirement 2: Run your model to see how it predicts future Earth health. You may need to estimate parameters that you would normally determine from data. (Remember, this is just to test and understand the elements of your model, not to use it for prediction or decision making.) What kinds of factors will your model produce? Could it predict state change or tipping points in Earth's condition? Could it provide warning about global consequences of changing local conditions? Could it inform decision makers on important policies? Do you take into account the human elements in your measures and network properties?Requirement 3: One of the powerful elements of using network modeling is the ability to analyze the network structure. Can network properties help identify critical nodes or relationships in your model? If so, perform such analysis. How sensitive is your model to missing links or changing relationships? Does your model use feedback loops or take into account uncertainties? What are the data collection issues? Does your model react to various government policies and could it thus help inform planning? Requirement 4: Write a 20-page report (summary sheet does not count in the 20 pages) that explains your model and its potential. Be sure to detail the strengths and weaknesses of the model. Your supervisor will use your report as a major theme in the upcoming workshop and, if it is appropriate and insightful to planetary health modeling, will ask you to present at the upcoming workshop. Good luck in your network modeling work!Potentially helpful references include:Anthony D. Barnosky, Elizabeth A. Hadly, Jordi Bascompte, Eric L. Berlow, James H. Brown, Mikael Fortelius, Wayne M. Getz, John Harte, Alan Hastings, Pablo A. Marquet, Neo D. Martinez, Arne Mooers, Peter Roopnarine, Geerat Vermeij, John W. Williams, Rosemary Gillespie, Justin Kitzes, Charles Marshall, Nicholas Matzke, David P. Mindell, Eloy Revilla, Adam B. Smith. "Approaching a state shift in Earth's biosphere,". Nature, 2012; 486 (7401): 52 DOI: 10.1038/nature11018Donella Meadows, Jorgen Randers, and Dennis Meadows. Limits to Growth: The 30-year update, 2004.Robert Watson and A.Hamid Zakri. UN Millennium Ecosystem Assessment Synthesis Report, United Nations Report, 2005.University of California - Berkeley. "Evidence of impending tipping point for Earth." ScienceDaily, 6 Jun. 2012. Web. 22 Oct. 2012.。

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。

【VIP专享】华南师范大学参加2013年美国数学建模竞赛获奖成绩

教练刘秀湘刘秀湘刘秀湘杨坦杨坦教练组教练组邓键邓键李湖南李湖南李湖南李湖南李湖南李湖南李湖南李庆峰刘秀湘
队员一黄嘉欣 2010 数师丁格曼 2010 物电林文钊 2009 物电叶馨蔚 2011 理综二冯星凯 2011 数勷勤宋帆 2010 光电陈国 2010 光电梁文玲 2010 数统计章瑜珍 2011 物电吴锐涛 2011 光勷勤胡长康 2010 物电梁婕 2011 物勷勤钟英文 2010 物电陈晓弘 2010 数金融姚献文 2010 数师何越 2010 数师黄锦东 2010 经管谢少茹 2011 理综一
一等奖 22308
一等奖ห้องสมุดไป่ตู้20506
一等奖 22082
一等奖 20879
一等奖 22480
二等奖 20382
二等奖 20749
二等奖 21061
二等奖 21593
二等奖 21691
二等奖 22266
二等奖 22398
二等奖 22532
二等奖 22791
二等奖 18193
二等奖 18560
控制号 18252
队员三黄培鸿 2010 计算机孙淼 2010 计算机林俊成 2009 物电区诵宜 2011 心理张树邦 2011 数勷勤张俊仪 2010 光电吴锐欢 2010 光电吴欣 2010 数统计黄迪 2011 物电彭冬雨 2011 理综一郑嘉鹏 2010 物电尹李明 2011 物勷勤陈祖杰 2010 物电程志南 2010 计算机关恩 2010 数师史华杰 2010 数师刘梓煊 2011 光电萧筠儒 2011 理综一
队员二蔡淡芸 2010 物电谢文欣 2010 数师林晓丹 2009 数师郑诗莹 2011 理综一余庆超 2011 数勷勤林琪璇 2010 光勷勤李泽曦 2010 光电胡嘉仪 2010 数统计许健 2011 物电陈彦霖 2011 光勷勤李奕华 2010 物电陈文创 2011 物勷勤黎智添 2010 物电谭卓荣 2010 数信计何威 2010 数师朱桂静 2010 数师邱柯蓝 2010 经管王碧霄 2011 理综一

