2008年数学建模竞赛题目(A题)

2008年全国大学生数学建模竞赛A题全国一等奖论文数码相机定位摘要本文通过对数码相机的靶标和像平面相互之间关系的分析，利用选取相关对应点和坐标转换的方法，确定靶标圆心在像平面的投影位置，进而完成了系统标定模型，解决了相机的单目定位问题。

对于问题1，为确定靶标上圆的圆心在一个相机像平面的像坐标，需要得到相机像平面中点与靶标上点的对应关系。

通过将相机外部参数和内部参数联立可以建立模型1。

对于问题2，内部参数通过焦距可以得到，而外部参数的获得则需要事先确定一组特殊点。

由于靶标上两条线的交点在像平面上的投影点即为这两条线在像平面上的投影图线的交点，因此我们首先对图像进行边缘提取和椭圆拟合，然后利用程序选择靶标上A 、C 两个圆的外公共切线的切点作为特殊点。

将对应特殊点带入（1）式，就可以求得外部参数。

最后利用几何关系得出靶标上圆心的坐标，带入得到它们在该相机像平面的坐标。

结果为：vA O （-4.4324，-6.7785，0）、vB O （-2.3，-6.4456，0）、vC O （3.39，-5.9757，0）、vD O （-4.5471，3.7096，0）、vE O （2.1965，3.2275，0）。

见图3。

对于问题3，为了检验模型，本文通过计算机模拟数据，可以得到一个内外参数都已知的图像。

进而可以确定这四个顶点在像平面的准确坐标。

根据（1）式可以得到这四个顶点的计算坐标，把计算坐标与准确坐标的距离为对角线的矩形面积称为误差面积，误差率=误差面积/相纸面积。

计算误差率分别为：0.017591%、0.01777%、0.01532%、0.01557%。

从而可知用此模型精确度高，稳定性强。

对于问题4，类似于问题3，进行计算机模拟，得到空间两不同角度拍摄图像，进而得到在此数码相机坐标系下的特殊点坐标。

由于在求像坐标时考虑到了数码相机的透视效应，也就是内部参数，而两个数码相机的空间位置关系仅仅是外部参数的关系，因此可以求得仅考虑外部参数时两个像平面上的坐标，进而做差求出两个数码相机的相对位置坐标。

R 旋转矩阵t 平移矢量K 相机标定矩阵AB（粗体）向量AB，粗体表示向量3尺寸、形状等信息，我们就可以确定靶标在三维空间中与相机的位置关系。

确定了靶标与相机的位置关系后，就可以很容易的将靶标上的点投射到靶标像平面上，当然也包括五个圆心。

在求取出五个圆心的空间坐标之后，将其投射到像平面上，就得到了第一问需要求的坐标。

在求解时涉及到一个问题，就是像平面上怎样确定切线。

因为像平面上的图形是不规则的，所以很难确定这些形状的切线。

因此我们考虑另外的方法，使用搜索的办法，利用模拟退火算法求解。

在如何检验模型的问题上，需要分两方面进行检验，一是精度，而是稳定性。

按照以上的方法求圆心在像平面上的坐标，并没有充分利用像平面上所有轮廓点的信息，因此可以利用这些点来检验模型的精度。

对于稳定性问题，可以采用计算机模拟的方法，随机修改图形的轮廓，并用以上的方法再次进行求解，通过比较修改前后的结果来分析模型的稳定性。

最后，考虑另外一台相机的定位相对位置问题。

根据前面模型，我们应能够对任意一台相机确定靶标相对它的位置，因此可以以这个靶标作为参照物，建立一个世界坐标系，将这两台相机的位置在这个坐标系里面表示出来，以此确定两台相机的相对位置。

四、模型假设1、假设靶标像的中心恰好在光轴上2、假设数码相机中图像平面与光轴垂直3、假设相机两个方向上焦距相等4、假设透镜的焦距很小，像距约等于焦距五、模型准备(一)靶图像矩阵表示首先将题目中的图片保存出来，得到的图像可以很方便的放到Matlab 里面进行处理。

但在处理之前还要进行进一步加工：(1) 将文件读入Matlab，使用imread()函数(2) 将矩阵变为0-1 矩阵对于用以上方式得到的矩阵，有两个值：0、15。

其中0 代表像素为白色的点，15代表像素为黑色的点。

为了方便下面处理，对需要把以上像素为15 的点值全部变为1。

以上两步的源代码见附录一。

4(二)图像轮廓的提取在提取图像轮廓时，首先要引入计算机图像处理技术中四邻域的概念。

具有仿射不变性的几何结构在相机定位中的应用摘要本文采用小孔成像的模型研究相机成像问题。

基于靶平面上的点与像平面上的点一一对应，本文研究了几种几何结构。

发现靶平面上两个圆的内公切线交点与两个圆心共线这种几何结构仿射到像平面上依然成立，即两个圆心和内公切线交点在像平面上的3个像点共线，并证明了这一结论。

本文提出一种运用0-1矩阵求公切线的算法，但在实际操作时采用作图法。

运用作图法可以在像平面上确定两个椭圆的内公切线交点，该交点为靶平面上两个圆的内公切线交点在像平面上所成的像。

靶平面上5个圆可以确定10个内公切线交点，这样用作图法就可以确定靶平面上10个内公切线交点在像平面上的10个像点。

在像平面上建立坐标，每个靶平面上的圆心的像用两个未知量表示，共有10个未知量。

根据已证明的结论可知，对于每个内公切线交点在像平面上的像点，都有相对应的两个圆心的像点与之共线，就可以得到共线所满足的方程。

10个内公切线交点的像点对应10个2次方程，10个未知量就可求出。

靶平面上的圆心的像就可以确定。

本文采用牛顿迭代法对2次方程组进行求解。

并研究了解的稳定性。

为了得到两部固定相机的相对位置，建立了2个像平面坐标系、2个相机坐标系和1个三维世界坐标系。

本文采用最小二乘法确定相机坐标系与三维世界坐标系的关系。

在具体算法中，并没有利用所求出来的靶平面上圆心以及它的像点的坐标求解，而是采用10个内公切线交点及其像点的坐标求解，这是因为圆心的像点是由内公切线交点的像点求出的，误差更大。

