2012美国大学生数学建模题目(英文原版加中文翻译)
美国大学生数学建模论文及其翻译31552
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Best all time college coach Summary
In order to select the “best all time college coach” in the last century fairly, We take selecting the best male basketball coach as an example, and establish the TOPSIS sort - Comprehensive Evaluation improved model based on entropy and Analytical Hierarchy Process. The model mainly analyzed such indicators as winning rate, coaching time, the time of winning the championship, the number of races and the ability to perceive .Firstly ， Analytical Hierarchy Process and Entropy are integratively utilized to determine the index weights of the selecting indicators Secondly，Standardized matrix and parameter matrix are combined to construct the weighted standardized decision matrix. Finally, we can get the college men's basketball com
美国数学建模题目2001至2012翻译
2001年A题（一）Choosing a Bicycle Wheel选择自行车车轮有不同类型的车轮可以让自行车手们用在自己的自行车上。
两种基本的车轮类型是分别用金属辐条和实体圆盘组装而成（见图1）。
辐条车轮较轻，但实体车轮更符合空气动力学原理。
对于一场公路竞赛，实体车轮从来不会用作自行车的前轮但可以用作后轮。
职业自行车手们审视竞赛路线，并且请一位识文断字的人推断应该使用哪种车轮。
选择决定是根据沿途山丘的数量和陡度，天气，风速，竞赛本身以及其他考虑作出的。
你所喜爱的参赛队的教练希望准备妥当一个较好的系统，并且对于给定的竞赛路线已经向你的参赛队索取有助于确定宜用哪种车轮的信息。
这位教练需要明确的信息来帮助作出决定，而且已经要求你的参赛队完成下面列出的各项任务。
对于每项任务都假定，同样的辐条车轮将总是装在前面，而装在后面的车轮是可以选择的。
任务1. 提供一个给出风速的表格，在这种速度下实体后轮所需要的体能少于辐条后轮。
这个表格应当包括相应于从百分之零到百分之十增量为百分之一的不同公路陡度的风速。
（公路陡度定义为一座山丘的总升高除以公路长度。
如果把山丘看作一个三角形，它的陡度是指山脚处倾角的正弦。
）一位骑手以初始速度45kph从山脚出发，他的减速度与公路陡度成正比。
对于百分之五的陡度，骑上100米车速要下降8kph左右。
任务2. 提供一个例证，说明这个表格怎样用于一条时间试验路线。
任务3. 请判明这个表格是不是一件决定车轮配置的适当工具，并且关于如何作出这个决定提出其他建议。
MCM2001B题Escaping a Hurricane's Wrath逃避飓风怒吼(一场恶风…)1999年，在Floyd飓风预报登陆之前，撤离南卡罗来纳州沿海地区的行动导致一场永垂青史的交通拥塞。
车水马龙停滞在州际公路I-26上，那是内陆上从Charleston通往该州中心Columbia相对安全处所的主要干线。
正常时轻松的两个小时驱车路要用上18个小时才能开到头。
2012年美国大学生数学建模竞赛B题特等奖文章翻译
We develop a model to schedule trips down the Big Long River. The goalComputing Along the Big Long RiverChip JacksonLucas BourneTravis PetersWesternWashington UniversityBellingham,WAAdvisor: Edoh Y. AmiranAbstractis to optimally plan boat trips of varying duration and propulsion so as tomaximize the number of trips over the six-month season.We model the process by which groups travel from campsite to campsite.Subject to the given constraints, our algorithm outputs the optimal dailyschedule for each group on the river. By studying the algorithm’s long-termbehavior, we can compute a maximum number of trips, which we define asthe river’s carrying capacity.We apply our algorithm to a case study of the Grand Canyon, which hasmany attributes in common with the Big Long River.Finally, we examine the carrying capacity’s sensitivity to changes in thedistribution of propulsion methods, distribution of trip duration, and thenumber of campsites on the river.IntroductionWe address scheduling recreational trips down the Big Long River so asto maximize the number of trips. From First Launch to Final Exit (225 mi),participants take either an oar-powered rubber raft or a motorized boat.Trips last between 6 and 18 nights, with participants camping at designatedcampsites along the river. To ensure an authentic wilderness experience,at most one group at a time may occupy a campsite. This constraint limitsthe number of possible trips during the park’s six-month season.We model the situation and then compare our results to rivers withsimilar attributes, thus verifying that our approach yields desirable results.Our model is easily adaptable to find optimal trip schedules for riversof varying length, numbers of campsites, trip durations, and boat speeds.No two groups can occupy the same campsite at the same time.Campsites are distributed uniformly along the river.Trips are scheduled during a six-month period of the year.Group trips range from 6 to 18 nights.Motorized boats travel 8 mph on average.Oar-powered rubber rafts travel 4 mph on average.There are only two types of boats: oar-powered rubber rafts and motorizedTrips begin at First Launch and end at Final Exit, 225 miles downstream.*simulates river-trip scheduling as a function of a distribution of trip*can be applied to real-world rivers with similar attributes (i.e., the Grand*is flexible enough to simulate a wide range of feasible inputs; andWhat is the carrying capacity of the riverÿhe maximum number ofHow many new groups can start a river trip on any given day?How should trips of varying length and propulsion be scheduled toDefining the Problemmaximize the number of trips possible over a six-month season?groups that can be sent down the river during its six-month season?Model OverviewWe design a model thatCanyon);lengths (either 6, 12, or 18 days), a varying distribution of propulsionspeeds, and a varying number of campsites.The model predicts the number of trips over a six-month season. It alsoanswers questions about the carrying capacity of the river, advantageousdistributions of propulsion speeds and trip lengths, how many groups canstart a river trip each day, and how to schedule trips.ConstraintsThe problem specifies the following constraints:boats.AssumptionsWe can prescribe the ratio of oar-powered river rafts to motorized boats that go onto the river each day.There can be problems if too many oar-powered boats are launched with short trip lengths.The duration of a trip is either 12 days or 18 days for oar-powered rafts, and either 6 days or 12 days for motorized boats.This simplification still allows our model to produce meaningful results while letting us compare the effect of varying trip lengths.There can only be one group per campsite per night.This agrees with the desires of the river manager.Each day, a group can only move downstream or remain in its current campsiteÿt cannot move back upstream.This restricts the flow of groups to a single direction, greatly simplifying how we can move groups from campsite to campsite.Groups can travel only between 8 a.m. and 6 p.m., a maximum of 9hours of travel per day (one hour is subtracted for breaks/lunch/etc.).This implies that per day, oar-powered rafts can travel at most 36 miles, and motorized boats at most 72 miles. This assumption allows us to determine which groups can reasonably reach a given campsite.Groups never travel farther than the distance that they can feasibly travelin a single day: 36 miles per day for oar-powered rafts and 72 miles per day for motorized boats.We ignore variables that could influence maximum daily travel distance, such as weather and river conditions.There is no way of accurately including these in the model.Campsites are distributed uniformly so that the distance between campsites is the length of the river divided by the number of campsites.We can thus represent the river as an array of equally-spaced campsites.A group must reach the end of the river on the final day of its trip:A group will not leave the river early even if able to.A group will not have a finish date past the desired trip length.This assumption fits what we believe is an important standard for theriver manager and for the quality of the trips.MethodsWe define some terms and phrases:Open campsite: Acampsite is open if there is no groupcurrently occupying it: Campsite cn is open if no group gi is assigned to cn.Moving to an open campsite: For a group gi, its campsite cn, moving to some other open campsite cm ÿ= cn is equivalent to assigning gi to the new campsite. Since a group can move only downstream, or remain at their current campsite, we must have m ÿ n.Waitlist: The waitlist for a given day is composed of the groups that are not yet on the river but will start their trip on the day when their ranking onthe waitlist and their ability to reach a campsite c includes them in theset Gc of groups that can reach campsite c, and the groups are deemed “the highest priority.” Waitlisted groups are initialized with a current campsite value of c0 (the zeroth campsite), and are assumed to have priority P = 1 until they are moved from the waitlist onto the river.Off the River: We consider the first space off of the river to be the “final campsite” cfinal, and it is always an open campsite (so that any number of groups can be assigned to it. This is consistent with the understanding that any number of groups can move off of the river in a single day.The Farthest Empty CampsiteOurscheduling algorithm uses an array as the data structure to represent the river, with each element of the array being a campsite. The algorithm begins each day by finding the open campsite c that is farthest down the river, then generates a set Gc of all groups that could potentially reach c that night. Thus,Gc = {gi | li +mi . c},where li is the groupÿs current location and mi is the maximum distance that the group can travel in one day.. The requirement that mi + li . c specifies that group gi must be able to reach campsite c in one day.. Gc can consist of groups on the river and groups on the waitlist.. If Gc = ., then we move to the next farthest empty campsite.located upstream, closer to the start of the river. The algorithm always runs from the end of the river up towards the start of the river.. IfGc ÿ= ., then the algorithm attempts tomovethe groupwith the highest priority to campsite c.The scheduling algorithm continues in this fashion until the farthestempty campsite is the zeroth campsite c0. At this point, every group that was able to move on the river that day has been moved to a campsite, and we start the algorithm again to simulate the next day.PriorityOnce a set Gc has been formed for a specific campsite c, the algorithm must decide which group to move to that campsite. The priority Pi is a measure of how far ahead or behind schedule group gi is:. Pi > 1: group gi is behind schedule;. Pi < 1: group gi is ahead of schedule;. Pi = 1: group gi is precisely on schedule.We attempt to move the group with the highest priority into c.Some examples of situations that arise, and how priority is used to resolve them, are outlined in Figures 1 and 2.Priorities and Other ConsiderationsOur algorithm always tries to move the group that is the most behind schedule, to try to ensure that each group is camped on the river for aFigure 1. The scheduling algorithm has found that the farthest open campsite is Campsite 6 and Groups A, B, and C can feasibly reach it. Group B has the highest priority, so we move Group B to Campsite 6.Figure 2. As the scheduling algorithm progresses past Campsite 6, it finds that the next farthest open campsite is Campsite 5. The algorithm has calculated that Groups A and C can feasibly reach it; since PA > PC, Group A is moved to Campsite 5.number of nights equal to its predetermined trip length. However, in someinstances it may not be ideal to move the group with highest priority tothe farthest feasible open campsite. Such is the case if the group with thehighest priority is ahead of schedule (P <1).We provide the following rules for handling group priorities:?If gi is behind schedule, i.e. Pi > 1, then move gi to c, its farthest reachableopen campsite.?If gi is ahead of schedule, i.e. Pi < 1, then calculate diai, the number ofnights that the group has already been on the river times the averagedistance per day that the group should travel to be on schedule. If theresult is greater than or equal (in miles) to the location of campsite c, thenmove gi to c. Doing so amounts to moving gi only in such a way that itis no longer ahead of schedule.?Regardless of Pi, if the chosen c = cfinal, then do not move gi unless ti =di. This feature ensures that giÿ trip will not end before its designatedend date.Theonecasewhere a groupÿ priority is disregardedisshownin Figure 3.Scheduling SimulationWe now demonstrate how our model could be used to schedule rivertrips.In the following example, we assume 50 campsites along the 225-mileriver, and we introduce 4 groups to the river each day. We project the tripFigure 3. The farthest open campsite is the campsite off the river. The algorithm finds that GroupD could move there, but GroupD has tD > dD.that is, GroupD is supposed to be on the river for12 nights but so far has spent only 11.so