2013年数学建模A优秀论文2

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2013年研究生数学建模优秀论文E2

二、模型假设
假设一：收入是划分中等收入的主要依据。但反映收入水平的具体指标可以有多种可能，如城镇居民人均年全部收入、人均年生活费收入、人均年可支配收入，农村居民可以有人均年纯收入等，因为虽然这些指标的口径不同，但并不影响反映居民收入水平的真实性只是在同一范围内应选用相同口径的指标进行分析
3
假设二：中等收入一个用区域值表示的数量，而不是一个确定的数值假设三：收入域值的界定具有一定的假定性从下面中等收入的界定方法中，可以看到最低收入水平和最高收入水平的界定都有假定条件假设四：收入域值的边界也是模糊的，可以有一定的上下浮动空间。
参赛密码（由组委会填写）
第十届华为杯全国研究生数学建模竞赛
学
校
贵州大学
参赛队号
10657008 1.秦书琳
队员姓名
2.王影 3.任丽
参赛密码（由组委会填写）
第十届华为杯全国研究生数学建模竞赛
题目
E:中等收入人口度量与实证研究
摘
要：
本文对中等收入人口进行了定量研究，并建立了有效的度量中等人口范围的数学模型。 Lorentz 曲线是研究人口收入的一个重要手段，为此，本文首先提出了一种拟合 Lorentz 曲线的模型，用 L( p) p (1 (1 p) ) , 1, 0 ， 0 1, 来拟合 Lorentz 曲线，得到了相当精确的结果（MSE 精确到 106 ）。该模型在此时成立，这是之前所没有的结论。为了度量中等收入人口，本文提出了衡量社会分配公平的局部公平指数
此模型有效地解决了传统“人口空间”模型对两极分化不敏感的问题。基于以上研究，本文提出了更为一般的中等收入人口的定义、原理、即经济学意义，并提出了测算方法。关键词：中等收入；Lorentz 曲线；分配公平度

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。

数学建模竞赛优秀大学生论文

数学建模竞赛优秀大学生论文随着科学技术的高速发展，数学的应用价值越来越得到众人的重视，因此数学建模也被逐渐的引起重视了。

下面是店铺为大家整理的数学建模优秀论文，供大家参考。

数学建模优秀论文篇一：《数学建模用于生物医学论文》1数学建模的过程1.1模型准备首先要了解实际背景，寻找内在规律，形成一个比较清晰的轮廓，提出问题。

1.2模型假设在明确目的、掌握资料的基础上，抓住问题的本质，舍弃次要因素，对实际问题做出合理的简化假设。

1.3模型建立在所作的假设条件下，用适当的数学方法去刻画变量之间的关系，得出一个数学结构，即数学模型。

原则上，在能够达到预期效果的基础上，选择的数学方法应越简单越好。

1.4模型求解建模后要对模型进行分析、求解，求解会涉及图解、定理证明及解方程等不同数学方法，有时还需用计算机求数值解。

1.5模型分析、检验、应用模型的结果应当能解释已存的现象，处理方法应该是最优的决策和控制方案，所以，对模型的解需要进行分析检验。

把求得的数学结果返回到实际问题中去，检验其合理性。

如果理论结果符合实际情况，那么就可以用它来指导实践，否则需再重新提出假设、建模、求解，直到模型结果与实际相符，才能进行实际应用。

总之，数学建模是一项富有创造性的工作，不可能用一些条条框框的规则规定的十分死板，只要是能够做到全面兼顾、能抓住问题的本质、最终检验结果合理，都是一个好的数学模型。

2数学建模在生物医学中的应用2.1DNA序列分类模型DNA分子是遗传信息存储的基本单位，许多生命科学中的重大问题都依赖于对这种特殊分子的深入了解。

因此，关于DNA分子结构与功能的问题，成为二十一世纪最重大的课题之一。

DNA序列分类问题是研究DNA分子结构的基础，它常用的方法是聚类分析法。

聚类分析是使用数据建模简化数据的一种方法，它将数据分成不同的类或者簇，同一个簇中的数据有很大的同质性，而不同的簇中的数据有很大的相异性。

在对DNA序列进行分类时，需首先引入样品变量，比如说单个碱基的丰度、两碱基丰度之比等;然后计算出每条DNA序列的样品变量值，存入到向量中;最后根据相似度度量原理，计算出所有序列两两之间的Lance与Williams距离，依据距离的远近进行分类。

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!。

2013全国数模竞赛A题优秀论文祥解

2013高教社杯全国大学生数学建模竞赛编号专用页赛区评阅编号（由赛区组委会评阅前进行编号）：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）：车道被占用对城市道路通行能力的影响摘要本文主要研究车道被占用对城市道路通行能力的影响并建立了相应的数学模型。

针对问题一，考虑到交通信号灯的周期，我们选择1分钟为周期，结合不同车辆的标准车当量的折算系数，求出每个采样点的交通量，通过MATLAB作图，从定性方面对道路通行能力进行分析，然后通过基本通行能力和4个修正系数建立动态通行能力的模型。

图像显示，事故发生后（采样点5附近），实际通行能力下降至一个较低水平，并且横断面处的实际能力变化过程呈先下后上的波形变化，在事故解决（第20个采样点）以后，由图像看出实际通行能力持续上升。

针对问题二，利用问题一建立的模型，结合视频二，比较交通事故所占不同车道时横断面的实际通行能力，可以发现二者实际通行能力变化趋势大致相同，但视频二实际通行能力大于视频一实际通行能力。

可见占用车流量大的车道使道路通行能力降低更多。

针对问题三，首先我们建立单车道排队车辆数目的积分模型，单个车道的滞留车辆为上游车流量和实际通行能力的差值。

我们以30s为一个时间段，对视频一中的车流量进行统计，得到横截面处每个监测段的实际通行能力。

本题要求考虑三车道，总体排队长度不容易通过积分模型确定，所以我们将队列长度问题转化为车辆数目问题，通过视频资料统计120米对应24辆车，据此关系转换，从而得到车辆排队长度与事故横断面实际通行能力、事故持续时间和上游车流量的关系。

针对问题四，在对问题3研究的基础上，根据问题3建立的数学模型，建立起某一段时间间隔车辆排队的长度，然后，通过求得的关系得到当排队长度为140m的时候所对应的时间段，由于每段时间间隔设为30s，因此，可以求得排队长度到达上游时用的时间为347.7273s。

关键词：交通事故车道占用通行能力排队论一、问题的重述车道被占用是指因交通事故、路边停车、占道施工等因素，导致车道或道路横断面通行能力在单位时间内降低的现象。

优秀的数学建模论文范文(通用8篇)