分别确定2个相机坐标系与三维世界坐标系的关系之后，就可以确定2个相机坐标系之间的关系。

最后，本文对模型进行了分析，对一些方法的精度进行了讨论。

关键词相机定位仿射不变性内公切线交点1 问题分析双目定位是用两部相机给物体拍照来定位。

对于物体上的一个特征点，用两部不同位置的照相机拍照，就获得该点在两个像平面上的坐标。

如果知道两部相机的相对位置，就可以知道该特征点的具体位置。

AbstractThe global temperature is rising rapidly today which has caused an extensive ice melt, so the study of predicting rising sea level because of ice melt in North Polar is essential. Our study will try to predict the impact to Florida from melting ice in North Polar .Our studies have three steps:●Predict the temperature: We did the prediction by Neural network and give thechange of temperature in 50 years, based on a large amount of data from the Intergovernmental Panel on Climate Change (IPCC);●Model the mass of ice melting and the sea level: The sea level model is mainlybased on the principle of Thermodynamics and iteration. The results demonstrate that the sea level will rise by 10.8 cm totally in 50 years. Prediction of our model can be proved to be credible by consulting the data from IPCC. We introduce a correct term αto modify our model. We can change αto simulate sea level rise in different temperature condition.●Analyze the impacts to Florida: We model the erosion of Florida’s shoreline tomake it clear that when sea level rise to a certain extent that Florida will face many serious problems such as flooding, destruction of biodiversity, (Health Care), Loss of agriculture production (salinization of soil) and so on over the next50 years.. We find the 17 cities or areas and 15 airports which are severelyimpacted by the rise of sea level.(Based on our results,) Without attaching more importance to solving the problem, shoreline of several coastal cities like Miami will be eroded seriously, and the lowest place−Key West will be disappeared. It will cost a huge financial loss, so further protection should be put into place.Content Introduction (3)Background (3)Our work (3)Study object: Ice cap in Greenland (3)Modeling the sea level (3)Analysis of Florida (3)Assumption (4)Model Ⅰ:Temperature Prediction (4)Grey Prediction Model: (4)Neural Network prediction Model: (4)Model Ⅱ:Melting ice and the rise of sea level (5)Model the rise of sea level (5)Heat from rise of temperature: (5)Mass of melting ice: (6)Design of Algorithms: (7)Model Results: Sea level will elevate by 10 centimeters in 50 years (7)Validation of our model: (8)Model ⅢAnalysis: The effects towards Florida (10)Major Cities Analysis (11)Miami (11)Tampa (11)Cape Coral (12)Key West (12)Other Impacts in Florida: (12)Recommendations to coastal Florida: (13)Judgments (13)Strengths (13)Weaknesses (13)Reference (14)IntroductionBackgroundGlobal Warming and sea level rise“Air temperatures at the top of the world continue to rise twice as fast as temperatures in lower latitudes, causing significant ice melt on land and sea” [Fears, December 17, 2014]. One of the serious consequences is that sea level will rise. Global average sea-level rose at an average rate of about 3.1[2.4 to 3.8] mm per year from 1993 to 2003[IPCC]. This information suggests that from 1993 to 2003 the sea-level rise by 3.1cm totally.Our workThe question requires us to predict the next 50 years’ condition of ice melting and analyze the effects on the Florida, especially some big cities. So we can separate this question into two parts:●How much and how fast will the see level rise within 50 years?●What are the effects on the Florida because of the rise of sea level, especiallysome big cities?Study object: Ice cap in GreenlandArctic mainly consists of Greenland, which occupies about 9% glaciers all over the world. Melting in Arctic is mainly due to Greenland, melting of floating ice can be ignored. So we can consider Greenland as study object.Modeling the sea levelWe develop a model for sea-level rise as the function of time. This model can predict sea-level rise in future.Analysis of FloridaAfter having calculated the increased sea level within next 50 years, we analyze the impact to the Florida.●Rising sea level can seriously threaten the development of cities. It has beenthreatening some islands and coastal cities. Over the next 18 years, about twothirds among 544 American towns will be twice as likely to face floods [Huang].More frequency hurricane will happen.●Sea water will corrode seacoast.● A large quantity of drinking water will be polluted.Assumption●Sea level rise is primarily due to the melting of ice cap in Green Land. We ignorethe other floating ice in the Northern Polar.●The increment of sea water from melting will flow over the oceans uniformly●Salt in the ice will not affects the procedure of melting.Model Ⅰ:Temperature PredictionGrey Prediction Model:The weakness of the grey prediction is that the result is increasing all the time. In other words, it cannot show the changes in detail.Neural Network prediction Model:Model Ⅱ:Melting ice and the rise of sea level Model the rise of sea levelThe main reason of the sea level rising is the melting of ice cap and the mass of melting ice is equal to the mass of sea water generated from melting. So, based on several physical principles, we model the rise of sea level by calculating the mass of melting ice. We assume that the increment of sea water from melting will flow over the world uniformly, which means the melting ice will contribute to the rise of sea level, divided by the area of the ocean.ρw V w=ρi V i=m i∆x=V w S oV w The increment of sea water from melting iceV i The total volume of melting ice capV m The total volume of water generated from melting∆x The sea level riseS o The overall ocean area: 361745300km2, this is 71 percent of earth’s total surface area (Wikipedia).Heat from rise of temperature:According to the principle of thermal transmission, heat will always be transmitted from high temperature to low temperature. So, the final state of stuff in the thermalcycling system will reach to a same temperature. So, we assume that the temperature of the whole ice cap will increase by ∆T when the world temperature rise by ∆T. However, it takes time for the ice cap to transmit the heat from the rise temperature. We use the ∆T every month to calculate the increase of melting ice in each month and get the total increment by accumulation, which means parts of the heat from temperature will be absorbed and used to melt ice. So, defining a coefficient(α)and we will have the heat which ice cap absorbs from the rising temperature in the n-th month:Q n=αc i(m c−∆m i(n−1))∆T nQ n The heat comes from rises of temperature in the n-th monthαThe coefficient of capacity of absorbing heat in one monthc i The specific heat capacity of icem c The mass of the whole ice cap 2.45×1016kg∆m in The mass of melting ice in the n-th month∆T n The change of temperature in the n-th monthWe try to find the α by calculating the mass of melting ice in known years. “Recently reported GrIS mass balance varies from near-balance to modest mass losses [47 to 97 gigatons (Gt) year−1] in the 1990s, increasing to a mass loss of 267 ± 38 Gt year−1 in 2007”(Michiel).α= 5.6735×10−3Mass of melting ice:The mass of melting ice in this month will depends on not only the rising temperature, but also the mass of melting ice in the last month.We divide the heat absorbed by the ice by the melting enthalpy of fusion for water to obtain the mass of extra melting ice resulted from the rise of temperature in this month.∆m in=∆m i(n−1)+Q n△fusHθmm i(k)=∑∆m in12kn=1∆m in The mass of melting ice in the n-th month△fusHθm: The melting enthalpy3.36×105J/kgm i(k)The total mass of melting ice in the next k yearsMoreover, “Since 2006, high summer melt rates have increased Greenland ice sheet mass loss to 273 gigatons per year” (Partitioning Recent Mass Loss). And the initial mass of melting ice in the first month will be calculated as follows:2.73×1014kg/12∆m i(−1)=m ii=2.73×101412kg∆x(k)=V wS o=m i(k)ρw S om ii: The mass of melting ice in the initial year.∆x(k)The total rise of sea level in the next k yearsDesign of Algorithms:Since we have the function ∆m in=f(∆m i(n−1),∆T n), we are able to calculate the mass of melting ice in the next n months using computer program with the data of temperatures and the initial amount of melting ice in the first month.This is easy to achieve using two linear arrays ∆m i[]and ∆T[]in MATLAB. Run a simple for loop from 2 to n and calculate ∆m in during each pass so that the whole ∆m i[]array can be found.So that the total amount of melting ice at n-th month is the summation from ∆m i[1] to ∆m i[n]. Also the total rise of sea level can be easily found.Model Results: Sea level will elevate by 10 centimeters in 50 years The solutions were coded using matlab:Figure 1: The mass of melting ice in the next 50 yearsFigure 2: The rise of sea level in the next 50 years The prediction about rise of sea level every decade in next 50 years:∆x(10)=0.9874 cm∆x(20)=2.5125 cm∆x(30)=4.6222 cm∆x(40)=7.3674 cm∆x(50)= 10.8037 cm Validation of our model:The results show that the sea level will elevate by approximately 10 centimeters totally and 2 millimeters/yr, which accord with the prediction in Relative Mean Sea Level trends from NOAA.Moreover, if we calculated the rise of sea level without considering the increase of temperature, which means sea level rise at the rate today in the next 50 years, the order of magnitudes is match up with our result. So, based on the analysis above, the results of our modelModel ⅢAnalysis: The effects towards FloridaBased on our results, the sea level will rise 10 centimeters in the following 50 years, which threaten Florida in the future and result in tremendous impacts. “Some 2.4 million people and 1.3 million homes, nearly half the risk nationwide, sit within 4 feet of the local high tide line. Sea level rise is more than doubling the risk of a storm surge at this level in South Florida.” (Florida and rising sea)Figure 3 the altitudes of Florida ()As we can see from this picture, most cities or counties in the southern Florida lie besides the coast. Statistic suggests that 17 counties with altitudes smaller than 3 feet will be threatened by the rise of sea level in 50 years. The counties and airports which will be involved are listed as follows:Cities: Miami, Homestead, Fort Lauderdale, West Palm Beach, Titusville, St Augustine, Clearwater, St Petersburg, Tampa, Brandon, Bradenton, Port Charlotte，Cape Coral, Bonita Springs, Naples, Marco Island, and Grand Isle.Airports: Miami International Airport, Fort Lauderdale International Airport, Central Florida Regional Airport, Cedar Knoll Flying Ranch, Daytona Beach Regional Airport, Craig Municipal Airport, Jacksonville International Airport, St George Island Airport.Major Cities AnalysisWe choose 4 metropolises which will be severely impacted to do some further analysis.MiamiMiami is the biggest city in Florida with the average elevation of 3 feet (0.9144m) [11ikipedia]. Also, it is a coastal city. So according to what we predict that sea level will rise 10.8037cm within 50 years, we can draw a conclusion that this city would be greatly influenced.The possibility of flooding would grow while the frequency of hurricane will increase. It is a big challenge to sewer system of Miami. According to our simulation, sea level rising would also threaten Miami International Airport. Another problem is that the sea water would gnaw at the shoreline. Many coastal man-made buildings are too closed to the sea which they would face a serious problem of being eroded. Aside from threatening of losing habitat, local drinking water would be polluted.TampaTampa is a city located on the west coast of Florida. It is the third largest city in Florida. It is famous because of tourism. Although the highest point in the city is only 48 feet (15 m) [wikipedia], rising sea level will do harm to its natural disaster. Tampais special because it has the Old Tampa Bay and Hillsborough Bay which is easy to be attacked by storm surge. Sea level rising would produce much more violent storm surge. The boundary of Tampa would also be lost. What’s more,Traffic facilities such as Tampa International Airport were under threatening of disappearing.Cape CoralThis city is famous because of its far-stretching beach and animated quay. Besides, it has more than 30 gardens and golf courses which attract many tourists. A variety of animals also promotes this city’s tourism. However, sea level rising would erode shoreline of Cape Coral. It would destroy natural environment of this area, and then damages biodiversity. And still worse, Pine Island would mostly disappear. So, the economic damage there will be hardly assessed.Key WestIt is an island of Florida, which have the lowest altitude. So if sea level rises to some extent, it would be the first to be under water.Other Impacts in Florida:Biodiversity: Wild life and rare animals in Florida will be impacted by loss of habitat and food. Moreover, it is hard for plants and animals in Florida to adapt the new climatic conditions and the increase of relative air humidity.Architecture:Sea level rise will cause salinization, which will impact the architectural production.Economy Pressure：More money will be put into the drainage systems and dam project, which means less city construction and business development.Health Care: Higher sea level will increase the risk of some disease like malaria.Recommendations to coastal Florida:●Build higher dams: In this way can cities hold back the rising flood waters.●Prepare for flooding: Complete supervisory control system. Guarantee thatcitizen can be evacuated in time.●Reduce carbon emission: More carbon emission means higher temperature, andthen lead to rising of sea level. So encourage citizen to live a low-carbon life.●Warn local citizen: Propagate relative knowledge of sea level rising to improvecitizen’s sense of self-protection when facing natural disaster. JudgmentsStrengths●We use Neural Network to predict temperature in future with a large amount ofreliable data. So our prediction of temperature in future is accurate relatively.●Our model can predict the sea level rising in different conditions, such asdifferent temperature.●Our model is relatively simple so it will take a little time to simulate. We caneasily get the result.Weaknesses●We ignore the areal variation about depth, salinity and temperature of the sea forsimplifying the model;●We neglect the floating ice which will bring some error;●We neglect the thermal expansion;The mode is only the function of time, so it can’t simulate unusual situations.ReferenceBen Strauss,Florida and rising sea,/news/floria-and-the-rising-seaFears, Darryl, Huang, Ming. “Rising sea level threatens millions of American and causes huge economic loss”. Souhu, March 16, 2012, 09:48 AM. Web. IPCC, http://www.ipcc.chMichiel van den Broeke，Partitioning Recent Mass Loss, Science 13 November 2009 “National Snow and Ice Data Center”, January 7, 2015。