Group D remains on the river, at some campsite between 171 and 224 inclusive.schedules of the four specific groups that we introduce to the river on day25. We choose a midseason day to demonstrate our modelÿs stability overtime. The characteristics of the four groups are:. g1: motorized, t1 = 6;. g2: oar-powered, t2 = 18;. g3: motorized, t3 = 12;. g4: oar-powered, t4 = 12.Figure 5 shows each groupÿs campsite number and priority value foreach night spent on the river. For instance, the column labeled g2 givescampsite numbers for each of the nights of g2ÿs trip. We find that each giis off the river after spending exactly ti nights camping, and that P ÿ 1as di ÿ ti, showing that as time passes our algorithm attempts to get (andkeep) groups on schedule. Figures 6 and 7 display our results graphically.These findings are consistent with the intention of our method; we see inthis small-scale simulation that our algorithm produces desirable results.Case StudyThe Grand CanyonThe Grand Canyon is an ideal case study for our model, since it sharesmany characteristics with the Big Long River. The Canyonÿs primary riverrafting stretch is 226 miles, it has 235 campsites, and it is open approximatelysix months of the year. It allows tourists to travel by motorized boat or byoar-powered river raft for a maximum of 12 or 18 days, respectively [Jalbertet al. 2006].Using the parameters of the Grand Canyon, we test our model by runninga number of simulations. We alter the number of groups placed on thewater each day, attempting to find the carrying capacity for the river.theFigure 7. Priority values of groups over the course of each trip. Values converge to P = 1 due to the algorithm’s attempt to keep groups on schedule.maximumnumber of possible trips over a six-month season. The main constraintis that each trip must last the group’s planned trip duration. Duringits summer season, the Grand Canyon typically places six new groups onthe water each day [Jalbert et al. 2006], so we use this value for our first simulation.In each simulation, we use an equal number of motorized boatsand oar-powered rafts, along with an equal distribution of trip lengths.Our model predicts the number of groups that make it off the river(completed trips), how many trips arrive past their desired end date (latetrips), and the number of groups that did not make it off the waitlist (totalleft on waitlist). These values change as we vary the number of new groupsplaced on the water each day (groups/day).Table 1 indicates that a maximum of 18 groups can be sent down theriver each day. Over the course of the six-month season, this amounts to nearly 3,000 trips. Increasing groups/day above 18 is likely to cause latetrips (some groups are still on the river when our simulation ends) and long waitlists. In Simulation 1, we send 1,080 groups down river (6 groups/day?80 days) but only 996 groups make it off; the other groups began near the end of the six-month period and did not reach the end of their trip beforethe end of the season. These groups have negligible impact on our results and we ignore them.Sensitivity Analysis of Carrying CapacityManagers of the Big Long River are faced with a similar task to that of the managers of the Grand Canyon. Therefore, by finding an optimal solutionfor the Grand Canyon, we may also have found an optimal solution forthe Big Long River. However, this optimal solution is based on two key assumptions:?Each day, we put approximately the same number of groups onto theriver; and?the river has about one campsite per mile.We can make these assumptions for the Grand Canyon because they are true for the Grand Canyon, but we do not know if they are true for the Big Long River.To deal with these unknowns,wecreate Table 3. Its values are generatedby fixing the number Y of campsites on the river and the ratio R of oarpowered rafts to motorized boats launched each day, and then increasingthe number of trips added to the river each day until the river reaches peak carrying capacity.The peak carrying capacities in Table 3 can be visualized as points ina three-dimensional space, and we can find a best-fit surface that passes (nearly) through the data points. This best-fit surface allows us to estimatethe peak carrying capacity M of the river for interpolated values. Essentially, it givesM as a function of Y and R and shows how sensitiveM is tochanges in Y and/or R. Figure 7 is a contour diagram of this surface.The ridge along the vertical line R = 1 : 1 predicts that for any givenvalue of Y between 100 and 300, the river will have an optimal value ofM when R = 1 : 1. Unfortunately, the formula for this best-fit surface is rather complex, and it doesn’t do an accurate job of extrapolating beyond the data of Table 3; so it is not a particularly useful tool for the peak carrying capacity for other values ofR. The best method to predict the peak carrying capacity is just to use our scheduling algorithm.Sensitivity Analysis of Carrying Capacity re R and DWe have treatedM as a function ofR and Y , but it is still unknown to us how M is affected by the mix of trip durations of groups on the river (D).For example, if we scheduled trips of either 6 or 12 days, how would this affect M? The river managers want to know what mix of trips of varying duration and speed will utilize the river in the best way possible.We use our scheduling algorithm to attempt to answer this question.We fix the number of campsites at 200 and determine the peak carrying capacity for values of R andD. The results of this simulation are displayed in Table 4.Table 4 is intended to address the question of what mix of trip durations and speeds will yield a maximum carrying capacity. For example: If the river managers are currently scheduling trips of length?6, 12, or 18: Capacity could be increased either by increasing R to be closer to 1:1 or by decreasing D to be closer to ? or 12.?12 or 18: Decrease D to be closer to ? or 12.?6 or 12: Increase R to be closer to 4:1.ConclusionThe river managers have asked how many more trips can be added tothe Big Long Riverÿ season. Without knowing the specifics ofhowthe river is currently being managed, we cannot give an exact answer. However, by applying our modelto a study of the GrandCanyon,wefound results which could be extrapolated to the context of the Big Long River. Specifically, the managers of the Big Long River could add approximately (3,000 - X) groups to the rafting season, where X is the current number of trips and 3,000 is the capacity predicted by our scheduling algorithm. Additionally, we modeled how certain variables are related to each other; M, D, R, and Y . River managers could refer to our figures and tables to see how they could change their current values of D, R, and Y to achieve a greater carrying capacity for the Big Long River.We also addressed scheduling campsite placement for groups moving down the Big Long River through an algorithm which uses priority values to move groups downstream in an orderly manner.Limitations and Error AnalysisCarrying Capacity OverestimationOur model has several limitations. It assumes that the capacity of theriver is constrained only by the number of campsites, the trip durations,and the transportation methods. We maximize the river’s carrying capacity, even if this means that nearly every campsite is occupied each night.This may not be ideal, potentially leading to congestion or environmental degradation of the river. Because of this, our model may overestimate the maximum number of trips possible over long periods of time. Environmental ConcernsOur case study of the Grand Canyon is evidence that our model omits variables. We are confident that the Grand Canyon could provide enough campsites for 3,000 trips over a six-month period, as predicted by our algorithm. However, since the actual figure is around 1,000 trips [Jalbert et al.2006], the error is likely due to factors outside of campsite capacity, perhaps environmental concerns.Neglect of River SpeedAnother variable that our model ignores is the speed of the river. Riverspeed increases with the depth and slope of the river channel, makingour assumption of constant maximum daily travel distance impossible [Wikipedia 2012]. When a river experiences high flow, river speeds can double, and entire campsites can end up under water [National Park Service 2008]. Again, the results of our model don’t reflect these issues. ReferencesC.U. Boulder Dept. of Applied Mathematics. n.d. Fitting a surface to scatteredx-y-z data points. /computing/Mathematica/Fit/ .Jalbert, Linda, Lenore Grover-Bullington, and Lori Crystal, et al. 2006. Colorado River management plan. 2006./grca/parkmgmt/upload/CRMPIF_s.pdf .National Park Service. 2008. Grand Canyon National Park. High flowriver permit information. /grca/naturescience/high_flow2008-permit.htm .Sullivan, Steve. 2011. Grand Canyon River Statistics Calendar Year 2010./grca/planyourvisit/upload/Calendar_Year_2010_River_Statistics.pdf .Wikipedia. 2012. River. /wiki/River .Memo to Managers of the Big Long RiverIn response to your questions regarding trip scheduling and river capacity,we are writing to inform you of our findings.Our primary accomplishment is the development of a scheduling algorithm.If implemented at Big Long River, it could advise park rangerson how to optimally schedule trips of varying length and propulsion. Theoptimal schedule will maximize the number of trips possible over the sixmonth season.Our algorithm is flexible, taking a variety of different inputs. Theseinclude the number and availability of campsites, and parameters associatedwith each tour group. Given the necessary inputs, we can output adaily schedule. In essence, our algorithm does this by using the state of theriver from the previous day. Schedules consist of campsite assignments foreach group on the river, as well those waiting to begin their trip. Given knowledge of future waitlists, our algorithm can output schedules monthsin advance, allowing managementto schedule the precise campsite locationof any group on any future date.Sparing you the mathematical details, allow us to say simply that ouralgorithm uses a priority system. It prioritizes groups who are behindschedule by allowing them to move to further campsites, and holds backgroups who are ahead of schedule. In this way, it ensures that all trips willbe completed in precisely the length of time the passenger had planned for.But scheduling is only part of what our algorithm can do. It can alsocompute a maximum number of possible trips over the six-month season.We call this the carrying capacity of the river. If we find we are below ourcarrying capacity, our algorithm can tell us how many more groups wecould be adding to the water each day. Conversely, if we are experiencingriver congestion, we can determine how many fewer groups we should beadding each day to get things running smoothly again.An interesting finding of our algorithm is how the ratio of oar-poweredriver rafts to motorized boats affects the number of trips we can send downstream. When dealing with an even distribution of trip durations (from 6 to18 days), we recommend a 1:1 ratio to maximize the river’s carrying capacity.If the distribution is skewed towards shorter trip durations, then ourmodel predicts that increasing towards a 4:1 ratio will cause the carryingcapacity to increase. If the distribution is skewed the opposite way, towards longer trip durations, then the carrying capacity of the river will always beless than in the previous two cases—so this is not recommended.Our algorithm has been thoroughly tested, and we believe that it isa powerful tool for determining the river’s carrying capacity, optimizing daily schedules, and ensuring that people will be able to complete their trip as planned while enjoying a true wilderness experience.Sincerely yours,Team 13955。
2012年 AMC8 美国数学竞赛试题 答案(英文版)
The Mathematical Association of America November 13,2012
1. Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighborhood picnic?
