2008美国数学建模真题论文

合集下载

2008年数学建模A题_数码相机定位【一等奖】

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。

(完整版)数模美赛08年A题

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。

2008年数学建模A题论文

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。

2008AMC8试题答案

This Solutions Pamphlet gives at least one solution for each problem on this year’s exam and shows that all the problems can be solved using material normally as-sociated with the mathematics curriculum for students in eighth grade or below. These solutions are by no means the only ones possible, nor are they necessarily superior to others the reader may devise.We hope that teachers will share these solutions with their students. However, the publication, reproduction, or communication of the problems or solutions of the AMC 8 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, e-mail, World Wide Web or media of any type is a violation of the competition rules.Correspondence about the problems and solutions should be addressed to:Ms. Bonnie Leitch , AMC 8 Chair / bleitch@548 Hill Avenue, New Braunfels, TX 78130Orders for prior year Exam questions and solutions Pamphlets should be addressed to:Attn: Publications American Mathematics Competitions University of Nebraska-Lincoln P .O. Box 81606Lincoln, NE 68501-1606Copyright © 2008, The Mathematical Association of AmericaT he M aTheMaTical a ssociaTion of a MericaAmerican Mathematics Competitions 24th AnnualAMC 8(American Mathematics Contest 8)Solutions PamphletT uesday, NOvEMBER 18, 20081.Answer(B):Susan spent2×12=$24on rides,so she had50−12−24=$14to spend.2.Answer(A):Because the key to the code starts with zero,all the lettersrepresent numbers that are one less than their ing the key,C is 9−1=8,and similarly L is6,U is7,and E is1.BEST OF LUCK0123456789CLUE=86713.Answer(A):A week before the13th is the6th,which is theﬁrst Friday of themonth.Counting back from that,the5th is a Thursday,the4th is a Wednesday, the3rd is a Tuesday,the2nd is a Monday,and the1st is a Sunday.ORCounting forward by sevens,February1occurs on the same day of the week as February8and February15.Because February13is a Friday,February15is a Sunday,and so is February1.4.Answer(C):The area of the outer triangle with the inner triangle removedis16−1=15,the total area of the three congruent trapezoids.Each trapezoid =5.has area1535.Answer(E):Barney rides1661−1441=220miles in10hours,so his average=22miles per hour.speed is220106.Answer(D):After subdividing the central gray square as shown,6of the16congruent squares are gray and10are white.Therefore,the ratio of the area of the gray squares to the area of the white squares is6:10or3:5.7.Answer (E):Note that M45=35=3·95·9=2745,so M =27.Similarly,60N =35=3·205·20=60100,so N =100.The sum M +N =27+100=127.ORNote that M45=35,so M =35·45=27.Also 60N =35,so N60=53,andN =53·60=100.The sum M +N =27+100=127.8.Answer (D):The sales in the 4months were $100,$60,$40and $120.The average sales were 100+60+40+1204=3204=$80.ORIn terms of the $20intervals,the sales were 5,3,2and 6on the chart.Their sum is 5+3+2+6=16and the average is 164=4.The average sales were4·$20=$80.9.Answer (D):At the end of the ﬁrst year,Tammy’s investment was 85%of the original amount,or $85.At the end of the second year,she had 120%of her ﬁrst year’s ﬁnal amount,or 120%of $85=1.2($85)=$102.Over the two-year period,Tammy’s investment changed from $100to $102,so she gained 2%.10.Answer (D):The sum of the ages of the 6people in Room A is 6×40=240.The sum of the ages of the 4people in Room B is 4×25=100.The sum of the ages of the 10people in the combined group is 100+240=340,so the average age of all the people is 34010=34.11.Answer (A):The number of cat owners plus the number of dog owners is20+26=46.Because there are only 39students in the class,there are 46−39=7students who have both.ORBecause each student has at least a cat or a dog,there are 39−20=19students with a cat but no dog,and 39−26=13students with a dog but no cat.So there are 39−13−19=7students with both a cat and a dog.13197DogCat12.Answer (C):The table gives the height of each bounce.Bounce 12345Height 23·2=23·43=23·89=23·1627=in Meters 2438916273281Because 1627>1632=12and 3281<3264=12,the ball ﬁrst rises to less than 0.5meters on the ﬁfth bounce.Note:Because all the fractions have odd denominators,it is easier to doublethe numerators than to halve the denominators.So compare 1627and 3281to their numerators’fractional equivalents of 12,1632and 3264.13.Answer (C):Because each box is weighed two times,once with each ofthe other two boxes,the total 122+125+127=374poundsis twice the combined weight of thethree boxes.The combined weight is 3742=187pounds.14.Answer (C):There are only two possible spaces for the B in row 1and onlytwo possible spaces for the A in row 2.Once these are placed,the entries in theremaining spaces are determined.The four arrangementsare:ORThe As can be placed eitheror In each case,the letter next to the top A can be B or C.At that point the rest of the grid is completely determined.So there are 2+2=4possible arrangements.15.Answer(B):The sum of the points Theresa scored in theﬁrst8games is37.After the ninth game,her point total must be a multiple of9between37and 37+9=46,inclusive.The only such point total is45=37+8,so in the ninth game she scored8points.Similarly,the next point total must be a multiple of 10between45and45+9=54.The only such point total is50=45+5,so in the tenth game she scored5points.The product of the number of points scored in Theresa’s ninth and tenth games is8·5=40.16.Answer(D):The volume is7×1=7cubic units.Six of the cubes have5square faces exposed.The middle cube has no face exposed.So the total surface area of theﬁgure is5×6=30square units.The ratio of the volume to the surface area is7:30.ORThe volume is7×1=7cubic units.There areﬁve unit squares facing each of six directions:front,back,top,bottom,left and right,for a total of30square units of surface area.The ratio of the volume to the surface area is7:30. 17.Answer(D):The formula for the perimeter of a rectangle is2l+2w,so2l+2w=50,and l+w=25.Make a chart of the possible widths,lengths,and areas,assuming all the widths are shorter than all the lengths.Width123456789101112Length242322212019181716151413Area24466684100114126136144150154156 The largest possible area is13×12=156and the smallest is1×24=24,for a diﬀerence of156−24=132square units.Note:The product of two numbers with aﬁxed sum increases as the numbers get closer together.That means,given the same perimeter,the square has a larger area than any rectangle,and a rectangle with a shape closest to a square will have a larger area than other rectangles with equal perimeters.18.Answer(E):The length ofﬁrst leg of the aardvark’s trip is14(2π×20)=10πmeters.The third andﬁfth legs are each14(2π×10)=5πmeters long.Thesecond and sixth legs are each10meters long,and the length of the fourth leg is 20meters.The length of the total trip is10π+5π+5π+10+10+20=20π+40 meters.19.Answer(B):Choose two points.Any of the8points can be theﬁrst choice,and any of the7other points can be the second choice.So there are8×7=56ways of choosing the points in order.But each pair of points is counted twice,so there are 562=28possible pairs.A B CDEF G H Label the eight points as shown.Only segments AB ,BC ,CD ,DE ,EF ,F G ,GH and HA are 1unit long.So 8of the 28possible segments are 1unit long,and the probability that the points are one unit apart is 828=27.ORPick the two points,one at a time.No matter how the ﬁrst point is chosen,exactly 2of the remaining 7points are 1unit from this point.So the probabilityof the second point being 1unit from the ﬁrst is 27.20.Answer (B):Because 23of the boys passed,the number of boys in the class is a multiple of 3.Because 34of the girls passed,the number of girls in the classis a multiple of 4.Set up a chart and compare the number of boys who passed with the number of girls who passed to ﬁnd when they are equal.Total boysBoys passed 326496Total girls Girls passed 4386The ﬁrst time the number of boys who passed equals the number of girls who passed is when they are both 6.The minimum possible number of students is 9+8=17.ORBecause 23of the boys passed,the number of boys who passed must be a multiple of 2.Because 34of the girls passed,the number of girls who passed must be a multiple of 3.Because the same number of boys and girls passed,the smallestpossible number is 6,the least common multiple of 2and 3.If 6of 9boys and 6of 8girls passed,there are 17students in the class,and that is the minimum number possible.ORLet G =the number of girls and B =the number of boys.Then 23B =34G ,so 8B =9G .Because 8and 9are relatively prime,the minimum number of boysand girls is 9boys and 8girls,for a total of 9+8=17students.21.Answer (C):Using the formula for the volume of a cylinder,the bologna hasvolume πr 2h =π×42×6=96π.The cut divides the bologna in half.Thehalf-cylinder will have volume 96π2=48π≈151cm 3.Note:The value of πis slightly greater than 3,so to estimate the volume multiply 48(3)=144cm 3.The product is slightly less than and closer to answer C than any other answer.22.Answer (A):Because n3is at least 100and is an integer,n is at least 300andis a multiple of 3.Because 3n is at most 999,n is at most 333.The possible values of n are 300,303,306,...,333=3·100,3·101,3·102,...,3·111,so the number of possible values is 111−100+1=12.E F 323.Answer (C):Because the answer is a ratio,it doesnot depend on the side length of the square.Let AF =2and F E =1.That means square ABCE has side length 3and area 32=9square units.The area of BAF is equal to the area of BCD =12·3·2=3square units.Triangle DEF is an isosceles right triangle with leg lengths DE =F E =1.The area of DEF is 12·1·1=12square units.The area of BF D is equal to the area of the square minusthe areas of the three right triangles:9−(3+3+12)=52.So the ratio of the area of BF D to the area of square ABCE is 529=518.24.Answer (C):There are 10×6=60possible pairs.The squares less than60are 1,4,9,16,25,36and 49.The possible pairs with products equal to the given squares are (1,1),(2,2),(1,4),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6)and (9,4).So the probability is 1160.。