优秀的数学建模论文范文第1篇摘要：将数学建模思想融入高等数学的教学中来，是目前大学数学教育的重要教学方式。

建模思想的有效应用，不仅显著提高了学生应用数学模式解决实际问题的能力，还在培养大学生发散思维能力和综合素质方面起到重要作用。

本文试从当前高等数学教学现状着手，分析在高等数学中融入建模思想的重要性，并从教学实践中给出相应的教学方法，以期能给同行教师们一些帮助。

关键词：数学建模；高等数学；教学研究一、引言建模思想使高等数学教育的基础与本质。

从目前情况来看，将数学建模思想融入高等教学中的趋势越来越明显。

但是在实际的教学过程中，大部分高校的数学教育仍处在传统的理论知识简单传授阶段。

其教学成果与社会实践还是有脱节的现象存在，难以让学生学以致用，感受到应用数学在现实生活中的魅力，这种教学方式需要亟待改善。

二、高等数学教学现状高等数学是现在大学数学教育中的基础课程，也是一门必修的课程。

他能为其他理工科专业的学生提供很多种解题方式与解题思路，是很多专业，如自动化工程、机械工程、计算机、电气化等必不可少的基础课程。

同时，现实生活中也有很多方面都涉及高数的运算，如，银行理财基金的使用问题、彩票的概率计算问题等，从这些方面都可以看出人们不能仅仅把高数看成是一门学科而已，它还与日常生活各个方面有重要的联系。

但现在很多学校仍以应试教育为主，采取填鸭式教学方式，加上高数的教材并没有与时俱进，将其与生活的关系融入教材内，使学生无法意识到高数的重要性以及高数在日常生活中的魅力，因此产生排斥甚至对抗的心理，只是在临考前突击而已。

因此，对高数进行教学改革是十分有必要的，而且怎么改，怎么让学生发现高数的魅力，并积极主动学习高数也是作为教师所面临的一个重大问题。

三、将数学建模思想融入高等数学的重要性第一，能够激发学生学习高数的兴趣。

建模思想实际上是使用数学语言来对生活中的实际现象进行描述的过程。

把建模思想应用到高等数学的学习中，能够让学生们在日常生活中理解数学的实际应用状况与解决日常生活问题的方便性，让学生们了解到高数并不只是一门课程，而是整个日常生活的基础。

2013年五一数学建模联赛 A题论文

图 1 男女大学生体重指数一览表
人数
体重指数一览表
800
712
700
600 500
475
400
300
200 100
10264 38
237 183157 26
0
7267 5
低体重者
正常体重者
超重者
四类人群
肥胖者
学生数男生数女生数
2
4.2.3 结果与分析
对照表 1 及相关的频数表，反映出大学生 BMI 指数基本正常，正常体重人群相对集
关键词：独立样本 t 检验方差检验 K-S 检验模糊综合评价 SPSS
一问题重述
据数据显示，近年来中国大学生的体质健康水平呈下降趋势。学生或者过重或者过瘦，对大学生体质健康的评价问题对于如何提高大学生的体质健康水平具有指导意义。
根据对某高校大一新生 36 个班级共 1000 多名学生进行的体质与健康测试，得到了一组相关资料，由于测试过程中学生未能按照要求规范测试，导致测量结果中出现一些偏差，进而影响了体质健康的测试，请结合各项测试评分标准，回答下列问题：问题 1：影响大一新生的体质健康状况的因素很多，体重是体现体质健康状况的重要指标，分析体重对体质健康的影响；在体质健康测试中，测试结果可能存在误差，在附表 1 中，有些测量资料不能反映同学的真实水平，根据附表 1 数据，请建立数学模型检验测试结果的正确性和准确性，找出附表 1 中 1、2、3 班同学的可能偏差测试结果，并说明理由。问题 2：生源地是影响体质健康状况的因素，请在不同生源地选取适当的样本，试检验不同地区学生的体能健康是否具有显著差别。问题 3：目前，我国体能测试主要采用《国家体质健康标准》对学生体质进行评价，根据附表 2 中（男生：sheet1；女生：sheet2）项目评价标准，试建立体质健康评价模型，评价该校学生的体质健康状况，并对 1 班的 30 名同学进行体质健康评价。问题 4：我国大多数高校学生体质健康合格率未达到国家要求，对于未达标的大一新生来说，就如何让学生在在校期间提高自身的体质健康写一份建议报告书，其中包括提高体质健康水平的措施和手段，如何量化提高体质健康指标等问题。

2013美国数学建模A题优秀论文

终极布朗尼烤烤盘一、摘要根据题意，我们把把要解决的分成三个问题；第一个就是建立一个模型来表示整个烤盘的外边缘热量的分布。

第二个就是优化组合题目中条件1和条件2，使得权重p和(1- p)能够描述随着W/L和p值的改变，最佳的烤烤盘形状和热量分布情况是如何改变的第三个问题就是为布朗尼美食家杂志准备一到两页的宣传广告，需要突出设计和结果。

对于第一个问题，我们结合傅里叶定律构建了二维热传导模型；然后通过模型中的S来限定范围得到六种不同形状烤盘对应的热传导偏微分方程。

然后对模型赋值和第二类边界条件（Neumann边界条件）下，应用comsol得出六种烤盘稳定热量分布图像和烤盘外边缘热量分布图像。

通过输出的图像，我们得出结论：矩形四角处温度较高，圆形外边缘热量分布比较均匀；随着烤盘边数的增加，烤盘外边缘热量分布愈加均匀，但在角处温度仍然会高一些对于问题二对于问题三关键词：二、问题重述当用一个长方形的平底烤盘（盘）烘烤时，热量被集中在4个角，在角落处，食物可能被烤焦了，而边缘处烤的不够熟。

在一个圆形的平底烤盘（盘）热量被均匀地分布在整个外边缘，在边缘处食物不会被烤焦。

但是，大多数的烤箱的形状是矩形的，采用了圆形的烤盘（盘）相对于烤箱的使用空间而言效率不高。

为所有形状的烤盘（盘）----包括从矩形到圆形以及中间的形状，建立一个模型来表示整个烤盘（盘）的外边缘热量的分布。

假设：1. 形状是矩形的烤箱宽长比为W/L；2. 每个烤烤盘（盘）的面积为A；3. 每个烤箱最初只有两个均匀放置的烤架。

根据以下条件，建立一个能使用的最佳类型或形状的烤烤盘（盘）：1.放入烤箱里的烤烤盘（盘）数量的最大值为(N)；2.烤烤盘（盘）的平均分布热量最大值为(H)；3.优化组合条件1和条件2，使得权重p和(1- p)能够描述随着W/L和p值的改变，最佳的烤烤盘形状和热量分布情况是如何改变的。

除了完成规定的解决方案，为布朗尼美食家杂志准备一到两页的宣传广告，需要突出你的设计和结果。

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高教社杯全国大学生数学建模竞赛A题论文.