9. The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds? (A) 61 (B) 122 (C) 139 (D) 150 (E) 161
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(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
6. A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures 8 inches high and 10 inches wide. What is the area of the border, in square inches? (A) 36 (B) 40 (C) 64 (D) 72 (E) 88
2012美国大学生数学建模竞赛_MCM_A题
SummaryMany scholars conclude that leaf shape is highly related with the veins. Based on this theory, we assume the leaf growth in each direction satisfies a function. For the leaves in the same tree, the parameters are different; for those of separate trees, the function mode is different. Thus the shape of leaf differs from that of another. In the end of section 3, we simulate one growing period and depict the leaf shape.Through thousands of years of evolution, the leaves find various wa ys to make a full use of natural resources, including minimizing overlapping individual shadows. In order to find the main factors promoting the evolution of leaves, we analyze the distribution of adjacent leaves and the equilibrium point of photosynthesis and respiration. Besides, we also make a coronary hierarchical model and transmission model of the solar radiation to analyze the influence of the branches.As to the tree structure and the leaf shape, first we consider one species. Different tree shapes have different space which is built up by the branch quality and angle, effect light distribution, ventilation and humidity and concentration of CO2 in the tree crown. These are the factors which affect the leaf shape according to the model in section 1. Here we analyze three typical tree shapes: Small canopy shape, Open center shape and Freedom spindle shape, which can be described by BP network and fractional dimension model. We find that the factors mainly affect the function of Sthat affects the additional leaf area. Factors are assembled in different ways to create different leaf shapes. So that the relationship between leaf shape and tree profile/branching structure is proved.Finally we develop a model to calculate the leaf mass from the basic formula of . By adjusting the crown of a tree to a half ellipsoid, we first define thefunction of related factors,such as the leaf density and the effective ratio of leaf area. Then we develop the model using calculus. With this model, weapproximately evaluate the leaf mass of a middle-sized tree is 141kg.Dear editor,How much the leaves on a tree weigh is the focus of discussion all the time. Our team study on the theme following the current trend and we find something interesting in the process.The tree itself is component by many major elements. In our findings, we analyze the leaf mass with complicated ones, like leaf shape, tree structure and branch characteristics, which interlace with each other.With the theory that leaf shape is highly related with the veins, we assume theleaf growth in each direction satisfies a function . For the leaves in thesame tree, the parameters are different; for those of separate trees, the function mode is di fferent. That’s why no leaf shares the same shape. Also, we simulate one growing period and depict the leaf shape.In order to find the main factors promoting the evolution of leaves, we analyze the distribution of adjacent leaves and the equilibrium point of photosynthesis and respiration. Besides, we also make a coronary hierarchical model and transmission model of the solar radiation to analyze the influence of the branches.As to the tree structure and the leaf shape, different tree shapes have differe nt space which is built up by the branch quality and angle, effect light distribution, ventilation and humidity and concentration of CO2 in the tree crown that affect leaf shapes. Here we analyze three typical tree shapes which can be described by BP network and fractional dimension model. We find that the factors mainly affect the function of Sthat affects the additional leaf area. Factors are assembled in different ways to create different leaf shapes. So that the relationship between leaf shape and tree profile or branching structure is proved.Finally we develop the significant model to calculate the leaf mass from thebasic formula of . By adjusting the crown of a tree to a half ellipsoid,we first define the function of related factors and then we develop the model using calculus. With this model, we approximately evaluate the leaf mass of a middle-sized tree is 141kg.We are greatly appreciated that if you can take our findings into consideration. Thank you very much for your precious time for reading our letter.Yours sincerely,Team #14749Contents1. Introduction (4)2. Parameters (4)3. Leaves have their own shapes (5)3.1 Photosynthesis is important to plants (5)3.2 How leaves grow? (6)3.3 Build our model (7)3.4 A simulation of the model (10)4. Do the shapes maximize exposure? (14)4.1 The optimum solution of reducing overlapping shadows (14)4.1.1 The distribution of adjacent leaves (14)4.1.2 Equilibrium point of photosynthesis and respiration (15)4.2 The influence of the “volume” of a tree and its branches (17)4.2.1 The coronary hierarchical model (17)4.2.2 Spatial distribution model of canopy leaf area (18)4.2.3 Transmission model of the solar radiation (19)5. Is leaf shape related to tree structure? (20)5.1 The experiment for one species (21)5.2 Different tree shapes affect the leaf shapes (23)5.2.1 The light distribution in different shapes (23)5.2.2 Wind speed and humidity in the canopy (23)5.2.3 The concentration of carbon dioxide (24)5.3 Conclusion and promotion (25)6. Calculus model for leaf mass (26)6.1 How to estimate the leaf mass? (26)6.2 A simulation of the model (28)7. Strengths and Weakness (29)7.1 Strengths (29)7.2 Weaknesses (30)8. Reference (30)1. IntroductionHow much do the leaves on a tree weigh? Why do leaves have the various shapes that they have? How might one estimate the actual weight of the leaves? How might one classify leaves?We human-beings have never stopped our steps on exploring the natural world. But, as a matter of fact, the answer to those questions is still unresolved. Many scientists continue to study on this area. Recently , Dr. Benjamin Blonder (2010) achieved a new breakthrough on the venation networks and the origin of the leaf econo mics spectrum. They defined a standardized set of traits – density , distance and loopiness and developed a novel quantitative model that uses these venation traits to model leaf-level physiology .Now, it is commonly thought that there are four key leaf functional traits related to leaf economics: net carbon assimilation rate, life span, leaf mass per area ratio and nitrogen content.2. Parametersthe area a leaf grows decided by photosynthesisthe additional leaf area in one growing periodthe leaf growing obliquity Pthe total photosynthetic rate 0d R the dark respiration rate of leavesn P the net photosynthetic rateh the height of the canopyd the distance between two branchesdi the illumination intensity of scattered light from a given directionthe solar zenith angleh the truck highh the crown high13. Leaves have their own shapes3.1 Photosynthesis is important to plantsIt is widely accepted that two leaves are different, no matter where they are chosen from; even they are from the very tree. To understand how leaves grow is helpful to answer why leaves have the various shapes that they have.The canopy photosynthesis and respiration are the central parts of most biophysical crop and pasture simulation models. In most models, the acclamatory responses of protein and the environmental conditions, such as light, temperature and CO2 concentration, are concerned[1].In 1980, Farquhar et al developed a model named FvCB model to describe photosynthesis[2]:The FvCB model predicts the net assimilation rate by choosing the minimum between the Rubisco-limited net photosynthetic rate and the electron transport-limited net photosynthetic rate.Assume A n, A c, A j are the symbols for net assimilation rate, the Rubisco-limited net photosynthetic rate and the electron transport-limited net photosynthetic rate respectively, and the function can be described as:(1)(2)where and are the intercellular partial pressures of CO2 and O2,respectively, and are the Michaelis–Menten coefficients ofRubisco for CO2and O2, respectively, is the CO2compensation point inthe absence of (day respiration in andis the photosystem II electron transport rate that is used for CO2fixation and photorespiration[3][4].We apply the results of this model to build the relationship between the photosynthesis and the area a leaf grows during a period of time. It can be released as:(3)and are the area of the target leaf and the period of time it grows.is a function which can transfer the amount ofCO2into the area the leaf grows and the are parameters which affect S p. S p can be used as a constraint condition in our model.3.2 How leaves grow?As the collocation of computer hardware and software develops, people can refer to bridging biology, morphogenesis, applied mathematics and computer graphics to simulate living organisms[5], thus how to model leaves is of great challenge. In 2001, Dengler and Kang[6]brought up the thought that leaf shape is highly related to venation patterns. Recently, Runions[7] brought up a method to portray the leaf shape by analyzing venation patterns. Together with the Lindenmayer system (L-system), an advanced venation model can adjust the growth better that it solved the problem occurred in the previous model that the secondary veins are retarded.We knowleaves have various shapes.For example, leaves can be classified in to simple leaves which have an undivided blade and compound leaves whose blade is divided into two or more distinctleaflets such as the Fabaceae. As to the shape of a leaf, it may have marginal dentations of the leaf blades or not, and like a palm with various fingers or an elliptical cake. Judd et al defined a set of terms which describe the shape of leaves as follows [8]:We chose entire leaves to produce this model as a simplification. What’s more, they confirmed again that the growth of venations relates with that of the leaf.To disclose this relationship, Relative Elementary Rate of Growth (RERG) can be introduced to depict leaves growth [9]. RERG is defined as the growth rate per distance, in the definitive direction l at a point p of the growing object, yielding(4)Considered RERG , the growth patterns of leaves are also different. Roth-Nebelsick et al brought up four styles in their paper [10]:3.3 Build our modelWe chose marginal growth to build our model. Amid all above-mentioned studies, weFigure 3.1 Terms pertinent to the description of leaf shapes.Figure 3.2 A sample leaf (a) and the results of its: (b) marginal growth, (c) uniform isotropic (isogonic) growth, (d) uniform anisotropic growth, and (e) non-uniform anisotropic growth.Figure 3.3 The half of a leaf is settled in x-y plane like this with primary vein overlapping x-axis. The leaf grows in the direction of .assume that the leaf produce materials it needs to grow by photosynthesis to expand its leaf area from its border and this process is only affected by what we have discussed in the previous section about photosynthesis. The border can be infinitesimally divided into points. Set as the angle between the x-axis and thestraight line connecting the grid origin and one point on the curve,andas the growth distance in the direction of . To simplify the model, we assume that the leaf grows symmetrical. We put half of the leaf into the x-y plane and make the primary vein overlap x-axis.This is how we assume the leaf grows.In one circle of leaf growth, anything that photosynthesis provided transfers into theadditional leaf area, which can be described as:(5)while in the figure.In this case, we can simulate leaf growth thus define the leaf shape by using iterative operations the times N a leaf grow in its entire circle ①.① For instance, if the vegetative circle of a leaf is 20 weeks on average, the times of iterative operations N can beset as 20 when we calculate on a weekly basis.Figure 3.4 The curves of the adjacent growing period and their relationship.First, we pre-establish the border shape of a leaf in the x-y plane, yielding . where , the relevant satisfies:(6)(7) In the first growing period, assume , the growth distance in the direction of ,satisfies:(8)(9) It releases the relationship of the coordinates in the adjacent growing period. In this case, we can use eq.(9)to predict the new border of the leaf after one period of growth②:(10)And the average simple recursions are③:②That means, in the end of period 1.③As we both change the x coordinate and the y coordinate, in the new period, these two figures relate through those in the last period in the functions.(11)After simulate the leaf borders of the interactive periods, use definite integral ④ to settle parametersin the eq.(8) then can be calculated in each direction of , thus the exact shape of a leaf in the next period is visible.When the number of times N the leaf grows in its life circle applies above-mentioned recursions to iterate N times and the final leaf shape can be settled.By this model, we can draw conclusions about why leaves have different shapes. For the leaves on the same tree, they share the same method of expansion which can be described as the same type of function as Eq.(8). The reason why they are different, not only in a sense of big or small, is that in each growing period they acquire different amount of materials used to expand its own area. In a word, the parameters in the fixedly formed eq.(8) are different for any individual leaf on the same tree. For the leaves of different tree species, the corresponding forms of eq.(8) are dissimilar. Some are linear, some are logarithmic, some are exponential or mixtures of that, which settle the totally different expansion way of leaf, are related with the veins. On that condition, the characters can be divided by a more general concept such as entire or toothed.3.4 A simulation of the modelWe set the related parameters by ourselves to simulate the shape of a leaf and to express the model better.First we initialize the leaf shape by simulating the function of a leaf border at the④The relationship must meet eq.