当前位置：搜档网 › 2013年美国大学生数学建模大赛B题获奖论文

2013年美国大学生数学建模大赛B题获奖论文

For office use only

T1________________ T2________________ T3________________ T4________________Team Control Number

22599

Problem Chosen

For office use only

F1________________

F2________________

F3________________

F4________________ 2013

Mathematical Contest in Modeling(MCM/ICM)Summary Sheet

(Attach a copy of this page to your solution paper.)

Heat Radiation in The Oven

Heat distribution of pans in the oven is quite different from each other,which depends on their shapes.Thus,our model aims at two goals.One is to analyze the heat distri-bution in different ovens based on the locations of electrical heating cubes.Further-more,a series of heat distribution which varies from circular pans to rectangular pans could be got easily.The other is to optimize the pans placing,in order to choose a best way to maximize the even heat and the number of pans at the same time.

Mathematically speaking,our solution consists of two models,analyzing and optimi-zing.In part one,our whole-local approach shows the heat distribution of every pan.Firstly,we use the Stefan-Boltzmann law and Fourier theorem to describe the heat distribution in the air around the electrical heating tube.And then, based on plane in-tercept method and simplified Monte Carlo method,the heat distribution of different shapes of pans is obtained.Finally,we explain the phenomenon that the corners of a pan always get over heated with water waves stirring by analogy.In part two,our discretize-convert approach optimizes the shape and number of the pans.Above all,we discre-tize the side length of the oven, so that the number and the average heat of the pans vary linearly.In the end,the abstract weight P is converted into a specific length,in order to reach a compromise between the two factors.

Specially,we create a unique method to convert the variables from the whole space to the local section.The special method allows us to draw the heat distribution of every single section in the oven.The algorithm we create does a great job in flexibility,which can be applied to all shapes of pans.

Type a summary of your results on this page.Do not include

the name of your school,advisor,or team members on this page.

Heat Radiation in The Oven

Summary

Mathematically speaking,our solution consists of two models,analyzing and optimi-zing.In part one,our whole-local approach shows the heat distribution of every pan. Firstly,we use the Stefan-Boltzmann law and Fourier theorem to describe the heat distribution in the air around the electrical heating tube.And then,based on plane in-tercept method and simplified Monte Carlo method,the heat distribution of different shapes of pans is obtained.Finally,we explain the phenomenon that the corners of a pan always get over heated with water waves stirring by analogy.In part two,our discretize-convert approach optimizes the shape and number of the pans.Above all, we discre-tize the side length of the oven,so that the number and the average heat of the pans vary linearly.In the end,the abstract weight P is converted into a specific length,in order to reach a compromise between the two factors.

Keywords:Monte Carlo thermal radiation section heat distribution discretization

Introduction

Many studies on heat conduction wasted plenty of time in solving the partial differential equations,since it’s difficult to solve even for computers.We turn to another way to work it out.Firstly,we study the heat radiation instead of heat conduction to keep away from the sophisticated partial differential equations.Then, we create a unique method to convert every variable from the whole space to section. In other words,we work everything out in heat radiation and convert them into heat contradiction.

Assumptions

We make the following assumptions about the distribution of heat in this paper.

·Initially two racks in the oven,evenly spaced.

·When heating the electrical heating tubes,the temperature of which changes from room temperature to the desired temperature.It takes such a short time that we can ignore it.

·Different pans are made in same material,so they have the same rate of heat conduction.

·The inner walls of the oven are blackbodies.The pan is a gray body.The inner walls of the oven absorb heat only and reflect no heat.

·The heat can only be reflected once when rebounded from the pan.

Heat Distribution Model

Our approach involves four steps:

·Use the Fourier theorem to calculate the loss energy when energy beams are spread in the medium.So we can get the heat distribution around each electrical heating tube.The heat distribution of the entire space could be go where the heat of two electrical heating tubes cross together.

·When different shapes of the pans are inserted into the oven,the heat map of the entire space is crossed by the section of the pan.Thus,the heat map of every single pan is obtained.

·Establish a suitable model to get the reflectivity of every single point on the pan with the simplified Monte Carlo method.And then,a final heat distribution map of the pan without reflection loss is obtained.

·A realistic conclusion is drawn due to the results of our model compared with water wave propagation phenomena.

First of all,the paper will give a description of the initial energy of the electrical hea-ting tube.We see it as a blackbody who reflects no heat at all.Electromagnetic know-ledge shows that wavelength of the heat rays ranges from um 110?to um 210as shown below[1]:

Figure 1.

Figure 2.

We apply the Stefan-Boltzmann’s law[2]whose solution is ()1

/512?=?T c b e c E λλλ(1)()λλλλλd e c d E E T c b b ∫∫∞?∞?==0

/51012(2)

Where b E means the ability of blackbody to radiate. 1c and 2c are constants.Obviously,,the initial energy of a black body is )(0122398.320m w e E b ×+=.Combine Figure 1with Figure 2,we integrate (1)from 1λto 2λto get the equation as follow:

λλλλλd E E b b ∫=?2121)((3)

Figure 3.

From Figure 3,it can be seen how the power of radiation varies with wavelength.Secondly,based on the Fourier theorem,the relation between heat and the distance from the electrical heating tubes is:

dt S Q λ?=(4)Where Q is the power of heat (W s J =/),S is the area where the energy beam

radiates (2m ),dx

dt represents the temperature gradient along the direction of energy beam.[3]

It is known that the energy becomes weaker as the distance becomes larger.According to the fact we know:

dQ =ρ(5)Where ρis the rate of energy changing.

We assume that the desired temperature of electrical heating tube is 500k.With the two equations,the distribution of heat is shown as follow:

Figure 4.(a)Figure 4.(b)

In order to draw the map of heat distribution in the oven,we use MATLAB to work on the complicated algorithm.The relation between the power of heat and the distance is shown in Figure4(a).The relation between temperature and distance is presented in Figure4(b).The spreading direction of energy beam is presented in Figure5.

Figure5.

The shape of electrical heating tube is irregular.The heat distribution of a single electrical heating tube can be draw in3D space with MATLAB.The picture is shown in Figure6.After superimposing,the total heat distribution of two tubes is shown below in Figure7and Figure8.

Figure6.Figure7.

Figure8.

The pictures above show the energy in an oven with no pan.We put in a rectangular pan whose area is A,and intercept the maps with MATLAB.The result is show in Figure9.

Figure9.

Figure10.