2013高教社杯全国大学生数学建模竞赛承诺书我们仔细阅读了《全国大学生数学建模竞赛章程》和《全国大学生数学建模竞赛参赛规则》（以下简称为“竞赛章程和参赛规则”，可从全国大学生数学建模竞赛网站下载）。

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章程和参赛规则的，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，严格遵守竞赛章程和参赛规则，以保证竞赛的公正、公平性。

如有违反竞赛章程和参赛规则的行为，我们将受到严肃处理。

我们授权全国大学生数学建模竞赛组委会，可将我们的论文以任何形式进行公开展示（包括进行网上公示，在书籍、期刊和其他媒体进行正式或非正式发表等）。

我们参赛选择的题号是（从A/B/C/D中选择一项填写）： A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）：吉林医药学院参赛队员(打印并签名) ：1. 于邦文2. 薛盈军3. 杨国庆指导教师或指导教师组负责人(打印并签名)：霍俊爽（论文纸质版与电子版中的以上信息必须一致，只是电子版中无需签名。

以上内容请仔细核对，提交后将不再允许做任何修改。

如填写错误，论文可能被取消评奖资格。

）日期： 2013 年 9 月 16 日赛区评阅编号（由赛区组委会评阅前进行编号）：2013高教社杯全国大学生数学建模竞赛编号专用页赛区评阅编号（由赛区组委会评阅前进行编号）：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）：车道被占用对城市道路通行能力的影响摘要本文通过对城市中车道因交通事故被占用问题的分析，探讨了事故所处道路横断面的实际通行能力的变化过程，并依据事故路段车辆排队长度与实际通行能力、事故持续时间、路段上游车辆流量之间的关系，最后针对各个问题建立模型并求解。

2013全国数学建模A题优秀作品

第六届“认证杯”数学中国数学建模网络挑战赛承诺书我们仔细阅读了第六届“认证杯”数学中国数学建模网络挑战赛的竞赛规则。

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。

我们允许数学中国网站()公布论文，以供网友之间学习交流，数学中国网站以非商业目的的论文交流不需要提前取得我们的同意。

我们的参赛队号为：2261参赛队员(签名) ：队员1：张述平队员2：魏方征队员3：乔赛参赛队教练员(签名)：参赛队伍组别：2261第六届“认证杯”数学中国数学建模网络挑战赛编号专用页参赛队伍的参赛队号：2261竞赛统一编号（由竞赛组委会送至评委团前编号）：竞赛评阅编号（由竞赛评委团评阅前进行编号）：2013年第六届“认证杯”数学中国数学建模网络挑战赛题目护岸框架减速效果的优化方案关键词护岸框架减速效果单因素方差分析最小二乘拟合回归分析摘要：四面六边透水框架是一种新型江河护岸工程技术，对于降低岸边流速、稳定河道、保护堤岸有显著的作用。

本文针对所引用参考文献中的图像、数据，从四面六边透水框架群框架尺寸、架空率和长度三方面出发，对框架群的水力特性及其影响因素进行分析，探讨三要素对减速效果的影响，建立三个模型，为这种“亲水”式生态防护技术在工程中推广运用提供参考依据。

模型一：架空率对减速效果影响的分析。

首先，利用单因素方差分析，推断出架空率对减速效果的影响较显著；其次，采取最小二乘法拟合曲线具体演示二者的发展趋势，并得到关系模型，依照此模型得出架空率对减速效果影响显著。

得出框架率ε=4.2～4.8 之间时，框架群的减速率比较高，能够使得框架群的阻水消能作用和“亲水”功能较好的结合起来。

2013电工杯数学建模A题论文(国家二等奖)

答卷编号：论文题目：A题：风电功率波动特性的分析姓名专业、班级有效联系电话参赛队员1参赛队员2参赛队员3指导教师：冯玉昌参赛学校：证书邮寄地址及收件人：答卷编号：阅卷专家1 阅卷专家2 阅卷专家3 专家签字风电功率波动特性的分析摘要本论文针对“风电功率波动特性的分析”问题，根据所给的风电机组功率数据建立风电功率波动特性的概率分布模型和灰色预测模型。

由此，进行相关的问题分析及解决。

对于问题一、二，借鉴分离min 级负荷的算法，采用滑动平均法分离s 级风电功率，且处理了丢失数据及错误数据，以此提高数据的准确性。

经过如此处理所给数据后，再采用Matlab 的概率密度拟合工具箱dfittool 得出五台风电机组的功率概率直方图及t location-scale 分布、正态分布、逻辑斯特分布的概率分布图。

发现t location-scale 分布比其他分布更适于拟合各风电场概率密度函数，并作相关分析及检验。

再用t location-scale 分布以每日为时间窗宽，对5个风电功率分别计算30个时段的概率分布参数并做出检验。

问题二与问题一方法一致，只需要从从上述5台机的风电功率数据中提取出间隔为1分钟的数据序列)(k m i t P ，重复问题一的方法即可。

对于问题三，由上可以分别获得采样间隔为5s 和1min 的相关波动特性参数，以此为依照进行波动特性与时间尺度关系的分析。

对于问题四，整合出20台机组的数据，再分别将采样间隔改为1min ，5min ，15min ，使用问题一中同样的方法分析处理即可。

对于问题五，预测未来四小时的风电场总功率，我们采用了灰色模型（Grey Model ，GM ），使用MATLAB 对灰色模型GM （1，1）编程得到预测值，残差，级比偏差等相关数据结果。

由于初步编程得出的预测模型为其累加后的方程，通过生成序列预测值及模型还原值之间的关系及之前所求的预测值模型易求的未来四小时风电场总功率预测模型。

2013国赛A题全国一等奖论文(英文版)