(5).Figure 3.5 and the recurrence relations.beginning of growing period 1 in the x-y plane. By observation, we assume that themovement of the initial leaf border satisfied:(12)Suppose the curve goes across the origin of coordinates, then the constraint conditionscan be:(13)Thus the solution to eq.(13) is:(14)By using Mathematica we calculatewhereandthe area of the half leaf is: (15)Wesettleaccording to the research by S. V . Archontoulisin et al [11] in eq.(3). On a weekly basis, theparameter. In eq.(15), we have .Figure 3.6When assume ,thus constraint condition eq.(5) becomes: (16)When eq.(8) is linear and after referring to Runions’s paper, we assume that Y -valuedecreases when X-value increases, which means the leaf grows faster at the end of theprimary vein. If the grow rate at the end of the primary vein is 0,as ,eq.(8) can be described as: (17)where b is decided by eq.(16).To settle the value of b , we calculate multiple sets of data by Excel then use a planecurve to trace them and get the approximation of b . In this method, we use grid toapproximate .Apparently,is monotone.When b is 2: The square of each square is 0.0139cm 2, and the total number of the squares in theadditional area is about 240. SoFigure 3.7When b is 2.5:Thesquare of each squareis 0.0240cm 2, and the total number of the squares in theadditional area is about 201. SoWhen b is 3:The square of each square is 0.0320cm 2, and the total number of the squares in the additional area is about 193. SoAfter comparing, we can draw a conclusion that fit eq.(16)best; accordingly, in period 1: (18)In each growing period, may be different for the amount of material produced isFigure 3.9Figure 3.8related with various factors, such as the change of relative location and CO 2 or O 2concentration, and other reasons. By using the same method, the leaf shape in period2 or other period can be generated on the basis of the previous growing period untilthe end of its life circle.What’s more, when eq.(8) is remodeled, the corresponding leaf shape can be changed.With , the following figure shows the transformationof a leaf shape in the firstgrowing period with the samesettled above and we can see that the shape will be dissected in the end.4. Do the shapes maximize exposure?4.1 The optimum solution of reducing overlapping shadows4.1.1 The distribution of adjacent leavesV ein is the foundation of the leaves. With the growing of veins, the leaves graduallyexpand around. The distribution of main and lateral veins plays an important decisiverole in the shapes of leaves. The scientists created a mathematical model which usesthree decisive factors - the relationship between the rate of photosynthesis, leaf life,carbon consumption or nitrogen consumption, to simulate the leaves’ shape. Becauseof carbon consumption is a constant for one tree, and we take the neighboring leavesin the same growth cycle to observe. So, we can only focus on one factor - the rate ofphotosynthesis.Figure 3.10The shape differs from figure 3.7-3.9, as the kind offunction of is different. In this case, it is a cubic model while a linear model in figure 3.7-3.9.Through the observation of dicotyledon, leaves on a branch will grow in a staggered way that can reduce the overlapping individual shadows of adjacent leaves and make them get more sunlight (Figure 4.1).Figure 4.1 The rotation distribution of leavesBase on the similar environment, we assume that adjacent leaves nearly have the same shape. From the perspective of looking down, the leaves grow from a point on the branch. So, we can simplify the vertical view of leaves as a circle of which the radius is the length of a vein which is represented with r. The width of the leaf is represented with w. The angle between two leaves is represented with β(Figure4.2).Figure 4.2 The vertical view of leavesThe leaves should use the space as much as possible, and for the leaf with one main vein, oval is the best choice. In general, βis between 15°and 90°. In this way, effectively reduce the direct overlapping area. According to the analysis of the first question, r and w are determined by the rate of photosynthesis and respiration. Besides, the width of leaf is becoming narrower when the main vein turns to be thinner.4.1.2 Equilibrium point of photosynthesis and respirationThe organism produced by photosynthesis firstly satisfies needs of leaf itself. Then the remaining organism delivered to the root to meet the growth needs of the tree. As we know, respiration needs to consume organism. If the light is not sufficient, organism produced by photosynthesis may no longer be able to afford the materials required for the growth of leaves. There should be an equilibrium point so as toprevent the leaf is behindhand in its circumstances.The formula [12] that describes the photosynthetic rate in response to light intensity with gradual exponential growth index can be expressed as:(19) Where P is total photosynthetic rate, m ax P is maximum photosynthetic rate ofleaves, a is initial solar energy utilization and I is photosynthetic photon quanta flux density .The respiration rate is affected by temperature ，using the formal of index to describe as follows:025102T d d R R -=⋅ (20)Where T is temperature and 0d R is the dark respiration rate of leaves, when the temperature is 25 degrees. As a model parameter, 0d R can be determined bynonlinear fitting. Therefore, the net photosynthetic rate can be expressed in the indexform as follows:(21)Where n P is the net photosynthetic rate, which does not include the concentration ofcarbon dioxide and other factors. The unit of n P is 21m ol m s μ--⋅⋅.We assume that the space for leaf growth is limited, the initial area of the leaf is 0S . At this point, the entire leaf happens to be capable of receiving sunlight. If the leaf continue to grow, some part of the leaf will be in the shadows and the area in the shadows is represented with x . Ignore the fluctuation cycle of photosynthesis and respiration, on average, the duration of photosynthesis is six hours per day . In the meanwhile, the respiration is ongoing all the time. When the area increased to S , we established an equation as follows:(22)where μ is the remaining organism created by the leaf per day . And from where 0μ=, which means the organism produced by photosynthesis has all been broken down completely in the respiration, we get the equilibrium point.4n n d S P x P R ⋅=+ (23) The proportion of the shaded area in the total area:4nn d P xS P R =+ (24)Cite an example of oak trees, we found the following data (Figure 4.3), which shows the relationship between net photosynthesis and dark respiration during 160 days. Thehabitat of the trees is affected by the semi-humid monsoon climate.Figure 4.3 The diurnal rate of net photosynthesis and dark respiration [13]In general, 217n P m ol m s μ--=⋅⋅，214d R m ol m s μ--=⋅⋅. Taking the given numbersinto the equation, we can get the result.0025xS =From the result, we can see that in the natural growth of leaves, with the weakening of photosynthesis, the leaves will naturally stop growing once they come across the blade between blocked. Therefore, the leaves can always keep overlapping individual shadows about 25%, so as to maximize exposure.4.2 The influence of the “volume” of a tree and its branches4.2.1 The coronary hierarchical modelMr. Bōken and Dr. J. Fischer (1987) found that in order to adequate lighting, leaves have different densities and the branches is distributed according to certain rules. They observed tropical plants in Miami, found that the ratio of main branch and two side branches is 1:0.94:0.87, and the angles between them are 24.4°and 36.9°. According to the computer simulation, the two angles can maximize the exposure of leaves.For simply, we use hemisphere to simulate the shape of the canopy. According to the light transmittance rate, we can divide the canopy into outer and inner two layers(Figure 4.3). The volume of the hemisphere depends on the size and distribution of branches.Figure 4.3 The coronary hierarchical modelThe angle between the branches will affect the depth of penetration of sun radiation and the distribution of leaves. Besides, thickness of the branches will affect the transfer of nutrients to the leaves.4.2.2 Spatial distribution model of canopy leaf areaAssume that the distribution of leaves is uniform in the section xz but not uniform in the y direction, as the figure 4.3. Then, we can get the formula of leaf area index (LAI) as follows [14]:/20/21(,)h d d dz x z dx LAI d -∂=⎰⎰(25)Where h is the height of the canopy , d is the distance between two branches, and (,)x z ∂ is the leaf area density function of the micro-body at the point (,)x z . x a is the function of leaf area which means the distribution of cross-section of the X direction, and Its value is a dimensionless, defined as follows: /2/21()d x d a x dx LAI d -=⎰ (26)()()x s a x d LAI C x =⋅⋅ (27)For the canopy which is not uniform in the horizontal direction, the distribution of its leaves is not entirely clear. We use s C to represent the distribution function of thedensity of leaves. s C can be expressed as a quadratic function or a Gaussiandistribution function.4.2.3 Transmission model of the solar radiationAssume that the attenuation of the solar radiation accords with the law of Beer-Lambert. The attenuation value at a point of canopy where the light arrives with a certain angle of incidence and azimuth is proportional to the length of the path, and the length can be calculated by Goudriaan function. Approximate function of the G function is shown as follows [15]：00(12)cos G G k G θ=+- (28)where k is a parameter decided by different plants.Take a micro unit in the canopy . Direct sunlight intercepted in this micro unit can be expressed in the following form:()(,)b b L dI I z G z dLAI =-Θ (29)Where b I means the direct sunlight, Θ is the solar zenith angle and L dL A I is the LAI on the path of light.Figure 4.4 Micro unit of the canopyThrough the integral and the chain rule, we can get the transmittance at the point (',')x z as follows:(,)[()]tan sin ()(',')exp()()h b b z G z a x z a z I x z I h ΘΘ∆Φ+=-⎰ (30)Average direct transmission rate:/2/21(')(',')'d b b d t z t x z dx d --=⎰(31)Different with the direct light, the scattered light in all directions is intercepted by theleaf surface from the upper hemisphere. The irradiance d dI of scattering at the point(',')x z can be expressed as follows:'cos d d b dI i t d ωθ=⋅⋅ (32) where d i is the illumination intensity of scattered light from a given direction.The transmission rate of the sun scattered light is as follows:2/2'00(',')1sin cos ()d b d I x z t d d I h ππθθθϕπ=⎰⎰ (33) Average scattering transmission rate:/2/21(')(',')'d d d d t z t x z dx d --=⎰(34)Global solar radiation reaching the canopy with a given depth z :(')()[(1)(')(')]d b d d I z I h k t z k t z ---=-+ (35)According to eq.(30) and eq.(33), the maximum depth the solar radiation can reach has a major link with h and θ. The shape of leaves have a relationship with photosynthesis. So, the h and θ of branches do have an influence on the leaves. Results showed that the light level and light utilization of high stem and open ce nter shape as well as small and sparse canopy shape were better than others ．Double canopy shape ，spindle shape and center shape took second place ，while big canopy shape had the lowest light distribution [16]．5. Is leaf shape related to tree structure?Due to internal and external factors, there are many kinds of tree shapes in the nature. For example, the apple's tree shape is semi-ellipsoidal, the willow's is hemispherical, the peach's likes a cup, the pine's likes cone and so on. The shape of their leaves varies. The leaf shape of apple is oval, pine's is needle, and the Indus's is palm. Is leaf shape related to tree shape? Even for the same species, there are many kinds of tree shapes. For instance, Small canopy shape, Open center shape, Freedom spindle shape, high stem and open center shape, Double canopy shape and so on. The sizes of leaves are different. Does the tree profile/branching structure effects the leaf shape?5.1 The experiment for one speciesTo solve this problem, we first consider the relationship of one tree species such as apple, which is semi-ellipsoidal. According to the second question, the hemispherical model is further extended to the semi-ellipsoidal model. Every tree individual shows irregularly because of natural and man-made factors, which reflects on the diversity of the crown in the canopy height direction of the changing relationship between the crown and canopy height. Figure 5.1 shows the relationship.Figure5.1 Schematic diagram of tree crown contourWhere 0h means the truck high, 1h means the crown high, 8710,,,,d d d d ⋅⋅⋅ meanscrown diameter. Tree shape with the scale change can be characterized in fractal dimension. Across the same scales, a fixed fractal dimension indicates the boundary shape's self-similarity; on different scales, the change of fractal dimension means that different processes or limiting factor has superiority. (Wiens, 1989)[17] According to the application of BP network and fractional dimension, we describe the tree shapes.[18]According to Li Guodong and Zhang Junke's work which detected and evaluated in different tree shapes of ‘Fuji’ apple. We choose three kinds of tree shapes to study; they are Small canopy shape, Open center shape and Freedom spindle shape as figure5.2 shows.。
2012年数学建模美赛B题
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Page 3 of 18
1 Introduction & Backgrounds
Tourism is the sum of relationships and the phenomena which happens in the process when people make the non resident trip in order to seek the spirit of the pleasant feeling. it is a kind of spiritual pursuit when people have met the material life. Every time when we plan to select a tourist destination, we are always considering many issues, such as: the safety, the quality, the time and so on. As the administrator of the tourism area, we hold the tourism resources, At the same time, in order to improve the safety of tourist and the quality of tourism, we must do our best to avoid too many tourists at the same time or in a given period of time to enter the scenic area. For hot spots and spots of limited opening hours, if conditions permit, improving the system to achieve the maximum utilization of resource is the problem to be solved urgently, in other words, as the managers, we must give access to more tourists in the case of quality assurance.