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。

美赛建议协议范文

美赛建议协议范文尊敬的美赛组委会：我写信给您是为了提出一些建议，希望能够改进美赛的组织和进行。

美赛是一个很好的机会，让学生在团队合作的环境中提高数学建模和解决问题的能力。

然而，有一些方面可以进一步改善，以使比赛更具价值和公正性。

首先，建议增加准备时间。

目前，每支队伍只有4个小时的时间来解决问题，这对于许多参赛者来说是非常有挑战性的。

尤其是对于非英语为母语的学生来说，理解文档和题目可能需要更多的时间。

因此，我建议将准备时间延长到6或8个小时，这样可以给学生更多的思考和准备问题的时间，从而更好地展示他们的能力。

其次，建议增加更多的题目选择。

当前的比赛只有3个问题供选择，这使得队伍可能会因为题目没有太多的选择而在解决问题时受限。

增加更多的题目选择可以使参赛队伍能够根据自己的优势选择更适合的问题进行解决，同时也能够更好地展示他们的能力和才华。

第三，建议增加与现实生活相关的问题。

尽管当前的题目也是很有挑战性和有趣的，但是它们对于现实世界的应用可能不够明显。

考虑到美赛的目标是培养学生解决实际问题的能力，我建议组委会增加与现实生活相关的问题，这样可以更好地培养学生的应用能力和创新思维。

此外，建议组委会加强评分流程的透明度。

比赛的评分是决定获胜队伍的关键因素。

但是，目前的评分过程对参赛队伍来说是不透明的，他们无法了解或理解评委的评分标准和决策过程。

这可能导致参赛队伍对评分结果的不满和投诉。

我建议组委会在评分过程中增加透明度，向参赛队伍提供更多关于评分标准和过程的信息，以便他们更好地理解评委的决策依据。

最后，建议组委会加强与参赛学校和导师的交流。

当前，很多学校和导师对于比赛的组织和进行缺乏深入了解。

我建议组委会加强与参赛学校和导师的交流，提供更多关于比赛的信息和资源，以便学校和导师能够更好地准备学生参赛，并为学生提供更好的指导和支持。

希望我的建议能够对美赛的组织和进行有所改进。

美赛是一个非常有价值的活动，可以为学生提供宝贵的学习和成长机会。

2013美国大学生数学建模论文终稿

本文建立了三个模型(model)，模型一（model 1 ）用于解释不同形状的pan（从矩形到圆形中的任一形状）在其外围边沿的热量分布（原文：the distribution of heat across the outer edge of a pan for pans of different shapes --rectangular to circular and other shapes in between ），模型二用于在一定条件下选取最优形状的pan（the best type of pan (shape)），第三个模型为对问题一、二的优化（optimize（ .））方案。

In this paper, we formulate 3 relevant models. Through model 1, we display the distribution of heat across the outer edge of a pan for pans of different shapes -rectangular to circular and other shapes in between. While in model 2, we can select out the best type of pan in certain condition. Optimize a combination of model 1 and 2, then we get model 3.首先对于模型一，我们将求解pan的外沿热量分布（the distribution of heat across the outer edge of a pan ）转化为求解pan所在平面的温度场（temperature field），根据热力学理论（thermodynamic theory ）写出温度分布方程（Temperature distribution function），同时由假设条件（assumed condition）确定方程的初始条件（initial condition）和边界条件（boundary condition），利用Matlab（软件）求其数值解（numerical solution）。