Put in a circular pan to intercept the maps,whose area is A,also.The distribution of heat is shown in Figure11.

Figure11.

When put in a pan in transition shape,which is neither rectangular nor circular.The area of it is A,also.The heat distribution on such a pan is shown as follow:

Figure12(a)

Figure12(b).

Figure13.

Next,learning from the Monte Carlo simulation[4],a model is established to get obtain the reflectivity.We generate a random number between0and1to determine if the energy beam on certain point is reflected.

?Firstly,to demonstrate the question better,we construct a simple model:

Figure 14.

Where θis the viewing angle from electrical heating tube to the pan.360

θ=R is the proportion of the beams radiated to the pan.

?What is more,we assume the total beam is 1M .Ideally,the number of absorption is

360

1θ×M .Then,each element of the pan is seen as a grid point.Each grid point can generate a-360

1θ×M -random-number vector between 0and 1in MATLAB.?After MATLAB simulating,the number of beams decreased by 2M ,due to the

reflection.So we define a probability θ

ρ12360M M ×=to describe the number of beams reflected.

The conclusion is :

?If R ≤ρ,the energy beam is absorbed.

?If R >ρ,the energy beam is reflected.[5]

Based on the analysis above,our model get a final result of heat-distribution on the pan as shown below:

Figure 15(a)

Figure15(b).

The conclusion is known that the closer the shape of pans is to circle,the more evenly the heat is distributed.Moreover,the phenomenon that the corners always get over heated can be explained by water wave propagation in different containers.

When there is a fluctuation in the center of the water,the ripples will fluctuate and spread in concentric circles,as shown in Figure16.The fluctuation stirs waves up when contacting the https://www.sodocs.net/doc/2612647993.html,pared with the waves with one boundary,the waves in corner make a higher amplitude.

The thermal conduction on the pan is exactly the inverse process of the waves propagation.The range of thermal motion is much smaller than it on the side.That’s why the corners is easy to get over heated.

In order to make the heat evenly distributed on the pan,the sides of the pan should be as few as possible.Therefore,if nothing is considered about the utilization of space,a circle pan is the best choice.

Figure16,the water waves propagation[6]

According to the analysis above and Figure7,the phenomenon shows that the heat conduction is similar to water waves propagation.So it is proved that heat

concentrates in the four corners of the rectangular pan.

The Super Pan Model

Assumptions

?The width of the oven(W)is mm

100,the length is L.

?There are three pans at most in vertical direction.

?Each pan’s area is A.

The first part.

Calculate the maximum number of pans in the oven.Different shapes of pan have different heat distribution which affects the number of pans,judging from the previous solution.According to the conclusion in the first model,the heat is distributed the most evenly on a circular pan rather than a rectangular one.However, the rectangular pans make fuller use of the space the space than circular ones.Both factors considered,a polygonal pan is chosen.

A circle can be regard as a polygon whose number of boundaries tends to infinity. Except for rectangle,only regular hexagon and equilateral triangle can be closely placed.Because of the edges of equilateral triangle,heat dissipation is worse than rectangle.So,hexagonal pans are adopted after all the discussion.

Considering the gaps near boundaries,we place the hexagonal pans closely attached each other on the long side L.There are two kinds of programs as shown below.

Program1.

Program2.

Obviously,Program2is better than Program1when considering space utilization.So scheme 1is adopted.

Then,design a size of each hexagonal pan to make the highest space utilization.With the aim of utilization,hexagonal pans has to be placed contact closely with each other on both sides.It is necessary to assume a aspect ratio of the oven to work out the number of pans(N ).

Assume that the side length of a regular hexagon is a ,the length-width ratio of the oven is λand L ?is the increment in discretization.Because the number of pans can not change continuously when ???=+∈3,2,1),1,(m m m n ,the equations would be as follows.

????????????

???????????????????????????====??+

k L W W L k L W L k L a L L k L L L W a

W 23,810;23105000λ(6)Result:??????????==?+=???=?=+??+=3,2,1,22

33,2,1,1212130201k k n n N N k k n n N N Where 1N represents the number of pans when n is odd,2N represents the number of pans when n is even.The specific number of pans is depended on the width-length ratio of oven.

The second part.

Maximum the heat distribution of the pans.We define the average heat(H )as the ratio of total heat and total area of the pans.Aiming to get the most average heat,we set the width-length ratio of the oven λ.Space utilization is not considered here.

A conclusion is easy to draw from Figure 8that a square area in the oven from 150mm to 350mm in length shares the most heat evenly.So the pans should be

placed mainly in this area.From model1we know that the corners of the oven are apt to gather heat.Besides,four more pans are added in the corners to absorb more heat. Because heat absorbing is the only aim,there is no need to consider space utilization. Circular pans can distribute heat more evenly than any other shape due to model1.So circular pans are used in Figure17.

Figure8.

Figure17.

We set the heat of the pans in the most heated area(the middle row)as Q.Pans in the corners receive more heat but uneven theoretically.And the square of the four pans in the corners is so small compared with the total square that we set the heat of the four as Q too.When the length of oven(L)increases,the number of pans increases too. It makes the square of the gaps between pans bigger,meanwhile.If each pan has a same radius(r)and square(A),the equation about average heat,length-width ratio and number of pans would be(7).

?????????????????????????=+=????????====??+

k n

N N r W L n W L N L k L W W L k L W L k L r L L k L L L W r A W r λπ(7)

Here we get the most average heat (H ):

29400W

Q H ?=πThe third part.

We discussed two different plans in the previous parts of the paper.One is aimed to get the most average heat,while the other aimed to place the most pans.The two plans are contradictory with each other,and can not be achieved together.

Firstly,the weight of plan 1is P and the weight of plan 2is P ?1.Obviously,this kind optimization has difficulty in solving and understanding.So we turn to another way to make it a easier and linear question.It has been set that the width of the oven is a constant W and there should be three pans at most in vertical direction.We make the weight P a proportion of the two plans.Thus the two plans could be achieved together due to proportion P and P ?1,as shown in Figure

18.

Figure 18.

As been told in model 1,the corners have a higher temperature than other parts of the oven.So plan 1is used in district 1(in Figure 10)and plan 2is used in district 2(in Figure 10).A better compromise could be reached in this way,as shown in Figure 19.

Figure 19.

Every pan has a square of A .Radius of circular ones is r .Side length of regular hexagon is a .