美国大学生数学建模竞赛试题AB题中文
A 题热水澡一个人进入浴缸洗澡放松。
浴缸的热水由一个水龙头放出。
然而浴缸不是一个可以水疗泡澡的缸，没有辅助加热系统和循环喷头，仅仅就是一个简单的盛水容器。
过一会，水温就会显著下降。
因此必须从热水龙头里面反复放水以加热水温。
浴缸的设计就是当水达到浴缸的最大容量，多余的水就会通过一个溢流口流出。
做一个有关浴缸水温的模型，从时间和地点两个方面来确定在浴缸中泡澡的人能采用的最佳策略，从而泡澡过程中能保持水温并在不浪费太多水的情况下使水温尽量接近最初的水温。
用你的模型来确定你的策略多大程度上依赖于浴缸的形状和容量，浴缸中的人的体型/体重/体温，以及这个人在浴缸中做出的动作。
如果这个人在最开始放水的时候加入了泡泡浴添加剂，这将会对你的模型结果有什么影响？要求提交一页MCM的总结，此外你的报告必须包括一页给浴缸用户看的非技术性的解释，其中描述了你的策略并解释了在泡澡过程中为什么保持平均的水温会非常困难。
B题太空垃圾地球轨道周围的小碎片的数量受到越来越多的关注。
据估计，目前大约有超过50万片太空碎片被视为是宇宙飞行器的潜在威胁并受到跟踪，这些碎片也叫轨道碎片。
2009年2月10号俄罗斯卫星科斯莫斯-2251与美国卫星iridium-33相撞的时候，这个问题在新闻媒体上就愈发受到广泛讨论。
已经提出了一些方法来清除这些碎片。
这些方法包括小型太空水流喷射器和高能量激光来瞄准具体的碎片，还有大型卫星来清扫碎片等等。
这些碎片数量和大小不一，有油漆脱离的碎片，也有废弃的卫星。
碎片高速转动使得定位清除变得困难。
建一个随时间变化的模型来确定一个最佳选择或组合的选择提供给一家私人公司让它以此为商业机遇来解决太空碎片问题。
你的模型应该包括对成本、风险、收益的定量和/或定性分析以及其他重要因素的分析。
你的模型应该既能够评估单个的选择也能够评估组合的选择，且能够探讨一些重要的”what if ”情景。
用你的模型来确定是否存在这样的机会，在经济上很有吸引力；或是根本不可能有这样的机会。
2012年 AMC8 美国数学竞赛试题+答案(英文版)
12. What is the units digit of 132012? (A) 1 (B) 3 (C) 5 (D) 7
(E) 9
13. Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?
(A) 5 : 10pm (B) 5 : 21pm (C) 5 : 41pm (D) 5 : 57pm (E) 6 : 03pm
4. Peter’s family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?
(A) 3 (B) 4 (C) 5 (D) 6 (E) 7
18. What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?
(A) 3127 (B) 3133 (C) 3137 (D) 3139 (E) 3149
22. Let R be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of R ?
数学建模美赛题目及翻译
PROBLEM A: The Keep-Right-Except-To-Pass Rule The Keep-Right-Except-To-Pass RuleIn countries where driving automobiles on the right is the rule (that is, USA, China and most other countries except for Great Britain, Australia, and some former British colonies), multi-lane freeways often employ a rule that requires drivers to drive in the right-most lane unless they are passing another vehicle, in which case they move one lane to the left, pass, and return to their former travel lane.Build and analyze a mathematical model to analyze the performance of this rule in light and heavy traffic. You may wish to examine tradeoffs between traffic flow and safety, the role of under- or over-posted speed limits (that is, speed limits that are too low or too high), and/or other factors that may not be explicitly called out in this problem statement. Is this rule effective in promoting better traffic flow? If not, suggest and analyze alternatives (to include possibly no rule of this kind at all) that might promote greater traffic flow, safety, and/or other factors that you deem important.In countries where driving automobiles on the left is the norm, argue whether or not your solution can be carried over with a simple change of orientation, or would additional requirementsbe needed.Lastly, the rule as stated above relies upon human judgment for compliance. If vehicle transportation on the same roadway was fully under the control of an intelligent system – either part ofthe road network or imbedded in the design of all vehicles using the roadway – to what extent would this change the results ofyour earlier analysis?问题A ：除非超车否则靠右行驶的交通规则在一些汽车靠右行驶的国家（比如美国，中国等等），多车道的高速公路常常遵循以下原则：司机必须在最右侧驾驶，除非他们正在超车，超车时必须先移到左侧车道在超车后再返回。
北美数学建模PROBLEMA、B、C、D题翻译-6页
PROBLEM A:The Ultimate Brownie Pan终极布朗尼潘When baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with respect to using the space in an oven. 当在一个矩形的锅热烘烤时，热量被浓缩在4个角落中，并在拐角处（以及在较小程度上在边缘处）：产品会过头。
在一个圆形盘中，热量被均匀地分布在整个外缘，并且在边缘处的产品不过头。
然而，因为大多数烤炉使用圆形平底锅，而形状是矩形的，这样是效率不高的相对于使用在烘箱中的空间。
Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes -rectangular to circular and other shapes in between. 开发一个模型来显示横跨平底锅平底锅不同形状-矩形之间的圆形和其他形状的外边缘的热量分布。
Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.1宽度与长度之比（W / L）的是矩形的烘箱。
历年美国大学生数学建模竞赛试题MCM.(翻译版)doc
1985 年美国大学生数学建模竞赛MCM 试题1985年MCM:动物种群选择合适的鱼类和哺乳动物数据准确模型。
模型动物的自然表达人口水平与环境相互作用的不同群体的环境的重要参数,然后调整账户获取表单模型符合实际的动物提取的方法。
包括任何食物或限制以外的空间限制,得到数据的支持。
考虑所涉及的各种数量的价值,收获数量和人口规模本身,为了设计一个数字量代表的整体价值收获。
找到一个收集政策的人口规模和时间优化的价值收获在很长一段时间。
检查政策优化价值在现实的环境条件。
1985年MCM B:战略储备管理钴、不产生在美国,许多行业至关重要。
(国防占17%的钴生产。
1979年)钴大部分来自非洲中部,一个政治上不稳定的地区。
1946年的战略和关键材料储备法案需要钴储备,将美国政府通过一项为期三年的战争。
建立了库存在1950年代,出售大部分在1970年代初,然后决定在1970年代末建立起来,与8540万磅。
大约一半的库存目标的储备已经在1982年收购了。
建立一个数学模型来管理储备的战略金属钴。
你需要考虑这样的问题:库存应该有多大?以什么速度应该被收购?一个合理的代价是什么金属?你也要考虑这样的问题:什么时候库存应该画下来吗?以什么速度应该是画下来吗?在金属价格是合理出售什么?它应该如何分配?有用的信息在钴政府计划在2500万年需要2500万磅的钴。
美国大约有1亿磅的钴矿床。
生产变得经济可行当价格达到22美元/磅(如发生在1981年)。
要花四年滚动操作,和thsn六百万英镑每年可以生产。
1980年,120万磅的钴回收,总消费的7%。
1986 年美国大学生数学建模竞赛MCM 试题1986年MCM A:水文数据下表给出了Z的水深度尺表面点的直角坐标X,Y在码(14数据点表省略)。
深度测量在退潮。
你的船有一个五英尺的草案。
你应该避免什么地区内的矩形(75200)X(-50、150)?1986年MCM B:Emergency-Facilities位置迄今为止,力拓的乡牧场没有自己的应急设施。
2012年 全国数学建模大赛A题获奖论文英文版English
The Evaluation of WinesSummaryThe general way is to employ a number of qualified wine critics to taste the wi nes when needing to determine the quality of the wines. But the grapes of making wines are able to influence the quality of the wines in acertain extent.In Task 1, we firstly solve the arithmetic mean to the two groups of the same s ample wine ratings of Appendix I of the wine critic Members , then prove and verif y two groups Tasting ratings results have the existence of significant differences using SPSS software by paired T test ,finally solve the second assessment wine grou p more reliable by analysis of variance method.In Task 2, by an accurate analysis of the impact of the physical and•chemic al indicators of wine grape and the quality of wine to wine grapes ,we extract the pri ncipal component that embodies the basic characteristics of the object of study, so w e can reduce redundancy, and reduce the dimension of the physical and chemical in dicators of wine grapes, which of the various samples conduct a comprehensive e valuation and ranking grapes .Red grapes into four categories on this basis, the thr ee levels of white grape.In Task 3, we analyze the correlation degree of both using the typical correlation, it is concluded that both has the very high correlation, that is, the better the quality of wine grape, the higher the quality of the wine.In Task 4,we again use SPSS software to visually show the correlation coefficient between the three study and concluded that the impact on wine quality is more than two, there are other factors not taken into account.Through the Third Schedule aromatic substances added argumentation analysis, we have confirmed the larger factors exist .Physical and chemical indicators of wine grapes and wine can not be very accurate assessment of the quality of the wine, you can consider the introduction of a sensory analysis of taste and smell. Keywords:Paired samples T-test Principal Component Analysis Canonical- correlation analysis Path AnalysisIntroductionThe general way is to employ a number of qualified wine critics to tast the wines when needing to determine the quality of the wines. First,each tasting member in the taste of the wine samples give rates in accordance with the classification index, then sum the total scores to determine the quality of the wines. Quality of wine grape has a direct bearing with the quality of the wines. the physical and chemical indicators of wine grape and the wine can reflect the quality of the wine and grape to some extent. Following issues need to be addressed :1. Analysis in Annex 1 two groups of evaluation of wine member of the evaluation results whether there were significant differences of both, which a set of results more reliable.2. According to the physical and chemical indicators of the wine grape and wine quality, how about were these wine grape classified ?3. Analyze the link between the physical and chemical indicators of wine grapes and wine.4. Analyze the physical and chemical index of the wine grape and wine to the influence on the quality of wine, and demonstrate the ability to use the physical and chemical indicators of grape and wine to evaluate the quality of the wine.The analysis of issueBackgroundThe high-quality wines are popular in 2012. It ’s seems to be urgent to study that the main raw materials - whether the quality of the red grapes and white grapes of the wine is good or bad a decisive role. Therefore analyzed the relationship between wines ’ and grapes ’ quality and physical and chemical indicators over thirty kinds of physical and chemical indicators of grape and wine.Assumptions1 Each tasting wine samples from an approximate normal distribution of the distribution of the overall ；2 Tasting members are normal senses, there is not much difference ；3 Annex all the physical and chemical indicators can be representative of the nature of the study, omission of the object of study have a significant impact on physical and chemical indicators ；Symbolic representationα： Significant parameters ；W ： Rejection region range ；m ： The number of indicator variables ； 12,,...,,m x x x ： Evaluation object ；i j a⋅ Standardized index value ； ,j j s μ： Sample mean and sample standard deviation of the j-th indicator ； R : Correlation coefficient matrix ； A : Standardized matrix; (1,2,,)i i m λ= Eigenvalue ；(1,2,,)i e i m = Eigenvectors ；1V Eigenvectors of the correlation coefficient matrix of red wine grapes; 1D Red wine grape correlation coefficient matrix eigenvalue;2V White wine grape correlation coefficient matrix of eigenvectors; 2D Eigenvalues of the correlation matrix of white wine grapes;12,,,p λλλ The characteristic value corresponding to the first, second ...... p maincomponen ；p The number of indicators in the wine ；q The number of indicators of the wine grape ；11R The coefficient matrix of the first set of variables ；22R The coefficient matrix of the second set of variables ；11'R 、22'R The correlation coefficient of the