2012年美国大学生数学建模C题参考论文
2012年美国数学建模最终版
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.Now that all our deciduous trees are in leaf, and roadside ditches and forest floors in the Georgia Piedmont are green with life, it is a marvelous time to take a walk and see how many different leaf shapes you can find. Many shapes are particular to a certain kind of tree, shrub, or vine; some, like that of the maple, have even been celebrated on a country's flag. There are the three jagged leaflets of poison ivy, and five of Virginia creeper; the many lobes (rounded or jagged) of the oaks; the tulip-shaped leaves of the tulip tree tree; the five-pointed star of the sweet gum; and the heart-shaped leaves of the wild yam. Some plants cannot settle upon one leaf shape, but instead have several. Leaves of the sassafras tree can be simple ovals, shaped like mittens, or have three broad, blunt lobes. As it is known to us, different countries and districts have different criterions for mercury toxicity. In our case, we adopt LD50 as the toxic criterions(LD50 is the dosage at which 50% of the humans exposed to a particular chemical will die. The LD50 for methylmercury is 50 mg/kg.). We speculate mercury toxicity has effect on the ability of eliminating mercury, therefore, we set up variable-elimination model at the basis of the first model. According to the first model, the amount of methylmercury in human body is 50 ug/kg, far less than 50 mg/kg, so we reach the conclusion that the fish consumption restrictions put forward by the reservoir advisories can protect the average adult. If the amount of methylmercury ingested increases, the amount of bioaccumulation will go up correspondingly. If 50 mg/kg is the maximum amount of methylmercury in human body, we can obtain the maximum number of fish that people consume safely permonth is 997.Keywords: methylmercury discrete dynamical system model variable-elimination modeldiscrete uniform random distribution model random-ingestion modelIntroductionIt is one of those wonders of the natural world that there is such diversity inthe shapes of leaves. As one website on leaf shapes remarks, "Plants have leaves in many different shapes - the thicker the book you refer to, the more leaf shapes they seem to find." The various classifications and permutations of shape form an arcane language, limited to a handful of botanists and elementary Montessori school students: terms such as runcinate, trifoliate, cordate, digitate, and deltoid. Although the words may be unfamilar, they describe the shapes of leaves encountered all around us: dandelion, clover, morning glory, maple, cottonwood. Beyond terms for general shape are further classifications for leaf form: toothed or untoothed, simple or compound, entire or lobed. Finally, even though one set of terms might be used to classify a leaf from a particular species of plant, some plants, particularly oaks, have leaves that vary considerably while keeping to generally the same overall shape.Problem Onediscrete dynamical system modelWhy? Why is there such diversity in leaves? What makes one leaf angular andanother rounded, one leaf wide and another narrow? These are the kinds of questions wondering children might ask, after being satisfied with a general explanation for the color of the sky and the forces causing the wind to blow. And, like most questions children generate, the answer is not an easy one. Indeed, there is no one clear explanation out there. Only a few days ago, a physics blog reported a new theoretical model that purports to explain all leaf shape variation as an incidental effect of the different patterns of veins in the leaf.Assumptions● The amount of methylmercury in fish is absorbed completely and instantly bypeople.