08--历年数学建模优秀论文大全

Can We Assess a Health Care System's Performance?参赛队员：董希望（自动化学院），刘琳燕（城环学院）刘福亮（软件学院）指导教师：肖剑参赛单位：重庆大学参赛时间：2008年2月15∼18日Can We Assess a Health Care System's Performance?1.BackgroundHealth systems consist of all the people and actions whose primary purpose is to improve health. They may be integrated and centrally directed, but often they are not. After centuries as small-scale, largely private or charitable, mostly ineffectual entities, they have grown explosively in this century as knowledge has been gained and applied. They have contributed enormously to better health, but their contribution could be greater still, especially for the poor. Failure to achieve that potential is due more to systemic failings than to technical limitations. It is therefore urgent to assess current performance and to judge how health systems can reach their potential.The World Health Organization (WHO) is a specialized agency of the United Nations (UN) that acts as a coordinating authority on international public health. Established on 7 April 1948, and headquartered in Geneva, Switzerland, the agency inherited the mandate and resources of its predecessor, the Health Organization, which had been an agency of the League of Nations.The WHO's constitution states that its objective "is the attainment by all peoples of the highest possible level of health." Its major task is to combat disease, especially key infectious diseases, and to promote the general health of the people of the world.As well as coordinating international efforts to monitor outbreaks of infectious diseases, such as SARS, malaria, and AIDS, the WHO also sponsors programs to prevent and treat such diseases. The WHO supports the development and distribution of safe and effective vaccines, pharmaceutical diagnostics, and drugs. The WHO also carries out various health-related campaigns — for example, to boost the consumption of fruits and vegetables worldwide and to discourage tobacco use.The annual World Health Report (http://www.who.int/whr/en/index.html) assesses global health factors and World Health Statistics provides health statistics for the countries in the UN. The production and dissemination of health statistics is a major function of the WHO. To many people, these data and the associated analyses are considered unbiased and very valuable to the world community.2. Basic Assumption and Hypotheses1.Assume that in a certain interval such as 5years the main components (metrics) ofthe health care system stays steady, that is to say the metric won’t change continually.2.Assume that all the statistics we get from the database of the WHO is authentic.3.Assume that the ranking of the world's health systems in 2000 made by WHO isscientific and dependable.4.During the data processing if a data little than x we can replace it with x.5.If there existing data missing for some year’s indicator we can value it with thecorresponding value of the near years.3. SymbolsSymbol Definition and Property'Z The matrix before standardizationZ The matrix after standardizationz j The statistic of the j indicator to each countryu The main component to be evaluatedu m The m th main component of the indicatorl ij The load of the original indicatorR1 The correlation matrix of Zc i The contribution rate of the i th main componentS i The summation of the front i main components’ contribution rateQ The integrated score of each countryX The project set ( the Member States)U The attribute set which also means main component seta ij The attribute value of x i in reference to u jA The decision making matrixa i The mean value of the line I in the primitive matrixb i The standard deviation of row I in the primitive matrix4. Problem AnalysisTo determine several important and viable metrics for assessing the performance of a health care system and comparing health care systems in different countries. We have to know what metrics or indicators are there in a health care system, as is shown in the problem we search the web of the WHO and get the database of the indicators. There exists statistics for 50 core indicators on mortality, morbidity, risk factors, service coverage, and health systems, which take on more than one hundred and fifty terms of raw indicators. We must use some data mining technology or method to distill the crucial metrics.Considering the data is promiscuous and inconsistent and not all the countries have the corresponding data to each indicator from the year 1960 to 2006, we first need to choose certain year’s data as our study object. Then to the mass actual statistical data we can’t expect all the indicators are complete so what to do with the incomplete data to make sure that all the indicators or all the data we used below are universal or effective is an inevitable problem. There are 159 raw indicators how could we select the most important ones and combine them scientifically to make them more useful in measuring quality is another basal problem. Then how could we accomplish this goal? The main components analysis method which we could use to devise our first model will help a lot.Furthermore how could we assess a country’s health care system and make some comparisons with the combined metrics? This situation much agrees with the multiple attribute decision problems. So we could solve this problem by ranking all the countries health care systems using this multiple attribute decision method.5. The Establishment of Model5.1 The Primary Data and Indicators ProcessingAccording to the above problem analysis part we know that we could obtain enough raw data for almost 159 indicators from 1960 to 2006. We first choose a year 2004 whose data is much completer than other years as our study object. Then if some of the indicators of certain country in 2004 have no value and the year close to 2004 such as the year of 2005 or 2003 has the corresponding value we treat this close value as the valve of the country in that indicator in 2004.Based on these we select the indicators that 95% of the country has the corresponding data for them from all the 159 raw indicators. By doing this primary selection we make sure that all the indicators or all the data we used below are universal or effective. After the primary selection we get 48 crucial indicators as our primary outcomes (metrics).5.2 Model 1 DesignFollowing the above analysis we utilize the main components analysis method to devise our first model.When it comes to main components analysis the biggest effect to it is the dimension of the data. So in the practical application we first should make standardization to the data.Assume that 'Z is the matrix before standardization Z is the matrix after standardization z j is the statistic of the j indicator to each country; u is the main component to be evaluated, so the objective function could be:11111221221122221122p p p p m m m mp p u l z l z l z u l z l z l z u l z l z l z =++⎧⎪=++⎪⎨⎪⎪=++⎩""""""" (1)Where u 1, u 2,… u m is called the 1st, 2nd, … mth main component of the indicator z 1, z 2, z p ; l ij is the load of the original indicator z j (j=1,2, …,p) in each main component.The detailed process of this solution is as follows:Step1: Evaluate the standardized matrix Z of the matrix'Z The standardization of the 'Z is just replace the (i=1,2, …,p) and the z 'i z ijof the matrix 'Z with z i (i=1,2, …,p) and with z ij respectively, which is shown in table1Step 2: Evaluate the correlation matrix R1 of matrix ZR1 could be evaluated by the following matrix:1112121222121p p p p pp r r r r r r R r r r ⎡⎤⎢⎥⎢=⎢⎢⎥⎢⎥⎣⎦""##"#"⎥⎥ (2) Where r ij (i, j =1, 2,…,p) is the original indicator z i and z j ’s correlation coefficient specially r ij =r ji . r ij could be derived by the following formula:ij r = (3) Step 3: By formula 1 we can compute the characteristic root and characteristic vector of matrix R1 then rank the characteristic values of R1 as expression 110I R λ−= (4)(5)0≥≥≥≥λλλ"p 21Step 4: Evaluate the contribution rate and the accumulative contribution rate according to formula 1and 1of the main components, determine the proper number of the main components.(6) λ∑==="1(1,2,,i i p k k c i λ)pWhere is the contribution rate of the i th main component.i c(7) S i λ∑p λ====∑"11(1,2,,)i k k i p k k Here is the summation of the front i main components’ contribution rate. If thevalue of the accumulative contribution rate reaches to more than 80% we can approbate the effect of the main components.i S Step 5: Compute the load of the main component l ij(y ,z )(,1,2,,)ij i j ij l p i j p ===" (8)Step 6: Sum up the above five steps get our objective function11111221221122221122............p p p p m m m mp u l z l z l z u l z l z l z u l z l z l z =+++⎧⎪=+++⎪⎨⎪⎪=+++⎩"""p Step 7: Evaluate the integrated value of each country and make a ranking of them with the formula 1.112211(m m m ii )u u λλλλ==+++∑"Q u (9) Q is the integrated score of each country.5.3 Model 2 Design5.3.1 The Description of the PrincipleThe method of the multiple attribute decision making based on dispersion maximization is used to solve the multiple attribute decision making problems with the uncertain weight attribute. We can use this method to make ranking and comparison between different projects with multiple attributes.In more details the smaller the difference between certain attribute for all the projects is the less affection it has on the decision making and ranking of the projects. On the contrary the bigger it is the more affection it has on the decision making and ranking. As a result in the view of ranking the bigger of one attribute’s deviation is the bigger weight of this attribute should be given. Especially if there is no deviation for certain attribute to all the projects which means that this attribute will have little affection on the ranking we can value a zero to its weight.5.3.2 Model DevelopmentStep 1.Structure and Normalize the Decision Making Matrix1.1 Structure the Decision Making MatrixAssume that:M={1,2,…,m},N={1,2,…,n} (10)The projects set which also is the set of the Member States in WHO is XX={x 1,x 2,…,x n } (11)The attribute set which means main component set here is UU={u 1,u 2,…,u m } (12) is the attribute value of in reference to so we obtain the decision making matrix ()ij n m A a ×=whose form is shown as table 2 Table2. The form of the decision making matrixu 1u 2… u m x 1a 11a 12… a 1m x 2a 21a 22… a 2m ## # # x na n1a n2… a nm5.4 Model 3 DesignModel 3 is our predictive model, from model 1 we can get the objective function with the data of that year.When it comes to predicating for the convenience of evaluating the main components we can change the main component which is expressed by the standardization indicator z i into the form that expressed by nonstandard indicator z i ’ to predicate the main components.''''''1111122110''''''2211222220''''''11220(13)............p p p p m m m mp p m u l z l z l z l u l z l z l z l u l z l z l z l ⎧=++++⎪=++++⎪⎨⎪⎪=++++⎩"""Here(14)'()/i i i z z a b =−i Substitute into formula 1 can we obtain the formula 2.'i z i a is the mean value of the line i in the primitive matrix; is the standard deviationof row i in the primitive matrix.i bThen utilize the formula 9 to compute the synthetic score and get variability ofthe system.Normalize the Decision Making MatrixThere are many types of attributes such as benefit type, cost type, fixation type, deviate type, interval type, deviate interval type etc. In our model all the attribute could be sorted to two types the benefit type and the cost type approximately. The benefit cost requires the value of the attribute as big as possible; the cost type requires the value of the attribute as small as possible.To eliminate the impact of the different dimensions to the decision making result we should normalize the decision making matrix A whose values could be obtained from the model 1.Assume that I i (i=1, 2) stands for the subscript set of the benefit type and cost type. If the attribute is benefit type we value i in I i as 1. If the attribute is cost type we value i in I i as 2.1min(),,max()min()ij ij i ij ij ij ii a a r i a a N j I −=−∈∈ (15)2max(),,max()min()ij ij i ij ij ij i i a a r i a a N j I −=−∈∈ (16)After this step we get the normalized matrix ()ij n m R r ×= whose form is the samewith the matrix A.Step 2: Calculus the optimization weight vector w11111,n n ji kj i k j m n nij kj j i k r r w r r =====−=−∑∑∑∑∑j M ∈ (17) Where w j is the j th main component’s weight.Step 3: Computer the synthetic attribute z i (w) (i ∈N) of project x i .1(),,mi ij j j z w r w i N j ==∈∑M ∈ (18)Step 4: Make ranking and comparison to the projects (countries) using z i (w)(i ∈N)6. Applying the Model1 and Model 26.1 Applying the Model 1 to the Statistics of the Year 20046.1.1 Data for Model 1 in the Year of 2004We first select the indicators that 95% of the countries own these indicators from all 159 indicators getting 28 indicators which could be seen in appendix Ⅰ. Then weselect the countries that have the data for all these 28 indicators from all 193 Member States getting 163 countries. By doing these we have made good preparation for our model 1’s solution.6.1.2 Solution of the Model 1 for the Year of 2004Based on the above data we solve our model 1 in matlab using the function of zscore to normalize the data, and then we calculate the characteristic roots and characteristic vector. The characteristic roots are shown in the table 3.Table3. Part Valves of Model 1From the table 3 we can see that the front six red colored components’ accumulative contribution rate reaches to 81.5% which means that most of the main components are involved, so these six components are just our combined metrics. We renamed these six combined indicators with A, B, C, D, E, F metrics all of which are constituted by several raw indicators and could reflect certain performance of a health care system.In more detail the metric A is much positively related with life expectancy, per capita total expenditure on health at international dollar rate etc and much negatively related with mortality rate, incidence of tuberculosis (per 100 000 population per year) etc. The visual relationship between the metric and the 28 indicators is shown in figure 1. The x axis is the order of the 28 indicators which maps to corresponding 28 indicators in appendix Ⅰ. The y axis is the affection of each of the 28 indicator on metric A. All the rest five figures follow this instruction so we won’t explain the rest five figures again.Figure1. The affection of the 28 indicators on metric A The metric B is much positively related with expenditure on health, disease detection rate etc and much negatively related with government expenditure on health, alcohol consumption etc.Figure2. The affection of the 28 indicators on metric B The metric C is much positively related with General government expenditure on health as percentage of total expenditure on health, immunized with disease etc and much negatively related with private expenditure on health as percentage of total expenditure on health, population (in thousands) total etc.Figure3. The affection of the 28 indicators on metric C The metric D is much positively related with private expenditure on health as percentage of total expenditure on health, immunized with disease etc and muchnegatively related with General government expenditure on health as percentage of total expenditure on health etc.Figure4. The affection of the 28 indicators on metric D The metric E is much positively related with external resources for health as percentage of total expenditure on health etc and much negatively related with tuberculosis: DOTS case detection rate, probability of dying (per 1 000 population) between 15 and 60 years etc.Figure5. The affection of the 28 indicators on metric E The metric F is much positively related with immunized with disease, out-of-pocket expenditure as percentage of private expenditure on health etc and much negatively related with general government expenditure on health as percentage of total government expenditure, population (in thousands) total etc.Figure6. The affection of the 28 indicators on metric FThe above descriptions show that our metrics is reasonable, moreover all the 28indicarors we selected could be found in 92% of all Member States, which means that our metrics could be easily collected.Furthermore we get the ranking for all the 163 countries that own orbicular and effective data. The front and the back 20 countries in our ranking and their scores calculated by our model 1 are listed as table 4:Table4. Part of our Ranking by Our Model1The whole ranking is shown in appendix Ⅱ.After obtaining the six metrics we treat the still missing value’s indicator as zero then recompute the ranking of the year 2004 with the model 1 and get another ranking for all the Member States which we list in appendix Ⅲ.In conclusion we put forward 28 important indicators from all the 159 indicators furthermore we combine the 28 important indicators getting 6 main components which we renamed as metric A, B, C, D, E and F. Then we assess the health care system with these six metrics and make a ranking of all the 163 countries.6.1.3 Applying model 1 to the Statistics of the Year 2000Using model 1 and the six metrics obtained from 4.11 we assess the health care system of each country in the year of 2000. This time we only utilize the data of this year, which means that we just substitute the data of 2004 with that of 2000.By doing this we get the ranking of this year as table 5 which just show out the front and the back 20 countries too.Table5. The Part Ranking of the Year 2000 by Model 1 (There are 194 Member States in 2000; the score here is just a relative value computed by our model; the whole ranking is shown in appendix Ⅳ)6.2 Applying the Model 2The six metrics obtained from model1 is ordered. Although the six metrics keep the same in the model 2 as what they are in model 1 according to our assumptions, there is no certain order among them in our model 2. The data processing methods for the raw data are the same with what we have described and used before.6.2.1 Applying the Model 2 to the year of 2004With the help of the software matlab we realize the algorithm of dispersion maximization computing the weight of the six metrics, and then we calculate the synthetic score of each of the 163 country that own holonomic statistics after our data mining process. After comparing the synthetic score of these countries we get the ranking of their health system as shown in table 6.Table6. Part of the ranking for the 163Member States(The score here is just a relative value computed by our model; the whole ranking is shown in appendix Ⅴ)6.2.2 Applying the Model 2 to the year of 2000Similar to 4.1.2 we just replace the data of 4.2.1 with the data of the year 2000 then compute the synthetic score of all the 194 Member States. After comparing the different countries we get the ranking as table7.Table7. Part of the ranking for 2000 by model 27. Comparisons7.1 Comparisons between Different RankingsFrom the above solution we obtain 4 different rankings. The precise clues of our models have already showed their validity. Besides we can load a ranking for all the 190 Member states in 2000 from the WHO’s official web which we list in appendix Ⅷ.By making comparisons between our two rankings with the official ranking of the year 2000 we can test the reliability and practicability of our model to a certain extent. The figure7 shows their relationship clearly.Figure7. The corresponding relationship between our rankings and the WHO’s We can see that the dots which stand for parts of the countries in the rankings match quite well with each other in the three polygonal lines. That means the model1 and model 2’s results not only agree with each other but also agree with the official results quite well. So we can conclude that our two models are practical and reasonable.Since the solution for the year of 2000 is dependable, we have reason enough to predicate that our solution for 2004 is authentic as the only difference between 2004 and 2000 is the substituted statistics and the data of 2004 is more holonomic than that of 2000.To make sure that both of our models’ results for the year 2004 are unitive we make a comparison between their rankings. We select some characteristic countries in both rankings and compare those countries rankings as shown in figure 8.Figure8. The comparison between the two rankings for the year 2004From the figure we find that the two rankings match quite well.In conclusion our models are scientific and our results are authentic.7.2 Comparisons between US and FranceIn the 2000’s ranking of WHO France takes the first place, which also could be seen clearly in the appendix. In the year 2004 there is no official ranking so we assess these two countries health care system with our model 1 to see which country has the better health care system then.The table8 shows their score according to our six metrics:Table8. The comparisons between US and France’s health system according to our metricsThe metric A, C, F belongs to benefit type and the rest belong to cost type. Base on this we can see that the health care system of US in 2004 is better than France in metric A, B, D. According to our table 1 we know that the synthetic score of US is better than France.7.3 Comparisons between US and IndiaIn the ranking of WHO the health system of US is better than India. With the help of our mode 1 we consider that India has the poor health care system in 2004, so we make a comparison between them.Table9.The comparisons between US and India’s health system according to our metricsSimilar to 7.2 we can see that the health care system of US is better than India in metric A, B, C, D, F. Also from the table 9 we know that the ranking of US is much better than India.8. Applying the Model 3Based on model 1 and model 2 with the help of the software matlab we realize the algorithm in model 3 and get the predictive function of the synthetic scores as follows:''''1234'7''''67891''''1112131410.0033810.00966920.0105590.00194680.000522790.00052406 3.06100.0782130.00421940.0003620.00873510.00956890.00984890.000392260.0043929Q z z z z z z z z z '5z z z z z z −=−++−−−−×−−−−++−+'5''''1617181920''''2122232425'''2627280.0053440.0257520.00377190.000143460.000182770.000109410.000136790.00534410.056440.00292340.000840420.029650.012447 3.8668''z z z z z z z z z z z z ++−++++−+−−++−z (19)Considering the affection of the weight on the synthetic score we could find that the bigger the absolute value of weight is the bigger the impact is on the synthetic score of the country. On the contrary if the absolute value of weight is small then the variation of the metric won’t produce big changes to the synthetic score. Then we take some indicators of the all 28 indicators as examples to discuss what affection it will has on the health care system if the various changes are occurred.'8z is the formula is the total fertility rate (per woman). It has a negative correlation with the synthetic score. What’s more it has a big affection on the score so this indicator should be as small as possible, which means that the government should take some measures to control the population within a proper range to improve the health care system of the nation.'24z is the total expenditure on health as percentage of gross domestic product. Itis an indicator that positively related with the synthetic score which means that the more it spend on the total expenditure on health as percentage of gross domestic product the better score it has in the system.'17z is the general government expenditure on health as percentage of totalgovernment expenditure. It is an indicator that positively related with the synthetic score which means that the bigger the general government expenditure on health as percentage of total government expenditure is the better score it has in the system'3z is the life expectancy at birth (years) males. It is an indicator that positivelyrelated with the synthetic score which means that the longer the life expectancy at birth (years) males is the better score it has in the system.'z stands for the neonatal mortality rate (per 1 000 live births). It has a negative 11correlation with the synthetic score. That’s to say the smaller the neonatal mortality rate (per 1 000 live births)is the better the health care system will become.9. The Strength and Weakness9.1The StrengthWe obtain the statistics directly from the raw database of the WHO’s official web not from the report of the WHO. We use some data mining technology to draw the available and effective data from thousands terms of data ourselves.We develop three different models to solve all the six parts of the problem, those models are built with precise logic, scientific principle which could solve the problems efficaciously.We don’t solve the problem part by part but solve them in our models’ development and solution process, which keeps the whole paper’s with a good continuity.We compare our result with the practical result, which tests our models’ practicability and validity greatly.Our models could be easily extended to other fields to solve the multiple attribute decision making problems.Our models are independent to the metric (indicators) to a certain extent as the algorithm of our models has the universal applications.9.2 The WeaknessThe raw data we get is the data from the real world, which means that there must be some imperfect data which do have some negative impact on our result.As there are so many indictors that it is hard to select proper metrics to assess the health system properly without some kind of error.Because the limitation of the time and resource it’s inevitable to have some imperfect aspects in our models, analysis and paper.10. References[1] Zeshui Xu, 8/2004, Uncertain Multiple Attribute Decision Making: Methods andApplications, Tsinghua University Press.[2] Qiyuan Jiang, Jinxing Xie, 12/2004, Mathematical Model, Higher Education Press[3] The World Health Report 2000 - Health systems: improving performance.http://www.who.int/whr/2000/en/whr00_en.pdf[4] World Health Organization, http://www.who.int/research/en/s[5] Principal Component Analysis,/jpkc/jldlx/admin/ewebeditor/UploadFile/200783101241734.ppt[6] /wiki/World_Health_Organisation,"World Health Organization"11. AppendixAppendixⅠ: The list of all the 28 indicators and their sequence numberAppendixⅡ: The Ranking of all 163 Countries in 2004 by model 1Appendix Ⅲ: The Ranking of all 194 Countries in 2004 by model 1。