靶标圆心像坐标确定与数码相机定位摘要数码相机实现定位功能，需确定靶标圆心的像坐标。

本文就如何确定靶标圆心像坐标展开了讨论，并给出了计算两部相机相对位置的模型。

在问题一中，我们采用坐标变换的方法建立确定靶标圆心像坐标的模型。

根据坐标系之间的关系，分别通过物坐标系的旋转、平移以及相机坐标系的缩放，引入绕物坐标系三坐标轴旋转的角度θξϕ，，以及物坐标系平移的量度321,,t t t 等参数确定出物坐标系到像坐标系变换的方程，由此即可得到求解靶标圆心像坐标的模型。

求解方程里面的参数时，考虑到计算的方便，我们选择两圆内公切线的交点作为标定点。

计算它们的物坐标与像坐标，代入上述方程即可求得参数的值。

对于问题二，根据圆的有关性质，两条内公切线的斜率(或斜率倒数)分别为连接对应两圆上任意两点连线斜率(或斜率倒数)的最大值和最小值。

基于此，容易求得像坐标系里面对应的内公切线的方程，它们的交点即为标定点的像坐标，对应的物坐标容易得到。

然后将这些标定点的坐标分别代入问题一建立的物坐标系到像坐标系变换的方程，求解得到相应的参数θξϕ，，，321,,t t t 的值。

最后再将各园圆心的物坐标代入上述方程，求得各圆圆心像坐标结果为：A(-49.8577，50.6559)，B(-24.5423，49.1824)，C(32.5168，48.5784)，D(18.3139，-30.6194)，E(-60.3038，-30.3856)。

在问题三中，我们选取物坐标系里面一条直线上的9个点，对它们对应的像坐标进行一元线性回归分析，对模型的精度进行检验；最终得到这9个点拟合优度为0.9096非常接近1，说明模型精度较高。

对于模型稳定性的分析，我们将各圆圆心的物坐标向左偏移1mm,考查对应的像坐标的变化；得到各圆心像坐标的偏移量的平均值与圆心物坐标的偏移量的相对误差是2.62%，说明模型稳定性较好。

最后我们对问题一、二中模型进行了检验，在A,C,D,E 四个圆上分别选取一些特定的点，利用它们的像坐标分别求出其对应的物坐标，找到这些物坐标与对应圆心物坐标之间的距离，比较这些距离同圆半径的实际值(即12mm)的差值，最终得到它们相对误差的平均值是1.66%，说明模型的可行性是较高的。

美国数学建模竞赛题目1985年：A题：动物群体的管理B题：战略物资储备的管理问题1986年：A题：海底地型测量问题B题：应急设施的优化选址问题1987年：A题：堆盐问题（盐堆稳定性问题）B题：停车场安排问题1988年：A题：确定毒品走私船位置B题：平板列车车厢的优化装载1989年：A题：蠓虫识别问题；最佳分类与隔离B题：飞机排队模型1990年：A题：脑中多巴胺的分布B题：铲雪车的路径与效率问题1991年：A题：估计水塔的水流量B题：通信网络费用问题1992年：A题：雷达系统的功率与设计式样B题：紧急修复系统的研制1993年：A题：堆肥问题B题：煤炭装卸场的最优操作1994年：A题：保温房屋设计问题B题：计算机网络的最小接通时间1996年：A题：大型水下物体的探测B题：快速遴选优胜者问题1997年：A题：恐龙捕食问题B题：会议混合安排问题1998年：A题：MRI图象处理问题B题：分数贬值问题1999年：A题：小星体撞击地球问题B题：公用设施的合法容量问题C题：确定环境污染的物质、位置、数量和时间的问题2000年：A题：空间交通管制B题：无线电信道分配C题：大象群落的兴衰2001年：A题：选择自行车车轮B题：逃避飓风怒吼C题：我们的水系-不确定的前景2002年：A题：风和喷水池B题：航空公司超员订票C题：如果我们过分扫荡自己的土地，将会失去各种各样的蜥蜴。

2003年：A题：特技演员B题：Gamma刀治疗方案C题：航空行李的扫描对策2004年：A题：指纹是独一无二的吗？B题：更快的快通系统C题：安全与否？2005年：A题：flood planningB题：tollboothsC题: Nonrenewable Resources2006年：A题：Positioning and Moving SprinklerSystems for IrrigationB题：Wheel Chair Access at AirportsC题：Trade-offs in the fight againstHIV/AIDS2007年：A题：GerrymanderingB题：The Airplane Seating ProblemC题：Organ Transplant: The Kidney Exchange Problem2008年：A题：Take a BathB题：Creating Sudoku PuzzlesC题：Finding the Good in Health Care Systems2009年：A题：Designing a Traffic CircleB题：Energy and the Cell PhoneC题：Creating Food Systems: Re-Balancing Human-Influenced Ecosystems。

2008年度全国大学生数学建模竞赛郑州轻工业学院选拔赛备选题目A. 电梯控制问题我校教三楼有四部电梯。

等电梯的人给出要上下的信号，电梯只有在空闲或同方向行进时才接受这个指令。

然而，电梯经常出现十分拥挤的状况，特别在上下课的时候，要等很长的时间，所以埋怨声很多。

请为电梯设计一个调度方案，减少大家的等待时间，减少师生的不满。

并分析说明你所设计方案的合理性和可操作性。

请你撰写一份800—2000字之间的建议书，说明你的方案使得管理者能够接受你的方案。

B. 汽车车库库存的数学模型某汽车制造厂有一大型仓库存放成品小型汽车，厂方希望将尽可能多的汽车贮存在车库内。

在满足一定要求的条件下，尽可能提高仓库的利用率。

设车库形状为200米╳300米的矩形，仓库只有一个门，位于矩形长边的正中央，门宽5米。

假设汽车形状只有两种形式，如下图所示：从网上查出以上两种型号汽车的形状尺寸。

要求：1、在任何时刻只有一辆汽车开出仓库大门，开出过程中不得有任何碰撞；2、摆放时任意两辆汽车之间至少保持40cm的间距，不重叠；3、出门时必须车头先出，不得使用任何其他辅助设备。

试建立合理的数学模型，解决以下问题。

1、在每辆车都可顺利开出车库的条件下，如何摆放，可提高车库利用率。

2、假设在车辆无法调出时，可以先将阻碍的车辆开出车库外，在这种情况下，给出车辆摆放的优化数学模型。

3、对问题2的车俩摆放模型，假定每辆汽车开出仓库时的速度均相同，且汽车前轮可以左右转动90度，给出将车库4个角落的汽车全部开出所需最少时间的调运方案。

C. 自习教室开放的优化管理近年来，大学用电浪费比较严重，集中体现在学生上晚自习上，一种情况是去某个教室上自习的人比较少，但是教室内的灯却全部打开，第二种情况是晚上上自习的总人数比较少，但是开放的教室比较多，这要求我们提供一种最节约、最合理的管理方法。

下面是某学校收集的部分数据，请完成以下问题.管理人员只需要每天晚上开一部分教室供学生上自习，每天晚上从7：00---10：00开放（如果哪个教室被开放，则假设此教室的所有灯管全部打开）。