● The elimination of mercury is proportional to the amount remaining.● People absorb fixed amount of methylmercury at fixed term per month. ● We assume the half-life of methylmercury in human body is 69.3 days. SolutionsLet 1α denote the proportion of eliminating methylmercury per month, 1β denote the accumulation proportion. As we know, methylmercury decays about 50 percent every 65 to 75 days, if no further methylmercury is ingested during that time. Consequently,111,βα=-69.3/3010.5.β=Through calculating, we get10.7408.β=L et’s define the following variables ：0ω denotes the amount of methylmercury at initial time,n denotes the number of month,n ω denotes the amount of methylmercury in human body at the moment people have just ingested the methylmercury in the month n ,1x denotes the amount of methylmercury that people ingest per month and113000.7910x ug ug =⨯=.Moreover, we assume0=0.ωThough,111,n n x ωωβ-=⋅+we get1011x ωωβ=⋅+2201111x x ωωββ=⋅+⋅+⋅⋅⋅10111111n n n x x x ωωβββ-=⋅+⋅+⋅⋅⋅+⋅+121111(1)n n n x ωβββ--=++⋅⋅⋅++⋅11111.1n n x βωβ--=- With the remaining amount of methylmercury increasing, the elimination of methylmercury is also going up. We know the amount of ingested methylmercury per mouth is a constant. Therefore, with time going by, there will be a balance between absorption and elimination. We can obtain the steady-state value of remaining methylmercury as n approaches infinity.1*1111111lim 3505.11n n n x x ug βωββ-→∞-===-- The value of n ωis shown by figure 1.Figure 1. merthylmercury completely coming from fish and ingested at fixed term per monthIf the difference of the remaining methylmercury between the month n and 1n - is less than five percent of the amount of methylmercury that people ingest per month, that is,115%.n n x ωω--<⋅Then we can getAt the same time, we can work out the time that people have taken to achieve 3380 ug is 11 months.From our model, we reach the conclusion that the maximum amount of methylmercury the average adult human will bioaccumulate in their lifetime is 3505 ug.If people ingest methylmercury every half of a month, however, the sum of methylmercury ingested per month is constant, consequently,11910405,0.86.2x ug β=== As a result, we obtain the maximun amount of methylmercury in human body is 3270ug. When the difference is within 5 %, we get the time people have taken to achieve it is 11 months.Similarly, if people ingest methylmercury per day, we get the maximum amount is 3050ug, and the time is 10 months.Revising ModelSome aspects of leaf shape have been explained based upon comparing tropical forest plants with temperate forest ones, such as those here in Georgia. A visit to a tropical rainforest, or a greenhouse full of tropical plants, will reveal that most tropical leaves are thicker than temperate ones, as well as more rounded and smooth-edged Assumptions● The amount of methylmercury in the seafood is absorbed completely andinstantly by people.● The elimination of methylmercury is proportional to the amount remaining. ● People ingest fixed amount of methylmercury from other seafood every day. ● We assume the half-life of methylmercury in human body is 69.3 days. SolutionsLet 0ωdenote the amount of methylmercury at initial time, t denote the number of days, t ω denote the remaining amount on the day t , and 2x denote the amountof methylmercury that people ingest per day. Moreover, we assume0=0.ωIn addition, we work out2x =50.4/30=1.68 ug.The proportion of remaining methylmercury each day is 2β, then69.320.5.β=Through calculating, we get20.99.β=Because of12221,1t t x βωβ--=- we obtain steady-state value of methylmercury1*2222211lim 168.11t t t x x ug βωββ-→∞-===-- If the difference of remaining methylmercury between the day t and 1t - is less than five percent of the amount of methylmercury that people ingest every day, that is,125%.t t x ωω--<⋅We have301= 160 ug.