1.1:23322=??=?r a a r π(8)

Based on the equation (8),if the pans are placed as shown in Figure 19,regular hexagons are placed full of district 1,the circular ones will be placed beyond the border line.If the circular ones are placed full of district 2,there will be more gaps in district 1,which will be wasted.So we change our plan of placing pans as Figure 20.

Figure 20.

The number of circular ones decreases by two,but the space in district 1is fully used,and no pan will be placed beyond the borderline.

We assume that P is bigger than P ?1,so that,the heat in district 1will be fully used.By simple calculating,we know that the ratio of the heat absorbed in circular pan (1H )and in regular hexagon (2H )is 1.2:1.

Figure21.

So,based on the pans placing plan,a equation on heat can be got as follow:

???

≤

≈

...

3,2,1

)

;

(9)

Resolution:?????????????????=??+?+??+=??+?+???+==?+?+=?=+??+?+=...3,2,1)24(2.1)325()24(2.1)3215()2(232)12(12132221212111121211201k A N n Q Q n H A N n Q Q n H k n n n N N k n n n N N (10)

1N and 1H means the number of pans and average heat absorbed when n is odd.2N and 2H means the number of pans and average heat absorbed when n is even.For example:

(1)When 37.0=λ,6.0=P :

16=N ,A

Q H 075.1=.The best placing plan is:(2)When 37.0=λ,7.0=P :

18=N ,A

Q H ?=044.1.The best placing plan

is:

(3)When 58.0=λ,6.0=P ：

12=N ,A

Q H ?=067.1.The best placing plan is:

A conclusion is easy to draw that when the ratio of width and length of the oven (λ)is a constant,the number of pans increases with an increasing P,but the average heat decreases (example (1)and (2)).When the weight P is a constant,the number of pans decreases with an increasing λ,and the average heat decreases also.So,the actual plan should be base on your specific needs.

Conclusion

In conclusion,our team is very certain that the method we came up with is effective in heat distribution analysis.Based on our model,the more edges the pan has,the more evenly the heat distribute on.With the discretize-convert approach,we know that when the ratio of width and length of the oven (γ)is a constant,the number of pans increases with an increasing P ,but the average heat decreases.When the weight P is a constant,the number of pans decreases with an increasing γ,and the average heat decreases also.So,the actual plan should be base on your specific needs.

Strengths &Weaknesses

Strengths

?Difficulties Avoided Avoided.

.In model 1,we turn to another way to work simulate the heat distribution instead of work on heat conduction directly.Firstly,we simulate heat radiation not heat conduction to keep away from the sophisticated partial differential equations.Then,we create a unique method to convert every variable from the whole space to section.In other words,we work everything out in heat radiation and convert them into heat contradiction.

?Close to Reality.

Our model considers both the thermal radiation and surface reflection,which is relatively close to the actual situation.

?Flexibility Provided.

Our algorithm does a great job in flexibility.The heat distribution map on sections are intercepted from the heat distribution maps of the entire space.All shapes of sections can be used in the algorithm.The heat distribution in the whole space is generated based on the location of the electrical heating tubes and the decay curve of the heat, which can be modified at any time.

?Innovation.

Based on our model,the space of an oven can be divided into six parts with different hear distribution.In order to make full use of the inner space,we invent a new pan which allows users to cook six different kinds of food at same time.An advertisement is published in the end of the paper.

Weaknesses

Pan’’s Thermal Conductivity Ignored.

?Pan

The heat comes from not only the electrical heating tubes,but also heat conduction of the pans themselves.But the pan’s thermal conduction is ignored in the model,which may cause little inaccuracy.

?Thermal Conductivity of Electrical Heating Tubes Ignored

Ignored..

it is assumed that there are two electrical heating tubes in the oven and placed in a specific location.The initial temperature of the tubes is a desired constant temperature. In other words,the time electrical heating tubes spend to heating themselves is ignored.The simplification can cause some inaccuracy.

simplification..

?Linear simplification

In model2,the length of the oven is discretized,so that the number of pans will changes linearly.calculating through simple integer linear method.This will lead to the result of our model is not accurate enough.

Application

We have discussed the heat distribution in the oven in model1.The heat distribution

is shown in figure1and figure2.

美国数学建模大赛比赛规则

数学中国MCM/ICM参赛指南翻译（2014版） MCM:The Mathematical Contest in Modeling MCM：数学建模竞赛 ICM:The InterdisciplinaryContest in Modeling ICM：交叉学科建模竞赛ContestRules, Registration and Instructions 比赛规则，比赛注册方式和参赛指南 (All rules and instructions apply to both ICM and MCMcontests, except where otherwisenoted.)（所有MCM的说明和规则除特别说明以外都适用于 ICM）每个MCM的参赛队需有一名所在单位的指导教师负责。指导老师：请认真阅读这些说明，确保完成了所有相关的步骤。每位指导教师的责任包括确保每个参赛队正确注册并正确完成参加MCM/ ICM所要求的相关步骤。请在比赛前做一份《参赛指南》的拷贝，以便在竞赛时和结束后作为参考。组委会很高兴宣布一个新的补充赛事（针对MCM/ICM 比赛的视频录制比赛）。点击这里阅读详情！ 1.竞赛前

A．注册 B．选好参赛队成员 2.竞赛开始之后 A．通过竞赛的网址查看题目 B．选题 C．参赛队准备解决方案 D．打印摘要和控制页面 3.竞赛结束之前 A.发送电子版论文。 4.竞赛结束的时候, A. 准备论文邮包 B．邮寄论文 5.竞赛结束之后 A. 确认论文收到 B．核实竞赛结果 C．发证书 D．颁奖 I. BEFORE THE CONTEST BEGINS:（竞赛前）A.注册所有的参赛队必须在美国东部时间2014年2月6号（星期四）下午2点前完成注册。届时，注册系统将会自动关闭，不再接受新的注册。任何未在规定时间