2008数学建模论文

高等教育学费标准探讨摘要本文针对高等教育的学费收费标准问题，在做出某些合理的假设下，建立了两个模型，分别得到各类地区各类学校和专业的学费收费标准。

在模型1中，通过分析影响高等教育价格的主要因素及高等教育本身的多元性，基于“谁受益，谁付费”的市场经济原则，首先利用已有数据和回归分析方法得到所有高校教育成本的均值，在此基础上再利用层次分析（AHP ）方法求得各类地区每类学校和专业的学费在教育总成本中所占的比重，再利用乘法原理，得到各类地区每类学校和专业的学生应交的学费，所得结果与当年的实际情况进行了比较分析。

如2006年来自2e 类地区的学生就读2a 类学校的1b 类专业查得其收费价格为5250元到5550元，与计算结果5280.23吻合的很好，由此可知该模型有很高的可行性，在模型1中，主要利用高校教育成本的均值来求得各类高校的学费标准。

在模型2中，我们利用建立数学规划方法各类地区每类学校和专业的学费收费标准。

考虑到总培养费用的来源与开销，以政府资助、学生学费、学校创收、社会捐赠为影响总培养费用来源的决定性因素，以教师工资、学校固定资产、基本开销为总培养费用的开销，通过搜集数据，针对不同地区的学生，利用MATLAB 数学软件作出政府资助、社会捐赠、教师工资、高校在校生人数、家庭收入、学校固定资产、毕业生工资等与时间的拟合曲线，得出拟合方程，用此模型即可预测出2009年不同地区的学费收费情况，其计算结果与实际数据基本符合。

最后对模型的合理性与实用性且推广难易程度进行了评价，并从中得知该模型若作出合理假设可改进为更优的模型。

1.问题的重述1.1 问题的提出高等教育事关高素质人才培养、国家创新能力增强、和谐社会建设的大局，因此受到党和政府及社会各方面的高度重视和广泛关注。

培养质量是高等教育的一个核心指标，不同的学科、专业在设定不同的培养目标后，其质量需要有相应的经费保障。

高等教育属于非义务教育，其经费在世界各国都由政府财政拨款、学校自筹、社会捐赠和学费收入等几部分组成。

2008数学建模国家一等奖论文(神经网络)

高等教育学费标准的研究摘要本文从搜集有关普通高等学校学费数据开始，从学生个人支付能力和学校办学利益获得能力两个主要方面出发，分别通过对这两个方面的深入研究从而制定出各自有关高等教育学费的标准，最后再综合考虑这两个主要因素，进一步深入并细化，从而求得最优解。

模块Ⅰ中，我们将焦点锁定在从学生个人支付能力角度制定合理的学费标准。

我们从选取的数据和相关资料出发，发现1996年《高等学校收费管理暂行办法》规定高等学校学费占生均教育培养的成本比例最高不得超过25%，而由数据得到图形可知，从2002年开始学费占教育经费的比例超过了25%，并且生均学费和人均GDP 的比例要远远超过美国的10%到15%。

由此可见，我国的学费的收取过高。

紧接着，我们从个人支付能力角度出发，研究GDP 和学费的关系。

并因此制定了修正参数，由此来获取生均学费的修正指标。

随后，我们分析了高校专业的相关系数，从个人支付能力角度，探讨高校收费与专业的关系，进一步得到了高校收费标准1i i y G R Q =。

在模块Ⅱ中，我们从学校办学利益获得能力出发，利用回归分析对学生应交的学杂费与教育经费总计、国家预算内教育经费、社会团体和公民个人办学经验、社会捐投资和其他费用的关系，发现学杂费与教育经费总计成正相关，与其他几项费用成负相关。