first set of variables and the second setof variables ；1Z Comprehensive evaluation function of the principal component of red grape wine grape ；2Z Comprehensive evaluation function of the principal components of the white grape wine grape ；,1,2,...,28j x j = Transverse section of physical and chemical indicators in accordance with Annex II to turn on behalf of the 27-level indicators of the total amino acids, proteins, VC, ......, as well as wine quality and wine quality ratings;,1,2,...,14,15;i y i = Wine quality, peel quality juice rate (%), respectively, in turn, said, stems ratio (%), one hundred quality / g, ear quality / g dry matter content g/100g solid acid than titratable acidity (g/l), PH value, soluble solids g/l, the reducing sugars g/L, total sugar g/L, and flavonols (mg/kg), resveratrol (mg/kg) and other physical and chemical indicators ；Model establishments and solutionsTask1：Analysis in Annex 1 two groups of evaluation of wine member of the evaluationresults whether there were significant differences of both, which a set of results more reliable.To review the wine member of the evaluation result, significant difference and credibility evaluation calculation methods are varied ，mainly including Sensory evaluation of significant differences, based on the evaluation of the credibility of the Analytic Hierarchy Process, discriminant analysis, T value analysis, F value analysis,etc.Firstly, in accordance with the principle of the score with the same sample that 10 Tasting 'average score obtained in Schedule I of the first and second sets each red and white wine sample tasting ratings. Are listed below:two tables of very difference.Establishments of Model 11-1 For the evaluation of red wine :First of all by the data observation, it is known that on the whole, in view of the same sample wine in the first group and the second group of score difference are more prominent，therefore the relationship between the two with a Broken line vividly expressed. From the sensory ,image display greater differences in two groups Tasting set of evaluation criteria, shown in the following figure line chart:Figure 1 the overall rating of FirstSet and SecondSet for each of the red wine samplesThen, the overall rating results of the two rating wine group in a significant level 0.05α= are made a significant difference test. Firstly, each wine sample is selected from the large number of the same kind of samples wine from testing samples ,sample population can be approximated as a normal distribution . Secondly, Of all samples tested constitute 27 paired samples tested overall. Therefore we paired samples T-test two samples. The results are as follows:Table 3 The paired samples T-test the first and second Sets(The red)Inspection objectthe difference with 95% confidence interval tNPLowerlimitCeilingFirst and Second0.41569 4.66579 2.458 26 0.021In fact, P<0.05α=and 1/2,0.975,262.458 2.0555n t t α->==, the result falls into Rejection region {}1/2W t t α-=≥.Therefore the overall evaluation criteria of the members of the twogroups of wine critic has a significant difference.Which is trustworthier : The smaller the v ariance,the trustworthier the group’s evaluate ,when we study a single kind of wine sample. In that ,we compare the variance of these two groups to define the trustworthier group which have a smaller variance .Table 4 The credibility test for red wineWine sample sFirstSet ’s Variance Second’sVarianceThetrustworthiergroupWine sample sFirstSet ’s Variance Second’sVarianceThetrustworthiergroup1 236.1 736.9 一 15 770.1 372.1 二2 358.1 146 二 16 112.9 180.9 一3 412.4 276.4 二 17 792.1 82.5 二 4 972.4 371.6 二 18 424.9 452.4 一which is the better,the ratio of the two groups for 8:19;therefore the Second is trustworthier.1 -1 For the evaluation of white wine :Firstly,in a word,only studying the same sample wine,two groups are in small diference., therefore the relationship between the two with a Broken line vividly expressed.Said out as follows:Figure 2 the overall rating of FirstSet and SecondSet for each of the white wine samples Because wo use the same way to answer about the white wines’ question , There is no longer the detailed solution process .In summary,the evaluation results of both show significant differences,and the Second is trustworthier wherever in the red wines or the white wines. At the same time ,we find a standard to measure the quality of wine—the tasing ratings of the trustworthier sommeliers. Task2: According to the physical and chemical indicators of the wine grape and wine quality, how about were these wine grape classified ?Analysis of Model 2:There are a lot of the physical and chemical indicators of wine grapes that include more than 30 level indicators and some secondary indicators in Annex 2.we hope that many high correlation variables in wine grape physicochemical indicators are converted into mutually independent or uncorrelated variables,to choose only a small amount of indicators that can reflect most of the nature of the object of study. There are so-called Principal component which can be used to explain the research object indicators.2-1 The Step of using Principal Component Analisis way:(1) Standardize raw dataThere are m indicator variables that can be used to Principal Component Analisis way,namedas 12,,...,.m x x x There are all (1n or )2n evaluation objects.In there ,128,27m n ==,and 228n =( 1n represents the red wine as a research object, 2nrepresents the white wine as aresearch object ). The value of the j th indicator of the ith evaluation object is i j a ⋅,thereforewe gain the initial matrix of the object of study:1111m n n m a a A a a ⋅⋅⋅⋅⎛⎫ ⎪= ⎪ ⎪⎝⎭ ，The value of each indicator is converted to a standardized indicator value.As follows:,1,2,...,;1,2,...,,i j i i j ja ai n j m s μ⋅⋅-=== Among ：11,1,2,...,,n i i j j i a s j m n μ⋅====∑At the same time,j and j srespectively is the sample’s mean and standard deviation Correspondingly ,1,2,...,j j j jx xj m s μ-==It ’s a standardized indicator variable.（2）Calculting the Correlation coefficient matrix R: the Correlation coefficient matrix()i j m n R r ⋅⨯=11122122212m m m m m m m r r r r r r R r r r ⋅⋅⋅⋅⋅⋅⋅⋅⋅⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦ TR A A =*，among ： A represents A’s standardized matrix .By nature that R is a realsymmetric matrix （j i i j r r ⋅⋅=r ij =r ji ），therefore we only need to calculate on the upper triangle element or lower triangular elements to seek R . The Correlation coefficient matrix of the red wine grapes is R1,and the Correlation coefficient matrix of the white is R2. Specific data, see Annex 1.0 and Annex 3.0.（3）Calculating eigenvalues and eigenvectorsFirstly , we have solved the characteristic equation 0=-R I λ. We usually obtain eigenvalues (1,2,,)i i m λ= by Jacobi method , then arrange its in order of size 12,0m λλλ≥≥≥≥Secondly ，we respectively obtain the eigenvector (1,2,,)i e i m = corresponding to theeigenvalue i λ. Here is a requirement ,means i e =1,211mijj e ==∑,of which ij e represents the j-th component of the vector i e .The eigenvector V1 and eigenvalue D1 of the red wine grapes’ Correlation coefficient matrix , see Annex 2.0.The Characteristic roots of red wine grapes D1 Sequence :-0.0000 0.0000 0.0016 0.0058 0.0112 0.0156 0.0299 0.0504 0.0652 0.0826 0.1764 0.2025 0.2275 0.2350 0.3019 0.3712 0.5499 0.6844 0.7307 0.8076 0.9969 1.2228 1.5217 1.9934 2.8682 3.2615 4.7702 6.8158The eigenvector V2 and eigenvalue D2 of the white wine grapes’ Correlation coefficient matrix , see Annex 4.0.The Characteristic roots of white wine grapes D2 Sequence :-0.0000 0.0008 0.0026 0.0060 0.0250 0.0441 0.0634 0.0790 0.0953 0.1352 0.2719 0.3142 0.3156 0.3798 0.4458 0.6326 0.7267 0.8919 0.9663 1.0753 1.2937 1.4996 1.6175 1.7869 2.0894 2.9810 4.6623 5.5981(4)Calculation of the contribution rate and cumulative contribution rate of the main componentsThe contribution rate of main component i z :1(1,2,,)imkk i m λλ==∑The cumulative contribution rate :11(1,2,,)ikk mkk i m λλ===∑∑We generally take the characteristic values of 85-95% of the cumulatiive contribution rate corresponding to 1-st,2-nd,…,p-th(p≤m) main cpmponent.2-2 The results of principal component analysis :The eigenvalues 123,,,.....p λλλλwhose accumulative contribution rate of 85-95% can be generally selected as principal component parameters . According to theconclusion of the Principal Component Analysis , whenever in red wine or the white wine, the quality of wine is regarded as the first principal component. This shows that the quality of the evaluation of the grade of wine grapes wine share an important role .In addition ,we also gain:1. In the system which the red wine grapes act as study ,the effect of the principal component analysis of the first eleven characteristic roots whose cumulative contribution rate achieves 90% above is very great . Therefore we choose the first fourteen principal components []121314,,...,,y y y yto run a comprehensive evaluation. The contributions of fourteen principal component variables are the weight tobuild principal component comprehensive evaluation model of the red wine grapes , namly:345678910111213112140.116482559y +0.10243608y 0.0711931110.054346623+0.043671585y +0.035603699y +0.02884296y +0.026096522y +0.024442944y +0.019639356y +0.0.243422980.17036401325719y 894y 0.010782181y Z y y y ++++=+Therefore, we put all the objects corresponding to factors of the various principal components into the model ,to gain the comprehensive evaluation and sort results of the red wine grapes.2-3 ConclutionFor the red wine grapes , there are quite difference to various samples of grapes , therefore the red wine grape samples are divided into four grades: "Premium", "Great", "Qualified" and "Bad" .No. 26 sample is classified as "Premium" level by visible quality and particularly high comprehensive evaluation ;No. 17,24,5, and 20 are classified as "Great" level by great quality and high comprehensive evaluation ;No. 23,25,10,12,18,27,6,8,14,9,19,and sample is classified as "Qualified" level ;And , No. 15,3,13,4,21,7,11,2,22,and 16 sample are only classified as "Bad" , because their evaluateons are less than 60.For the white wine grapes,…Task3:Analyze the link between the physical and chemical indicators of wine grapes and wine.3-1 Model preparation :Introduced the idea of canonical correlation analysis In task 3 , we study the correlation relationship between 27 physical and chemical indicators of wine grapes ,10 of the red wines and 9 of the white wines ; the method which is similar to the main component is used, to respectively find the linear combination of the two sets of variables.Can make the number of variables to simplify, and can achieve the purpose of analysis correlation.3-2 Modeling steps :一、According to the purpose of the analysis to establish the original matrixAmong: p is the number of indicators in the wines, q is the wine grape number of indicators11111111p q n np n nq x x y y x x y y ⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦；二、Standardization of the original data , and calculating a correlation coefficient matrix11212122R R R R R ⎡⎤=⎢⎥⎣⎦Among: 11R ，22R respectively is the correlation coefficient matrix of the first set ofvariables and a second set of variables .三、Seeking canonical correlation coefficient and canonical variablesCalculate the eigenvalues and eigenvectors of the matrix A and B ，to gain the canonical correlation coefficient and canonical variables.Among:1111122221A R R R R --= ,1122211112B R R R R --=.四、Making canonical aorrelation analysis by using SPSS,to analysis of the link between the physical and chemical indicators of the wine grape and wine.The first step, the original data entry wine grapes and wines .As follows : X1，X2， X3 ，X4 ，X5 respectively represents Anthocyanin ,Tannin, Total phenolic , Wine total flavonoids , DPPH half inhibition volume ; Y1 ，Y2 ，Y3 …Y16 respectively represents the original data of the wine grapes of Annex II (Vertical indicators from the total amino acids ). Due to the limited space here is only part of the data is given. Detailed data see Annex Table 5.0(redpu.xls).Run in SPSS results are as follows:between the various indicators are small.If the correlation coefficient of the two indicators, two indicators may reflect the same ,and you can consider the merger.Analyze the physical and chemical index of the wine grape and wine to the influence on the quality of wine, and demonstrate the ability to use the physical and chemical indicators of grape and wine to evaluate the quality of the wine.4-1 The idea of Model 4:In order to analysis of wine grape and wine physical and chemical index to the influence on the quality of wine , we introduce linear regression theory to realize size analysis method . Using the Path Analysis on the basis of multiple regression , correlation coefficient iy r is decomposed into the direct path coefficients and indirect path coefficients .Path Analysis of the theory has been proved that any simple Correlation coefficient (iy r)between an independent variable i x and dependent variable y= Direct path (iy P)coefficient between x and y+ Indirect path coefficients of all the i x and y;Indirect path coefficients ibetween an independent variable any i x and dependent variable any y=Correlation coefficient (iy r)×Path coefficients (jy p) .Making Path Analysis process, it is generally believed that the most difficult part to calculate is the path coefficients. In fact , the path coefficients that we need to can be obtained by liner regression calculation . Then Indirect path coefficients can be multiplied by the correlation coefficient .4-2Problem-solving steps :4.2-1Data entryStart SPSS program, data input SPSS，name of each variable, andset the variable label.Wine quality is acted as the dependent variable y, and the physical and chemical indicators of the wine grape and wine including Anthocyanins, Tannins, Total Phenols, Wine Total Flavonoids, Resveratrol,Trans polydatin,Cis polydatin,Trans-resveratrol,Cis-resveratrol,DPPH half inhibition volume,and Color were the independent variables x1, x2, x3,x4,x5,x6,x7,x8,x9,x10.4.2-2Normality test on the dependent variable yFigure 3y s Standard Q-Q diagramStrengths and Weaknesses1.The strengths of model(1) The model established by the questions one to four are in strict theory based on analysis derived，comparing theoretical calculations with actual background, we have amended ,therefore it makes model more reasonable.(2) The use of paired samples T-test makes the results more reliable . This paper originality is very strong，because of the most of the models in the article derived and established strictly.(3) All calculations are used to specialized mathematical software and processing large amounts of data, such conclusions credibility is higher.We quantitative analysis many influencing factors that model involves, to make the paper more persuasive.2.The weaknesses of modelIn questionI and II ,due to the choice of the numerical inevitably produce a slight error and the complexity of factors affect ,so the calculation results inevitably produce certain error. References[1] Dongyan Chen , Dongmei Li , Shuzhong Wang , Mathematical Model [M].Beijing:Science Press,2007[2] Fengqiu Liu , Shanqiang Li , Zuobao Cao, Mathematical experiment [M].Harbin: HarbinInstitute of Technology Press,2010[3] Shisong Mao,Yiming Cheng,Xiaolong Pu, Probability theory and mathematical statisticstutorial [M].Beijing: Higher Education Press,2004[4] Shoukui Si,Xi Sun, Algorithm and application of mathematical modeling [M].Beijing:National Defence Industry Press,2011[5] Zhixing Zhang, Design and Application of the MATLAB program [M].Beijing: TsinghuaUniversity Press,2002。
2012美国大学生国际建模B题
February 14, 2012
Summary
Visitors to Big Long River can enjoy the beautiful scenic views; while the park managers should consider the boat adding number and the carrying capacity of the river and then develop the best traveling schedule to fulfill their target on the condition of keeping the visitors' satisfaction. On the basis of consulting the relative datum and documents and analyzing and studying the river tourism, on the condition of the uniform distribution of campsites and passengers unsharing the same campsite, we organically bring in the Enumeration model and Genetic Algorithm model based on multilayer coding technology to solves trip build a simulating fusion mathematical model combined by Genetic Algorithm after Enumeration, and present the fixed camping trip in stages mode, which scheduling problem theoretically. By the contrast of the computer simulation and practical examples, the efficiency and reliability of the model can be proved, which provides a best trip schedule to the managers. And the details are as follows, Firstly, the tour season can be divided into off season, peak season and off season according to the analysis of the basic schedule. Regard the trip styles as setting problem of the balls in different colors in Y boxes ( Y can be the number of the campsites) and then do the research. The available trip styles can be m days camping by motorized boats, n camping by motorized boats and l days days by oar boats; and Y = m + n + l . Taking the
2012年美国数学建模MCM题目(中英对照版)
2012 Contest ProblemsMCM PROBLEMSPROBLEM A: The Leaves of a Tree"How much do the leaves on a tree weigh?" How might one estimate the actual weight of the leaves (or for that matter any other parts of the tree)? How might one classify leaves? Build a mathematical model to describe and classify leaves. Consider and answer the following:• Why do leaves have the various shapes that they have?• Do the shapes “minimize” overlapping individual shadows that are cast, so as to maximize exposure? Does the distribution of leaves within the “volume” of the tree and its branches effect the shape?• Speaking of profiles, is leaf shape (general characteristics) related to tree profile/branching structure?• How would you estimate the leaf mass of a tree? Is there a correlation between the leaf mass and the size characteristics of the tree (height, mass, volume defined by the profile)?In addition to your one page summary sheet prepare a one page letter to an editor of a scientific journal outlining your key findings.“一棵树的叶子有多重？”怎么能估计树的叶子（或者树的任何其它部分）的实际重量？怎样对叶子进行分类？建立一个数学模型来对叶子进行描述和分类。
2012年美国大学生数学建模竞赛MCM-A题-树与树叶-一等奖论文