AbstractThe global temperature is rising rapidly today which has caused an extensive ice melt, so the study of predicting rising sea level because of ice melt in North Polar is essential. Our study will try to predict the impact to Florida from melting ice in North Polar .Our studies have three steps:●Predict the temperature: We did the prediction by Neural network and give thechange of temperature in 50 years, based on a large amount of data from the Intergovernmental Panel on Climate Change (IPCC);●Model the mass of ice melting and the sea level: The sea level model is mainlybased on the principle of Thermodynamics and iteration. The results demonstrate that the sea level will rise by 10.8 cm totally in 50 years. Prediction of our model can be proved to be credible by consulting the data from IPCC. We introduce a correct term to modify our model. We can change to simulate sea level rise in different temperature condition.●Analyze the impacts to Florida: We model the erosion of Florida’s shoreline tomake it clear that when sea level rise to a certain extent that Florida will face many serious problems such as flooding, destruction of biodiversity, (Health Care), Loss of agriculture production (salinization of soil) and so on over the next50 years.. We find the 17 cities or areas and 15 airports which are severelyimpacted by the rise of sea level.(Based on our results,) Without attaching more importance to solving the problem, shoreline of several coastal cities like Miami will be eroded seriously, and the lowest place Key West will be disappeared. It will cost a huge financial loss, so further protection should be put into place.Content Introduction (3)Background (3)Our work (3)Study object: Ice cap in Greenland (3)Modeling the sea level (3)Analysis of Florida (3)Assumption (4)Model Ⅰ:Temperature Prediction (4)Grey Prediction Model: (4)Neural Network prediction Model: (4)Model Ⅱ:Melting ice and the rise of sea level (5)Model the rise of sea level (5)Heat from rise of temperature: (6)Mass of melting ice: (6)Design of Algorithms: (7)Model Results: Sea level will elevate by 10 centimeters in 50 years (8)Validation of our model: (9)Model ⅢAnalysis: The effects towards Florida (10)Major Cities Analysis (11)Miami (11)Tampa (11)Cape Coral (12)Key West (12)Other Impacts in Florida: (12)Recommendations to coastal Florida: (13)Judgments (13)Strengths (13)Weaknesses (13)Reference (14)IntroductionBackgroundGlobal Warming and sea level rise“Air temperatures at the top of the world continue to rise twice as fast as temperatures in lower latitudes, causing significant ice melt on land and sea” [Fears, December 17, 2014]. One of the serious consequences is that sea level will rise. Global average sea-level rose at an average rate of about 3.1[2.4 to 3.8] mm per year from 1993 to 2003[IPCC]. This information suggests that from 1993 to 2003 the sea-level rise by 3.1cm totally.Our workThe question requires us to predict the next 50 years’ conditi on of ice melting and analyze the effects on the Florida, especially some big cities. So we can separate this question into two parts:●How much and how fast will the see level rise within 50 years?●What are the effects on the Florida because of the rise of sea level, especiallysome big cities?Study object: Ice cap in GreenlandArctic mainly consists of Greenland, which occupies about 9% glaciers all over the world. Melting in Arctic is mainly due to Greenland, melting of floating ice can be ignored. So we can consider Greenland as study object.Modeling the sea levelWe develop a model for sea-level rise as the function of time. This model can predict sea-level rise in future.Analysis of FloridaAfter having calculated the increased sea level within next 50 years, we analyze the impact to the Florida.●Rising sea level can seriously threaten the development of cities. It has beenthreatening some islands and coastal cities. Over the next 18 years, about two thirds among 544 American towns will be twice as likely to face floods [Huang].More frequency hurricane will happen.●Sea water will corrode seacoast.● A large quantity of drinking water will be polluted.Assumption●Sea level rise is primarily due to the melting of ice cap in Green Land. We ignorethe other floating ice in the Northern Polar.●The increment of sea water from melting will flow over the oceans uniformly●Salt in the ice will not affects the procedure of melting.Model Ⅰ:Temperature PredictionGrey Prediction Model:The weakness of the grey prediction is that the result is increasing all the time. In other words, it cannot show the changes in detail.Neural Network prediction Model:Model Ⅱ:Melting ice and the rise of sea level Model the rise of sea levelThe main reason of the sea level rising is the melting of ice cap and the mass of melting ice is equal to the mass of sea water generated from melting. So, based on several physical principles, we model the rise of sea level by calculating the mass of melting ice. We assume that the increment of sea water from melting will flow over the world uniformly, which means the melting ice will contribute to the rise of sea level, divided by the area of the ocean.The increment of sea water from melting iceThe total volume of melting ice capThe total volume of water generated from meltingThe sea level riseThe overall ocean area: 361745300km2, this is 71 percent of earth’s total surface area (Wikipedia).Heat from rise of temperature:According to the principle of thermal transmission, heat will always be transmitted from high temperature to low temperature. So, the final state of stuff in the thermal cycling system will reach to a same temperature. So, we assume that the temperature of the whole ice cap will increase by when the world temperature rise by However, it takes time for the ice cap to transmit the heat from the rise temperature. We use the every month to calculate the increase of melting ice in each month and get the total increment by accumulation, which means parts of the heat from temperature will be absorbed and used to melt ice. So, defining a coefficient and we will have the heat which ice cap absorbs from the rising temperature in the n-th month:The heat comes from rises of temperature in the n-th monthThe coefficient of capacity of absorbing heat in one monthThe specific heat capacity of iceThe mass of the whole ice capThe mass of melting ice in the n-th monthThe change of temperature in the n-th monthWe try to find the by calculating the mass of melting ice in known years. “Recently reported GrIS mass balance varies from near-balance to modest mass losses [47 to 97 gigatons (Gt) year−1] in the 1990s, increasing to a mass loss of 267 ±38 Gt year−1in 2007”(Michiel).Mass of melting ice:The mass of melting ice in this month will depends on not only the rising temperature, but also the mass of melting ice in the last month.We divide the heat absorbed by the ice by the melting enthalpy of fusion for water to obtain the mass of extra melting ice resulted from the rise of temperature in this month.The mass of melting ice in the n-th month△fusHθm: The melting enthalpyThe total mass of melting ice in the next k yearsMoreover, “Since 2006, high summer melt rates have increased Greenland ice sheet mass loss to 273 gigatons per year” (Partitioning Recent Mass Loss). And the initial mass of melting ice in the first month will be calculated as follows:: The mass of melting ice in the initial year.The total rise of sea level in the next k yearsDesign of Algorithms:Since we have the function, we are able to calculate the mass of melting ice in the next n months using computer program with the data of temperatures and the initial amount of melting ice in the first month.This is easy to achieve using two linear arrays and in MATLAB. Run a simple for loop from 2 to n and calculate during each pass so that the whole array can be found.So that the total amount of melting ice at n-th month is the summation fromto . Also the total rise of sea level can be easily found.Model Results: Sea level will elevate by 10 centimeters in 50 yearsThe solutions were coded using matlab:Figure 1: The mass of melting ice in the next 50 yearsFigure 2: The rise of sea level in the next 50 yearsThe prediction about rise of sea level every decade in next 50 years:Validation of our model:The results show that the sea level will elevate by approximately 10 centimeters totally and 2 millimeters/yr, which accord with the prediction in Relative Mean Sea Level trends from NOAA.Moreover, if we calculated the rise of sea level without considering the increase of temperature, which means sea level rise at the rate today in the next 50 years, the order of magnitudes is match up with our result. So, based on the analysis above, the results of our modelModel ⅢAnalysis: The effects towards FloridaBased on our results, the sea level will rise 10 centimeters in the following 50 years, which threaten Florida in the future and result in tremendous impacts. “Some 2.4 million people and 1.3 million homes, nearly half the risk nationwide, sit within 4 feet of the local high tide line. Sea level rise is more than doubling the risk of a storm surge at this level in South Florida.” (Florida and rising sea)Figure 3 the altitudes of Florida ()As we can see from this picture, most cities or counties in the southern Florida lie besides the coast. Statistic suggests that 17 counties with altitudes smaller than 3 feetwill be threatened by the rise of sea level in 50 years. The counties and airports which will be involved are listed as follows:Cities: Miami, Homestead, Fort Lauderdale, West Palm Beach, Titusville, St Augustine, Clearwater, St Petersburg, Tampa, Brandon, Bradenton, Port Charlotte，Cape Coral, Bonita Springs, Naples, Marco Island, and Grand Isle.Airports: Miami International Airport, Fort Lauderdale International Airport, Central Florida Regional Airport, Cedar Knoll Flying Ranch, Daytona Beach Regional Airport, Craig Municipal Airport, Jacksonville International Airport, St George Island Airport.Major Cities AnalysisWe choose 4 metropolises which will be severely impacted to do some further analysis.MiamiMiami is the biggest city in Florida with the average elevation of 3 feet (0.9144m) [11ikipedia]. Also, it is a coastal city. So according to what we predict that sea level will rise 10.8037cm within 50 years, we can draw a conclusion that this city would be greatly influenced.The possibility of flooding would grow while the frequency of hurricane will increase. It is a big challenge to sewer system of Miami. According to our simulation, sea level rising would also threaten Miami International Airport. Another problem is that the sea water would gnaw at the shoreline. Many coastal man-made buildings are too closed to the sea which they would face a serious problem of being eroded. Aside from threatening of losing habitat, local drinking water would be polluted.TampaTampa is a city located on the west coast of Florida. It is the third largest city in Florida. It is famous because of tourism. Although the highest point in the city is only 48 feet (15 m) [wikipedia], rising sea level will do harm to its natural disaster. Tampa is special because it has the Old Tampa Bay and Hillsborough Bay which is easy to be attacked by storm surge. Sea level rising would produce much more violent storm surge. The boundary of Tampa would also be lost. What’s more,Traffic facilities such as Tampa International Airport were under threatening of disappearing.Cape CoralThis city is famous because of its far-stretching beach and animated quay. Besides, it has more than 30 gardens and golf courses which attract many tourists. A variety of animals also promotes this city’s tourism. However, sea level rising would erode shoreline of Cape Coral. It would destroy natural environment of this area, and then damages biodiversity. And still worse, Pine Island would mostly disappear. So, the economic damage there will be hardly assessed.Key WestIt is an island of Florida, which have the lowest altitude. So if sea level rises to some extent, it would be the first to be under water.Other Impacts in Florida:Biodiversity: Wild life and rare animals in Florida will be impacted by loss of habitat and food. Moreover, it is hard for plants and animals in Florida to adapt the new climatic conditions and the increase of relative air humidity.Architecture:Sea level rise will cause salinization, which will impact the architectural production.Economy Pressure：More money will be put into the drainage systems and dam project, which means less city construction and business development.Health Care: Higher sea level will increase the risk of some disease like malaria.Recommendations to coastal Florida:●Build higher dams: In this way can cities hold back the rising flood waters.●Prepare for flooding: Complete supervisory control system. Guarantee thatcitizen can be evacuated in time.●Reduce carbon emission: More carbon emission means higher temperature, andthen lead to rising of sea level. So encourage citizen to live a low-carbon life.●Warn local citizen: Propagate relative knowledge of sea level rising to improvecitizen’s sense of self-protection when facing natural disaster. JudgmentsStrengths●We use Neural Network to predict temperature in future with a large amount ofreliable data. So our prediction of temperature in future is accurate relatively.●Our model can predict the sea level rising in different conditions, such asdifferent temperature.●Our model is relatively simple so it will take a little time to simulate. We caneasily get the result.Weaknesses●We ignore the areal variation about depth, salinity and temperature of the sea forsimplifying the model;●We neglect the floating ice which will bring some error;●We neglect the thermal expansion;The mode is only the function of time, so it can’t simulate unusual situations.ReferenceBen Strauss,Florida and rising sea,/news/floria-and-the-rising-seaFears, Darryl, Huang, Ming. “Rising sea level threatens millions of American and causes hugeeconomic loss”. Souhu, March 16, 2012, 09:48 AM. Web.IPCC, http://www.ipcc.chMichiel van den Broeke，Partitioning Recent Mass Loss, Science 13 November 2009 “National Snow and Ice Data Center”, January 7, 2015怎样写作数学建模竞赛论文一如何建立数学模型—建立数学模型的涉骤和方法建立数学模型没有固定的模式，通常它与实际问题的性质、建模的目的等有关。