对此产生的数据验证分析符合标准。

然后，再根据专业相关系数来确定学校收取学费的标准。

从而，得到了学校办学利益的收费标准2i i i y y R =。

在模块Ⅲ中，为了获取最优解，我们综合了前面两个模块所制定的收费指标，并分别给予不同权系数，得到最终学费的表达式12i i C ay by =+。

然后，我们从学校收费指标的权系数b 考虑，利用神经网络得到的区域划分，根据不同区域而计算出的权系数b 的范围。

最终得到的表达式[]12345**(1)(1.0502 1.1959 1.3108 1.36360.7929)**b i i C R G Q b x x x x x R =-+----;由此便可得到综合学费标准C 的取值范围。

[C]美国数学建模比赛题1985-2009

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 (5)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 (6)MCM89问题-A 蠓的分类 (6)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 (7)MCM91问题-B 通讯网络的极小生成树 (7)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 (8)MCM93问题-B 倒煤台的操作方案 (8)MCM94问题-A 住宅的保温 (9)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 (10)MCM95题-B A1uacha Balaclava学院 (10)MCM96问题-A 噪音场中潜艇的探测 (11)MCM96问题-B 竞赛评判问题 (11)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 (12)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 (14)MCM2000问题-A空间交通管制 (14)MCM2000问题-B: 无线电信道分配 (14)MCM2001问题- A: 选择自行车车轮 (15)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 (16)MCM2002问题-A风和喷水池 (16)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 (18)MCM2003问题-B: Gamma刀治疗方案 (18)MCM2003问题-C航空行李的扫描对策 (19)MCM2004问题-A：指纹是独一无二的吗？ (19)MCM2004问题-B：更快的快通系统 (19)MCM2004问题-C安全与否？ (19)MCM2005问题A.水灾计划 (19)MCM2005B.Tollbooths (19)MCM2005问题C：不可再生的资源 (20)MCM2006问题A: 用于灌溉的自动洒水器的安置和移动调度 (20)MCM2006问题B: 通过机场的轮椅 (20)MCM2006问题C : 抗击艾滋病的协调 (21)MCM2008问题A:给大陆洗个澡 (24)MCM2008问题B：建立数独拼图游戏 (24)MCM2009 问题A:设计一个交通环岛 23 MCM 2009问题B：能源和手机 24 MCM 2009问题C : 构建食物系统: 重新平衡被人类影响的生态系统25MCM85问题-A 动物群体的管理在一个资源有限，即有限的食物、空间、水等等的环境里发现天然存在的动物群体。

AMC2008

AMC2008年真题Problem 1Susan had dollars to spend at the carnival. She spent dollars on food and twice as much on rides. How many dollars did she have left to spend?Problem 2The ten-letter code represents the ten digits , in order. What4-digit number is represented by the code word ?Problem 3If February is a month that contains Friday the , what day of the week is February 1?Problem 4In the figure, the outer equilateral triangle has area , the inner equilateral triangle has area , and the three trapezoids are congruent. What is the area of one of the trapezoids?Problem 5Barney Schwinn notices that the odometer on his bicycle reads , a palindrome, becauseit reads the same forward and backward. After riding more hours that day and the next,he notices that the odometer shows another palindrome, . What was his average speed in miles per hour?In the figure, what is the ratio of the area of the gray squares to the area of the white squares?Problem 7If , what is ?Problem 8Candy sales from the Boosters Club from January through April are shown. What were theaverage sales per month in dollars?Problem 9In Tycoon Tammy invested dollars for two years. During the the first year her investment suffered a loss, but during the second year the remaining investmentshowed a gain. Over the two-year period, what was the change in Tammy's investment?The average age of the people in Room A is . The average age of the people in RoomB is . If the two groups are combined, what is the average age of all the people?Problem 11Each of the students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and students have a cat. How manystudents have both a dog and a cat?Problem 12A ball is dropped from a height of meters. On its first bounce it rises to a height of meters.It keeps falling and bouncing to of the height it reached in the previous bounce. On which bounce will it not rise to a height of meters?Problem 13Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than pounds or more than pounds. So the boxes are weighed in pairs in every possible way. The results are, and pounds. What is the combined weight in pounds of the three boxes?Three , three , and three are placed in the nine spaces so that each row andcolumn contain one of each letter. If is placed in the upper left corner, how manyarrangements are possible?Problem 15In Theresa's first basketball games, she scored and points. In her ninthgame, she scored fewer than points and her points-per-game average for the nine gameswas an integer. Similarly in her tenth game, she scored fewer than points and herpoints-per-game average for the games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?Problem 16A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?Problem 18Two circles that share the same center have radii meters and meters. An aardvark runs along the path shown, starting at and ending at . How many meters does theaardvark run?Problem 19Eight points are spaced around at intervals of one unit around a square, as shown.Two of the points are chosen at random. What is the probability that the two points are oneunit apart?The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?Problem 21Jerry cuts a wedge from a -cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?Problem 22For how many positive integer values of are both and three-digit whole numbers?In square , and . What is the ratio of the area ofto the area of square ?Problem 24Ten tiles numbered through are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?Problem 25Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Approximately what percent of the design is black?。

美国中学生数学建模竞赛获奖论文

Abstract
In this paper, we undertake the search and find problem. In two parts of searching, we use different way to design the model, but we use the same algorithm to calculate the main solution. In Part 1, we assume that the possibilities of finding the ring in different paths are different. We give weight to each path according to the possibility of finding the ring in the path. Then we simplify the question as pass as more weight as possible in limited distance. To simplify the calculating, we use Greedy algorithm and approximate optimal solution, and we define the values of the paths(according to the weights of paths) in Greedy algorithm. We calculate the possibility according to the weight of the route and to total weights of paths in the map. In Part 2, firstly, we limit the moving area of the jogger according to the information in the map. Then we use Dijkstra arithmatic to analysis the specific area of the jogger may be in. At last, we use greedy algorithm and approximate optimal solution to get the solution.

2008年D题全国一等奖论文2

NBA赛程的分析与评价摘要本文首先综合考虑了NBA上个赛季的赛程、赛绩和本赛季的赛程确定出赛程对球队利弊的三个主要影响因素，并对其进行了定量分析。

其次利用偏大型柯西分部隶属函数确定主要影响因素的权值，给出了一个利弊的评价指标——利弊指数，并计算了各球队的利弊指数值。

从得到的结果看本次赛程对火箭队而言是比较有利的，其中最有利的球队是凯尔特人队，最不利的是快船队。

对于问题三，基于公平性和观赏性考虑，同部不同区球队实力尽可能悬殊的队尽可能少赛（赛3场）。

由此建立0-1规划模型，并利用LINDO软件求解出了赛3场球队的最优选取方案。

关键词：隶属函数利弊指数 0-1规划一.问题的重述NBA赛程的安排对球队实力的发挥和战绩存在着客观的影响，但编制一个完整的、对各球队尽可能公平的赛程是一件非常复杂的事情。

为了更直观的体现出这些客观因素的存在，利用数学建模方法对2008～2009年的赛季安排表进行定量的分析与评价：1）确定出赛程对某一支球队的利弊的主要影响因素，根据所确定的因素将赛程转换为便于进行数学处理的数字格式，同时给出评价赛程利弊的数量指标。

2）按照1）的结果计算、分析赛程对火箭队的利弊，并找出赛程对30支球队最有利和最不利的球队。

3）对2008～2009年的赛季安排表进行分析可以发现，每支球队与同区的每一支球队赛4场（主客各2场），与不同部的每一球队赛2场（主客各1场），与同部不同区的每一球队有赛4场和赛3场（2主1客或2客1主）两种情况，每支球队的主客场数量相同且同部3个区的球队间保持均衡。

试根据赛程找出与同部不同区球队比赛中，选取赛3场的球队的方法。

这种方法如何实现，对该方法给予评价，也可以给出认为合适的方法。

二.问题分析问题1首先应综合分析上一赛季的赛绩和本次赛季的赛程确定赛程对球队利弊的主要影响因素，其次要确定影响因素权值；根据本次赛场各球队的影响指标，对东西联盟的30支球队进行排序。

问题2根据上一问所得的结果，重点分析赛程对火箭队的利弊及赛程对那个队是最有利的，对那个队是最不利的。

2008年NBA建模获奖论文

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

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

我们郑重承诺，严格遵守竞赛规则，以保证竞赛的公正、公平性。

如有违反竞赛规则的行为，我们将受到严肃处理。

我们参赛选择的题号是（从A/B/C/D中选择一项填写）：我们的参赛报名号为（如果赛区设置报名号的话）：所属学校（请填写完整的全名）：参赛队员(打印并签名) ：1.2.3.指导教师或指导教师组负责人(打印并签名)：日期：年月日赛区评阅编号（由赛区组委会评阅前进行编号）：编号专用页赛区评阅编号（由赛区组委会评阅前进行编号）：赛区评阅记录（可供赛区评阅时使用）：评阅人评分备注全国统一编号（由赛区组委会送交全国前编号）：全国评阅编号（由全国组委会评阅前进行编号）：NBA赛程的分析与评价摘要论文对于已经制定好的NBA赛程进行了统计分析，对已给数据进行筛选和处理，更好地对NBA赛程进行了定量的分析与评估。

问题一，考虑NBA整体赛事运行以及影响球队发挥的关键因素，我们结合题目中给出的数据，找出影响赛程的弊端因素，我们主要考虑三种因素分别为：背靠背场次，连续客场3场及以上，对手的平均实力。

然后将赛程数字化，分析赛程对于各支球队的利弊，然后给出评价指标。

问题二，基于问题1的结论，求出赛程对每一支球队的利弊，在30支球队弊端因素值相比之下，得出2008-2009年度NBA常规赛安排，赛程对球队发挥最有利是：篮网；赛程对球队发挥最不利是：国王。

火箭队的弊端因素值为0.636，排在第18位，这样的赛程安排对于火箭较为不利。

问题三，首先我们从赛程中找出赛3场比赛的球队，可以得出，在每一球队在与同部不同区的比赛中，分别选取同部另外2个分区中，选择2只球队进行赛3场，这样可以保证每个赛区主客场数量相同，保持一定的平衡性；在这种情况下，我们考虑到赛3场（2主场1客场或者2客场1主场）的球队有一只球队拥有一个主场优势，实力相对较大的球队相互赛三场可以最大限度消除这一优势，建立线性规划模型，使得总的弊端因素值差取得最小值，通过lingo编程求解出同部不同区的球队比赛的场次，西部的见表（11），东部的见表（12）。

数模美国赛相关论文9

(h1')2 r12 r21 (h2')2 R'= . . . . rp1 rp2 … r1p … r2p … . … . … (hp')2
R'的前个特征根及其对应的单位化特征的前m个特征根及其对应的单位化特征的前向量就是主因子解. 向量就是主因子解.
迭代主因子法(iterated principal factor) )
Heywood现象
残差矩阵
五,因子旋转
目的:使因子负荷两极分化,要么目的:使因子负荷两极分化, 接近于0,要么接近于1. 接近于 ,要么接近于 . 常用的旋转方法: 常用的旋转方法:
(1)方差最大正交旋转
(varimax
orthogonal rotation)
基本思想:使公共因子的相对负荷基本思想: 的方差之和最大, (lij/hi2)的方差之和最大,且保持原公共因子的正交性和公共方差总和不变. 共因子的正交性和公共方差总和不变. 可使每个因子上的具有最大载荷的变量数最小,因此可以简化对因子的解释. 数最小,因此可以简化对因子的解释.
谢谢! 谢谢!
�
六,因子得分
Thomson法,即回归法回归法得分是由Bayes思想导出的,得思想导出的, 回归法得分是由思想导出的到的因子得分是有偏的, 到的因子得分是有偏的,但计算结果误差较小. 误差较小.
Bartlett法 Bartlett因子得分是极大似然估计,也因子得分是极大似然估计, 因子得分是极大似然估计是加权最小二乘回归, 是加权最小二乘回归,得到的因子得分是无偏的,但计算结果误差较大. 分是无偏的,但计算结果误差较大. 因子得分可用于模型诊断,也可用作因子得分可用于模型诊断, 进一步分析的原始资料. 进一步分析的原始资料.