2013年美赛a题h奖论文_大学论文

SummaryWe build two basic models for the two problems respectively: one is to show the distribution of heat across the outer edge of the pan for different shapes, rectangular, circular and the transition shape; another is to select the best shape for the pan under the condition of the optimization of combinations of maximal number of pans in the oven and the maximal even heat distribution of the heat for the pan.We first use finite-difference method to analyze the heat conduct and radiation problem and derive the heat distribution of the rectangular and the circular. In terms of our isothermal curve of the rectangular pan, we analyze the heat distribution of rounded rectangle thoroughly, using finite-element method. We then use nonlinear integer programming method to solve the maximal number of pans in the oven. In the even heat distribution, we define a function to show the degree of the even heat distribution. We use polynomial fitting with multiple variables to solve the objective function For the last problem, combining the results above, we analyze how results vary with the different values of width to length ratio W/L and the weight factor p. At last, we validate that our method is correct and robust by comparing and analyzing its sensitivity and strengths /weaknesses.Based on the work above, we ultimately put forward that the rounded rectangular shape is perfect considering optimal number of the pans and even heat distribution. And an advertisement is presented for the Brownie Gourmet Magazine.Contents1 Introduction (3)1.1Brownie pan (3)1.2Background (3)1.3Problem Description (3)2. Model for heat distribution (3)2.1 Problem analysis (3)2.2 Assumptions (4)2.3 Definitions (4)2.4 The model (4)3 Results of heat distribution (7)3.1 Basic results (7)3.2 Analysis (9)3.3 Analysis of the transition shape—rounded rectangular (9)4 Model to select the best shape (11)4.1 Assumptions (11)4.2 Definitions (11)4.3 The model (12)5 Comparision and Degree of fitting (19)6 Sensitivity (20)7 Strengths/weaknesses (21)8 Conclusions (21)9 Advertisement for new Brownie Magazine (23)10 References (24)1 Introduction1.1Brownie panThe Brownie Pan is used to make Brownies which are a kind of popular cakes in America. It usually has many lattices in it and is made of metal or other materials to conduct heat well. It is trivially 9×9 inch or 9×13 inch in size. One example of the concrete shape of Brownie pan is shown in Figure 1Figure 1 the shape of Brownie Pan (source: Google Image)1.2BackgroundBrownies are delicious but the Brownie Pan has a fetal drawback. When baking in a rectangular pan, the food can easily get overcooked in the 4 corners, which is very annoying for the greedy gourmets. In a round pan, the heat is evenly distributed over the entire outer edge but is not efficient with respect to using in the space in an oven, which most cake bakers would not like to see. So our goal is to address this problem.1.3Problem DescriptionFirstly, we are asked to develop a model to show the distribution of heat across the outer edge of a pan for different shapes, from rectangular to circular including the transition shapes; then we will build another model to select the best shape of the pan following the condition of the optimization of combinations of maximal number of pans in the oven and maximal even distribution of heat for the pan.2. Model for heat distribution2.1 Problem analysisHere we use a finite difference model to illustrate the distribution of heat, and it has been extensively used in modeling for its characteristic ability to handle irregular geometries and boundary conditions, spatial and temporal properties variations1. In literature 1, samples with a rectangular geometric form are difficult to heat uniformly,particularly at the corners and edges. They think microwave radiation in the oven can be crudely thought of as impinging on the sample from all, which we generally acknowledge. But they emphasize the rotation.Generally, when baking in the oven, the cakes absorb heat by three ways: thermal radiation of the pipes in the oven, heat conduction of the pan, and air convection in the oven. Considering that the influence of convection is small, we assume it negligible. So we only take thermal radiation and conduction into account. The heat is transferred from the outside to the inside while water in the cake is on the contrary. The temperature outside increase more rapidly than that inside. And the contact area between the pan and the outside cake is larger than that between the pan and the inside cakes, which illustrate why cakes in the corner get overcooked easily.2.2 Assumptions● We take the pan and cakes as black body, so the absorption of heat in eacharea unit and time unit is the same, which drastically simplifies ourcalculation.