ωSo we can reach the conclusion that the maximum amount of methylmercury the average adult human will bioaccumulate from seafood is 160 ug and the time that people take to achieve the maximum is 301 days.Let 1x denote the amount of methylmercury people ingest through bass at fixedterm per month, so the amount of methylmercury an average adult accumulate on the day t is1221221if t is a positive integer and not divisible by 30if t is a positive integer and divisible by 30.t t t t x x x ωωβωωβ--=⋅+⎧⎨=⋅++⎩ The value of t ωis shown by figure 2.Figure 2. merthylmercury coming from fish and other seafood and ingested at fixed term per dayThe change of t ω reflects the change of the amount of methylmercury in humanbody. Through revising model, we can figure out the maximum amount of methylmercury the average adult human will bioaccumulate in their lifetime is 3679 ug.Problem TwoRandom-ingestion modelAlthough people consume one fish per month, the consuming time has great randomness. We speculate the randomness has effect on the bioaccumulation of methylmercury, therefore, we construct a new model.Assumptions● The amount of methylmercury in fish is absorbed completely and instantly bypeople.● The elimination of methylmercury is proportional to the amount remaining. ● People consume one fish per month, but the consuming time has randomness.We assume the half-life of methylmercury in human body is 69.3 days.Let 0L denote the amount of methylmercury at initial time, n L denote theamount of methylmercury at the moment people have just ingested methylmercury in the month n , and x denote the amount of methylmercury that people absorb each time.We assume0=0.LWe have910.x ug =We define 1β the proportion of remaining methylmercury every day.Through69.310.5,β=we can get10.99.β=Let i obey discrete uniform random distribution with maximum 30 and minimum 1 and n t denote the number of days between the day 1n i - of the month 1n - and theday n i of the month n , then we have-130-,n n n t i i =+(1)1.n t n n L L x β-=⋅+The value of n L is shown by figure 3.Figure 3. merthylmercury completely coming from fish and ingested at random per monthFigure 3 shows the amount of methylmercury in human body has a great change due to the randomness of consuming time. Through the computer simulation, if we have numberless samples, n L will achieve the maximum value.That is,4261.n L ug =Revising modelUltimately, we simply do not know why there are so many leaf shapes are out there, and how (if at all) a star-shaped leaf might better serve a sweet gum tree's needs than the compound structure of ovate leaflets of a hickory. It is humbling and perhaps even a bit comforting, though, that something so commonplace as the formsIn that situation, we have1212.30(-1)30(-1)n n n nn n L L x if n n i L L x x if n n i ββ=⋅+≠⨯+⎧⎨=⋅++=⨯+⎩ Through the computer simulation, we can get a set of data about n L shown by figure4.Figure 4. remaining merthylmercury coming from fish consumed at random per month and otherfood consumed at fixed term per dayThough the revised model, we reach the conclusion that if we have numberless samples, n L will achieve the maximum value.That is,4420.n L ug =Variable-eliminateion modelAs a matter of fact, the state of human health can affect metabolice rate so that the ability of eliminating methylmercury is not constant. We have koown the amount of methylmercury in human body will affect human health. So we can draw the conclusion that the amount of methylmercury in human body will affect the abilitity of eliminating methylmercury.Assumptions● The amount of methylmercury in fish is absorbed completely and instantly bypeople.