08年数学建模B题

眼科病床的合理安排摘要本文结合眼科医疗特点，全面分析了各类眼疾对应床位安排的要求限制，引入了优先权重系数矩阵，构建了基于贪心算法的病床安排模型。

问题一中，我们设立了一个含有两类指标的评价体系。

第一类：针对五种类型的眼疾，以病人等待入院时间为决策变量，分别构造了五大满意度指标，并综合得到了满意度均值的评价标准。

第二类指标是61天内出院的病人数。

问题二中，我们首先建立了以等待住院总时间最短为目标的动态规划模型，并在该基础上采用排队模式简化得到静态模型，最后利用遗传算法对优先权重系数矩阵进行改进，最终求得：在7月13日至9月11日的时间段内，此方案比FCFS规则增加了45个出院病人。

问题三对于满足非正态分布的等待入院时间数据，采用大样本区间估计法，和等待手术2队列（针对白内障双眼）的队列，并且对白内障手术时间做出了相应的调整。

模型求解后得到规定白内障手术在周三和周五的方案是最合理的。

问题五中，我们首先建立了以总逗留时间最短为目标的动态规划模型。

为改进模型，我们利用贪心算法，以逗留时间最短为目标，分配达到极限情况（即无关键词：病床安排贪心算法优先权重遗传算法大样本法一、问题重述我们考虑某医院眼科病床的合理安排问题。

该医院眼科门诊每天开放，住院部共有病床79张。

该医院眼科手术主要分四大类：白内障、视网膜疾病、青光眼和外伤。

白内障手术较简单，而且没有急症。

目前该院是每周一、三做白内障手术，此类病人的术前准备时间只需1、2天。

做两只眼的病人比做一只眼的要多一些，大约占到60%。

如果要做双眼是周一先做一只，周三再做另一只。

外伤疾病通常属于急症，病床有空时立即安排住院，住院后第二天便会安排手术。

其他眼科疾病比较复杂，有各种不同情况，但大致住院以后2-3天内就可以接受手术，主要是术后的观察时间较长。

这类疾病手术时间可根据需要安排，一般不安排在周一、周三，且不考虑急诊。

通常情况下白内障手术与其他眼科手术（急症除外）不安排在同一天做。

美国大学生数学建模竞赛二等奖论文

美国⼤学⽣数学建模竞赛⼆等奖论⽂The P roblem of R epeater C oordination SummaryThis paper mainly focuses on exploring an optimization scheme to serve all the users in a certain area with the least repeaters.The model is optimized better through changing the power of a repeater and distributing PL tones,frequency pairs /doc/d7df31738e9951e79b8927b4.html ing symmetry principle of Graph Theory and maximum coverage principle,we get the most reasonable scheme.This scheme can help us solve the problem that where we should put the repeaters in general cases.It can be suitable for the problem of irrigation,the location of lights in a square and so on.We construct two mathematical models(a basic model and an improve model)to get the scheme based on the relationship between variables.In the basic model,we set a function model to solve the problem under a condition that assumed.There are two variables:‘p’(standing for the power of the signals that a repeater transmits)and‘µ’(standing for the density of users of the area)in the function model.Assume‘p’fixed in the basic one.And in this situation,we change the function model to a geometric one to solve this problem.Based on the basic model,considering the two variables in the improve model is more reasonable to most situations.Then the conclusion can be drawn through calculation and MATLAB programming.We analysis and discuss what we can do if we build repeaters in mountainous areas further.Finally,we discuss strengths and weaknesses of our models and make necessary recommendations.Key words:repeater maximum coverage density PL tones MATLABContents1.Introduction (3)2.The Description of the Problem (3)2.1What problems we are confronting (3)2.2What we do to solve these problems (3)3.Models (4)3.1Basic model (4)3.1.1Terms,Definitions,and Symbols (4)3.1.2Assumptions (4)3.1.3The Foundation of Model (4)3.1.4Solution and Result (5)3.1.5Analysis of the Result (8)3.1.6Strength and Weakness (8)3.1.7Some Improvement (9)3.2Improve Model (9)3.2.1Extra Symbols (10)Assumptions (10)3.2.2AdditionalAdditionalAssumptions3.2.3The Foundation of Model (10)3.2.4Solution and Result (10)3.2.5Analysis of the Result (13)3.2.6Strength and Weakness (14)4.Conclusions (14)4.1Conclusions of the problem (14)4.2Methods used in our models (14)4.3Application of our models (14)5.Future Work (14)6.References (17)7.Appendix (17)Ⅰ.IntroductionIn order to indicate the origin of the repeater coordination problem,the following background is worth mentioning.With the development of technology and society,communications technology has become much more important,more and more people are involved in this.In order to ensure the quality of the signals of communication,we need to build repeaters which pick up weak signals,amplify them,and retransmit them on a different frequency.But the price of a repeater is very high.And the unnecessary repeaters will cause not only the waste of money and resources,but also the difficulty of maintenance.So there comes a problem that how to reduce the number of unnecessary repeaters in a region.We try to explore an optimized model in this paper.Ⅱ.The Description of the Problem2.1What problems we are confrontingThe signals transmit in the way of line-of-sight as a result of reducing the loss of the energy. As a result of the obstacles they meet and the natural attenuation itself,the signals will become unavailable.So a repeater which just picks up weak signals,amplifies them,and retransmits them on a different frequency is needed.However,repeaters can interfere with one another unless they are far enough apart or transmit on sufficiently separated frequencies.In addition to geographical separation,the“continuous tone-coded squelch system”(CTCSS),sometimes nicknamed“private line”(PL),technology can be used to mitigate interference.This system associates to each repeater a separate PL tone that is transmitted by all users who wish to communicate through that repeater. The PL tone is like a kind of password.Then determine a user according to the so called password and the specific frequency,in other words a user corresponds a PL tone(password)and a specific frequency.Defects in line-of-sight propagation caused by mountainous areas can also influence the radius.2.2What we do to solve these problemsConsidering the problem we are confronting,the spectrum available is145to148MHz,the transmitter frequency in a repeater is either600kHz above or600kHz below the receiver frequency.That is only5users can communicate with others without interferences when there’s noPL.The situation will be much better once we have PL.However the number of users that a repeater can serve is limited.In addition,in a flat area ,the obstacles such as mountains ,buildings don’t need to be taken into account.Taking the natural attenuation itself is reasonable.Now the most important is the radius that the signals transmit.Reducing the radius is a good way once there are more users.With MATLAB and the method of the coverage in Graph Theory,we solve this problem as follows in this paper.Ⅲ.Models3.1Basic model3.1.1Terms,Definitions,and Symbols3.1.2Assumptions●A user corresponds a PLz tone (password)and a specific frequency.●The users in the area are fixed and they are uniform distribution.●The area that a repeater covers is a regular hexagon.The repeater is in the center of the regular hexagon.●In a flat area ,the obstacles such as mountains ,buildings don’t need to be taken into account.We just take the natural attenuation itself into account.●The power of a repeater is fixed.3.1.3The Foundation of ModelAs the number of PLz tones (password)and frequencies is fixed,and a user corresponds a PLz tone (password)and a specific frequency,we can draw the conclusion that a repeater can serve the limited number of users.Thus it is clear that the number of repeaters we need relates to the density symboldescriptionLfsdfminrpµloss of transmission the distance of transmission operating frequency the number of repeaters that we need the power of the signals that a repeater transmits the density of users of the areaof users of the area.The radius of the area that a repeater covers is also related to the ratio of d and the radius of the circular area.And d is related to the power of a repeater.So we get the model of function()min ,r f p µ=If we ignore the density of users,we can get a Geometric model as follows:In a plane which is extended by regular hexagons whose side length are determined,we move a circle until it covers the least regular hexagons.3.1.4Solution and ResultCalculating the relationship between the radius of the circle and the side length of the regular hexagon.[]()()32.4420lg ()20lg Lfs dB d km f MHz =++In the above formula the unit of ’’is .Lfs dB The unit of ’’is .d Km The unit of ‘‘is .f MHz We can conclude that the loss of transmission of radio is decided by operating frequency and the distance of transmission.When or is as times as its former data,will increase f d 2[]Lfs .6dB Then we will solve the problem by using the formula mentioned above.We have already known the operating frequency is to .According to the 145MHz 148MHz actual situation and some authority material ,we assume a system whose transmit power is and receiver sensitivity is .Thus we can conclude that ()1010dBm mW +106.85dBm ?=.Substituting and to the above formula,we can get the Lfs 106.85dBm ?145MHz 148MHz average distance of transmission .()6.4d km =4mile We can learn the radius of the circle is 40mile .So we can conclude the relationship between the circle and the side length of regular hexagon isR=10d.1)The solution of the modelIn order to cover a certain plane with the least regular hexagons,we connect each regular hexagon as the honeycomb.We use A(standing for a figure)covers B(standing for another figure), only when As don’t overlap each other,the number of As we use is the smallest.Figure1According to the Principle of maximum flow of Graph Theory,the better of the symmetry ofthe honeycomb,the bigger area that it covers(Fig1).When the geometric centers of the circle andthe honeycomb which can extend are at one point,extend the honeycomb.Then we can get Fig2,Fig4:Figure2Fig3demos the evenly distribution of users.Figure4Now prove the circle covers the least regular hexagons.Look at Fig5.If we move the circle slightly as the picture,you can see three more regular hexagons are needed.Figure 52)ResultsThe average distance of transmission of the signals that a repeater transmit is 4miles.1000users can be satisfied with 37repeaters founded.3.1.5Analysis of the Result1)The largest number of users that a repeater can serveA user corresponds a PL and a specific frequency.There are 5wave bands and 54different PL tones available.If we call a code include a PL and a specific frequency,there are 54*5=270codes.However each code in two adjacent regular hexagons shouldn’t be the same in case of interfering with each other.In order to have more code available ,we can distribute every3adjacent regular hexagons 90codes each.And that’s the most optimized,because once any of the three regular hexagons have more codes,it will interfere another one in other regular hexagon.2)Identify the rationality of the basic modelNow we considering the influence of the density of users,according to 1),90*37=3330>1000,so here the number of users have no influence on our model.Our model is rationality.3.1.6Strength and Weakness●Strength:In this paper,we use the model of honeycomb-hexagon structure can maximize the use of resources,avoiding some unnecessary interference effectively.It is much more intuitive once we change the function model to the geometric model.●Weakness:Since each hexagon get too close to another one.Once there are somebuildingsor terrain fluctuations between two repeaters,it can lead to the phenomenon that certain areas will have no signals.In addition,users are distributed evenly is not reasonable.The users are moving,for example some people may get a party.3.1.7Some ImprovementAs we all know,the absolute evenly distribution is not exist.So it is necessary to say something about the normal distribution model.The maximum accommodate number of a repeater is 5*54=270.As for the first model,it is impossible that 270users are communicating in a same repeater.Look at Fig 6.If there are N people in the area 1,the maximum number of the area 2to area 7is 3*(270-N).As 37*90=3330is much larger than 1000,our solution is still reasonable to this model.Figure 63.2Improve Model3.2.1Extra SymbolsSigns and definitions indicated above are still valid.Here are some extra signs and definitions.symboldescription Ra the radius of the circular flat area the side length of a regular hexagon3.2.2Additional AdditionalAssumptionsAssumptions ●The radius that of a repeater covers is adjustable here.●In some limited situations,curved shape is equal to straight line.●Assumptions concerning the anterior process are the same as the Basic Model3.2.3The Foundation of ModelThe same as the Basic Model except that:We only consider one variable(p)in the function model of the basic model ;In this model,we consider two varibles(p and µ)of the function model.3.2.4Solution and Result1)SolutionIf there are 10,000users,the number of regular hexagons that we need is at least ,thus according to the the Principle of maximum flow of Graph Theory,the 10000111.1190=result that we draw needed to be extended further.When the side length of the figure is equal to 7Figure 7regular hexagons,there are 127regular hexagons (Fig 7).Assuming the side length of a regular hexagon is ,then the area of a regular hexagon is a .The area of regular hexagons is equal to a circlewhose radiusis 22a =1000090R.Then according to the formula below:.221000090a R π=We can get.9.5858R a =Mapping with MATLAB as below (Fig 8):Figure 82）Improve the model appropriatelyEnlarge two part of the figure above,we can get two figures below (Fig 9and Fig 10):Figure 9AREAFigure 10Look at the figure above,approximatingAREA a rectangle,then obtaining its area to getthe number of users..The length of the rectangle is approximately equal to the side length of the regular hexagon ,athe width of the rectangle is ,thus the area of AREA is ,then R ?*R awe can get the number of users in AREA is()，2**10000 2.06R a R π=????????9.5858R a =As 2.06<<10,000,2.06can be ignored ,so there is no need to set up a repeater in.There are 6suchareas(92,98,104,110,116,122)that can be ignored.At last,the number of repeaters we should set up is,1276121?=2)Get the side length of the regular hexagon of the improved modelThus we can getmile=km 40 4.1729.5858a == 1.6* 6.675a =3)Calculate the power of a repeaterAccording to the formula[]()()32.4420lg ()20lg Lfs dB d km f MHz =++We get32.4420lg 6.67520lg14592.156Los =++=32.4420lg 6.67520lg14892.334Los =++=So we get106.85-92.156=14.694106.85-92.334=14.516As the result in the basic model,we can get the conclusion the power of a repeater is from 14.694mW to 14.516mW.3.2.5Analysis of the ResultAs 10,000users are much more than 1000users,the distribution of the users is more close toevenly distribution.Thus the model is more reasonable than the basic one.More repeaters are built,the utilization of the outside regular hexagon are higher than the former one.3.2.6Strength and Weakness●Strength:The model is more reasonable than the basic one.●Weakness:Repeaters don’t cover all the area,some places may not receive signals.And thefoundation of this model is based on the evenly distribution of the users in the area,if the situation couldn’t be satisfied,the interference of signals will come out.Ⅳ.Conclusions4.1Conclusions of the problem●Generally speaking,the radius of the area that a repeater covers is4miles in our basic model.●Using the model of honeycomb-hexagon structure can maximize the use of resources,avoiding some unnecessary interference effectively.●The minimum number of repeaters necessary to accommodate1,000simultaneous users is37.The minimum number of repeaters necessary to accommodate10,000simultaneoususers is121.●A repeater's coverage radius relates to external environment such as the density of users andobstacles,and it is also determined by the power of the repeater.4.2Methods used in our models●Analysis the problem with MATLAB●the method of the coverage in Graph Theory4.3Application of our models●Choose the ideal address where we set repeater of the mobile phones.●How to irrigate reasonably in agriculture.●How to distribute the lights and the speakers in squares more reasonably.Ⅴ.Future WorkHow we will do if the area is mountainous?5.1The best position of a repeater is the top of the mountain.As the signals are line-of-sight transmission and reception.We must find a place where the signals can transmit from the repeater to users directly.So the top of the mountain is a good place.5.2In mountainous areas,we must increase the number of repeaters.There are three reasons for this problem.One reason is that there will be more obstacles in the mountainous areas. The signals will be attenuated much more quickly than they transmit in flat area.Another reason is that the signals are line-of-sight transmission and reception,we need more repeaters to satisfy this condition.Then look at Fig11and Fig12,and you will know the third reason.It can be clearly seen that hypotenuse is larger than right-angleFig11edge(R>r).Thus the radius will become smaller.In this case more repeaters are needed.Fig125.3In mountainous areas,people may mainly settle in the flat area,so the distribution of users isn’t uniform.5.4There are different altitudes in the mountainous areas.So in order to increase the rate of resources utilization,we can set up the repeaters in different altitudes.5.5However,if there are more repeaters,and some of them are on mountains,more money will be/doc/d7df31738e9951e79b8927b4.html munication companies will need a lot of money to build them,repair them when they don’t work well and so on.As a result,the communication costs will be high.What’s worse,there are places where there are many mountains but few persons. Communication companies reluctant to build repeaters there.But unexpected things often happen in these places.When people are in trouble,they couldn’t communicate well with the outside.So in my opinion,the government should take some measures to solve this problem.5.6Another new method is described as follows(Fig13):since the repeater on high mountains can beFig13Seen easily by people,so the tower which used to transmit and receive signals can be shorter.That is to say,the tower on flat areas can be a little taller..Ⅵ.References[1]YU Fei,YANG Lv-xi,"Effective cooperative scheme based on relay selection",SoutheastUniversity,Nanjing,210096,China[2]YANG Ming,ZHAO Xiao-bo,DI Wei-guo,NAN Bing-xin,"Call Admission Control Policy based on Microcellular",College of Electical and Electronic Engineering,Shijiazhuang Railway Institute,Shijiazhuang Heibei050043,China[3]TIAN Zhisheng,"Analysis of Mechanism of CTCSS Modulation",Shenzhen HYT Co,Shenzhen,518057,China[4]SHANGGUAN Shi-qing,XIN Hao-ran,"Mathematical Modeling in Bass Station Site Selectionwith Lingo Software",China University of Mining And Technology SRES,Xuzhou;Shandong Finance Institute,Jinan Shandon,250014[5]Leif J.Harcke,Kenneth S.Dueker,and David B.Leeson,"Frequency Coordination in the AmateurRadio Emergency ServiceⅦ.AppendixWe use MATLAB to get these pictures,the code is as follows:1-clc;clear all;2-r=1;3-rc=0.7;4-figure;5-axis square6-hold on;7-A=pi/3*[0:6];8-aa=linspace(0,pi*2,80);9-plot(r*exp(i*A),'k','linewidth',2);10-g1=fill(real(r*exp(i*A)),imag(r*exp(i*A)),'k');11-set(g1,'FaceColor',[1,0.5,0])12-g2=fill(real(rc*exp(i*aa)),imag(rc*exp(i*aa)),'k');13-set(g2,'FaceColor',[1,0.5,0],'edgecolor',[1,0.5,0],'EraseMode','x0r')14-text(0,0,'1','fontsize',10);15-Z=0;16-At=pi/6;17-RA=-pi/2;18-N=1;At=-pi/2-pi/3*[0:6];19-for k=1:2;20-Z=Z+sqrt(3)*r*exp(i*pi/6);21-for pp=1:6;22-for p=1:k;23-N=N+1;24-zp=Z+r*exp(i*A);25-zr=Z+rc*exp(i*aa);26-g1=fill(real(zp),imag(zp),'k');27-set(g1,'FaceColor',[1,0.5,0],'edgecolor',[1,0,0]);28-g2=fill(real(zr),imag(zr),'k');29-set(g2,'FaceColor',[1,0.5,0],'edgecolor',[1,0.5,0],'EraseMode',xor';30-text(real(Z),imag(Z),num2str(N),'fontsize',10);31-Z=Z+sqrt(3)*r*exp(i*At(pp));32-end33-end34-end35-ezplot('x^2+y^2=25',[-5,5]);%This is the circular flat area of radius40miles radius 36-xlim([-6,6]*r) 37-ylim([-6.1,6.1]*r)38-axis off;Then change number19”for k=1:2;”to“for k=1:3;”,then we get another picture:Change the original programme number19“for k=1:2;”to“for k=1:4;”,then we get another picture:。