● We assume the air convection negligible, considering its complexity and thesmall influence on the temperature increase .● We neglect the evaporation of water inside the cake, which may impede theincrease of temperature of cakes.● We ignore the thickness of cakes and the pan, so the model we build istwo-dimensional.2.3 DefinitionsΦ: heat flows into the nodeQ: the heat taken in by cakes or pans from the heat pipesc E ∆: energy increase of each cake unitp E ∆: energy increase of the pan unit,i m n t : temperature at moment i and point (m,n)C 1: the specific heat capacity of the cakeC 2: the specific heat capacity of the panipan t : temperature of the pan at moment iT 1: temperature in the oven, which we assume is a constant2.4 The modelHere we use finite-difference method to derive the relationship of temperatures at time i-1 and time i at different place and the relationship of temperatures between the pan and the cake.First we divide a cake into small units, which can be expressed by a metric. In the following section, we will discuss the cake unit in different places of the pan.Step 1；temperatures of cakes interior(m,n+1)x△Figure 2 heat flow According to energy conversation principle, we can get 0up down left right c Q E Φ+Φ+Φ+Φ+-∆=(2.4.1) Considering Fourier Law and △x=△y, we get1,,1,,1,,1,,,1,,1,,1,,1,()()()()i i m n m n i i left m n m n i i m n m n i i right m n m n i i m n m n i i up m n m n i i m n m ni i down m n m n t t y t t xt t y t t x t t x t t y t t x t t y λλλλλλλλ--++++---Φ=-∆=-∆-Φ=∆=-∆-Φ=∆=-∆-Φ=∆=-∆ (2.4.2)According to Stefan-Boltzman Law,441,[()]i m n Q Ac T t σ=- (2.4.3)Where A is the area contacting, c is the heat conductance.σis the Stefan-Boltzmann constant, and equals 5.73×108 Jm -2s -1k -4.21,,()i i c m n m n E cm t t -∆=-(2.4.4) Substituting (2.4.2)-(2.4.4) into (2.4.1), we get441,1,,1,1,1,1,,4[()]()0i i i i i i m n m n m n m n m n m n i i m n m n Ac t t t t t T t cm t t σλλ-++--+++-+---=This equation demonstrates the relationship of temperature at moment i and moment i-1 as well as the relationship of temperature at (m,n) and its surrounding points.Step 2: temperature of the cake outer and the pan● For the 4 cornerscakeFigure 3 the relative position of the cake and the pan in the first cornerBecause the contacting area is two times, we get4411,1,122[()]2()i i i i m n pan pan A c T t t c M t t σλ---=-● For every edgecakeFigure 4 the relative position of the cake and the pan at the edgeSimilarly, we derive4411,,2[()]()i i i i m n m n pan pan Ac T t t c M t t σλ---=-Now that we have derived the express of temperatures of cakes both temporally and spatially, we can use iteration to get the curve of temperature with the variables, time and location.3 Results of heat distribution3.1 Basic resultsRectangularPreliminarily, we focus on one corner only. After running the programme, we obtain the following figure.Figure 5 heat distribution at one cornerFigure 5 demonstrates the temperature at the corner is higher than its surrounding points, that’s why food at corners get overcooked easily.Then we iterate globally, and get Figure 6.Figure 6 heat distribution in the rectangular panFigure 6 can intuitively illustrates the temperature at corners is the highest, and temperature on the edge is less higher than that at corners, but is much higher than that at interior points, which successfully explains the problem “products get overcooked at the corners but to a lesser extent on the edge”.After drawing the heat distribution in two dimensions, we sample some points from the inside to the outside in a rectangular and obtain the relationship between temperature and iteration times, which is shown in Figure 7Figure 7From Figure 7, the temperatures go up with time going and then keep nearly parallel to the x-axis. On the other hand, temperature at the center ascends the slowest, then edge and corner, which means given cooking time, food at the center of the pan is cooked just well while food at the corner of the pan has already get overcooked, but a lesser extent to the edge.RoundWe use our model to analyze the heat distribution in a round, just adapting the rectangular units into small annuluses, by running our programme, we get the following figure.Figure 8 the heat distribution in the circle panFigure 8 shows heat distribution in circle area is even, the products at the edge are cooked to the same extent approximately.3.2 AnalysisFinally, we draw the isothermal curve of the pan.