全国数学建模竞赛一等奖论文

交巡警服务平台的设置与调度摘要由于警务资源有限，需要根据城市的实际情况与需求建立数学模型来合理地确定交巡警服务平台数目与位置、分配各平台的管辖范围、调度警务资源。设置平台的基本原则是尽量使平台出警次数均衡，缩短出警时间。用出警次数标准差衡量其均衡性，平台与节点的最短路衡量出警时间。对问题一，首先以出警时间最短和出警次数尽量均衡为约束条件，利用无向图上任意两点最短路径模型得到平台管辖范围，并运用上下界网络流模型优化解,得到A区平台管辖范围分配方案。发现有6个路口不能在3分钟内被任意平台到达，最长出警时间为5.7分钟。其次，利用二分图的完美匹配模型得出20个平台封锁13个路口的最佳调度方案，要完全封锁13个路口最快需要8.0分钟。最后，以平台出警次数均衡和出警时间长短为指标对方案优劣进行评价。建立基于不同权重的平台调整评价模型，以对出警次数均衡的权重u和对最远出警距离的权重v 为参数，得到最优的增加平台方案。此模型可根据实际需求任意设定权重参数和平台增数，由此得到增加的平台位置，权重参数可反映不同的实际情况和需求。如确定增加4个平台，令u=0.6，v=0.4，则增加的平台位置位于21、27、46、64号节点处。对问题二，首先利用各区平台出警次数的标准差和各区节点的超距比例分析评价六区现有方案的合理性，利用模糊加权分析模型以城区的面积、人口、总发案次数为因素来确定平台增加或改变数目。得出B、C区各需改变2个平台的位置，新方案与现状比较，表明新方案比现状更合理。D、E、F区分别需新增4、2、2个平台。利用问题一的基于不同权重的平台调整评价模型确定改变或新增平台的位置。其次，先利用二分图的完美匹配模型给出80个平台对17个出入口的最优围堵方案，最长出警时间12.7分钟。在保证能够成功围堵的前提下，若考虑节省警力资源，分析全市六区交通网络与平台设置的特点，我们给出了分阶段围堵方案，方案由三阶段构成。最多需调动三组警力，前后总共需要29.2分钟可将全市路口完全封锁。此方案在保证成功围堵嫌疑人的前提下，若在前面阶段堵到罪犯，则可以减少警力资源调度，节省资源。【关键字】：不同权重的平台调整评价模糊加权分析最短路二分图匹配

全国大学生数学建模竞赛论文--范例

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）：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）：

眼科病床的合理安排摘要病床是医院的重要卫生资源，其使用情况是反映医院工作效率的重要指标，合理分配床位、提高病床使用率对于充分利用医疗资源、提高医院的两个效益有着十分重要的意义。本题针对某医院眼科病床分配中存在的不合理现象，让我们建立一个合理的病床安排模型，以解决病床的最优分配问题，从而提高对医院资源的有效利用。针对问题一，本文制定的指标评价体系包括门诊相关指标集（病人平均等待时间、门诊等待平均队长、病人平均满意度）和病床相关指标集（出院者平均住院日数、病床平均工作日、病床平均周转率、实际病床利用率）。为了能够全面地评价出模型的优劣，本文采用目前普遍使用的密切值法、TOPSIS法和RSR法等综合评价方法，并对应建立了三个评价模型，以得出更为科学合理的结论。针对问题二，本文建立了以病床需求数为状态转移变量、以各类病人的病床安排数为决策变量的动态规划模型。模型中，充分考虑了观测期内病人平均等待时间、病床平均周转率、病床利用率和潜在流失率等指标，且在制定寻优策略时，引入了病人满意度量化函数和优先级函数，使得模型更加合理。通过Matlab 对该模型求解，得出了次日病床安排方案（结果见表4）。综合评价模型时，以该医院目前的病床安排方案和我国医院通用的病床安排方法为比较对象，借助上述三种评价方法和模型，进行了综合评价比较，从综合评价结果来看，本文的模型相对较优（评价结果见表9）。针对问题三，本文既充分考虑了如何缩短病人平均等待时间和提高病床利用率，又兼顾了公平原则，根据病症的不同和就诊病人到院的顺序制订了优先服务策略，给出了每个病人相应的入住时间区间（见P18）。针对问题四，由于住院部周六和周日不安排手术，对某些类型病人的病床安排产生了一定的影响，因此我们对问题二中模型的优先级函数进行了相应的调整，并利用Matlab进行了求解（结果见表10）。为了判断手术安排时间是否改变，本文根据问题一的评价方法和模型对修改后的模型进行了综合评价，从评价结果得知，手术安排时间应该做相应的调整。针对问题五，为了使所有病人在系统内的平均逗留时间（含等待入院及住院时间）最短，本文建立了以其为目标函数且带约束条件的非线性规划模型，并利用了Lingo 软件对其进行求解，得出的结论是：分配给外伤、白内障（双眼）、白内障（单眼）、青光眼、视网膜疾病等各类型病人的床位数依次为：8、16、12、21、22，分别占总床数的比例为：10.13%、20.25%、15.19%、26.58%、27.85%。最后，本文对所建模型的优点和缺点进行了客观的评价，认为本文研究的结果在实际医院病床安排中有一定的参考价值。关键词：病人平均等待时间；实际病床利用率；RSR 法；满意度量化函数；动态规划模型；非线性规划 1.问题重述医院就医排队是大家都非常熟悉的现象，它以这样或那样的形式出现在我们面前，例如，

美国数学建模比赛题目及翻译

PROBLEM A: The Ultimate Brownie Pan When baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven. Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes - rectangular to circular and other shapes in between. Assume 1. 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)