历年美赛数学建模优秀论文大全

2008国际大学生数学建模比赛参赛作品---------WHO所属成员国卫生系统绩效评估作品名称：Less Resources, more outcomes参赛单位：重庆大学参赛时间：2008年2月15日至19日指导老师：何仁斌参赛队员：舒强机械工程学院05级罗双才自动化学院05级黎璨计算机学院05级ContentLess Resources, More Outcomes (4)1. Summary (4)2. Introduction (5)3. Key Terminology (5)4. Choosing output metrics for measuring health care system (5)4.1 Goals of Health Care System (6)4.2 Characteristics of a good health care system (6)4.3 Output metrics for measuring health care system (6)5. Determining the weight of the metrics and data processing (8)5.1 Weights from statistical data (8)5.2 Data processing (9)6. Input and Output of Health Care System (9)6.1 Aspects of Input (10)6.2 Aspects of Output (11)7. Evaluation System I : Absolute Effectiveness of HCS (11)7.1Background (11)7.2Assumptions (11)7.3Two approaches for evaluation (11)1. Approach A : Weighted Average Evaluation Based Model (11)2. Approach B: Fuzzy Comprehensive Evaluation Based Model [19][20] (12)7.4 Applying the Evaluation of Absolute Effectiveness Method (14)8. Evaluation system II: Relative Effectiveness of HCS (16)8.1 Only output doesn’t work (16)8.2 Assumptions (16)8.3 Constructing the Model (16)8.4 Applying the Evaluation of Relative Effectiveness Method (17)9. EAE VS ERE: which is better? (17)9.1 USA VS Norway (18)9.2 USA VS Pakistan (18)10. Less Resources, more outcomes (19)10.1Multiple Logistic Regression Model (19)10.1.1 Output as function of Input (19)10.1.2Assumptions (19)10.1.3Constructing the model (19)10.1.4. Estimation of parameters (20)10.1.5How the six metrics influence the outcomes? (20)10.2 Taking USA into consideration (22)10.2.1Assumptions (22)10.2.2 Allocation Coefficient (22)10.3 Scenario 1: Less expenditure to achieve the same goal (24)10.3.1 Objective function： (24)10.3.2 Constraints (25)10.3.3 Optimization model 1 (25)10.3.4 Solutions of the model (25)10.4. Scenario2: More outcomes with the same expenditure (26)10.4.1Objective function (26)10.4.2Constraints (26)10.4.3 Optimization model 2 (26)10.4.4Solutions to the model (27)15. Strengths and Weaknesses (27)Strengths (27)Weaknesses (27)16. References (28)Less Resources, More Outcomes1. SummaryIn this paper, we regard the health care system (HCS) as a system with input and output, representing total expenditure on health and its goal attainment respectively. Our goal is to minimize the total expenditure on health to archive the same or maximize the attainment under given expenditure.First, five output metrics and six input metrics are specified. Output metrics are overall level of health, distribution of health in the population,etc. Input metrics are physician density per 1000 population, private prepaid plans as % private expenditure on health, etc.Second, to evaluate the effectiveness of HCS, two evaluation systems are employed in this paper:●Evaluation of Absolute Effectiveness(EAE)This evaluation system only deals with the output of HCS，and we define Absolute Total Score (ATS) to quantify the effectiveness. During the evaluation process, weighted average sum of the five output metrics is defined as ATS, and the fuzzy theory is also employed to help assess HCS.●Evaluation of Relative Effectiveness(ERE)This evaluation system deals with the output as well as its input, and also we define Relative Total Score (RTS) to quantify the effectiveness. The measurement to ATS is units of output produced by unit of input.Applying the two kinds of evaluation system to evaluate HCS of 34 countries (USA included), we can find some countries which rank in a higher position in EAE get a relatively lower rank in ERE, such as Norway and USA, indicating that their HCS should have been able to archive more under their current resources .Therefore, taking USA into consideration, we try to explore how the input influences the output and archive the goal: less input, more output. Then three models are constructed to our goal:●Multiple Logistic RegressionWe model the output as function of input by the logistic equation. In more detains, we model ATS (output) as the function of total expenditure on health system. By curve fitting, we estimate the parameters in logistic equation, and statistical test presents us a satisfactory result.●Linear Optimization Model on minimizing the total expenditure on healthWe try to minimize the total expenditure and at the same time archive the same, that is to get a ATS of 0.8116. We employ software to solve the model, and by the analysis of the results. We cut it to 2023.2 billion dollars, compared to the original data 2109.8 billion dollars.●Linear Optimization Model on maximizing the attainment. We try to maximize the attainment (absolute total score) under the same total expenditure in2007.And we optimize the ATS to 0.8823, compared to the original data 0.8116.Finally, we discuss strengths and weaknesses of our models and make necessary recommendations to the policy-makers。