●RectangularFigure 9 the isothermal curve of the rectangular panFigure 9 demenstrates the isothermal lines are almost concentric circles in the center of the pan and become rounded rectangles outer, which provides theory support for following analysis..●CircularFigure 10 the isothermal curve of circularThe isothermal curves of the circular are series of concentric circles, demonstrating that the heat is even distributed.3.3 Analysis of the transition shape—rounded rectangularFrom the above analysis, we find that the isothermal curve are nearly rounded rectangulars in the rectangular pan, so we perspective the transition shape between rectangular and circular is rounded rectangular, considering the efficiency of using space of the oven and the even heat distribution. In the following section, we will analyze the heat distribution in rounded rectangular pan using finite element approach.During the cooking process, the temperature goes up gradually. But at a certain moment, the temperature can be assumed a constant. So the boundary condition yields Dirichlet boundary condition. And the differential equation is:22220T T x y∂∂+=∂∂ Where T is the temperature, and x, y is the abscissa and the ordinate.And the boundary condition is T=constant.After running the programme, we get the heat distribution in a rounded rectangular pan, the results is in the following.Figure 11 the heat distribution in a rounded rectangular pan From 11, we can see the temperature of the edge and the corner is almost the same, so the food won ’t get overcooked at corners. We can assume the heat in a rounded rectangular is distributed uniformly. We then draw the isothermal curve of the rounded rectangular pan.Figure 12 the isothermal curve of the rounded rectangular pan To show the heat distribution more intuitively, we also draw the vertical view of the heat distribution in a rounded rectangular pan.Figure 13 the vertical view of the heat distribution on a rounded rectangular platform4 Model to select the best shape4.1 AssumptionsBesides the assumptions given, we also make several other necessary assumptions.●The area of the even equals the area of the pan with small lattices in it.●There is no space between lattices or small pans on the pan.4.2 DefinitionsS: the area of the ovenk : the width of the external rectangular of the rounded rectangularh :the length of the external rectangular of the rounded rectangulara1: the ratio of the width and length of the external rectangular of the rounded rectangular, equals k/ha2: the ratio of the width and length of the external rectangular of the rounded rectangular, equals W/Ln: the amounts of the rounded rectangular in each rowm: the amounts of the rounded rectangular in each columnr: the radius of the rounded rectangularIn order to illustrate more clearly, we draw the following sketch.Figure 144.3 The modelProblem ⅠWe use nonlinear integer programming to solve the problem.Figure 15 the configuration of the rounded rectangular and the pan:max objective function N n m =⋅222:,004subject to n k m h k h A r r A r r Sππ⋅≤⋅≤==≥≥≤-+≤Where n, m are integers.Both a1 and a2 are variables, we can study the relationship of a1 and N at a given a2. Here we set a2=0.8.Considering the area of the oven S and the area of the small pan A are unknown, we collect some data online, which is shown in the following table.Table 1(source: / )Then we calculate the average of S and A , respectively 169.27 inch 2 and 16.25 inch 2. After running the programme, we get the following results.Figure 16 the relationship of the maximal N and the radius of the rounded rectangular r From the above figure, we know as r increases, the optimal number of pans decreases. For the data collected can not represent the whole features, the relationship is not obvious .Then we change the area of the oven and the area of each pan, we draw another figure.Figure 17This figure intuitively shows the relationship of r and N.Appearently, the ratio of the width to length of the rounded rectangular has a big influence on the optimization of N. In the following section, we will study this aspect.Figure 18 the relationship of N and a1 Figure 18 illustrates only when the ratio of the width to the length of the rounded rectangular a1 equals the ratio of the width and length of the