Effect of Lane Occupied on Traffic CapacityAbstract:The article aims at the problem how the lane occupation affects the traffic capacity. By the means of wavelet analysis and time series analysis method, we study on the periodicity and stability of the actual capacity which varies with time going. Meanwhile, we get the expression of the vehicle equivalent queue length based on the Two Fluid Theory.For the question 1: Firstly, considering that the number of vehicles which arrive at the cross-section in the accident obeying periodical distribution, we choose the beginning of green signal time as a starting point and calculate the traffic density every 15 seconds at the cross-section. (Average traffic density within the range from the accident site to the point 120 meters away instead). Secondly, according to the relationship among the three parameters of the traffic flow, we get saturated traffic flow at each time, which represents the actual traffic capacity at the moment that the accident happened. Finally, we analyze the periodicity and stability of the actual traffic capacity by utilizing wavelet analysis and SAS software tools, which result in the conclusion that the actual capacity decays periodically and transits from a stable state to another one.For the question 2: We obtain the actual capacity at the same cross-section when different kinds of accidents happen in video 2. Through the analysis, it is easy to find that both variation laws are basically the same, and the stable value is also basically the same, but the latter decays faster than the former. Therefore, we establish explicit equation about the real traffic capacity and draw the following conclusion: When the road capacity is not saturated, the outer one of lanes blocked by the accident has the greater influence on the actual traffic capacity than the inner one but the influence will disappear when the road cannot bear more cars.For the question 3: we can introduce the concept of the equivalent queue length to deal with the real line length which may be quite complicated. And then we classify all the vehicle running states into two equivalently based on the Two Fluid Theory. This processing method helps us achieve the expression of the equivalent queue length. Here we list the expression '00()()()()t t o o m j m N q t dt W t dt k LM L t M k k +--=-⎰⎰. At last we examine our model by the means of thethat the model we established can preliminarily explain the relationship amongst the four parameters.For the question 4: We put the data into the expression in question 3. And then we get that the vehicle queue length will reach the upstream intersection after 3.34 minutes.Keywords:Wavelet Analysis, Time Series Analysis, Computer Simulation, the Two Fluid Theory, the Equivalent Queue Length.ContentEffect of Lane Occupied on Traffic Capacity (1)Abstract: (1)1. Description of the Problem (3)2. Assumptions (3)3. Definition (3)3.1 Description of symbols (3)3.2 Definition of terms (4)4. Primary Analysis (4)5. Preparations for the model (5)5.1 Determining the cross section of the actual traffic capacity (5)5.2 Basic relationships of three traffic flow parameters (6)6. Foundation and solution of model (6)6.2 Establishing and solving of the model for problem two (10)6.3 Establishment and solution to the model of Question 3 (13)6.4Establishment and Solution of the Model for the Forth Question .. 187. Model Test (19)8 Advantages and Disadvantages of the model (19)9. Reference (20)Appendix (21)1. Description of the ProblemThe phenomenon of lane occupied will lead to lower lane or cross section traffic capacity. A city’s traf fic flow could have large road density and strong continuity, which will reduce the road capacity even if only one lane is occupied. Even though the time is short, it may also cause a queue of vehicles even a traffic jam. And if handled inappropriately, it will lead to emergent regional congestion.The variety of lane occupied is complicated though, it will properly offer the theory basis for the traffic management if we are able to estimate the impact of lane occupied on the road traffic capacity our city.According to the different situation where one or two lanes are blocked shown in the video 1 and video 2, you need to solve the questions below.1: Describe the changing process of the actual capacity at the cross-section during period from traffic accident happens to the end.2: Explain the difference when one or two lane is occupied based on the conclusion drawn from the question 1and the video 2.3: Construct a mathematical model to analyze the relationship among vehicle queue length, lane traffic capacity, the ongoing time of the accident and the upstream traffic flow.4: Try to estimate the duration from the beginning of the accident to the moment the queue line reaches the upstream junction. Assuming that the distance between the traffic accident site and the junction becomes 140 meters, the downstream flow demand is unchanged, road traffic upstream flow is 1500pcu / h, the initial queue length is zero when accident happens and the accident lasts for enough long time.2. AssumptionsThe event that vehicles arrived at the accident site is independent.The standard car equivalent number through the upstream cross-sectional obeys the Poisson Distribution.Ignoring the influence on the actual traffic flow statistics which arises from non-standard vehicle on the road;The vehicles reach into the section where the accident happens at a constant speed.3. Definition3.1 Description of symbols3.2 Definition of terms1. The basic capacity W: the maximum number of standard vehicles which pass through cross-sectional lane during the unit time at the situation where road, traffic and environment are ideal.2. The saturation capacity Q: the maximum number of standard vehicles which pass through cross-sectional lane during the unit time at the situation where road, traffic and environment are ideal and traffic flow is continuous.3. Chain block: referring to the situation where the next cyclical vehicles arrive before the former vehicles entirely pass the accident site.4. Primary AnalysisThis problem is a comprehensive problem, including the changes of the actual capacity at the cross-section, the influence of different lanes occupied, the queue length on the road and the relationship among them. It may involve the Poisson Distribution, the Computer Simulations and other knowledge.Firstly, we need to extract and process the data. After researching the video 1 and video 2, we record the actual number of vehicles through the accident site before, during and after car the accident. And then we convert it into the standard vehicle equivalent number.Question 1 is to describe the changing process of the actual traffic capacity in the accident cross-section during the car accident. The key to solve the question is how todetermine the dynamic process of the vehicle flow. After analysis, we find that the actual road traffic capacity can be valued by referring to the correlation about the traffic flow, vehicle speed and the traffic density. In such way, we can draw a diagram to vividly illustrate the dynamic changing process of the flow.Question 2 is a problem that aims at comparing the difference. Firstly, we need to draw the graphs respectively about the flow density at the two different situations. And then we can tell the similarities and differences between the two cases. Secondly, We need to find the reason causing this phenomenon. Through the analysis, we should divide the blockage into two steps from the moment accident happens to evacuates, which are chain blockage and partial blockage. During the chain blockage, in video 1W o is a constant value the same in video 2. However, it’s complicated in partial blockage. Finally, we build a function model using the vehicle flow proportional coefficient as variable. Obviously we can obtain the different influence on actual traffic capacity as a result of different occupied lanes.For the question 4: It’s vital for us to build an expression about the equivalent queue length and time. According to the analysis of question 3, what we need is just make some appreciate changes for model of question 3. Then we can get the answer of question 4.For the modified model, we analyze the value of each parameter and ultimately solve the equation based on the assumption of the title. The final solution is the unknown quantity t. Finally, we tested the rationality and validity of the model by comparing the actual recorded data with the data of theoretical analysis.5. Preparations for the model5.1 Determining the cross section of the actual traffic capacityConsidering the actual traffic flow, the road is controlled by the traffic light at the upstream intersection. If vehicles are released by the first phase stopped by the second phase, we can see the road’s traffic appears as cyclical trends. Besides both the time of first and second phase are 30 seconds, therefore 30 seconds can be taken as a time interval. Through the analysis of video 1 and 2, we recorded every 30s of actual traffic volume to obtain the relevant data traffic from two minutes before accident happened to two minutes after accident withdrawing. Due to different vehicles having different effects on the ability of W and Q,we introduce standard car equivalent number C. namely the vehicles which are the four-wheeled vehicles and more vehicles or electric vehicles amount to equivalent vehicles. Specific data tables are in Appendix 1 1-1. Conversation coefficients [ 2 ] is shown in Table 1.Table 1 the traffic investigation vehicle type and vehicle conversion coefficient5.2 Basic relationships of three traffic flow parametersAccording to Green Hilts flow density model, the relation between the parameters of the model provisions comply with the following formula, we can see:Q VK =... (1)According to Green Hilts speed density model , we can see:(1)f fk V V k =- (2) According to Green Hilts flow density relationship model, we can see:(1)f fk Q kV k =-…………(3) The expression (3) shows the average flow rate and an average density of a quadratic function.6. Foundation and solution of modelIn view of question 1, we are asked to build a model to describe the changing process of actual capacity from the time that the accident happens to evacuation. Apparently, this question is a dynamic analysis problem. First we need draw a picture to describe the changing process visually. Then we can use Matlab and SAS software to analyze qualitatively the periodicity of the actual capacity change process and the stability of the stability of time series.