美国大学生数学建模大赛模拟1论文

又实际中影响生猪存栏量和猪肉价格的因素复杂多变因此我们的预测模型只能进行短期预成艳聂锐灰色预测模型gm11在江苏省能源消费量预测中的应用中国矿业大学管理学院江苏徐州221116我国生猪生产的波动性研究辽宁农业职业技术学院辽宁营口1150093盛建华程功鹏金喜新任广智高腾云李晓峰河南省猪肉价格变化规律及其预测研究河南农业大学学报1315生猪年末存栏量及猪肉价格周期性波动研究附录
生猪年末存栏量及猪肉价格周期性波动研究
摘要
本文是关于时间序列数据的预测探究的，我们利用灰色系统理论和自己建立的模拟函数对生猪的年末存栏量和 36 个大中城市的猪肉价格进行了模拟和预测。在对第一问生猪年末存栏量的预测中，由于 1997 年以前的统计数据不真实，从而导致可用信息贫乏，据此特点我们采用了灰色系统理论中的 GM(1,1)模型。通过对已知数据和生猪养殖业的分析我们发现生猪养殖随供求关系的变化呈周期性的波动，波动周期为 3 至 4 年，又鉴于我国经济社会和人民生活水平的不断发展，生猪养殖业总体上呈现上升趋势。GM(1,1)并无对周期性变化数据的预测能力，但对单调趋势的小信息量数据有较好的预测能力，据此我们剔除数据中的波动成分，即选取原始序列时每隔 3 年取一点，得出 2010 的生猪存栏量为 44770.6 （万头）。最后对预测结果的检验发现其与原数据的偏差不会超过 3.03%，从而保证了预测结果的可靠性。利用生猪年末存栏量随时间增长但呈现周期性波动的特性，我们还建立了与之适应的模拟函数模型，模拟函数包括由一次函数表征的增长部分、由正弦函数表征的波动部分和由正态函数表征的冲击部分（如 03 年和 06 年生猪养殖业遭到了重大变故，反应在存栏量上有很大波动）。以最小二乘法确定此模拟函数的各项参数，进而预测出 2010 年末的生猪存栏量为 46813 万头。由于模拟函数是在机理分析的基础上建立的，所以很好的吻合了已知数据，从而对未来数据的预测也就有了保证。两种预测模型预测结果还是有较大差异的，但我们认为机理分析在预测中是更有效的方式，所以更倾向于把 46813 万头作为最终预测结果。第二问要求对 36 个大中城市的猪肉价格做出预测。通过对猪肉市场和已有数据的分析研究，我们发现猪肉价格在总体上仍然呈现 3 到 4 年为一个周期的波动，而在每一年中，猪肉价格受节假日、经济波动和病疫等诸多因素的影响，呈现小幅度短周期的波动情况。鉴于此变化规律，我们构造的模拟函数包括表示初始价格的常数、表示总体波动的周期为 4 年的正弦函数和表示众多繁杂因素影响的傅里叶级数（出于计算考虑取级数前 50 项）。通过最小二乘法确定各参数，得出的模拟函数对已知数据完全贴合。在模型检验中，我们只用部分已知数据来确定模拟函数，发现模拟结果与未使用的数据也有很好的吻合程度。为了进一步检验模型我们查阅了新的猪肉价格信息发现其与预测价格有相近的走势。这些都验证了模拟方法的正确性和模拟函数的有效性。

数模美赛08年A题

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 networkand give the change 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 levelmodel is mainly based 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 termαto 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’sshoreline to make 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 next 50 years.. We find the 17 cities or areas and 15 airports which are severely impacted 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.ContentIntroduction (4)Background (4)Our work (4)Study object: Ice cap in Greenland (4)Modeling the sea level (5)Analysis of Florida (5)Assumption (5)Model Ⅰ:Temperature Prediction (6)Grey Prediction Model: (6)Neural Network prediction Model: (6)Model Ⅱ:Melting ice and the rise of sea level (6)Model the rise of sea level (7)Heat from rise of temperature: (7)Mass of melting ice: (8)Design of Algorithms: (9)Model Results: Sea level will elevate by 10 centimeters in 50 years (10)Validation of our model: (11)Model ⅢAnalysis: The effects towards Florida (13)Major Cities Analysis (14)Miami (14)Tampa (15)Cape Coral (15)Key West (16)Other Impacts in Florida: (16)Recommendations to coastal Florida: (18)Judgments (18)Strengths (18)Weaknesses (19)Reference (20)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,especially some 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 studyobject.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. Ithas been threatening some islands and coastal cities. Over the next18 years, about two thirds among 544 American towns will be twiceas 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 ignore the other floating ice in the Northern Polar.●The increment of sea water from melting will flow over the oceansuniformly●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 sealevelModel 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=ρiV i=m i ∆x=V woV 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 thermal cycling system will reach to a same temperature. So, we assume that the temperature of the whole ice capwill 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.Wedivide 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 wo=m i(k)ρwS 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 computerprogram 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 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:∆x(10)=0.9874 cm∆x(20)=2.5125 cm∆x(30)=4.6222 cm∆x(40)=7.3674 cm∆x(50)= 10.8037 cmValidation 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) [15ikipedia]. 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 floodwaters.●Prepare for flooding: Complete supervisory control system.Guarantee that citizen can be evacuated in time.●Reduce carbon emission: More carbon emission means highertemperature, and then lead to rising of sea level. So encouragecitizen to live a low-carbon life.●Warn local citizen: Propagate relative knowledge of sea level risingto improve citizen’s sense of self-protection when facing natural disaster.JudgmentsStrengths●We use Neural Network to predict temperature in future with a largeamount of reliable data. So our prediction of temperature in future is accurate relatively.●Our model can predict the sea level rising in different conditions,such as different temperature.●Our model is relatively simple so it will take a little time to simulate.We can easily get the result.Weaknesses●We ignore the areal variation about depth, salinity and temperature ofthe sea for simplifying 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, 09:48AM. 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。

2008年数学建模C题

2r 40 66.67 秒 v 2 0 .6 1 1 ） 20.71 秒 0.6 1.2
（ 2 1 ）（ “扫角” ： t 2 20
“转向” ： t 3 直走+扫角 66.67 20.71 87.38 秒 5.1.2
800 800 网格
类似的，队列搜索时也会出现以上两种搜索方式。其中“转向”又分为两组情况，如图 4：
因为队列在移动过程中形成一条宽为 800 m 的搜索带，所以我们可以将整个搜索区域划分 14 9 个 800 800 的小正方形。搜索路线图如下：
图 2 搜索路线图
-4-
注：1 号为图中粗线，2 号为图中细线 4.2 问题二的分析
为了加快速度，搜索队伍有 50 人，拥有 3 台卫星电话，分成 3 组进行搜索。我们可以充分利用第一问的搜索方案，将 50 人分为 20， 20， 10 这三个小组。 20 人的小组直接可以在第一问中 800 800 的网格中制定搜索路线，10 人小组则在 400 400 的网格中制定搜索路线。 4.1.4 是否要增加队员
40m 就能将
一个 40 40 的网格搜索完全，不留死角。 “转向” ：这种搜索方式是在队列变动搜索方向是采用的，但是按照 90 角改变方向时会留下一个死角。所以搜索队员在每一次“转向”时都要进行“扫角” 。
图3
搜索方式示意图
-5-
下面我们具体计算一下两种搜索方式所用的时间： “直穿” ： t1
【摘要】：本文针对地震发生后地面搜索特定区域的最佳路径为方向，以搜索
整个区域所用时间最少的角度来研究。在小队搜索完整个区域后，以小队的用时最少为目标建立优化模型。对问题一，首先分析各个影响因素，找出其关键因素---路径最短，从中找到出发点，即尽量走直线、不留死角。搜索队员一字排开，组长在第十一位，每个队员间隔 40 m ，横向搜索，形成一条搜索宽度为 800 m 的搜索带。根据队员的搜索半径是 20 m ，我们巧妙的将整个搜索区域划分为 50400 个 40 40 的网格。在网格中制定搜索路线，直观明了。对问题二，利用问题一中的解题思路，让三组队员分区域搜索，在分配三组人员的问题上利用问题一中的 20 人一组的地毯式搜索方法，经过合理的推断把 50 名队员分成 20，20，10 三个小组，然后算出搜索完整个区域所用的时间。

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。

Take a bath for mainlandAbstract：全世界约有三分之一的人口，包括许多大城市位于居海岸线60km的范围内。

在美国，相对海平面上升100cm，按现在经济发展水平和价格估算，仅保护发达地区与滨海旅游区所需费用和欠发达地区的受淹损失，累计经济损失就将高达2700-4500亿美元。

由全球气候变暖所导致的海平面上升已经被世界各国所关注，海平面上升可能带来的影响包括淹没低地、加剧海岸侵蚀、增加风暴潮的发生频率、盐水入侵等。

本文主要研究由于全球气温升高造成的北极冰帽融化对大陆的影响。

以佛罗CO排放量上升引起气温升高，气温升高导里达州为例，用灰色模型对近几十年2致冰川融化引起海平面上升，以及海平面上升造成的陆地面积减少分别建立灰色模型、线性模型。

就陆地面积减少而言预测了今后50年陆地面积因海平面上升的减少量。

Keywords:气温升高，海平面上升，灰色模型，线性拟合一、引言CO、CH、N O等温室气体含量急剧增大量研究表明，人类活动造成的大气中24X加所引起的气候与环境效应，将对下一个世纪人类的生存与发展构成极大威胁，其中最严重的威胁之一是气候变暖导致的冰川融化使全球性的海平面的加速上升。

在过去100年中，全球海平面平均已上升了10-20cm，上升速率约为1-2mm/a，大多数研究者认为，随着全球变暖，下一个世纪海平面上升速度将明显加快，年上升速率可能将达过去100年来平均上升速率的3-8倍，在地壳显著沉降地区，其相对海平面上升速率将高于全球平均值。

而在沿海地区，因受地质等多种因素影响，其地壳垂直沉降运动频率远大于上升运动频率。

再加上人为大量使用地下资源加剧的地面下沉，从而使世界各地的相对海平面呈明显的加速上升趋势。

自1978年以来，北极地带冰帽正以每10年减少3%至4%的速度消融。

在过去20年间，北极地带冰帽缩小了大约100万平方公里。

佛罗里达州（英文：State of Florida）是美国南部的一个州，亦属于墨西哥湾沿岸地区，是美国人口第四多的州。

本文主要研究由于全球气温升高造成的北极冰帽融化对大陆的影响，以佛罗里达州为例，为佛罗里达海岸今后50年，每十年由于融化造成的影响建立模型。

二、问题重现研究一下由于全球气温升高造成的北极冰帽融化对大陆的影响。

具体来说，为佛罗里达海岸今后50年，每十年由于融化造成的影响建立模型，要特别注意大型都市区。

提出一些适当的回应来处理这些问题。

对所使用的数据的详细讨论是答案的一个重要部分。

三、假设条件1、全球气候变暖是由温室效应引起的，温室效应是由全球二氧化碳的排放引起的；2、一定时期内没有重大灾害发生；3、人类活动对自然造成影响无太大波动；4、冰川融化是平均融化；5、城市的海拔高度是平滑递增；四、模型的建立与求解近100多年以来，全球平均气温经历了冷暖-冷暖两次波动，总体为上升趋势。