oven a2, can N be optimized.Finally, we take both a1 and a2 as variables and study the relationship of N and a1, a2. The result is as follows.Figure 19 the relationship of N and a1, a2 Problem ⅡTo solve this problem, we first define a function u(r) to show the degree of the even heat distribution for different shapes. ()s u r AWhere s is the area surrounded by the closed isothermal curve most external of the pan.Now we think the rationality of the function. The temperature of the same isothermal curve is equivalent. We assume the temperature of the closed isothermal curve most external of the pan is t 0 , for the unclosed isothermal, the temperature is higher than t 0, and the temperature inside is lower than t 0. So we count the number of the pixel points inside the closed isothermal curve most external of the pan num 1 and the number of the whole pixel points num 2. Consequently, u(r)=num 1/num 2. To illustrate more clearly, we draw the following figure.The following table shows the relationship of u and r. And we set a=1Figure 21 the scatter diagram of u and rFirst, we consider u and r is linear, and by data fitting we deriveu r r=+⨯()0.85110.021Then, we consider u and r is second -order relationship, and we derive anothercurve.Figure 22 another relationship of u and r2 =+⨯-⨯u r r r()0.85110.02880.001According to the points we count, we can get the following figure, demonstrating the relationship u and r, a.Figure 23 the relationship of u and r, We aFrom the analysis above, we find with r increasing, u increases, which means when the radius of the rounded rectangular r increases, the degree of the even heat distribution. Given one extreme circumstance, when r gets its maximal value, the pan becomes a circular, and the degree of the even heat distribution is also the most.Finally, we can derive that the degree of the even heat distribution increases from the rectangular, rounded rectangular with smaller r, the rounded rectangular with bigger r and the circular.Figure 24 the comparision of degree of even heat distribution for different shapes Problem ⅢIn this section, we add a weight factor p to analyze the results with the varying values of W/L and p.:max (1)(,)objective function pN p u r a +-222:,004subject to n k m h k h A r r A r r Sππ⋅≤⋅≤==≥≥≤-+≤01p ≤≤In this problem, there are three variables , a, r and p .what we need to do is to select the best shape for the pan, namely, to select a and r. Firstly, we set a and get the relationship of objective function, y and r, p.Figure 25 the relationship of y and r, p From the figure, we can see with p decreasing and r increasing, which means the degree of the even heat distribution is larger, y increases.Then we set p, and get the relationship of the y and r, a.Figure 26 the relationship of y and a, rThis figure shows that the value a has little influence on the objective function.5 Comparision and Degree of fittingComparisionWhen solving the problem to select the best shape of the pan, we only take the rounded rectangular into account and ignore other shapes, Here, we concentrate on one of the polygon—the regular hexagon as an example to demonstrate our model is correct and retional.We use finite element method to derive the heat distribution, just like analyzing the rounded rectangular.Figure 27 the heat distribution of the regular hexagon panFigure 28 the heat distribution of the regular hexagon pan in two dimensions From the above two pictures, we can see the temperature of the corners is much higher than that in other sections of the pan. So the food at corners gets overcooked more easily. In fact, the longer the distance from the center to the corner is , the higher the temperature becomes.● Degree of fittingIn the second problem of selecting the best shapes of the pan, we get two equations of u and r by data fitting. Now, we will analyze the degree of fitting, which is expressed by the residual errors.To calculate the residual errors, we use the following equations.1ˆ()ˆˆT T Y X X X X Y YX βεββ-=+== Finally, we can get the residual errors by2ˆ()L i i iS y y=-∑ Where i is the number of the data sampled, here i is 11.S 1=5.021 and S 2=0.018. Obviously, the second equation is more accuracy.6 Sensitivity● From figure 16 and 17, we know that the larger the ratio of the area of theoven and the area of the pan is, the better our model fits.● We differentiate the equation derived by figure 22,and find that with rincreasing, the value of the differential goes down, which means u will become steady as r increases.