本文件夹中共有7个m文件，分别为：m1.m:用边缘检测算法检测的圆的像的边缘点，并用多元线性回归求出椭圆的方程，然后按照我们的模型中提供的方法，求出每个圆心的像的位置moni.m:用计算机模拟的方法检验相机和标靶的距离对我们模型带来的影响，即统计不同距离下的误差值moni2.m：用计算机模拟的方法检验α的变化对我们模型带来的影响，即统计不同α下的误差值moni3.m：用计算机模拟的方法检验β的变化对我们模型带来的影响，即统计不同β下的误差值gongqiexian.m：计算圆的像的的公切线并根据这个算出圆心的像的坐标jiaodian.m：函数，用于计算分别由两个点确定的两条直线的交点jiaodian2.m：函数，用于计算由8个点确定的4条直线的交点连线的交点坐标说明：1.m1.m，moni.m，moni2.m，moni3.m，gongqiexian.m为可执行代码，直接运行即得结果（当然必须保证代码即本文件中的图片pictrue.bmp在matlab的当前工作区）2.执行m1.m时程序会将每个圆的像的边缘点以excel文件保存在D盘，而执行gongqiexian.m 需要这些文件文件顺序为：format long;[p1,txt1,raw1]=xlsread('d:/1.xls');[p2,txt2,raw2]=xlsread('d:/2.xls');[p3,txt3,raw3]=xlsread('d:/3.xls');[p4,txt4,raw4]=xlsread('d:/4.xls');[p5,txt5,raw5]=xlsread('d:/5.xls');qd=ones(6,4);P=imread('picture.bmp');imshow(P);hold on;%第1组圆1,3[min1 i]=min(p1(:,1));qd(1,1)=i;[min3 i]=min(p3(:,1));qd(1,2)=i;size1=size(p1);size3=size(p3);flag=1;while flag>0flag=0;for i=1:size1(1)ifp1(i,1)<(p1(qd(1,1),1)-p3(qd(1,2),1))/(p1(qd(1,1),2)-p3(qd(1,2),2))*(p1(i,2)-(p1(qd(1,1),2)))+p1(q d(1,1),1)qd(1,1)=i;flag=1;endendfor i=1:size3(1)ifp3(i,1)<(p1(qd(1,1),1)-p3(qd(1,2),1))/(p1(qd(1,1),2)-p3(qd(1,2),2))*(p3(i,2)-(p1(qd(1,1),2)))+p1(q d(1,1),1)qd(1,2)=i;flag=1;endendendfor i=1:1024y(i)=(p1(qd(1,1),1)-p3(qd(1,2),1))/(p1(qd(1,1),2)-p3(qd(1,2),2))*(i-(p1(qd(1,1),2)))+p1(qd(1,1),1) ;endplot([1:1024],y);hold on;%另一条切线[max1 i]=min(p1(:,1));qd(1,3)=i;[max3 i]=min(p3(:,1));qd(1,4)=i;size1=size(p1);size3=size(p3);flag=1;while flag>0flag=0;for i=1:size1(1)ifp1(i,1)>(p1(qd(1,3),1)-p3(qd(1,4),1))/(p1(qd(1,3),2)-p3(qd(1,4),2))*(p1(i,2)-(p1(qd(1,3),2)))+p1(q d(1,3),1)qd(1,3)=i;flag=1;endendfor i=1:size3(1)ifp3(i,1)>(p1(qd(1,3),1)-p3(qd(1,4),1))/(p1(qd(1,3),2)-p3(qd(1,4),2))*(p3(i,2)-(p1(qd(1,3),2)))+p1(q d(1,3),1)qd(1,4)=i;flag=1;endendendfor i=1:1024y(i)=(p1(qd(1,3),1)-p3(qd(1,4),1))/(p1(qd(1,3),2)-p3(qd(1,4),2))*(i-(p1(qd(1,3),2)))+p1(qd(1,3),1) ;endplot([1:1024],y);hold on;%第2组圆4,5[min4 i]=min(p4(:,1));qd(2,1)=i;[min5 i]=min(p5(:,1));qd(2,2)=i;size4=size(p4);size5=size(p5);flag=1;while flag>0flag=0;for i=1:size4(1)ifp4(i,1)<(p4(qd(2,1),1)-p5(qd(2,2),1))/(p4(qd(2,1),2)-p5(qd(2,2),2))*(p4(i,2)-(p4(qd(2,1),2)))+p4(q d(2,1),1)qd(2,1)=i;flag=1;endendfor i=1:size5(1)ifp5(i,1)<(p4(qd(2,1),1)-p5(qd(2,2),1))/(p4(qd(2,1),2)-p5(qd(2,2),2))*(p5(i,2)-(p4(qd(2,1),2)))+p4(q d(2,1),1)qd(2,2)=i;flag=1;endendendfor i=1:1024y(i)=(p4(qd(2,1),1)-p5(qd(2,2),1))/(p4(qd(2,1),2)-p5(qd(2,2),2))*(i-(p4(qd(2,1),2)))+p4(qd(2,1),1) ;endplot([1:1024],y);hold on;%另一条切线[max4 i]=min(p4(:,1));qd(2,3)=i;[max5 i]=min(p5(:,1));qd(2,4)=i;size4=size(p4);size5=size(p5);flag=1;while flag>0flag=0;for i=1:size4(1)ifp4(i,1)>(p4(qd(2,3),1)-p5(qd(2,4),1))/(p4(qd(2,3),2)-p5(qd(2,4),2))*(p4(i,2)-(p4(qd(2,3),2)))+p4(q d(2,3),1)qd(2,3)=i;flag=1;endendfor i=1:size5(1)ifp5(i,1)>(p4(qd(2,3),1)-p5(qd(2,4),1))/(p4(qd(2,3),2)-p5(qd(2,4),2))*(p5(i,2)-(p4(qd(2,3),2)))+p4(q d(2,3),1)qd(2,4)=i;flag=1;endendendfor i=1:1024y(i)=(p4(qd(2,3),1)-p5(qd(2,4),1))/(p4(qd(2,3),2)-p5(qd(2,4),2))*(i-(p4(qd(2,3),2)))+p4(qd(2,3),1) ;endplot([1:1024],y);hold on;%第3组圆1,4[min1 i]=min(p1(:,2));qd(3,1)=i;[min4 i]=min(p4(:,2));qd(3,2)=i;size1=size(p1);size4=size(p4);flag=1;while flag>0flag=0;for i=1:size1(1)ifp1(i,2)<(p1(qd(3,1),2)-p4(qd(3,2),2))/(p1(qd(3,1),1)-p4(qd(3,2),1))*(p1(i,1)-p1(qd(3,1),1))+p1(qd (3,1),2)qd(3,1)=i;flag=1;endendfor i=1:size4(1)ifp4(i,2)<(p1(qd(3,1),2)-p4(qd(3,2),2))/(p1(qd(3,1),1)-p4(qd(3,2),1))*(p4(i,1)-p1(qd(3,1),1))+p1(qd (3,1),2)qd(3,2)=i;flag=1;endendendfor i=1:1024y(i)=(p1(qd(3,1),1)-p4(qd(3,2),1))/(p1(qd(3,1),2)-p4(qd(3,2),2))*(i-p1(qd(3,1),2))+p1(qd(3,1),1); endplot([1:1024],y);hold on;%另一条切线[max1 i]=max(p1(:,2));qd(3,3)=i;[max4 i]=max(p4(:,2));qd(3,4)=i;size1=size(p1);size4=size(p4);flag=1;while flag>0flag=0;for i=1:size1(1)ifp1(i,2)>(p1(qd(3,3),2)-p4(qd(3,4),2))/(p1(qd(3,1),1)-p4(qd(3,4),1))*(p1(i,1)-p1(qd(3,3),1))+p1(qd (3,3),2)qd(3,3)=i;flag=1;endendfor i=1:size4(1)ifp4(i,2)>(p1(qd(3,3),2)-p4(qd(3,4),2))/(p1(qd(3,1),1)-p4(qd(3,4),1))*(p4(i,1)-p1(qd(3,3),1))+p1(qd (3,3),2)qd(3,4)=i;flag=1;endendendfor i=1:1024y(i)=(p1(qd(3,3),1)-p4(qd(3,4),1))/(p1(qd(3,3),2)-p4(qd(3,4),2))*(i-p1(qd(3,3),2))+p1(qd(3,3),1);endplot([1:1024],y);hold on;%第4组圆2,4[min2 i]=min(p2(:,2));qd(4,1)=i;[min4 i]=min(p4(:,2));qd(4,2)=i;size2=size(p2);size4=size(p4);flag=1;while flag>0flag=0;for