进入二十世纪八十年代，全球气温明显上升，1982～1992年全球平均气温比100年前上升了0.48℃，导致全球变暖的主要原因是人类活动排放出大量的CO等多种温室气体。

全球变暖的后果，会使全球降水量重新分配，冰川和冻土消融，海平2面上升等。

下图为1850～2000年全球平均温度、全球平均海平面高度和北半球积雪面积变化趋势图，图1-全球温度、海平面和北半球积雪变化CO的排放量决定了未来温度上升的速率，尽管其他气体的排放量温室气体2CO的过量也会对升温和降温有一定的影响，但整个地球上的温度升高，主要是2排放和累积造成的。

温度的升高导致冰川融化而使海平面升高。

模型一：2CO 排放量的估计灰色模型有严格的理论基础，最大的优点是实用，用灰色模型预测的结果比较稳定，不仅适用于大量数据的预测，在数据量较少时预测结果依然较准确。

已知2003～2009年2CO 含量（单位：ppm ），以2003年为基准第一年，用灰色模型预测得到2CO 含量与时间的关系：0.00473932c()83671.445679041.5234t CO e =-用得到的模型对2010以后50年2CO 含量进行预测，见下表：年份2010 2020 2030 2040 2050 2060 2CO 含量（单位：ppm ）389.9995 408.9277428.7745449.5846469.1758491.9468模型二：海平面平均高度与时间的模型我们采用的数据为佛罗里达州的key west 观测点1926年到2000年的观测数据。

随着温室效应的加剧，全球的气温越来越高，冰川的进一步融化导致海平面越来越高，如图1所示海平面高度呈上升趋势。

图2-海平面深度随季度变化通过观测，我们认为海平面的平均高度与时间之间呈线性关系，通过Matlab 线性拟合得到海平面高度与时间的拟合函数：0.0023t-2.9846h =其中：h 表示海平面高度，t 表示时间年份。

表1对拟合函数模型进行了验证，拟合得到的海平面高度与实际的高度相比差值均在0.1以下，残差小于0.5，说明了模型的可信度，可以应用于实际的预测。

根据上述得到的模型，佛罗里达海岸今后50年，每隔十年海平面平均高度预测数据如表2所示。

模型三：海平面上升造成的影响海平面上升的影响，是指海平面上升引起的生态、环境与资源的变化及其所造成的经济损失。

目前，研究较多的是海平面加速上升最敏感地区（如地势低平的海岸平原和大河口三角洲地区等）的海岸灾害和低洼地洪涝灾害，海平面上升将加剧这些灾害，造成更大的损失。

其他方面的影响研究较之灾害影响研究要少得多。

一般来讲，海平面加速上升影响研究内容主要包括自然环境影响、社会经济影响以及区域综合评价与对策研究三部分。

现就此三部分内容对佛罗里达州作简要分析，并提出相应的处理对策。

3.1自然环境影响（1）生态影响：主要研究海平面加速上升对潮间带生态系统、湿地生态系统和河口生态系统的组成、结构和功能的影响。

如海平面上升引起的潮间带生物与环境及其相互关系的变化；各类湿地生态系统的演替；可口潮流、盐淡水混合等环境要素的变化及其对河口生物种类的组成和分布的影响等。

（2）环境影响：主要为海平面加速上升对海洋与海岸带环境的物理、化学过程的影响。

如海平面上升引起的沿海潮汐、波浪、海流等水环境要素的变化。

（3）灾害影响：海平面加速上升影响最明显、危害最严重的是对沿海各种自然灾害演变趋势的影响。

主要研究海平面上升引起的海岸侵蚀、风暴潮与洪水、盐水入侵以及低洼地洪、涝等各种自然灾害的强度和频度变化。

3.2社会经济影响佛罗里达州是美国人口第四多的州，人口之多有较高的生产力，海平面加速上升引起的各种自然环境要素变化，均不同程度直接或间接影响沿海地区社会经济发展和人民生活。

具体影响主要有：（1）对资源及其利用的影响：海平面上升对沿海地区蕴藏的土地、水、旅游、生物和矿产等资源数量、质量及开发利用条件的影响。

如海岸侵蚀、潮滩湿地及沿海低地淹没对土地、旅游和生物资源的影响，盐水入侵对供水的影响等。

（2）对社会发展的影响：海平面上升给社会发展和人民生活带来的各种无法用经济价值衡量的有形或无形的影响。

如增加的风险性对沿海地区投资行为、城市与工业发展方向与规模、资源开发利用强度的影响以及因旅游、休息地消失、各种资源数量减少等引起的居民生活质量下降。

（3）经济损失与预防措施价值的估算：由海平面上升引起的各种灾害等造成的直接或间接的经济损失。

3.3模型的建立以佛罗里达州陆地面积的减少量为例来说明海平面上升造成的影响，结合佛罗里达州的地形和海拔高度，大陆部分割补法处理后可以抽象成一个四棱台（EFGH CBAD -），如下图所示，OP 为佛罗里达州海拔高度，12∠∠和分别为大陆东西和南北坡度。

由模型二可知海平面每十年上升的高度为2.3cm ，则大陆每十年减少的垂直高度为2.3cm ，以2010年大陆面积为基准面（即面CBAD ），基准面数据已知，由此可算出以后每十年大陆面积的减少量。

由于拟合的海平面高度随时间的关系是线性函数，且大陆的坡度认为是不变的，因此每十年大陆面积的减少量是相等的（已知=35OP m ，225AB km =，625BC km =）。

经计算得到每十年大陆面积的减少量为187.75平方公里，截至2060年大陆面积的减少量相对2010年总比减少938.75平方公里。

图3五、海平面加速上升对佛罗里达州的影响分析5.1佛罗里达州沿岸城市简介佛罗里达洲沿海的主要城市有彭沙科拉、巴拿马城、圣彼得斯堡、迈阿密和圣奥古斯丁等著名城市。

（1）彭沙科拉：彭沙科拉是一座历史古城，她将美丽的海滩、军事战略、艺术、自然和悠久的历史尽收其中，它的多重魅力让每一个人都能在这里找到一个令自己心驰神往的目的地。

佛罗里达州立博物馆是一扇通往市中心彭沙科拉历史村落的“门户”，这里坐落着不同时期风格迥异的房屋和建筑。

彭沙科拉位于深海湾口，这一极佳的战略位置从一开始便决定了彭沙科拉成为重要军事要塞的地位。

（2）巴拿马城：巴拿马城原是印第安人的一个渔村，现分为古城、旧城和新城三部分，人口83万，是一座历史悠久的著名城市。

同世界上其他国家的首都相比较，巴拿马城是一座以商业和服务业为主的城市。

由于市区工业不多，不存在令人头痛的环境污染问题，城市上空总是蔚蓝一片，碧空万里。

（3）圣彼得斯堡：美国佛罗里达州西部城市。

临墨西哥湾的坦帕湾。

人口24.1万。

全年有360天晴日，有“阳光城”之称。

因气候温暖，美国北部和中西部退休者移居至此甚众，老龄人口多。

有小舟、用汽车牵引的旅行者的房屋、水泥等制,造业和水果、水产等加工业。

（4）迈阿密：迈阿密是国际性的大都市，在金融、商业、媒体、娱乐、艺术和国际贸易等方面拥有重要的地位，是许多公司、银行和电视台的总部所在。

迈阿密还被认为是文化的大熔炉，受庞大的拉丁美洲族群和加勒比海岛国居民的影响很大（当地居民多使用西班牙语和海地克里奥尔语），与北美洲、南美洲、中美洲以及加勒比海地区在文化和语言上关系密切，因此有时还被称为“美洲的首都”。

5.2佛罗里达洲地形和工业分布图4-佛罗里达州的经济分布5.4处理对策（1）分区对策：根据以上简介各市各地区的不同经济状况等因素因地制宜的采取防御措施，如建造人工海滩等。

（2）建立海岸信息系统CIS数据库的基本数据包括自然环境要素（如海岸类型、相对海平面上升幅度、洪涝淹没面积以及潮位、波高等等各种海洋水文要素的特征值）以及社会经济要素（如人口密度、经济密度、城市化程度、平均灾害经济损失）等。

（3）海平面上升综合影响等级评价与制图：包括确定综合评价与分区的原则与方法；评价各岸段受海平面上升影响的风险性以及海平面上升的综合影响等级；划分海平面上升综合影响分区并编制综合影响等级及分区图。

参考文献[1]林玎，安宁，孙继龙，等.大气变暖对气候状况影响的分析及预测[J].数学的实践与认识，2009，39（3）：87-92.[2]赵静.北极冰帽融化影响佛罗里达海岸问题分析.人文与科学.[3]杨桂山，朱季文.全球海平面上升影响研究的内容、方法与问题[J].地球科学进展，1993，8（3）：70-76.[4]江泽民.对中国能源问题的思考[J].上海交通大学学报，2008，42（3）：345-358.[5]任美锷.海平面研究的最近进展[J].南京大学学报（自然科学），2000，36（3）：270-279.[6]何朗，周薇，张应碧，等.未来海平面上升的预测及其影响[J].数学杂志，2008.[7]卓金武，秦健.Matlab在数学建模中的应用[M].北京：北京航空航天大学出版社，2011.4,30-31.[8]孔兰，陈晓宏，杜建，等.基于数学模型的海平面上升对咸潮上溯的影响[J].自然资源学报，2010,25（7）：1097-1103.附录附录一：灰色模型预测c=[a b]';A=[375.5,377.5,378.9,381.0,382.5,384.0,386.8];B=cumsum(A);n=length(A);for i=1:(n-1)C(i)=(B(i)+B(i+1))/2;endD=A;D(1)=[];D=D';E=[-C;ones(1,n-1)];c=inv(E*E')*E*D;c=c';a=c(1);b=c(2);F=[];F(1)=A(1);for i=2:(n+10)F(i)=(A(1)-b/a)/exp(a*(i-1))+b/a;endG=[];G(1)=A(1);for i=2:(n+10)G(i)=F(i)-F(i-1);endt1=2003:2009;t2=2003:2009;Gplot(t1,A,'*',t2,G)附录二：海平面平均高度与时间拟合程序>> x=[1926 1927192819291930193119321933193419351936193719381939194019411942194319441945194619471948194919501951195219531954196919701971197219731974197519761977197819791980 1981198219831984198519861987198819891990199119921993199419951997199819992000];>> y=[1.477914 1.5077911.4882691.5031581.5121951.4698141.5152481.5471141.5040911.5337821.5571611.574881.5344961.5447871.5146631.5483581.573091.5689991.5664891.5094591.6115051.614505 1.642872 1.581321 1.57388 1.566032 1.5791 1.651319 1.571958 1.612367 1.585661 1.588044 1.630194 1.669661 1.630793 1.643673 1.582375 1.607486 1.62669 1.613029 1.63321 1.627565 1.639148 1.62787 1.641353 1.644845 1.680172 1.655087 1.635879 1.618864 1.635689 1.700405 1.684063 1.675821 1.676809 1.702479 1.663453 1.659057 1.712288 1.691365 ];>> plot(x,y,'*') >> plot(x,y)>> polyfit(x,y,1) ans =0.0023 -2.9846。