如何准备美国大学生数学建模比赛

如何准备美赛数学模型：数学模型的功能大致有三种：评价、优化、预测。几乎所有模型都是围绕这三种功能来做的。比如，2012年美赛A题树叶分类属于评价模型，B题漂流露营安排则属于优化模型。对于不同功能的模型有不同的方法，例如评价模型方法有层次分析、模糊综合评价、熵值法等；优化模型方法有启发式算法（模拟退火、遗传算法等）、仿真方法（蒙特卡洛、元胞自动机等）；预测模型方法有灰色预测、神经网络、马尔科夫链等。在数学中国、数学建模网站上有许多关于这些方法的相关介绍与文献。软件与书籍：软件一般三款足够：Matlab、SPSS、Lingo，学好一个即可。书籍方面，推荐三本，一本入门，一本进级，一本参考，这三本足够：《数学模型》姜启源谢金星叶俊高等教育出版社《数学建模方法与分析》Mark M. Meerschaert 机械工业出版社《数学建模算法与程序》司守奎国防工业出版社入门的《数学模型》看一遍即可，对数学模型有一个初步的认识与把握，国赛前看完这本再练习几篇文章就差不多了。另外，关于入门，韩中庚的《数学建模方法及其应用》也是不错的，两本书选一本阅读即可。如果参加美赛的话，进级的《数学建模方法与分析》要仔细研究，这本书写的非常好，可以算是所有数模书籍中最好的了，没有之一，建议大家去买一本。这本书中开篇指出的最优化模型五步方法非常不错，后面的方法介绍的动态模型与概率模型也非常到位。参考书目《数学建模算法与程序》详细的介绍了多种建模方法，适合用来理解模型思想，参考自学。分工合作：数模团队三个人，一般是分别负责建模、编程、写作。当然编程的可以建模，建模的也可以写作。这个要视具体情况来定，但这三样必须要有人擅长，这样才能保证团队最大发挥出潜能。这三个人中负责建模的人是核心，要起主导作用，因为建模的人决定了整篇论文的思路与结构，尤其是模型的选择直接关系到了论文的结果与质量。对于建模的人，首先要去大量的阅读文献，要见识尽可能多的模型，这样拿到一道题就能迅速反应到是哪一方面的模型，确定题目的整体思路。其次是接口的制作，这是体现建模人水平的地方。所谓接口的制作就是把死的方法应用到具体问题上的过程，即用怎样的表达完成程序设计来实现模型。比如说遗传算法的方法步骤大家都知道，但是应用到具体问题上，编码、交换、变异等等怎么去做就是接口的制作。往往对于一道题目大家都能想到某种方法，可就是做不出来，这其实是因为接口不对导致的。做接口的技巧只能从不断地实践中习得，所以说建模的人任重道远。另外，在平时训练时，团队讨论可以激烈一些，甚至可以吵架，但比赛时，一定要保持心平气和，不必激烈争论，大家各让3分，用最平和的方法讨论问题，往往能取得效果并且不耽误时间。经常有队伍在比赛期间发生不愉快，导致最后的失败，这是不应该发生的，毕竟大家为了一个共同的目标而奋斗，这种经历是很难得的。所以一定要协调好队员们之间的关系，这样才能保证正常发挥，顺利进行比赛。美赛特点：一般人都认为美赛比国赛要难，这种难在思维上，美赛题目往往很新颖，一时间想不出用什么模型来解。这些题目发散性很强，需要查找大量文献来确定题目的真正意图，美赛更为注重思想，对结果的要求却不是很严格，如果你能做出一个很优秀的模型，也许结果并不理想也可能获得高奖。另外，美赛还难在它的实现，很多东西想到了，但实现起来非常困难，这需要较高的编程水平。除了以上的差异，在实践过程中，美赛和国赛最大的区别有两点：第一点区别当然是美赛要用英文写作，而且要阅读很多英文文献。对于文献阅读，可以安装有道词典，

2012美国大学生数学建模题目(英文原版加中文翻译)

2012 MCM Problems PROBLEM A:The Leaves of a Tree "How much do the leaves on a tree weigh?" How might one estimate the actual weight of the leaves (or for that matter any other parts of the tree)? How might one classify leaves? Build a mathematical mode l to describe and classify leaves. Consider and answer the following: ? Why do leaves have the various shapes that they have? ? Do the shapes “minimize” overlapping individual shadows that are cast, so as to maximize exposure? Does the distribution of leaves within the “volume” of the tree and its branches effect the shape? ? Speaking of profiles, is leaf shape (general characteristics) related to tree profile/branching structure? ? How would you estimate the leaf mass of a tree? Is there a correlation between the leaf mass and the size characteristics of the tree (height, mass, volume defined by the profile)? In addition to your one page summary sheet prepare a one page letter to an editor of a scientific journal outlining your key findings. “多少钱树的叶子有多重？”怎么可能估计的叶子（或树为此事的任何其他部分）的实际重量？会如何分类的叶子吗？建立了一个数学模型来描述和分类的叶子。考虑并回答下列问题：?为什么叶片有，他们有各种形状？?请勿形状的“最小化”个人投阴影重叠，以便最大限度地曝光吗？树叶树及其分支机构在“量”的分布效应的形状？说起型材，叶形（一般特征）有关的文件树/分支结构？你将如何估计树的叶质量？有叶的质量和树的大小特性（配置文件中定义的高度，质量，体积）之间的关系吗？除了你一个页面的汇总表，准备一页纸的信中列出您的主要结果的一个科学杂志的编辑. PROBLEM B:Camping along the Big Long River Visitors to the Big Long River (225 miles) can enjoy scenic views and exciting whi t e water rapids. The river is inaccessible to hikers, so the only way to enjoy i t is to take a river trip that requires several days of camping. River trips all start at First Launch and exi t the river at Final Exit, 225 miles downstream. Passengers take either oar- powered rubber rafts, which travel on average 4 mph or motorized boats, which travel on average 8 mph. The trips range from 6 to 18 nights of camping on the river, start to finish.. The government agency responsible for managing this river wants every trip to enjoy a wilderness experience, with minimal contact wi t h other groups of boats on the river. Currently, X trips travel down the Big Long River each year during a six month period (the rest of the year it is too cold for river trips). There are Y camp sites on the Big Long River, distributed fairly uniformly throughout the river corridor. Given the rise in popularity of river rafting, the park managers have been asked to allow more trips to travel down the river. They want to determine how they might schedule an optimal mix of trips, of varying duration (measured in nights on the river) and propulsion (motor or oar) that will utilize the campsites in the best way possible. In other words, how many more boat trips could be added to the Big Long River’s rafting season? The river managers have hired you to advise them on ways in which to develop the best schedule

美国大学生数学建模竞赛优秀论文翻译

优化和评价的收费亭的数量景区简介由於公路出来的第一千九百三十，至今发展十分迅速在全世界逐渐成为骨架的运输系统，以其高速度，承载能力大，运输成本低，具有吸引力的旅游方便，减少交通堵塞。以下的快速传播的公路，相应的管理收费站设置支付和公路条件的改善公路和收费广场。然而，随着越来越多的人口密度和产业基地，公路如花园州公园大道的经验严重交通挤塞收费广场在高峰时间。事实上，这是共同经历长时间的延误甚至在非赶这两小时收费广场。在进入收费广场的车流量，球迷的较大的收费亭的数量，而当离开收费广场，川流不息的车辆需挤缩到的车道数的数量相等的车道收费广场前。因此，当交通繁忙时，拥堵现象发生在从收费广场。当交通非常拥挤，阻塞也会在进入收费广场因为所需要的时间为每个车辆付通行费。因此，这是可取的，以尽量减少车辆烦恼限制数额收费广场引起的交通混乱。良好的设计，这些系统可以产生重大影响的有效利用的基础设施，并有助于提高居民的生活水平。通常，一个更大的收费亭的数量提供的数量比进入收费广场的道路。事实上，高速公路收费广场和停车场出入口广场构成了一个独特的类型的运输系统，需要具体分析时，试图了解他们的工作和他们之间的互动与其他巷道组成部分。一方面，这些设施是一个最有效的手段收集用户收费或者停车服务或对道路，桥梁，隧道。另一方面，收费广场产生不利影响的吞吐量或设施的服务能力。收费广场的不利影响是特别明显时，通常是重交通。其目标模式是保证收费广场可以处理交通流没有任何问题。车辆安全通行费广场也是一个重要的问题，如无障碍的收费广场。封锁交通流应尽量避免。模型的目标是确定最优的收费亭的数量的基础上进行合理的优化准则。主要原因是拥挤的