● We analyze figure 23, and find the higher the value of r is, the moreobvious the influence a has on u.●We draw another figure as follows in comparision with figure 25, and getwhen the area of the oven S is very small, the relationship of y and r, p andbe seen more apparently.Figure 29 the relationship of y and r, p7 Strengths/weaknessesStrengths:●We use different methods, infinite difference and infinite element, to buildthe model, and the conclusions are consistent with each other.●We compare the heat distribution of the rounded rectangular and the regularhexagon and then calculate the residual errors, validating our model iscorrect.●The results generated by our model agree with empirical results.●Our model is straight, common and easy to understand.Weaknesses:●We didn’t give an analytic solution for the optimal number of the pans in theoven, N.●The model doesn’t take into account detailed things, like the air convectionin the oven.8 ConclusionsWe propose several models to solve the problem of the heat distribution and the optimization of the pan’s shape combining the maximal number of pans in the ovenand the maximal even heat distribution. After detailed analysis, we can get the following conclusions:●Rectangular can best fit in the oven considering the best efficiency of usingspace in an oven.●To get the maximal number of the pans in an oven, we should set the ratio ofthe width to the length of the oven equals that of the small pans.●Generally, the heat distribution of the circular is the most even. But when itcomes to the combination of the efficiency of the using space and the evenheat distribution, the rounded rectangular fits well. And with the radiusincreasing, the degree of even heat distribution increases, resulting in lesserefficiency of using space.9 Advertisement for new Brownie MagazineLove Brownies? Of course, follow us to see our new-designed ultimate Brownie Pan!Almost every Brownie gourmet may encounter the same problems when baking Brownies, cakes or other gourmets. And the most annoying thing may lie in the uneven cooked gourmets. For the heat is distributed uneven in the present pans, after baked, the cakes often can’t be get out of the pan easily or the edge is always difficult to cut because it is too filmsy. What’s worse, the overcooked food taste bad and become unhealthy containing bad things. But now, things are different. All of these trouble problems will disappear for we have ultimate pans. After careful calculation and analysis, we designed a new Brownie Pan—the rounded rectangular shape pan. We study the heat distribution thoroughly of different shapes of pans and find that, the rounded rectangular shape is almost perfect in terms of even heat distribution. The cakes at the corner of the pan will never be overcooked as long as you set the temperature appropriately. And you can get out of your edge-crisp and chewy-inside cake whenever you want. So, is it wonderful?Another troublesome thing is that, the traditional pan usually can’t get clean easily for its straight angle., which bring about many complaints from customers at American Amazon online shop. But for our rounded rectangular pan, you won’t worry about this trifle! We guarantee our ultimate Brownie pan is simple and time-saving to clean. And we recommend aluminum as the material of the pan, for it’s light and portable.In our model, we optimize the number of pans in each oven and derive the relationship of the number N and the radius r , the ratio of the width to the length of the pan . So given the ratio of the width to the length, we can get a certain r, and the number is also determined. Or given the radius r, we can also design the pan. This wonderful because different people have different demands for the number of the pans in each oven. Considering a family party or doing baby food, we need different number, of course.I believe the following merits may be attractive for most manufactures. We guarantee the rounded rectangular shape pan can save many materials. And the simple style confirms to the values of beauty. The environment-friendly, low –carbon style pan bring a new try for customers.Bring our ultimate Brownie pan to your home, you will find more surprises!10 References1Shixiong Liu, Mika Fukuoka, Noboru Sakai, A finite element model for simulating temperature distributions in rotating food during microwave heating, Journal of food engineering, V olume 115, issue 1, March 2013 Page 49-62 2Heat transfer theory /unitoperations/httrtheory.htm。