i=1:size2(1)ifp2(i,2)<(p2(qd(4,1),2)-p4(qd(4,2),2))/(p2(qd(4,1),1)-p4(qd(4,2),1))*(p2(i,1)-p2(qd(4,1),1))+p2(qd (4,1),2)qd(4,1)=i;flag=1;endendfor i=1:size4(1)ifp4(i,2)<(p2(qd(4,1),2)-p4(qd(4,2),2))/(p2(qd(4,1),1)-p4(qd(4,2),1))*(p4(i,1)-p2(qd(4,1),1))+p2(qd (4,1),2)qd(4,2)=i;flag=1;endendendfor i=1:1024y(i)=(p2(qd(4,1),1)-p4(qd(4,2),1))/(p2(qd(4,1),2)-p4(qd(4,2),2))*(i-p2(qd(4,1),2))+p2(qd(4,1),1); endplot([1:1024],y);hold on;%另一条切线[max2 i]=max(p2(:,2));qd(4,3)=i;[max4 i]=max(p4(:,2));flag=1;while flag>0flag=0;for i=1:size4(1)ifp4(i,2)>(p2(qd(4,3),2)-p4(qd(4,4),2))/(p2(qd(4,1),1)-p4(qd(4,4),1))*(p4(i,1)-p2(qd(4,3),1))+p2(qd (4,3),2)qd(4,4)=i;flag=1;endendfor i=1:size2(1)ifp2(i,2)>(p2(qd(4,3),2)-p4(qd(4,4),2))/(p2(qd(4,1),1)-p4(qd(4,4),1))*(p2(i,1)-p2(qd(4,3),1))+p2(qd (4,3),2)qd(4,3)=i;flag=1;endendendfor i=1:1024y(i)=(p2(qd(4,3),1)-p4(qd(4,4),1))/(p2(qd(4,3),2)-p4(qd(4,4),2))*(i-p2(qd(4,3),2))+p2(qd(4,3),1); endplot([1:1024],y);hold on;%第5组圆2,5[min2 i]=min(p2(:,2));qd(5,1)=i;[min5 i]=min(p5(:,2));qd(5,2)=i;size2=size(p2);size5=size(p5);flag=1;while flag>0flag=0;for i=1:size2(1)p2(i,2)<(p2(qd(5,1),2)-p5(qd(5,2),2))/(p2(qd(5,1),1)-p5(qd(5,2),1))*(p2(i,1)-p2(qd(5,1),1))+p2(qd (5,1),2)qd(5,1)=i;flag=1;endendfor i=1:size5(1)ifp5(i,2)<(p2(qd(5,1),2)-p5(qd(5,2),2))/(p2(qd(5,1),1)-p5(qd(5,2),1))*(p5(i,1)-p2(qd(5,1),1))+p2(qd (5,1),2)qd(5,2)=i;flag=1;endendendfor i=1:1024y(i)=(p2(qd(5,1),1)-p5(qd(5,2),1))/(p2(qd(5,1),2)-p5(qd(5,2),2))*(i-p2(qd(5,1),2))+p2(qd(5,1),1); endplot([1:1024],y);hold on;%另一条切线[max2 i]=max(p2(:,2));qd(5,3)=i;[max5 i]=max(p5(:,2));qd(5,4)=i;flag=1;while flag>0flag=0;for i=1:size2(1)ifp2(i,2)>(p2(qd(5,3),2)-p5(qd(5,4),2))/(p2(qd(5,1),1)-p5(qd(5,4),1))*(p2(i,1)-p2(qd(5,3),1))+p2(qd (5,3),2)qd(5,3)=i;flag=1;endendfor i=1:size5(1)ifp5(i,2)>(p2(qd(5,3),2)-p5(qd(5,4),2))/(p2(qd(5,1),1)-p5(qd(5,4),1))*(p5(i,1)-p2(qd(5,3),1))+p2(qd (5,3),2)qd(5,4)=i;flag=1;endendendfor i=1:1024y(i)=(p2(qd(5,3),1)-p5(qd(5,4),1))/(p2(qd(5,3),2)-p5(qd(5,4),2))*(i-p2(qd(5,3),2))+p2(qd(5,3),1); endplot([1:1024],y);hold on;%第6组圆3,5[min3 i]=min(p3(:,2));qd(6,1)=i;[min5 i]=min(p5(:,2));qd(6,2)=i;size3=size(p3);size5=size(p5);flag=1;while flag>0flag=0;for i=1:size3(1)ifp3(i,2)<(p3(qd(6,1),2)-p5(qd(6,2),2))/(p3(qd(6,1),1)-p5(qd(6,2),1))*(p3(i,1)-p3(qd(6,1),1))+p3(qd (6,1),2)qd(6,1)=i;flag=1;endendfor i=1:size5(1)ifp5(i,2)<(p3(qd(6,1),2)-p5(qd(6,2),2))/(p3(qd(6,1),1)-p5(qd(6,2),1))*(p5(i,1)-p3(qd(6,1),1))+p3(qd (6,1),2)qd(6,2)=i;flag=1;endendfor i=1:1024y(i)=(p3(qd(6,1),1)-p5(qd(6,2),1))/(p3(qd(6,1),2)-p5(qd(6,2),2))*(i-p3(qd(6,1),2))+p3(qd(6,1),1); endplot([1:1024],y);hold on;%另一条切线[max2 i]=max(p3(:,2));qd(6,3)=i;[max5 i]=max(p5(:,2));qd(6,4)=i;flag=1;while flag>0flag=0;for i=1:size3(1)ifp3(i,2)>(p3(qd(6,3),2)-p5(qd(6,4),2))/(p3(qd(6,1),1)-p5(qd(6,4),1))*(p3(i,1)-p3(qd(6,3),1))+p3(qd (6,3),2)qd(6,3)=i;flag=1;endendfor i=1:size5(1)ifp5(i,2)>(p3(qd(6,3),2)-p5(qd(6,4),2))/(p3(qd(6,1),1)-p5(qd(6,4),1))*(p5(i,1)-p3(qd(6,3),1))+p3(qd (6,3),2)qd(6,4)=i;flag=1;endendendfor i=1:1024y(i)=(p3(qd(6,3),1)-p5(qd(6,4),1))/(p3(qd(6,3),2)-p5(qd(6,4),2))*(i-p3(qd(6,3),2))+p3(qd(6,3),1); endplot([1:1024],y);hold on;%求圆1的圆心X=jiaodian2(p1(qd(1,1),1),p1(qd(1,1),2),p3(qd(1,2),1),p3(qd(1,2),2),p1(qd(3,1),1),p1(qd(3,1),2),p 4(qd(3,2),1),p4(qd(3,2),2),p1(qd(1,3),1),p1(qd(1,3),2),p3(qd(1,4),1),p3(qd(1,4),2),p1(qd(3,3),1),p 1(qd(3,3),2),p4(qd(3,4),1),p4(qd(3,4),2))x=(X(2)-512)/3.78y=(384-X(1))/3.78%求圆2的圆心circle=2X=jiaodian2(p2(qd(4,1),1),p2(qd(4,1),2),p4(qd(4,2),1),p4(qd(4,2),2),p2(qd(5,1),1),p2(qd(5,1),2),p 5(qd(5,2),1),p5(qd(5,2),2),p2(qd(4,3),1),p2(qd(4,3),2),p4(qd(4,4),1),p4(qd(4,4),2),p2(qd(5,3),1),p 2(qd(5,3),2),p5(qd(5,4),1),p5(qd(5,4),2))x=(X(2)-512)/3.78y=(384-X(1))/3.78%求圆3的圆心circle=3X=jiaodian2(p1(qd(1,1),1),p1(qd(1,1),2),p3(qd(1,2),1),p3(qd(1,2),2),p3(qd(6,1),1),p3(qd(6,1),2),p 5(qd(6,2),1),p5(qd(6,2),2),p1(qd(1,3),1),p1(qd(1,3),2),p3(qd(1,4),1),p3(qd(1,4),2),p3(qd(6,3),1),p 3(qd(6,3),2),p5(qd(6,4),1),p5(qd(6,4),2))x=(X(2)-512)/3.78y=(384-X(1))/3.78%求圆4的圆心circle=4X=jiaodian2(p4(qd(2,1),1),p4(qd(2,1),2),p5(qd(2,1),1),p5(qd(2,1),2),p1(qd(3,1),1),p1(qd(3,1),2),p 4(qd(3,2),1),p4(qd(3,2),2),p4(qd(2,3),1),p4(qd(2,3),2),p5(qd(2,4),1),p5(qd(2,4),2),p1(qd(3,3),1),p 1(qd(3,3),2),p4(qd(3,4),1),p4(qd(3,4),2))x=(X(2)-512)/3.78y=(384-X(1))/3.78%求圆5的圆心circle=4X=jiaodian2(p4(qd(2,1),1),p4(qd(2,1),2),p5(qd(2,1),1),p5(qd(2,1),2),p3(qd(6,1),1),p3(qd(6,1),2),p 