● the elimination of methylmercury is not only proportional to the amountremaining, but also affected by the change of human health which are caused by the amount of methylmercury.● People absorb fixed amount of methylmercury at fixed term per month throughconsuming bass.● We assume the half-life of methylmercury in human body is 69.3 days.● In condition that no further methylmercury is ingested during a period of time, welet χ denote the eliminating proportion per month. We have known methylmercury decays about 50 percent every other day 5 to a turn 5 days, so we determine the half-life of methylmercury in human body is 69.3 days. Then we have69.3/301(1)0.5χ⋅-=.By calculating, we getχ=0.2592.We adopt LD50 as the toxic criterions, then we get the maximum amount of methylmercury in human body is 63.510⨯ ug.L et’s define the following variables ：0ω denotes the amount of methylmercury at initial time,n denotes the number of month,n ω denotes the amount of methylmercury in human body at the moment peoplehave just ingested the methylmercury in the month n ,n χ denotes the ability of eliminating methylmercury in the month n .γ denotes the effect on human health caused by methylmercury toxicity.1161 3.510r n n ωχχ-⎛⎫⎡⎤=⋅- ⎪⎢⎥ ⎪⨯⎣⎦⎝⎭ 1(1)n n n ωωχϕ-=⋅-+Hence, we have101(1)ωωχϕ=⋅-+20212(1)(1)(1)ωωχχϕχϕ=⋅-⋅-+⋅-+[]01233(1)...(1)(1)(1)...(1)(1)...(1)...(1)1n n n n n ωωχχϕχχχχχχ=⋅--+⋅-⋅--+--++-+ We define the value of γ is 0.5, then we get the maximum amount of maximum in human body is3567 ug, that is,*=3567 ug n ωNot taking the effect on the ability of eliminating maximum caused by methylmercury toxicity into account in model one,we obtain the maximum amount is 3510 ug. The difffference proves methylmercury toxicity has effect on eliminating methylmercury. We find out through calculating when r increases, the amount of methylmercury go up correspondingly. The reason for it is that methylmercury toxicity rises as a result of r increasing. Correspondingly, the effect on human health will increase, which is in accordance with fact.Problem ThreeAccording to the first model revised, we can get the maximum amount of bioaccumulation methylmercury is 3679 ug. We assume the average weight of an adult is 70 kg and the amount of methylmercury in human body is 53 ug/kg, far less than 50 mg/kg. Therefore, according to our model, the fish consumption restrictions put forward by the reservoir advisories can protect the average adult from reaching the LD50(LD50 is the dosage at which 50% of the humans exposed to a particular chemical will die. The LD50 for methylmercury is 50 mg/kg).We assume the lethal dosage of methylmercury is not gradually increasing. If the amount of methylmercury people ingests goes up rapidly, the bioaccumulation amount will reach to a higher value. Moreover, the value probably endangers human safety. Let LD50 be the maximum amount of methylmercury in human body, that is,*n =50 mg/kg 70 kg=3500 mg.ω⨯Let 1x denote the amount of methylmercury people ingest per month. According to the first model,1*1111111lim .11n n n x x βωββ-→∞-==-- We can figure out1 x =907.2 mg.We know the mean value of methylmercury in bass samples is 1.3 mg/kg, hence, we can obtain the maximum amount of fish that people consume safely per month is1max 698.1.3x M kg =≈ The maximum number of fish is 698/0.7=997.ConclusionSome aspects of leaf shape have been explained based upon comparing tropical forest plants with temperate forest ones, such as those here in Georgia. A visit to a tropical rainforest, or a greenhouse full of tropical plants, will reveal that most tropical leaves are thicker than temperate ones, as well as more rounded and smooth-edged (untoothed). Tropical plants retain their leaves for years, while deciduous plants in