(1) Method to find the actual capacity:Step 1: The Extraction of raw data In the Video 1, we record a value of the traffic density every 15 minutes from the time that green signal just bright to the evacuation of the accident. Generally speaking, it’s thought that vehicles which are below 120 meters away from the accident site will be influenced and meanwhile the average velocity of vehicles with this range approximately is equal to the speed of vehicles in the accident site. For simplifying the model, it’s reasonable to think the average traffic flow density as the traffic flow density in the accident site. Step 2: Formula for Solving the Actual CapacityWhen the upstream intersection can provide unlimited traffic , the value of Q in the formula ( 3) can be considered as the actual capacity under different speed.In accordance with the relevant literature [1],when the actual traffic flow density is greater than half of the blocking traffic density, it can also be thought that road is congested, otherwise road is in a smooth state. This can be described byk (1),k 2k ,k 42j j f j j f j k k V k W k V ⎧->⎪⎪=⎨⎪<⎪⎩ (4) Step 3: Estimation of values tor j k and f VIn order to facilitate the calculation below, we estimate the best traffic flow density at the same.Calculations of values for j k and f V :According to the traffic flow theory [6], the relationship between traffic density and flow can be described with many models. We use the classical inverted U-shaped model to go on a qualitative description of the position where j kappears.Figure 1 is an inverted U-shaped flow-density diagram:Figure1 Inverted U-shaped flow —Schematic density relationshipFigure 1 shows that j k is the maximum density when the velocity of the traffic obstruction is 0;m k is traffic density when road traffic flow reaches the maximum density. According to the related literature and requirements of road vehicle safetydistance, j k and m k are related to road conditions, the driver’s literacy and other factors, so in practice the two values should be calibrated according to survey data at the road section. Here, we take the data in Video 1 as standard value. Through observation, we can find that before the traffic accident, the road traffic flow density corresponding to the traffic flow is the optical traffic flow density with its value being 48.45,that is to say, m k =48.45 .After the traffic accident, the traffic jam will appear at the cross-sectional area of the accident. We identify several time points of the most serious blockage and calculate the traffic density corresponding to the time point, of which the average value is 287.5, that is to say,j k =287.5.m k and j k here are the sum of the traffic flow density of three lanes,j k and m k are determined by a weighted average of traffic flow density ratio of the downstream direction. Calculating the value of f V : During the time when accident has not happened, weselect multiple free vehicles and calculate the time of driving by using the following formula:1n i f i i S V t ==∑（5）Where:i S is the journey that the first I vehicle travelled. i t is the time that the first I vehicle spent.At this time, the actual capacity is obtained. The specific values have been givenin Schedule 1 to 3of Appendix 1. The waveform is shown in Figure 2.Figure 2 accident cross-sectional actual capacity changes over time graph in Video 1To the next, we made up a Matlab program in order to go on the periodical analysis of wavelet data. Taking the characteristic of the actual data into account, we choose db4 wavelet function to decompose the data signal into 10 layers, then we reconstruct 1 to 10 layers detail signal.Therefore, we get the second layer detail signal d2, as shown in Figure 3 (The program is shown in program2-1 of Appendix 2):Figure 3 Trend of the detail signal of the actual capacityFrom Figure 3, we can find that the actual capacity shows a trend of cyclical fluctuation.Next, we get the autocorrelation function diagram of the actual capacity by using SAS software to analyze the stability of the data, which is shown in Figure 4:Figure 4 The autocorrelation function diagram of the actual capacity According to the knowledge of time series theory, it is clear that this time series has the stability [5] if the autocorrelation coefficient of a time series whose expectation is zero shows the tendency of rapid attenuation to the range within the double standard deviation. Analyzing Figure 4, we can find that the autocorrelation coefficients are within the range of two times of the standard deviation. Therefore the trends of road capacity can be considered as stable.With comprehensive analysis of Figure 2, Figure 3 and Figure 4, the following conclusions can be drawn:Before the accident, namely it is 16:40:15—16:41:45, the actualcapacityW shows periodic fluctuation. By further analysis, we find that the oroad is open before the accident. So the actual road capacity is mainlycontrolled by traffic lights. In the second phase, vehicles from the upstreamare prohibited. It can be seen that these vehicles which pass by the pointwhere an accident happens mainly comes from vehicles of the first phase inthe previous stage, so the value of o W is smaller. However in the secondphase, vehicles from the upstream part start to move forward. Due to themore vehicle sources, capacity gets even bigger here. From what has beendiscussed above, the changing trend of actual traffic is periodic.During the accident, it is f16:42:30—16:55:30 (corresponding to [4,25] of xaxis in Figure 1),we can find that the actual capacity decreases over time andtends to be a stable value in the end except the abnormal part of time missingin the video.With deep analysis of the underlying causes, it can be found that when the traffic accident happens, two lanes are occupied, so all vehicles can only choose the only lane to travel through. In the sphere of influence at accident section, the speed of the car need to decrease, and then it makes the actual capacity to be reduced to a constant value. Along with the subsequent vehicles to arrive, the actual demand also increases. It is consistent with of conclusions of the stability analysis .Therefore , the actual demand is close to even greater than saturated traffic Q. According to the Theorem 1, traffic jams occur in line with the actual situation in the Video 1.At the same time, considering the impact of traffic lights alone, the traffic volume is also periodical and it’s period is about 1 minute in line with the periodic of the wavelet analysis and it is close to the period of the traffic light signal.To sum up: After a traffic accident, actual capacity at the cross-section where an accident happened cyclically shocks and decays over time besides transiting from a stable state to another one.6.2 Establishing and solving of the model for problem two(1) Analysis of problem twoFor problem two, it asks us to analyze and illustrate different influences in the actual capacity, with the same section of traffic accident and different occupied lane.Firstly, we need get original data of the actual capacity of various time points, so we can select data obtained in appendix1. Secondly, we build a model to describethe influences on the actual capacity at the same section of traffic accident and different occupied lane through a comprehensive comparison of video 1 and 2.