谢金星告数学建模爱好者的一封信

告数学建模爱好者的一封信清华大学数学科学系谢金星Email: jxie@一年一度的美国大学生数学建模竞赛即将开赛，应数学中国网站马壮站长之约，我通过这个网络平台和各位数学建模爱好者就大学生数学建模竞赛中的纪律问题进行一下交流、沟通。

大家知道，数学建模竞赛对促进数学教育改革、丰富学生课外科技活动起到了很好的作用，是培养大学生竞争意识和团队精神、提高大学生创新能力和综合素质的一个具体的、重要的载体。

二十多年来，在广大数学建模爱好者的大力支持和积极参与下，大学生数学建模竞赛活动在我国得到了快速的发展，取得了巨大的成绩：我国的竞赛规模长期居于世界最一，受益面很大；近几年我国参赛队已经占到美国赛参赛总队数的90%左右（还有继续上升的势头），并取得了较好的竞赛成绩，展示了中国大学生的能力与风采，也在一定程度上显示了中国高等教育的成绩。

所以很多人说，数学建模竞赛是在美国诞生，在中国开花、结果的。

成绩、优点有目共睹，就不多谈了。

我这里想说的主要是存在的问题和不足，特别是竞赛纪律问题。

虽然竞赛是通讯赛的形式，没有专人监督，但参赛队应该自律，自觉遵守竞赛规则（中美竞赛的规则基本是类似的，我们这里不加区分）。

具体来讲，按照竞赛规则（摘自[1]），“参赛队可以利用任何非生命提供的数据和资料——包括计算机，软件，参考书目，网站，书籍等，但是所有引用的资料必须注明出处，如有参赛队未注明引用的内容的出处，将被取消参赛资格。

”这一条实际上是关于写作规范的要求，近年来通过各类赛前培训活动和指导教师的努力，已经反复向参赛学生强调，多数参赛队已经认识到其重要性。

之所以大家对此很重视，是因为如果引用不规范，很容易被认定为剽窃，从而失去评奖资格；此外，这种写作规范实际上比较容易通过写作技巧的训练达到所希望的效果。

但竞赛还有另一条如下的重要规则（摘自[1]）：“参赛队成员不允许向指导教师或者除了本团队成员以外的其他人寻求帮助或讨论问题。

与除本团队成员以外的人的任何形式的接触都是严格禁止的。

格式模板2

承诺书我们仔细阅读了中国大学生数学建模竞赛的竞赛规则.我们完全明白，在竞赛开始后参赛队员不能以任何方式（包括电话、电子邮件、网上咨询等）与队外的任何人（包括指导教师）研究、讨论与赛题有关的问题。

我们知道，抄袭别人的成果是违反竞赛规则的, 如果引用别人的成果或其他公开的资料（包括网上查到的资料），必须按照规定的参考文献的表述方式在正文引用处和参考文献中明确列出。

我们郑重承诺，严格遵守竞赛规则，以保证竞赛的公正、公平性。

如有违反竞赛规则的行为，我们将受到严肃处理。

我们参赛选择的题号是（从A/B/C/D中选择一项填写）：我们的参赛报名号为（如果赛区设置报名号的话）：所属学校（请填写完整的全名）：参赛队员(打印并签名) ：1.2.3.指导教师或指导教师组负责人(打印并签名)：日期：年月日赛区评阅编号（由赛区组委会评阅前进行编号）：编号专用页赛区评阅编号（由赛区组委会评阅前进行编号）：赛区评阅记录（可供赛区评阅时使用）：评阅人评分备注全国统一编号（由赛区组委会送交全国前编号）：全国评阅编号（由全国组委会评阅前进行编号）：论文题目摘要一、问题重述正文部分二、问题的初步分析（1）正文部分（2）正文部分（3）正文部分三、基本假设1、一级标题（1）正文部分（2）正文部分（3）正文部分2、一级标题（1）正文部分3、一级标题（1）正文部分四、符号说明p：a:i符号全部用符号编辑器编辑？五、问题的分析和解答1、一级标题(1)正文部分(2)正文部分(3)正文部分2、一级标题（1）正文部分3、一级标题（1）正文部分六、模型的评价与改进1、模型的评价正文部分2、模型的改进正文部分参考文献书籍的表述方式为：[编号] 作者，书名，出版地：出版社，出版年参考文献中期刊杂志论文的表述方式为：[编号] 作者，论文名，杂志名，卷期号：起止页码，出版年参考文献中网上资源的表述方式为：[编号] 作者，资源标题，网址，访问时间（年月日）。