美国大学生数学建模竞赛赛题翻译

2015年美国大学生数学建模竞赛赛题翻译 2015年美国大学生数学竞赛正在进行，比赛时间为北京时间：20１5年2月６日(星期五)上午9点—２月10日上午9点.竞赛以三人(本科生）为一组，在四天时间内,就指定的问题，完成该实际问题的数学建模的全过程,并就问题的重述、简化和假设及其合理性的论述、数学模型的建立和求解(及软件)、检验和改进、模型的优缺点及其可能的应用范围的自我评述等内容写出论文。 2０15 ＭＣM/ICM Ｐｒｏblems 总计4题，参赛者可从MＣM Pｒoｂlｅm A, MＣM Problem B，ICM Pｒoblem C oｒICM Ｐｒｏbｌem D等四道赛题中自由选择。 201５Contest Proｂｌems MCM ＰROBLEMS PROBLEM A: Erａdiｃating Ebｏla The wｏrｌｄmedical assoｃiatiｏn has ａnnounced that ｔheｉｒneｗmｅdiｃatioｎcouｌd stop Ｅbola anｄcｕｒｅｐａtｉents ｗhose dｉsease ｉs not aｄｖanced. Ｂuild a realistｉc, sensｉblｅ, ａnｄusefｕｌmodel thaｔｃｏnsiders not onlｙｔhｅsｐｒead of the dｉsease，ｔｈｅqｕaｎtｉｔy of ｔhｅmedｉcine needed，ｐｏｓsｉble ｆeasible delivｅｒy sysｔemｓ（seｎdiｎg ｔhe meｄiｃine to where iｔis neｅdｅｄ), （gｅｏgrａphiｃal）loｃatioｎs of ｄelｉvery，ｓｐeed of mａｎufacturing of tｈe ｖa ｃcine ｏｒｄruｇ, but also ａｎy oｔheｒcriticａl factors your ｔeam cｏnｓidｅrs ｎecessaｒｙas ｐarｔｏf thｅｍodel to ｏpｔimize thｅerａdicaｔioｎoｆEｂola，oｒａt lｅａｓt ｉts current sｔrain. Iｎadd ｉtion ｔo your ｍodｅling apprｏach fｏr thｅcontest, prepare ａ１—2 page ｎon-teｃｈnical ｌettｅr for ｔhe wｏrld ｍｅｄicaｌａｓsocｉation ｔｏuse ｉｎtheir announｃｅmｅｎt．中文翻译：问题一：根除埃博拉病毒世界医学协会已经宣布他们的新药物能阻止埃博拉病毒并且可以治愈一些处于非晚期疾病患者。建立一个现实的,合理的并且有用的模型，该模型不仅考虑了疾病的蔓延,需要药物的量,可能可行的输送系统,输送的位置,疫苗或药物的生产速度，而且也要考虑其他重要的因素，诸如你的团队认为有必要作为模型的一部分来进行优化而使埃博拉病毒根除的一些因素，或者至少考虑当前的状态。除了你的用于比赛的建模方法外，为世界医学协会准备一份１-２页的非技术性的信,方便其在公告中使用。 PＲＯBLEＭB: Sｅａrchinｇfoｒａloｓt pｌａne Recａll the loｓｔMalａysian flight MH370．Build ａgeneriｃmaｔheｍatｉcaｌmｏｄel ｔｈat coｕld assist "seａrｃhers" ｉn plaｎninｇa useful seａrch ｆｏｒ a loｓt plaｎｅfeａred to hａｖe cｒａshed in ｏpｅn ｗatｅr sucｈas the Ａｔlantic, Paｃific，Iｎdian, Sｏuthｅrｎ，or Arcｔic Oceａn ｗｈil ｅｆlyinｇfrom PoinｔA to Poiｎt B. Ａｓsume thａt there are no sigｎals ｆroｍthe ｄowｎｅd pｌane。Your moｄel should recogｎiｚe thaｔｔherｅarｅmaｎy dｉfｆerｅnt ｔyｐes ｏf plaｎｅs ｆoｒｗ

第十一届五一数学建模联赛A优秀论文

2014年第十一届五一数学建模联赛承诺书我们仔细阅读了五一数学建模联赛的竞赛规则。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）与本队以外的任何人（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它公开的资料（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，我们愿意承担由此引起的一切后果。我们授权五一数学建模联赛组委会，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为（从A/B/C中选择一项填写）： A 我们的参赛报名号为：参赛组别（研究生或本科或专科）：所属学校（请填写完整的全名）参赛队员 (打印并签名) ：1. 2. 3. 日期：年月日获奖证书邮寄地址：邮政编码 2014年第十一届五一数学建模联赛编号专用页

竞赛评阅编号（由竞赛评委会评阅前进行编号）：评阅记录

裁剪线裁剪线裁剪线竞赛评阅编号（由竞赛评委会评阅前进行编号）：参赛队伍的参赛号码：（请各参赛队提前填写好）： 2014年第十一届五一数学建模联赛题目对黑匣子落水点的分析和预测摘要本文通过对飞机以及黑匣子受力情况进行分析，构建正交分解模型，得出飞机的坠落轨迹和黑匣子的落水点，及黑匣子在水中的移动情况。