(2)Establishing and solving of the modelBefore the accident , comparing to the value named o W of video 1(specific datacan be seen in the Appendix 1 ), we find that the actual capacity o W of Video 1 and Video 2 are all within the range from 1100 to 1400. Because the study is the actual capacity where are the same lanes before these lanes are occupied, the actual capacity in Video 1 and Video 2 is equal before an accident apparently.In order to research on changes of the data in video 2, during the accident occurring till evacuating, we described it by drawing it into graphs as the following graph 4 shows:Graph 4The diagram showing the capability of transportation during the accident 2 occurring till evacuating by comparing graph 1 with graph 4, we might notice visibly that the actual capability of transportation display.The oscillation-decaying tendency and after several cycles of the signal, the decay of video 1 and 2 are almost keeping on the line about 10,but the rate and amplitude of them have some major differences. Considering the situation above, we will divide the whole process into two parts.First part: before turning to stability, we call it partial jam stage.Based on the conclusion which is qualitatively analyzed above, we build the mathematics model and quantize the index to further clarify the difference of influence on the actual capacity of transportation by traffic accidents' occupation on different roads.Through the following ideas we build the model:For accident 2, the lane 1 and lane 2 are occupied, so we analyze the partial jam situation of the road first. To simplify the model, we think cars on road 3 directly drive across the intersecting surface to the accident site along the road and vehicles on road 2 drive to the intersecting surface of the accident earlier and then turn to road 3. Of course vehicles on road 1 also drive to the intersecting surface of the accident, stride road 2 and turn to road 3 to drive out.We define a set{}123,,V v v v =,iv is a0-1variable which indicates whether theroad is occupied. ‘1’ means road has been occupied and ‘0’ means contrarily. So the Set V is a collection of situations to road's occupation.We can know the accident 2 by the definition above{}1,1,0V =We can know t car flux of the road by the analysis above'i i Q Q α= (6)The time that the flow of traffics going straight needs at Lane 3 is '3(b d )Q μ+.The time that the traffic flow turns at Lane 2 is2Q t∆.The time that the flow of traffics turning and changing the lane at Lane 1 spendsis 13Q d μ .Consequentially, the total time of traffics passing is'312(b d )Q 3Q dt Q t μμ+=+∆+(7)In view of the simplified model that have been discussed earlier, we make a correction and unification according to the practical problems, then the total time that all of the traffics require to pass the place where accident happens is'312(b d )Q 3i i ii i i i Q dv i t t Q v t μμ=-+==+∆+∑. (8)According to this model, traffic capacity of a lane in unit time is'O Q W t=(9)So we can put Formula (1) and Formula (2) to the Formula (3), then we will get aresult thatis'123(b d )3k t (3t )k O W d μμμ=++∆+∆+ (10), andandare respectively the coefficient of traffic flow at the nearestoccupied lane and the farthest occupied lane.1) In Formula (9), traffic flow of a lane in unit time decreases with the increaseofandare respectively replaced with 21% and 35% in Video 1 and Video 2.A conclusion can be drawn that Lane 3 has more impact on traffic capability than Lane 1, which makes the amplitudeofchange more largely and quickly,according to the situation Figure 2 has more quick attenuation of the amplitude than Figure 1.2)According to 2o W k ∂∂＜1OW k ∂∂＜0,we know that the farthest occupied lane hasgreater effecton ,in other word ,the occupied Lane 1 or Lane 3 makes thedecreasing of amplitude of larger than what the Lane 2 does.From what has been discussed above, the degree of effect that different occupied lane have on practical traffic capacity can be ordered from big to small. Occupied Lane3 an Lane2 have the biggest effect, while occupied Lane1 and Lane2 have the least.So the result above indicates that during the accident occurring till evacuating in the road partial jam stage, if the traffic accident occupies the road with more car flux, the rate of oscillation decaying of the road's intersecting surface is higher and altitude is greater.After decaying to the stable entire jam stage, occupation of different roads of the same intersecting surface doesn't influence more on the actual capability of the intersecting surface.6.3 Establishment and solution to the model of Question 3（1）establishment of the model of question 3For question 3, we are required to establish mathematical model between the vehicle-queue length of the accident-affecting road, the actual transportation capability of intersecting surface of the accident, duration time of the accident and upstream car flux of the road. Firstly, we should define the queue length on concept in order to quantify it. so we bring in the equivalent L--queue length. Then we can establish a function relationship equation among these variables on the Two Fluid Theory.1）Introduction of the Equivalent Queue LengthIt is necessary to determine a new definition on the queue length that may differ from the additional one, which is also what we prefer to take advantage of to describe the situation showed in the video. Thus we introduce the concept of the equivalent queue length in our model. The new concept, in other words, is the ideal length in the Two Fluid theory in the traffic flow which not only reflects the stationary vehicles’influence on the queue length, but also takes the effect that the length has on the moving vehicles into consideration. And it will definitely produce more accurate result by combining the two dimensions together. The actual traffic flow operation state and the two fluid operation state diagrams are shown in the Figure 5 and Figure 6 below.Fig. 5 the actual running state of traffic flow on the road during the accidentFig. 6 accidents to evacuate Two Fluid operation state during the road traffic flowThe traffic flow in the actual running state can be divided into three conditions-the stagnation traffic flow Z, the uniform traffic flow Y and the transition traffic flow G , while in the two fluid theory, it only needs to be divided into two categories-the moving ones 'Z and the stationary ones 'Y . That is to say, in consideration of the gradual change and indeterminacy of the transition flow, in the theory, we can simply our model by putting the transition flow into two parts, each of which can be reckoned as the condition where the running state is the most familiar .Therefore, the theoretical situation itself includes the real three conditions, and it indicates that the theory is more than accuracy. We can draw a conclusion from the above analysis that the expression for the equivalent queue length is 'L Z Z =+.2) Establishment of the modelIn this part, our goal is to establish an useful expression that can best describe the variation of the equivalent queue length L when the real road traffic capacity 0W , accident duration t or the upstream flow changes (t)N based on the two fluid theory. First of all, let us figure out the comparatively easier situation where only the single lane is taken into account. And then we make some adjustments to apply it to the multiple lanes.i ） Single LaneAccording to the flow conservation principle, we get the equation of traffic flow on the road:N (t)N (t)(t)o u d N N +=+∆(11)Where :o N is the number of the vehicles between the accident site and the upstream intersection when the accident just happens.(t)u N is the accumulation of the upstream flow on the road at moment of t.(t)d N is the sum of vehicles which pass by the accident site at the moment of t.(t)N ∆is the number of vehicles between the accident site and the upstreamintersection at the moment of t.Combining the two fluid theory and the figure 6，we get'()()()j m N t k L t k L L t ⎡⎤∆=+-⎣⎦(12)Where :()L t is the equivalent queue length between the accident site and the upstreamintersection at the moment of t.'L is the extent of the road, here it is 240m.m k and j k respectively are the optical density of traffic flow and the traffic jam density between the accident site and the upstream intersection. They are constants.Simultaneous equations (11) and (12) can solve the model of equivalent length on the single lane:'()()()o u d m j mN N t N t k L L t k k +--=-(13) ii ）Adjustments applying for multiple lanesFor the situation of multiple lanes, the equivalent queue length on a lane is totally different when flow rate changes or the cars switch the lane. To describe the extent of vehicle line with a variable, we could introduce a method to measure the whole situation. And the approach is to get the average of the three lanes. After that,we get the model of the equivalent length on multiple lanes:'()()()()tto o m j m N q t dt W t dt k LML t M k k +--=-⎰⎰(14)Where:01()(,)Mtu i q t dt N i t ==∑⎰,1()(,)Mtodi W t dt Ni t ==∑⎰and the number of the lanesM=3.（2）Solution of the model for question 3 1）Deterioration of parameters in the average equivalent queue length on multiple lanesStep 1：o N is the number of the vehicles between the accident site and the upstream intersection when the accident just happens. We assume the happening time is 16:42:30, and we can count the number of vehicles in the video. The result is016N =.Step 2：Determine the value of1(,)Mui Ni t =∑, the expression means theaccumulation of the upstream cars flow at the moment of t;Through the analysis, we know that the upstream flowq is unpredictable, thuswe are capable of adopting the Poisson Distribution to describe it. The next is the。