5(qd(6,2),1),p5(qd(6,2),2),p4(qd(2,3),1),p4(qd(2,3),2),p5(qd(2,4),1),p5(qd(2,4),2),p3(qd(6,3),1),p 3(qd(6,3),2),p5(qd(6,4),1),p5(qd(6,4),2))x=(X(2)-512)/3.78y=(384-X(1))/3.78function y = jiaodian(m1,n1,m2,n2,m3,n3,m4,n4)A=[n1-n2 m2-m1;n3-n4 m4-m3];B=[(n1-n2)*m1-(m1-m2)*n1;(n3-n4)*m3-(m3-m4)*n3];y=A\B;function y = jiaodian2(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x7,y7,x8,y8) d1=jiaodian(x1,y1,x2,y2,x3,y3,x4,y4);d2=jiaodian(x5,y5,x6,y6,x7,y7,x8,y8);d3=jiaodian(x1,y1,x2,y2,x7,y7,x8,y8);d4=jiaodian(x5,y5,x6,y6,x3,y3,x4,y4);y=jiaodian(d1(1),d1(2),d2(1),d2(2),d3(1),d3(2),d4(1),d4(2));format longP=imread('picture.bmp');lefttop=[100 240142 244170 258450 545471 543];rightbutton=[200 368375 466591 685200 330537 624];for circle=1:5circleif(circle~=1)clear u;clear v;endk=1;for i=lefttop(circle,1):lefttop(circle,2)flag=0;for j=rightbutton(circle,1):rightbutton(circle,2) if flag==0 & P(i,j)<10u(k)=i;v(k)=j;k=k+1;flag=1;elseif flag==1 & P(i,j)<10 & P(i,j+1)>200u(k)=i;v(k)=j;k=k+1;endendendfor j=rightbutton(circle,1):rightbutton(circle,2) flag=0;for i=lefttop(circle,1):lefttop(circle,2)if flag==0 & P(i,j)<10f1=0;for temp=1:k-1if u(temp)==i & v(temp)==jf1=1;endendif f1==0u(k)=i;v(k)=j;k=k+1;flag=1;endelseif flag==1 & P(i,j)<10 & P(i+1,j)>200f1=0;for temp=1:k-1if u(temp)==i & v(temp)==jf1=1;endendif f1==0u(k)=i;v(k)=j;k=k+1;flag=1;endendendendaxis equal;axis([1 768 1 1024]);plot(u,v,'.');grid on ;hold on;x=ones(k-1,5);y=ones(k-1,1);for i=1:k-1x(i,:)=[1 v(i) u(i) u(i)*v(i) v(i)*v(i) ];y(i)=-u(i)*u(i);end[b,bint,r,rint,stats]=regress(y,x);bbintstatsA=[2 b(4);b(4) 2*b(5)];B=[-b(3);-b(2)];X=A\Bxo=(X(2)-512)/3.78yo=(384-X(1))/3.78xlswrite(strcat('d:/',num2str(circle),'.xls'),[u;v]'); endformat longclear;aa=140;bb=1000count=1;for t=aa:bbb=-0.7;r=0;a=0;L=3.78;f=1577/3.78;Tx=-t;Ty=0;Tz=-t;R1=[L*f 0 0 0;0 L*f 0 0;0 0 0 1];R2=[cos(b)*cos(r) sin(a)*sin(b)*cos(r)+cos(a)*sin(r) -cos(a)*sin(b)*cos(r)+sin(a)*sin(r) Tx;-cos(b)*sin(r) -sin(a)*sin(b)*sin(r)+cos(a)*cos(r) cos(a)*sin(b)*sin(r)+sin(a)*cos(r) Ty;sin(b) -sin(a)*cos(b) cos(a)*cos(b)Tz;0 0 01;];M=R1*R2;k=1;for w=0:2*pi/100:2*piX=12*cos(w);Y=12*sin(w);Zc=R2(3,1:3)*[X Y 0]'-Tz;u(k)=1/Zc*M(1,:)*[X Y 0 1]';v(k)=1/Zc*M(2,:)*[X Y 0 1]';k=k+1;endx=ones(k-1,5);y=ones(k-1,1);for i=1:k-1x(i,:)=[1 v(i) u(i) u(i)*v(i) v(i)*v(i) ];y(i)=-u(i)*u(i);end[b,bint,r,rint,stats]=regress(y,x);A=[2 b(4);b(4) 2*b(5)];B=[-b(3);-b(2)];X=A\B;M=M/(-Tz);F(count)=sqrt((X(1)-M(1,4))*(X(1)-M(1,4))+(X(2)-M(2,4))*(X(2)-M(2,4)));count=count+1;endplot(aa:bb,F);xlabel('k');ylabel('D')grid on;format longclear;count=1;for t=-pi/2:0.05:pi/2b=-0.7;r=0;a=t;L=3.78;f=1577/3.78;Tx=-350;Ty=0;Tz=-350;R1=[L*f 0 0 0;0 L*f 0 0;0 0 0 1];R2=[cos(b)*cos(r) sin(a)*sin(b)*cos(r)+cos(a)*sin(r) -cos(a)*sin(b)*cos(r)+sin(a)*sin(r) Tx;-cos(b)*sin(r) -sin(a)*sin(b)*sin(r)+cos(a)*cos(r) cos(a)*sin(b)*sin(r)+sin(a)*cos(r) Ty;sin(b) -sin(a)*cos(b) cos(a)*cos(b)Tz;0 0 01;];M=R1*R2;k=1;for w=0:2*pi/200:2*piX=12*cos(w);Y=12*sin(w);Zc=R2(3,1:3)*[X Y 0]'-Tz;u(k)=1/Zc*M(1,:)*[X Y 0 1]';v(k)=1/Zc*M(2,:)*[X Y 0 1]';k=k+1;endx=ones(k-1,5);y=ones(k-1,1);for i=1:k-1x(i,:)=[1 v(i) u(i) u(i)*v(i) v(i)*v(i) ];y(i)=-u(i)*u(i);end[b,bint,r,rint,stats]=regress(y,x);A=[2 b(4);b(4) 2*b(5)];B=[-b(3);-b(2)];X=A\B;M=M/(-Tz);F(count)=sqrt((X(1)-M(1,4))*(X(1)-M(1,4))+(X(2)-M(2,4))*(X(2)-M(2,4)));count=count+1;endplot(-pi/2:0.05:pi/2,F);xlabel('β');ylabel('D')grid on;format longclear;count=1;for t=-pi/2:0.05:pi/2b=t;r=0;a=0;L=3.78;f=1577/3.78;Tx=-350;Ty=0;Tz=-350;R1=[L*f 0 0 0;0 L*f 0 0;0 0 0 1];R2=[cos(b)*cos(r) sin(a)*sin(b)*cos(r)+cos(a)*sin(r) -cos(a)*sin(b)*cos(r)+sin(a)*sin(r) Tx;-cos(b)*sin(r) -sin(a)*sin(b)*sin(r)+cos(a)*cos(r) cos(a)*sin(b)*sin(r)+sin(a)*cos(r)Ty;sin(b) -sin(a)*cos(b) cos(a)*cos(b)Tz;0 0 01;];M=R1*R2;k=1;for w=0:2*pi/200:2*piX=12*cos(w);Y=12*sin(w);Zc=R2(3,1:3)*[X Y 0]'-Tz;u(k)=1/Zc*M(1,:)*[X Y 0 1]';v(k)=1/Zc*M(2,:)*[X Y 0 1]';k=k+1;endx=ones(k-1,5);y=ones(k-1,1);for i=1:k-1x(i,:)=[1 v(i) u(i) u(i)*v(i) v(i)*v(i) ];y(i)=-u(i)*u(i);end[b,bint,r,rint,stats]=regress(y,x);A=[2 b(4);b(4) 2*b(5)];B=[-b(3);-b(2)];X=A\B;M=M/(-Tz);F(count)=sqrt((X(1)-M(1,4))*(X(1)-M(1,4))+(X(2)-M(2,4))*(X(2)-M(2,4)));count=count+1;endplot(-pi/2:0.05:pi/2,F);xlabel('β');ylabel('D')grid on;。