2012年美国大学生数学建模竞赛MCM-A题-树与树叶-一等奖论文Leaf-Mass & Leaves ClassificationFebruary 14, 2012AbstractIn this paper, we present a statistical model to estimate the leaf mass and classify various leaf shapes based on the data obtained through tree harvest and leaf removal of 21 urban trees. We first explain the diversification of leaf shapes and describe the relationship between leaf shapes and tree profile in order to define the correlation between tree profile and leaf shapes. In order to estimate leaf mass, we apply the regression analysis with four observed independent variables, which allows us to estimate the leaf-mass of a whole-tree. In order to classify leaves, we use hierarchical clustering method according to the dry leaf-mass by Ward's Method. In this method we use the standard Euclidean distance as our principle of classification. Finally we obtain a regression model with vary high coefficient of determination and a good way to classify leaves.For office use onlyT1 ________________T2 ________________T3 ________________T4 ________________ Team Control NumberProblem Chosen AFor office use only F1 ________________ F2 ________________ F3 ________________ F4 ________________2012 Mathematical Contest in Modeling (MCM) Summary Sheet(Attach a copy of this page to each copy of your solution paper.)Type a summary of your results on this page. Do not include the name of your school, advisor, or team members on this page.Dear Editors,By operating and analyzing the experimental data, we have set up an optimizedregression model to estimate leaf mass, and have classified leaves into several specificgroups depending on a reasonable criterion. More importantly, we have noticed somereasonable results.To begin with, a classic regression model is cited to uncover the ambiguousrelationship between tree forms and leaf shapes. As a matter of fact, the leaf shapesare negatively correlated to the tree forms, and the result enables to expose the truthwhy narrower leaves on lower-crown trees are comparatively prolific.Moreover, compared with other regression models, an optimal regression modelestimating leaf mass was proposed. Under the specific significance level, tree mass isclosely relevant to some tree characteristics including crown height (m), crown radius(m) and leaf mass per area of crown projection (2/m g ). Therefore, if theseparameters are measured, the actual tree mass can beestimated.In addition, hierarchical clustering method is applied in leaf classification. In this way,leaves can be reasonably classified relying on the different dry mass.Contents1 Introduction (3)1.1 Outline of Our Approach (3)2 Leaves and the Leaf Shapes (4)2.1 The Diversification of Leaf Shapes (4)2.2 The Relationship between Leaf Shapes and Tree Profile (5)2.3 Leaf Mosaic (7)3 Estimating Leaf Mass (8)3.1 Assumptions of Regression Analysis (8)3.2 Basic Regression Model (8)3.2.1 Independent and Dependent Variables (9)3.2.2 Regression Equation (9)3.2.3 Correlation Matrix and Multicollinearity (9)3.2.4 A Global Test on the Set of Independent Variables (10)3.2.5 Checking the Significance of Each Independent Variable andANOVA Table (11)3.2.6 Diagnostic Plots for the Regression Model (12)3.3 Improved Regression Model (16)3.3.1 Solution and Results Analysis (17)4 Hierarchical Clustering in Leaves (19)4.1 Data Description (19)4.2 Methodology (20)4.3 Results of Clustering (21)4.4 Conclusion (21)5 References (22)。
2012年全国大学生数学建模竞赛A题“特等奖”译文
2012年美国大学生数学建模竞赛A题题目翻译：一棵树的叶子。
一棵树的叶子有多重？怎么能估计树的叶子（或者树的任何其它部分）的实际重量？怎样对叶子进行分类？建立一个数学模型来对叶子进行描述和分类。
摘要我们构建了四个模型来研究叶的分类，叶子形状和叶片分布之间的关系，叶子的形状和树的轮廓的关联，以及一棵树的树叶总质量。
模型1处理叶的分类。
我们主要侧重于最显着的叶片的特征，即，形状。
我们创建七个几何参数以量化叶的形状。
然后，我们选择了六种常见的类型的叶子构造一个数据库。
通过计算的样品的与这些典型叶子的偏差指数，我们可以将叶子进行分类。
为了说明这个分类过程中，我们使用了枫叶作为测试。
模型2研究叶的形状和叶片分布之间的关系。
首先，我们将一棵树化为理想模型，然后介绍太阳高度的概念。
通过考虑叶片长度和节间在不同太阳高度下的关系，分析重叠的片叶阴影，我们发现树叶的形状和分布随着太阳高度进行优化以最大化接受阳光照射。
我们将该模型应用到三类试验树木。
模型3讨论树的轮廓和叶的形状之间可能存在关联。
基于叶脉和树的分支结构的相似性，我们认为，叶子的形状在二维上相似于一个树的轮廓。
采用模型1的方法，我们设置了几个参数来反映每棵树的大概形状，并通过它们的叶子将其进行比较。
在统计工具的帮助下，我们展示了一个树的轮廓与其叶子形状之间的粗略的关联。
模型4通那过给定的树的大小特征，估计了一棵树叶子的总质量。
并引入固碳率和树龄来建立叶的总质量和树的大小之间的联系。
由于单位质量的一个叶以一个恒等的速率固碳，固碳率与树龄之间是一个二次关系，并且树的年龄关系经历逻辑斯蒂增长。
介绍：我们解决四个主要子问题：•分类的叶子，•叶分布和叶片形状之间的关系，•树的轮廓和叶的形状的关系，和•一棵树叶片总质量的计算。
要解决的第一个问题，我们选择一组参数来量化叶片的形状特征和使用叶的形状作为我们的分类过程的主要标准。
对于第二个问题，由于叶子的形状影响着叶之间的重叠，我们将一片叶子直接投射到它下面的一片叶子的阴影重叠区域作为叶片分布和叶片形状之间的一个联系，我们假设叶片分布总是趋于尽量减少重叠的区域。
2012美赛ICM C题题目翻译
星际犯罪塑型（ICM）正在调查串谋犯犯罪行为。
调查是非常有信心，他们知道的几名成员的阴谋，但他们进行逮捕之前，希望能找出其他成员和领导人。
阴谋和所有可能涉嫌同谋为同一家公司在一个大办公室复杂的工作。
“公司一直快速增长，并为自己的名称，开发和销售计算机银行和信用卡公司的软件。
ICM最近发现了一个消息从一个小集组82个工人，他们认为在公司将帮助他们找到最有可能的候选人身份不明的同谋者和未知的领导人。
由于信息流量是所有的办公在该公司的工人，它很可能是一些（或许很多）在确定的传播者消息流量不涉及阴谋。
事实上，他们是一定的，他们知道有些人谁是不是在阴谋。
建模工作的目标将是确定人们在办公室复杂谁是最有可能的同谋。
一个优先列表将是理想的，因此ICM可以调查，监视之下的地方，和/或询问最有可能的候选人。
一个判别线分离从非同谋的同谋也将是有益的，以明显的分类，在每个人组。
这也将是有用的模型，如果提名的阴谋领导人检察官办公室。
在当前情况下的数据是给你的犯罪建模团队，你的上司给你以下情形（称为调查的EZ），她曾在几年前在另一座城市。
甚至虽然她是她对简易案件的工作感到非常自豪，她说，这是一个非常小的，简单的例子，但它可以帮助你了解自己的任务。
她的数据如下：她考虑为同谋的十人分别为：安妮＃，鲍勃，卡罗尔，大卫*，艾伦，弗雷德，乔治·哈利，伊内兹和JAYE＃的。
（*表示已知的同谋者，＃表示事先已知nonconspirators）28消息，她为她的案件有编号为每个主题年表的消息，她分配的基础上分析她的消息：安妮对鲍勃：你为什么今天迟到了吗？（1）鲍勃对卡罗尔：这该死的安妮总是看着我。
我是不是晚了。
（1）卡罗尔对戴夫：安娜和鲍勃再次战斗是鲍勃的迟到。
（1）戴夫对艾伦：我要看看你今天早晨。
当你能来吗？带来的预算文件。
（2）戴夫对弗雷德：我能来，随时随地今天看到你。
让我知道什么时候是一个好时机。
我应该带来的预算文件吗？（2）戴夫对乔治：我会看到你以后---说不完的话。
浅析2012年美国大学生数学建模大赛 B题
确定过夜天数与露营点
规定过夜天数与露营点 失败，全随机 失败，与浆和马达正相关（从资料），解决。
解决一切你们需要讨论和解决的问题。
沿着“大长河”露营
理解题目（美？中？） 解题方向与思路（normal？ special？） 资料与信息收集 头脑风暴（brain storm） 算法分析与模型建立 论文写作 建议与心得 Q&A
• Given the rise in popularity of river rafting, the park managers have been asked to allow more trips to travel down the river. They want to PROBLEM B: Camping determine how they might schedule an optimal mix of trips, of varying along the Big Long duration (measured in nights on the river) and propulsion (motor or River oar) that will utilize the campsites in the best way possible. In other words, how many more boat trips could be added to the Big Long River’s rafting season? The river managers have hired you to advise them on ways in which to develop the best schedule and on ways in which to determine the carrying capacity of the river, remembering that no two sets of campers can occupy the same site at the same time. In addition to your one page summary sheet, prepare a one page memo to the managers of the river describing your key findings.
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2012 MCM Problems
PROBLEM A:The Leaves of a Tree
"How much do the leaves on a tree weigh?" How might one estimate the actual weight of the leaves (or for that matter any other parts of the tree)? How might one classify leaves? Build a mathematical mode l to describe and classify leaves. Consider and answer the following:
• Why do leaves have the various shapes that they have?
• Do the shapes “minimize” overlapping individual shadows that are cast, so as to maximize exposure? Does the distribution of leaves within the “volume” of the tree and its branches effect the shape?
• Speaking of profiles, is leaf shape (general characteristics) related to tree profile/branching structure?
• How would you estimate the leaf mass of a tree? Is there a correlation between the leaf mass and the size characteristics of the tree (height, mass, volume defined by the profile)?
In addition to your one page summary sheet prepare a one page letter to an editor of a scientific journal outlining your key findings.
“多少钱树的叶子有多重？”怎么可能估计的叶子（或树为此事的任何其他部分）的实际重量？会如何分类的叶子吗？建立了一个数学模型来描述和分类的叶子。
考虑并回答下列问题：•为什么叶片有，他们有各种形状？•请勿形状的“最小化”个人投阴影重叠，以便最大限度地曝光吗？树叶树及其分支机构在“量”的分布效应的形状？说起型材，叶形（一般特征）有关的文件树/分支结构？你将如何估计树的叶质量？有叶的质量和树的大小特性（配置文件中定义的高度，质量，体积）之间的关系吗？除了你一个页面的汇总表，准备一页纸的信中列出您的主要结果的一个科学杂志的编辑.
PROBLEM B:Camping along the Big Long River
Visitors to the Big Long River (225 miles) can enjoy scenic views and exciting whi t e water rapids. The river is inaccessible to hikers, so the only way to enjoy i t is to take a river trip that requires several days of camping. River trips all start at First Launch and exi t the river at Final Exit, 225 miles downstream. Passengers take either oar- powered rubber rafts, which travel on average 4 mph or motorized boats, which travel on average 8 mph. The trips range from 6 to 18 nights of camping on the river, start to finish.. The government agency responsible for managing this river wants every trip to enjoy a wilderness experience, with minimal contact wi t h other groups of boats on the river. Currently, X trips travel down the Big Long River each year during a six month period (the rest of the year it is too cold for river trips). There are Y camp sites on the Big Long River, distributed fairly uniformly throughout the river corridor. Given the rise in popularity of river rafting, the park managers have been asked to allow more trips to travel down the river. They want to determine how they might schedule an optimal mix of trips, of varying duration (measured in nights on the river) and propulsion (motor or oar) that will utilize the campsites in the best way possible. In other words, how many more boat trips could be added to the Big Long River’s rafting season? The river managers have hired you to advise them on ways in which to develop the best schedule
and on ways in which to determine the carrying capacity of the river, remembering that no two sets of campers can occupy the same site at the same time. In addi t ion to your one page summary sheet, prepare a one page memo to the managers of the river describing your key findings.
大隆河（225公里）的游客可以欣赏优美的景色和令人兴奋的白色水流湍急。
这条河是无法去踏青，那么享受它的唯一途径是采取一河之旅，需要数天的露营。
所有河流的旅行开始首次启动和退出在最后的出口，下游225英里的河流。
乘客采取要么桨为动力的胶筏，平均4英里或机动船，其中8英里每小时平均旅行旅行。
TRIPS协定的范围从河上露营的6至18晚，开始到结束...... 政府机构，负责管理这条河要享受旷野的经验，与其他团体的河上的船最小的接触，每一个行程。
目前，X人次前往大长江每年在六个月内（在今年余下时间，实在是太寒冷的河流人次）。
有Ÿ营地相当均匀地分布在整个河流廊道，大朗河。
鉴于漂流在河的知名度上升，公园管理人员已被要求让更多人次前往顺流而下。
他们希望以确定它们如何可能的时间长短不一（在河上的晚来衡量）和推进器（马达或桨）在尽可能最好的方式，将利用该营地，安排行程的最佳组合。
换句话说，有多少可以添加到更多乘船游览大朗河的漂流季节？河经理已经聘请你，向他们就如何在其中发展最好的时间表，并就如何在确定河流的承载能力，记住，没有两套营员们可以在同一时间占据同一站点。
另外给你一个页面汇总表，准备一个页的备忘录描述你的主要结果的河流的管理者。
2012 ICM Problem
PROBLEM C: Modeling for Crime Busting
Your ICM submission should consist of a 1 page Summary Sheet and your solution cannot
exceed 20 pages for a maximum of 21 pages.
*As m odelers, you have to deal with the dat a you have and through valid assumptions decide what to do with holes, irregularities, redundancies, and errors.
Click the ti tle below to download a ZIP file containing the 2012 ICM Problem.
Modeling for Crime Busting。