Georgia forests all keep their leaves only for one season. As a result, tropical leaves are sturdier than temperate forest ones, and therefore thicker. Thinner leaves require less energy to produce, and are more effective at gas exchange needed for photosynthesis. However, there is a cost of having thinner leaves: they are not as sturdy, particularly in areas distant from the major leaf veins (which provide structural support for the leaf). So those distant areas are simply done away with, resulting in lobed leaves like those of the white oak. The lobes (or teeth) of many leaves of Georgia plants also help to reduce wind resistance (and the damage that could result from it). Also, botanists have noticed that lobed or toothed leaves can permit sunlight to reach leaves beneath them, so perhaps the indentations help the plants filter sunlight down to leaves on their lower branches or stems.References[1] Dr.D.N.Rahni, PHD. Airborne Mercury Contamination and theNeversinkReservoir./dnabirahni/rahnidocs/Envsc/Airborne%20Mercury%20C ontamination%20and%20the%20Neversink%20Reservoir.doc[2] Hu Dong Bai Ke. Bass. /wiki%E9%B2%88%E9%B1%BC .[3] Centre for Food Safety Food and Environmental Hygiene Department TheGovernment of the Hong Kong Special Administrative Region. Mercury in Fish and Food Safety..hk/english/Programmme/programme_rafs/Programme_rafs_fc _01_19_mercury_in_fish.html.。
2012年美国数学建模竞赛
2012年美国数学建模竞赛（MCM/ICM）将于2.10（正月十九）—2.13（共4天）举行，我们邀请冯国灿教授介绍如何准备美国数学建模竞赛，内容包括:1、报名注册注意事项2、写作技巧3、数学建模知识的准备4、典型案例分析欢迎备战参赛的同学参加！时间：2011年12月21日（周三）晚上7：30地点：数学楼202【嘉宾简介】冯国灿：教授、博士生导师、中山大学数学与计算科学学院数学系副主任、广东省大学生数学建模竞赛组委会委员。
是十几年的数学建模竞赛的资深教练，经常参加全国、广东省大学生数学建模竞赛论文的评阅工作，指导数模美赛、国赛，获奖无数。
2012年美国（国际）大学生数学建模竞赛官方网站：/undergraduate/contests比赛时间：美国东部时间：2012年2月9日（星期四）下午8点-2月13日下午8点（共4天）北京时间：2012年2月10日（星期五）上午9点-2月14日上午9点农历：正月十九-正月二十三比赛之前注册报名1. 注意所有时间都是美国东部时间。
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2012年数学建模C题论文
承诺书我们仔细阅读了中国大学生数学建模竞赛的竞赛规则.我们完全明白，在竞赛开始后参赛队员不能以任何方式（包括电话、电子邮件、网上咨询等）与队外的任何人（包括指导教师）研究、讨论与赛题有关的问题。
我们知道，抄袭别人的成果是违反竞赛规则的, 如果引用别人的成果或其他公开的资料（包括网上查到的资料），必须按照规定的参考文献的表述方式在正文引用处和参考文献中明确列出。
我们郑重承诺，严格遵守竞赛规则，以保证竞赛的公正、公平性。
如有违反竞赛规则的行为，我们将受到严肃处理。
我们参赛选择的题号是（从A/B/C/D中选择一项填写）： C我们的参赛报名号为（如果赛区设置报名号的话）：所属学校（请填写完整的全名）：贵州电力职业技术学院参赛队员(打印并签名) ：1. 樊清松2. 温淑敏3. 陈慧婷指导教师或指导教师组负责人(打印并签名)：朱卫红日期： 2012 年 09 月 09日赛区评阅编号（由赛区组委会评阅前进行编号）：编号专用页赛区评阅编号（由赛区组委会评阅前进行编号）：全国统一编号（由赛区组委会送交全国前编号）：全国评阅编号（由全国组委会评阅前进行编号）：脑卒中发病环境因素分析及干预摘要为了探索脑卒中（俗称脑中风）的发病情况、分布规律以及对高危人群提出预警和干预的建议方案，分析脑卒中高危人群发病的环境因素，通过建立数学模型，进而研究脑卒中与气温、气压、相对湿度之间的关系，降低脑卒中高危人群的发病率，并帮助健康人了解自己得脑卒中的风险程度。
通过应用回归分析法、图表分析法和统计方法把发病人群职业类型人数、各年龄段人数、2007年月份发病人数、发病性别比例人数排列在工作表的列或行中，其数据可以绘制到柱形图中。
柱形图用于显示一段时间内的数据变化或显示各项数据之间的比较情况，可以直观地看出发病人群的发病情况以及分布规律。
综上所述，男性的老年人发病率较高，是潜在的脑卒中高危人群。
从职业上来看，脑卒中高危人群为60岁以上的农民和退休人员，且发病率随年龄的增大而上升。
2012美国数学建模大赛二等奖论文及格式——英文版
Dedicated Pipeline for Trip ArrangementSummaryIn the problem of camping, we should set reasonable schedule which can not only increase the utilization of campsites but also meet people's needs. Meanwhile, the carrying capacity of the river is also required. To solve the problem, this thesis will build optimization model with maximum campsite's utilization and river trips as the model's target function.The specific steps are as follows:step1: Determine the number of campsites Y. We use Computer Emulation Simulation to solve this problem by making full use of the given conditions that trips will spend6 to 18 nights on the river and the river is 225 miles long. We get 29 sets of data through programming, then curve fitting them by SPSS software. By comparing the value of sig. and adjusting R square and so on, the ideal number of the campsites is got .Step2: By using the number of the campsites 39 as well as the goal programming equation built in the first step, we get the number