三峡大学2008年数学建模竞赛试题A题自行车租赁策略很多旅游景点的面积都比较大，游客采用步行游览通常会感觉到比较疲惫，常见的旅游景点交通工具如电瓶车等由于游览路线比较固定，且不能中途停车，通常不适合一些乐于拍照的游客。

最近很多旅游景点推出了自行车租赁游览的方式，比较适合这些旅客的需要。

旅客可以在任意一个租赁点租到自行车，并且可以在任意一个租赁点归还自行车。

假设旅游景点如图2所示，该旅游景点每天的游客人数大约1万人，现有自行车500辆，图中蓝线所示的路线为游客必经的路线，游览苏堤时，从南向北游览的游客约占游客总数的60%。

问题一：考虑在图中的A，B两个位置建立租赁点，设计一套租赁方案，即每天早晨各个租赁点分别存放多少辆自行车。

尽量使得每个租赁点的自行车数量保持动态平衡。

问题二：考虑在图中的C，D两个位置新建立两个租赁点，重新设计一套租赁方案，尽量使得每个租赁点的自行车数量保持动态平衡。

图2 景点示意图B题奥运会奖牌预测2008年8月第29届奥运会将在中国北京举行，届时中国体育健儿将和来自世界各地的选手一同争夺奥运奖牌。

（1）请你设计一个数学模型预测中国队在本届奥运会上将获得多少枚奖牌。

附录中给出了88年，92年，96年，00年，04年各国家和地区所获金、银，铜牌数目。

（2）你认为衡量一个国家的体育实力应该以什么为标准，试建立相应的模型，并且从网上收集数据对模型进行求解。

附录：第24---28届奥运会各代表团奖牌榜第二十四届汉城奥运会（１９８８年）国家金牌银牌铜牌国家金牌银牌铜牌苏联55 31 46 捷克斯洛伐克3 3 2民主德国37 35 30新西兰3 2 8美国36 31 27加拿大3 2 5韩国12 10 11 波兰 2 5 9西德11 14 15 挪威 2 3 -匈牙利11 6 5 荷兰 2 2 5保加利亚10 12 13 丹麦 2 2 5罗马尼亚7 11 6 巴西 1 2 3法国 6 4 6芬兰 1 1 2意大利6 4 4 西班牙1 1 2中国 5 11 12 土耳其1 1 -英国 5 9 10 摩洛哥1 2 -肯尼亚5 2 2 苏里南1 - -日本 4 3 7 葡萄牙1 - -澳大利亚 3 6 5奥地利1 - -----------------------------精品word文档值得下载值得拥有----------------------------------------------南斯拉夫 3 4 5第二十五届巴塞罗那奥运会（１９９２年）国家金牌银牌铜牌国家金牌银牌铜牌独联体45 38 29 波兰 3 6 10美国37 34 37荷兰2 6 7德国33 21 28 肯尼亚2 4 2中国16 22 16挪威2 4 1古巴14 6 11 土耳其2 2 2西班牙13 7 2 印度尼西亚 2 2 1韩国12 5 12巴西 2 1 -匈牙利11 12 7 希腊 2 - -法国8 5 16瑞典 1 7 4澳大利亚7 9 11芬兰 1 2 2意大利6 5 7 丹麦 1 1 4加拿大6 5 7 摩洛哥1 1 1英国 5 3 12爱尔兰1 1 -罗马尼亚 4 2 1 埃赛俄比亚 1 - 2捷克斯洛伐克 4 2 1阿尔及利亚 1 - 1朝鲜 4 - 5 爱沙尼亚1 - 1日本 3 8 11立陶宛1 - 1保加利亚 3 7 6瑞士 1 - -第二十六届亚特兰大奥运会（１９９６年）国家金牌银牌铜牌国家金牌银牌铜牌美国44 32 25 爱尔兰3 0 1俄罗斯26 21 16 瑞典 2 4 2德国20 18 27 挪威 2 2 3中国16 22 12 比利时2 2 2法国15 7 15 尼日利亚2 1 3意大利13 10 12 朝鲜 2 1 2澳大利亚9 9 23 埃塞俄比亚 2 - 1古巴9 8 8 阿尔及利亚 2 - 1乌克兰9 2 12 英国 1 8 6韩国7 15 5 白俄罗斯1 6 8波兰7 5 5 肯尼亚1 4 3匈牙利7 4 10 牙买加1 3 2西班牙5 6 6 芬兰 1 2 1罗马尼亚 4 7 9 印尼 1 1 2荷兰 4 5 10 南斯拉夫1 1 2希腊 4 4 - 伊朗 1 1 1捷克 4 3 4 斯洛伐克1 1 1瑞士 4 3 - 亚美尼亚1 1 -----------------------------精品word文档值得下载值得拥有----------------------------------------------丹麦 4 1 1 克罗地亚1 1 -土耳其4 1 1 葡萄牙1 - 1加拿大3 11 8 泰国 1 - 1保加利亚 3 7 5 布隆迪1 - -日本 3 6 5 香港 1 - -哈萨克3 4 4 哥斯达尼加 1 - -巴西 3 3 9 叙利亚1 - -新西兰3 2 1 厄瓜多尔1 - -南非 3 1 1第二十七届悉尼奥运会(2000年)名次代表团金牌银牌铜牌总数名次代表团金牌银牌铜牌总数1 美国40 24 33 97 31 芬兰2 1 1 42 俄罗斯32 28 28 88 32 奥地利2 1 0 33 中国28 16 15 59 33 立陶宛2 0 3 54 澳大利亚16 25 17 58 34 阿塞拜疆2 0 1 35 德国13 17 26 56 35 斯洛文尼亚2 0 0 26 法国13 14 11 38 36 瑞士1 6 2 97 意大利13 8 13 34 37 印度尼西亚1 3 2 68 荷兰12 9 4 25 38 斯洛伐克1 3 1 59 古巴11 11 7 29 39 墨西哥1 2 3 610 英国11 10 7 28 40 阿尔及利亚1 1 3 511 罗马尼亚11 6 8 25 41 乌兹别克斯坦1 1 2 412 韩国8 10 10 28 42 南斯拉夫1 1 1 313 匈牙利8 6 3 17 43 拉托维亚1 1 1 314 波兰6 5 3 14 44 巴哈马1 1 0 215 日本5 8 5 18 45 新西兰1 0 3 416 保加利亚5 6 2 13 46 泰国1 0 2 317 希腊4 6 3 13 47 爱沙尼亚1 0 2 318 瑞典4 5 3 12 48 克罗地亚1 0 1 219 挪威4 3 3 10 49 喀麦隆1 0 0 120 埃塞俄比亚4 1 3 8 50 哥伦比亚1 0 0 121 乌克兰3 10 10 23 51 莫桑比克1 0 0 122 哈萨克斯坦3 4 0 7 52 巴西0 6 6 1223 白俄罗斯3 3 11 17 53 牙买加0 4 3 724 加拿大3 3 8 14 54 尼日利亚0 3 0 325 西班牙3 3 5 11 55 南非0 2 3 526 土耳其3 0 2 5 56 比利时0 2 3 527 伊朗3 0 1 4 57 阿根廷0 2 2 428 捷克共和国2 3 3 8 58 中国台北0 1 4 529 肯尼亚2 3 2 7 59 摩洛哥0 1 4 530 丹麦2 3 1 6 60 朝鲜0 1 3 4第二十八届雅典奥运会(2004年)----------------------------精品word文档值得下载值得拥有----------------------------------------------名次代表团金牌银牌铜牌总数名次代表团金牌银牌铜牌总数1 美国35 39 29 103 31 中国台北2 2 1 52 中国32 17 14 63 32 格鲁吉亚2 2 0 43 俄罗斯27 27 38 92 33 保加利亚2 1 9 124 澳大利亚17 16 16 49 34 牙买加2 1 2 55 日本16 9 12 37 34 乌兹别克斯坦2 1 2 56 德国14 16 18 48 36 摩洛哥2 1 0 37 法国11 9 13 33 37 丹麦2 0 6 88 意大利10 11 11 32 38 阿根廷2 0 4 69 韩国9 12 9 30 39 智利2 0 1 310 英国9 9 12 30 40 哈萨克斯坦1 4 3 811 古巴9 7 11 27 41 肯尼亚1 4 2 712 乌克兰9 5 9 23 42 捷克共和国1 3 4 813 匈牙利8 6 3 17 43 南非1 3 2 614 罗马尼亚8 5 6 19 44 克罗地亚1 2 2 515 希腊6 6 4 16 45 立陶宛1 2 0 316 挪威5 0 1 6 46 埃及1 1 3 517 荷兰4 9 9 22 46 瑞士1 1 3 518 巴西4 3 3 10 48 印度尼西亚1 1 2 419 瑞典4 1 2 7 49 津巴布韦1 1 1 320 西班牙3 11 5 19 50 阿塞拜疆1 0 4 521 加拿大3 6 3 12 51 比利时1 0 2 322 土耳其3 3 4 10 52 巴哈马1 0 1 223 波兰3 2 5 10 52 以色列1 0 1 224 新西兰3 2 0 5 54 喀麦隆1 0 0 125 泰国3 1 4 8 54 多米尼加共和国1 0 0 126 白俄罗斯2 6 7 15 54 爱尔兰1 0 0 127 奥地利2 4 1 7 54 阿拉伯联合酋长国1 0 0 128 埃塞俄比亚2 3 2 7 58 朝鲜0 4 1 529 伊朗2 2 2 6 59 拉托维亚0 4 0 429 斯洛伐克2 2 2 6 60 墨西哥0 3 1 4----------------------------精品word文档值得下载值得拥有----------------------------------------------。