2013年研究生数学建模竞赛A题全国三等奖论文

参赛密码（由组委会填写）第十届华为杯全国研究生数学建模竞赛学校参赛队号队员姓名1.2.3.参赛密码（由组委会填写）第十届华为杯全国研究生数学建模竞赛题目变循环发动机部件法建模及优化摘要变循环发动机可以实现超音速时的高推力与亚音速时的低油耗，其内在性能的优势引起了各航空强国的高度重视，因此是目前航空发动机的重要研究方向。

研究的重中之重是燃气涡轮发动机的特性，本文在了解变循环发动机的构造及工作原理的基础上，运用线性插值、遗传算法、多目标规划等方法给出各部件的出口总温、总压、流量和功率的算法、程序，进一步给出了平衡方程组成的非线性方程组的算法及其有效性分析，得到了发动机性能最优的相关约束条件，进而解决了发动机性能最优时的参数取值。

针对问题一，运用Matlab软件进行线性插值得到相应换算转速下增压比、流量、效率。

1）对附录4中风扇特性数据表中增压比进行标准化处理得到对应压比值，根据附录1中2.2.2得到相应换算转速下的流量，通过Matlab软件中Spline插值方法画出了相应换算转速下流量随压比函数值变化的图形。

2）根据附录1中2.1.2得到了进气道出口总温、总压，再由压气机（风扇、CDFS）进口总温、总压和物理转速得到的换算转速，利用线性插值得到对应增压比、流量、效率，再由附录1中2.2.2求出了压气机（风扇、CDFS）的出口总温、总压和流量。

针对问题二，给出了求解由发动机7个平衡方程组成的非线性方程组的遗传算法，进一步运用Matlab软件编写了子程序并进行有效性分析。

对于发动机7个平衡方程涉及的变量进行分类处理，结合附录1中变循环发动机各部件的计算公式给出对应类的算法及程序。

再将变量代入相应平衡方程，得到关于发动机参数的非线性方程组。

针对问题三，依据附录1给出的发动机性能参数：推力、单位推力和耗油率，建立了多目标规划模型。

根据所建模型，找出了发动机CDFS导叶角度、低压涡轮导叶角度和喷管喉道面积3个量与推力、单位推力和耗油率之间的关系，进而通过Lingo求解出3个量为多少时，发动机的性能最优。

2013高教杯数学建模竞赛A题论文二等奖

表二车道宽度和侧向净空修正系数表而驾驶员条件影响修正系数，一般取值在 0.9-1 之间，主要用来反映周末休闲出现活动或工作日正午时段驾驶人对设施的影响，因为这里我们研究的问题与其无关，故我们在研究过程中，固定取值为 1. 而对于大型车混行影响的修正系数，则需通过具体的计算实际中车辆通行情况来得到。通过交通资料，我们可以知道，大型车混行修影响正系数 f HW 的计算
关键词：修正系数
交通波
VISSIM 仿真
分段讨论
车辆排队长度
1
1、问题重述
车道被占用是指因交通事故、路边停车、占道施工等因素，导致车道或道路横断面通行能力在单位时间内降低的现象。由于城市道路具有交通流密度大、连续性强等特点，一条车道被占用，也可能降低路段所有车道的通行能力，即使时间短，也可能引起车辆排队，出现交通阻塞。如处理不当，甚至出现区域性拥堵。车道被占用的情况种类繁多、复杂，正确估算车道被占用对城市道路通行能力的影响程度，将为交通管理部门正确引导车辆行驶、审批占道施工、设计道路渠化方案、设置路边停车位和设置非港湾式公交车站等提供理论依据。视频 1（附件 1）和视频 2（附件 2）中的两个交通事故处于同一路段的同一横断面，且完全占用两条车道。请研究以下问题： 1. 根据视频 1（附件 1），描述视频中交通事故发生至撤离期间,事故所处横断面实际通行能力的变化过程。 2. 根据问题 1 所得结论，结合视频 2（附件 2），分析说明同一横断面交通事故所占车道不同对该横断面实际通行能力影响的差异。 3. 构建数学模型，分析视频 1（附件 1）中交通事故所影响的路段车辆排队长度与事故横断面实际通行能力、事故持续时间、路段上游车流量间的关系。 4. 假如视频 1 （附件 1）中的交通事故所处横断面距离上游路口变为 140 米，路段下游方向需求不变，路段上游车流量为 1500pcu/h,事故发生时车辆初始排队长度为零，且事故持续不撤离。请估算，从事故发生开始，经过多长时间，车辆排队长度将到达上游路口。