问题一要求在考虑空气气流影响的前提下，建立数学模型，描述飞机坠落轨迹并推测黑匣子的落水点。本文对飞机失去动力后的全过程建立动力学方程： 22d r m mg f dt =-+ 然后对动力学方程进行正交分解，在水平和竖直方向上分别进行分析，根据伯努利方程求得升力的计算公式，得出飞机在刚刚失去动力时，升力大于重力，所以飞机会先上升一段距离，随着水平速度的减小，升力也逐渐减小，然后飞机再下降，通过模拟计算可以得出当飞机坠落至失事点下10000m 时，飞机坠入海面，其飞行速度为m s ，飞机向东北方向飞行了28697m 。问题二要求建立数学模型，描述黑匣子在水中沉降过程轨迹，并指出它沉在海底的位置所在的区域范围。由于不用考虑洋流，黑匣子所受到的力中仅有水的阻力是变化的，其重力和浮力始终保持恒定，根据黑匣子的移动速度，得出相应的阻力和加速度。在不同的速度范围内，使用不同的阻力公式，计算出相应的移动距离并作出轨迹图。发现在水平方向仅漂出 m ，速度几乎为零，因此黑匣子在I 区域内。问题三要求描述黑匣子沉降轨迹方程，并求解出黑匣子沉入水下1000m ，2000m 和3000m 时离落水点的方位。根据问题一中得出的结果，可以大致判断出黑匣子的经纬度，查得当地的洋流为南赤道暖流，为风海

认证杯数学建模竞赛获奖论文

第七届“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七届“认证杯”数学中国数学建模网络挑战赛的竞赛规则。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，我们接受相应处理结果。我们允许数学中国网站(https://www.sodocs.net/doc/2612647993.html,)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：2900 参赛队员(签名) ：队员1：张安成队员2：勾旭东队员3：郑子嫣参赛队教练员(签名)：李石涛参赛队伍组别：本科组

第七届“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：（请各个参赛队提前填写好）：2900 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）：

2014年第七届“认证杯”数学中国数学建模网络挑战赛第一阶段论文题目土地储备方案的风险评估关键词土地储备主成分分析层次分析法风险函数风险评估摘要：本文针对土地在收储过程中存在一定的金融风险这一问题，综合运用了单一变量分析、多元统计等方法，建立了主成分分析模型和层次分析法模型，利用EXCEL和MATLAB 软件及C语言程序进行求解，进而构造了土地储备方案的风险函数，并利用该风险函数，分别对附件二中给出的数据进行统计分析，从而找出10个风险最大的项目，并给出了造成这 10个项目风险较大的原因。首先，我们分析了单一变量对土地储备方案的风险的影响。我们利用EXCEL软件，分别绘制了74组方案当中的收购储备面积、动态回收周期、总收储成本、预期收益的对比图，粗略的得出与土地储备方案风险有关的主要因素，即收购储备面积，总收储成本以及预期收益。得出影响风险的主要因素后，我们继续利用EXCEL软件，计算得出单位储备面积的成本以及收益，进而绘制出单位储备面积内，收益与成本的对比图，得出74组方案的收益与成本的差值范围。其次，我们建立主成分分析模型，利用MATLAB软件进行相关系数的计算，相对准确的找出与土地储备方案的风险有关的主要因素，并构造了土地储备方案的风险函数，即 S=0.264X1+0.422X2-0.313X3 其中，S表示加权之和，即风险总值，X1表示收购储备面积，X2表示总收储成本，X3表示预期收益。随后，我们根据构造的土地储备方案的风险函数S=0.264X1+0.422X2-0.313X3，将74组方案当中的相关数据代入，得出74组风险函数值，我们将74组数据值输入到C 语言程序中，设计从大到小顺序排列的程序，得出前10名的数据，与之对应的项目即为10个风险最大的项目，方案序号分别为10、37、47、50、51、57、60、64、66、74，而造成风险较大的原因大都是收购储备面积过大，总存储成本过高或预期收益较小。最后，为了保证所建立模型的可行性以及计算结果的可靠性，我们对所建立的模型进行了检验。因为土地储备方案风险评估是一个决策问题，所以也可用层次分析法进行求解。于是我们又建立了层次分析法模型，利用MATLAB软件进行求解，得出影响土地储备风险的三个主要因素分别为收购储备面积，总收储成本和预期收益，这一结果验证了主成分分析模型的可行性和风险函数计算结果的可靠性。参赛队号: 2900 Array 所选题目: C 题

数学建模竞赛如何写一篇能拿奖的论文

全国大学生数学建模竞赛如何写一篇能拿奖的论文 1.开篇数学建模竞赛实则为一种竞技比赛，则竞技比赛只要把握要应对技巧，渣渣队伍获奖可能性也会大大增加。作为一名过来人，除了参加过多次数学建模竞赛，同时跟评委老师有所沟通，大致可以得出这么一个定理：摘要箩筐判别法则：由于竞赛过程中，老师的数量是有限的，同时查阅论文的压力也是巨大的，时间的压迫及数量的追求，导致论文在查阅过程中无法非常详细地进行查阅。而在查阅过程中，摘要作为首要展示，也自然作为最重要的评判标准。也就出现了，摘要过拿省三，摘要挂回家睡，即使模型再怎么完美，摘要的撰写出现问题，在评分上也会受到很大的限制！假如把论文当做人来看，摘要就是人的脸，而在颜值当道的社会下，颜值不高从最开始就少了很多机会，所以写好摘要，为论文的脸认真化妆，这是在论文撰写过程中极为重要的！ 2.摘要（1）用1、2句话说明原题中要求解决的问题；（2）建立了什么模型（在数学上属于什么类型）建模的思想（思路），模型特点；（3）算法思想（求解思路），特色；（4）主要结果（数值结果，结论）；（5）模型优点，模型检验，灵敏度分析，有无改进、推广。 ·特色和创新之处必须在这里强调（稍夸张地）。 ·长度：理想长度很难说，必须包括上述要点，但简洁也非常重要。一般掌握在半页至2/3页左右。 ·摘要是文章最重要的部分。要保证准确、简明、条理清晰，突出特色和创新点。注：全国评阅时将首先根据摘要和论文整体结构及概貌对论文优劣进行初步筛选。 3.问题重述 ·不是题目的完整拷贝 ·根据自己的理解，用自己的语言清楚简明地阐述背景、条件和要求。注：有些同学提问可不可以直接复制问题，其实目前并没有明确不能直接复制，但通过自己的理解撰写出来的问题重述，一般都能为论文争取多一点分数。 4.模型假设假设要合理且全面，但不欣赏罗列大量无关紧要的假设，关键性假设不能缺。根据全国组委会确定的评阅原则，基本假设的合理性很重要。（1）根据题目中条件作出假设；（2）根据题目中要求作出假设；

数学建模美国赛历年试题

历年美国大学生数学建模赛题目录 MCM85问题-A 动物群体的管理 (3) MCM85问题-B 战购物资储备的管理 (3) MCM86问题-A 水道测量数据 (4) MCM86问题-B 应急设施的位置 (4) MCM87问题-A 盐的存贮 (4) MCM87问题-B 停车场 (5) MCM88问题-A 确定毒品走私船的位置 (5) MCM88问题-B 两辆铁路平板车的装货问题 (5) MCM89问题-A 蠓的分类 (5) MCM89问题-B 飞机排队 (6) MCM90-A 药物在脑内的分布 (6) MCM90问题-B 扫雪问题 (6) MCM91问题-B 通讯网络的极小生成树 (6) MCM 91问题-A 估计水塔的水流量 (7) MCM92问题-A 空中交通控制雷达的功率问题 (7) MCM 92问题-B 应急电力修复系统的修复计划 (7) MCM93问题-A 加速餐厅剩菜堆肥的生成 (7) MCM93问题-B 倒煤台的操作方案 (8) MCM94问题-A 住宅的保温 (8) MCM 94问题-B 计算机网络的最短传输时间 (9) MCM-95问题-A 单一螺旋线 (9) MCM95题-B A1uacha Balaclava学院 (10) MCM96问题-A 噪音场中潜艇的探测 (10) MCM96问题-B 竞赛评判问题 (10) MCM97问题-A Velociraptor(疾走龙属)问题 (11) MCM97问题-B为取得富有成果的讨论怎样搭配与会成员 (11) MCM98问题-A 磁共振成像扫描仪 (12) MCM98问题-B 成绩给分的通胀 (13) MCM99问题-A 大碰撞 (13) MCM99问题-B “非法”聚会 (13) MCM2000问题-A空间交通管制 (13) MCM2000问题-B: 无线电信道分配 (14) MCM2001问题- A: 选择自行车车轮 (14) MCM2001问题-B 逃避飓风怒吼（一场恶风...） .. (15) MCM2001问题-C我们的水系-不确定的前景 (15) MCM2002问题-A风和喷水池 (15) MCM2002问题-B航空公司超员订票 (16) MCM2002问题-C (16) MCM2003问题-A: 特技演员 (17) MCM2003问题-B: Gamma刀治疗方案 (18) MCM2003问题-C航空行李的扫描对策 (18)

美国数学建模经验

1, V olterra predator-prey model 沃尔泰拉捕食模型 2， a competition model 竞争模型 3，the Analytic Hierarchy Process 层次分析过程 4，Abstract 摘要 5，actual demographic data 真实人口统计数据relevant data 6，problem background 问题背景 7，assumptions 模型假定assume，suppose假定hypothetical假定的，假想的 8，have a negligible impact on 对。。。的影响很小 9，probability 概率，可能性the probability of ....的概率 10，extrapolate推断 11，simulate 模拟 12，the basic component of our model 13，attributes 属性，特性 14，model n/v塑造模型 15，计算，估计calculate 16,formula 公式，方程式 17,linear interpolation 线性插值 18，linear regression 线性回归 19，technological data 技术资料 20，optimal approximation最佳逼近 21，conclusion 总结 22，strengths 优点weaknesses缺点 23，bottom-up approach 倒置方法 24，infrastructure基础设施 25，base stations 基地 26,conservation of mass质量守恒 27，conservation of momentum 动量守恒 28，conservation of angular momentum 角动量守恒 29，coefficient of restitution 回弹系数 30，vibration震动，一次性往复震动 31，flexible object 弹性物体 32，orthogonality正交性 33，wave propagation 波的传播 34，impact 碰撞 35，waveform波形 36，parameter参数 37，angular velocity角速度 38，hysteresis curve滞后曲线 39，coefficient of restitution回弹系数 40，density 密度论文格式摘要Abstract (解决的问题，方法，结果)，关键字key words（可有，可无），目录（可有，

美国数学建模竞赛优秀论文阅读报告

2.优秀论文一具体要求：1月28日上午汇报 1）论文主要内容、具体模型和求解算法（针对摘要和全文进行概括）； In the part1, we will design a schedule with fixed trip dates and types and also routes. In the part2, we design a schedule with fixed trip dates and types but unrestrained routes. In the part3, we design a schedule with fixed trip dates but unrestrained types and routes. In part 1, passengers have to travel along the rigid route set by river agency, so the problem should be to come up with the schedule to arrange for the maximum number of trips without occurrence of two different trips occupying the same campsite on the same day. In part 2, passengers have the freedom to choose which campsites to stop at, therefore the mathematical description of their actions inevitably involve randomness and probability, and we actually use a probability model. The next campsite passengers choose at a current given campsite is subject to a certain distribution, and we describe events of two trips occupying the same campsite y probability. Note in probability model it is no longer appropriate to say that two trips do not meet at a campsite with certainty; instead, we regard events as impossible if their probabilities are below an adequately small number. Then we try to find the optimal schedule. In part 3, passengers have the freedom to choose both the type and route of the trip; therefore a probability model is also necessary. We continue to adopt the probability description as in part 2 and then try to find the optimal schedule. In part 1, we find the schedule of trips with fixed dates, types (propulsion and duration) and routes (which campsites the trip stops at), and to achieve this we use a rather novel method. The key idea is to divide campsites into different “orbits”that only allows some certain trip types to travel in, therefore the problem turns into several separate small problem to allocate fewer trip types, and the discussion of orbits allowing one, two, three trip types lead to general result which can deal with any value of Y. Particularly, we let Y=150, a rather realistic number of campsites, to demonstrate a concrete schedule and the carrying capacity of the river is 2340 trips. In part 2, we find the schedule of trips with fixed dates, types but unrestrained routes. To better describe the behavior of tourists, we need to use a stochastic model（随机模型）. We assume a classical probability model and also use the upper limit value of small probability to define an event as not happening. Then we use Greedy algorithm to choose the trips added and recursive algorithm together with Jordan Formula to calculate the probability of two trips simultaneously occupying the same campsites. The carrying capacity of the river by this method is 500 trips. This method can easily find the optimal schedule with X given trips, no matter these X trips are with fixed routes or not. In part 3, we find the optimal schedule of trips with fixed dates and unrestrained types and routes. This is based on the probability model developed in part 2 and we assign the choice of trip types of the tourists with a uniform distribution to describe their freedom