2013年数学建模A题优秀论文

2
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2013年美国数学建模A题论文中文版

4.2.2 热量的均匀分布
4.2.3(二) 如何得到最佳烤盘从上边两种情况可以分别得到不同形状的烤盘排列在烤炉中时的空间利用率，以及它们各自在达到平衡时的热量分布，同时我们由温度的方差得出温度的分布均匀性。考虑到实际情况中，我们常常既想追求温度的最均匀分布，因为此时得到的蛋糕品味最佳，又想使空间得到最有效利用，不致于资源浪费，显然这两者不能同时满足。这时我们应该考虑针对不同的需要，使用不同形状的烤盘，下面我们来解决这个问题。就像在招聘员工考虑不同因素来为求职者打分一样，我们从温度的均匀分布和空间利用率两方面考察一个烤盘的性能时，可以给以不同的权重 p 与 1-p，我们将在不同的 p 值下考察不同形状烤盘的性能。
占用率 Q=
NA 84 % . WL
WL 2(2 3 ) X 2
]=2[
L2
2(2 3 ) A
],
由此，我们比较四种不同图形在烤箱平面中数目和占用率，可以容易的得知：矩形（以正方形为例）烤盘在烤箱平面中占用率最高，为 100%，而圆形盘占用率最小仅为 84%。矩形盘和圆盘分为两个极端，中间分布着其他图形的数目和占用率。当选用烤箱平面中烤盘的最大数量时，显然矩形(正方形)最大。
为了减少由于方差 S 过大而引起的影响，我们引入了参数 U，以正方形的温度分布为基准，表示温度分布的相对不均匀度 Ui =
Si S1
由此得到不同情况下温度分布的相对不均匀度从而， U1 = 1， U2 = 0.526， U3 = 0.427， U4 = 0.397 U1 , U2 , U3 , U4 分别对应于正方形，正六边形，正八边形，圆形。我们再引入一个参数 R 来反映烤盘的相对综合性能 R=p∗Q− 1−p ∗U 这里的 R 相当于烤盘的性能得分，R 的值越大，则说明其性能越好。注意，这里对 U 的处理比较特殊，因为其值越小反映的烤盘的温度分布越均匀，故这里要用减号。（1） p 一定时，不同烤盘的性能比较

全国A题优秀论文

【关键词】GM（1，1）灰色预测 Logistic 阻滞增长模型 Matlab
常微分方程
SPSS
时间推移模型
一、问题重述
搜集某个具体地区（国家、省、市、县等）的人口普查数据，分析影响人口增长的因素（人口年龄分布，城镇化，出生率，妇女生育率），试回答下列问题。（1）合理建立相应一段时间内该地区的人口发展的常微分方程模型。（2）舍去最近 2 次数据，根据剩余数据利用最小二乘拟合法拟合模型参数。（3）计算最近 2 次数据对应的模型理论值，比较误差大小，决定是否需要修正模型。（4）根据（修正）模型对未来 50 年分年人口情况作预测。（5）试建立数学模型说明或是论证微分方程（修正）模型参数设定的合理性和准确性。
1
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该地区样本总人口数出生率固有增长率
四、模型的假设
1、所有人口都是自然出生和死亡； 2、死亡率不随医疗水等因素的改变而改变； 3、出生率取决于育龄妇女的的生育率及育龄妇女在总人口所占得比例； 4、根据所给出数据进行分析，我们假设人口增长因素只与乡村人口城市化、市、镇、乡的出生人口性别比、市、镇、乡育龄妇女育龄率和人口的老龄化有关； 5、假设该地区是个封闭的系统，不考虑移民情况。
同理，再利用此模型中灰色系统对出生率进行预测，进而建立该地区人口发展的常微分方程模型。问题二：根据问题一所建立的模型，把 2004 年和 2005 年最近的数据舍去，利用最小二乘拟合法拟合模型参数（以死亡率为例）：
^ ⎡0.0046 ⎤ a = ( BT B) −1 BT y = ⎢ ⎥ ⎣ 6.6581⎦
49
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数学建模A优秀论文

数学建模A优秀论文数学建模A优秀论文在日常学习、工作生活中，大家都接触过论文吧，论文是对某些学术问题进行研究的手段。

一篇什么样的论文才能称为优秀论文呢？以下是小编为大家收集的数学建模A优秀论文，供大家参考借鉴，希望可以帮助到有需要的朋友。

数学建模A优秀论文11. 问题重述：（略）2. 问题背景：交待问题背景，说明处理此问题的意义和必要性。

优点：叙述详尽，条理清楚，论证充分缺点：前两段过于冗长，可作适当删节3. 问题分析：进一步阐述解决此问题的意义所在，分析了问题，简述要解决此问题需要哪些条件和大体的解决途径优点：条理比较清晰，论述符合逻辑，表达清楚缺点：似乎不够详细，尤其是第三段有些过于概括。

4. 模型的假设与约定：共有8条比较合理的假设优点：假设有依据，合情合理。

比如第3条对上座率的假设，参考了上届奥运会的情况并充分考虑了我国国情，客观真实。

第8条假设用了分块规划和割补的方法，估计面积形状比较合理，而且达到了充分花剑问题的作用。

缺点：有些假设阐述不太清楚也存在不合理之处，第4条假设中面积在50-100之间，下面的假设应该是介于50-100之间的数，假设为最小的50平方米，有失一般性。

第6条假设中，假设MS最大营业额为20万，没有说明是多长时间内的，而且此处没有对下文提到的LMS作以说明。

5. 符号说明及名词定义优点：比较详细清楚，考虑周全，而且较合理地将定性指标数量化。

缺点：有些地方没有标注量纲，比如A和B的量纲不明确。

6. 模型建立与求解6.1问题一：对所给数据惊醒处理和统计，得出规律，找到联系。

优点：统计方法合理，所统计数据对解决问题确实必不可少，而且用图表和条形图的方式反映不同量的变化趋势，图文并茂，叙述清楚而且简明扼要，除了对数据统计情况进行报告以外，还就他们之间相关量之间的关系进行了详细阐述，使数据统计更具实效性。

6.2问题二：6.2.1最短路的确定为确定最短路径又提出了一系列假设并阐述了理由，在这些假设下规定了最短路径优点：假设有根据，理由合情合理缺点：第4条中假设观众消费是单向的，虽然简化了问题但有失一般性，事实上观众往返经过商业区消费的概率是相差比较大的，我认为应改为假设观众在往返过程中消费且仅消费一次。