of river trips that are allowed to enter, namely the carrying capacity of the river.Step3: By using the campsites 39, we adjust the campsites of different camping program and then divide them into 4 kinds through clustering analysis using SPSS. Then we select representatives in various types of camping programs according to repetition rate and the average transfer rate. So we streamline the camping programs into the problem of goal programming for 6, 8, 11, 12, 16 nights.Step4: In those five camping programs, 39 campsites which will not repeat are distributed in 3 dedicated pipelines . The first line accounts for 12 campsites and can only be available for 6 or 12 nights trip. Each day, a couple of 6 nights trips are distributed, and the starting trip camps the campsites in turn according to the even number of the pipeline while the secondary trip camps in turn according to the odd number. The second pipeline accounting for 16 campsites is arranged just as the first one .Under the premise of guaranteeing the variety of camping project, trips start as a pipeline to make the total number of trips camping in this line the biggest and the utilization of the campsites maximum. There are 11 campsites in the third pipeline which are available for 6 to 11 nights trip.According to the above analysis, the carrying capacity of Dedicated Pipeline, namely C_line, is less than that of the river, namely C_river, within 180 days. the park managers need to grasp passenger flow(P) of the river in the following period(T) and calculate P/( C_line/T)The best distribution program: the best utirlization of campsites is P/( C_line/T) in one period.According to the best utilization of campsites, The best distribution program can be got.Key words: Cluster Analysis, Bus Rapid, Transit Pipeline System, Curve Fitting , Computer Emulation SimulationContentsI. Introduction (3)1.1Restatement of the problem (3)1.2 Theory knowledge introduction (3)II. Definitions and Key terms (4)1，The conditions given (4)2，Symbol definition (4)III. General Assumptions (4)IV Model Design (5)4.1Model Establishment (5)4.2 Model Solution (6)4.2.1.To determine Y (6)4.2.1 To Determine the Camping program (11)4.2.3 To find capacity of the river (15)4.2.3 Determine Dedicated Pipeline (15)4.3 Strength and Weakness............................................................. 错误！未定义书签。
2012年美赛C题论文
1.2 Methods of the Analysis
As investigators, we now know well about the 83nodes, 400 links over 21,000 words of message traffic, 15 topics (three have been deemed to be suspicious), 7 known conspirators, and 8known non-conspirators. We now call the known information about the company INTELLIGENCE. From the practice, the method which is always used in the intelligence work is a combination of qualitative reasoning and quantified reasoning, that is called Analytic Hierarchy Process (called AHP in the later part). At the same time, we also use Structural Model Analysis (SMA).
1.3 Assumptions
There is only linguistic communication between people. Body language never exists. People can talk to each other freely, without the liof known conspiractors and non-conspiractors is correct. The talk happens only between two people. Talks between 3 or over 3 members never exists.
数学建模美赛2012年C题
问题分析：首先通过阅读题目，抽象问题。
交代了问题的背景——舆情分析。
获得一些消息，消息是一群人的谈话的网络，要求从消息中找到同谋者和组织领导人。
1.仔细阅读题目，看看出题人要求你做到什么程度。
要得到的最终的结论：1. 一个优先级列表 2. 一个判别是否为同谋人的分界线要求这个模型具有一般性，而不是解决这一个特殊问题，更具有普遍性。
2.最后确定所有需要或者能够做的思路3.明确符合条件的思路4.建立简单模型5.逐渐增加复杂度，建立精细化模型6.论文完善论文读书报告Abstract首先介绍问题，然后简洁的解释自己的模型，侧重于建立模型之后得到了什么样的效果，或者结论，论文分为三个模型，这三个模型都过有简单到复杂的原则进行展开，首先模型1建立单个节点的判断方式，模型2建立节点与相邻节点共同作用确定结论，模型3则是通过整个信息网络来确定为同某者的概率，最后介绍自己模型具有很好的应用价值，可以适用于数据量庞大的事件。
Declaration of the given data对于数据量比较大的问题，这一步还是有必要的，对数据进行进一步的说明或者说是预处理。
1．对奇异数据（如果是冗余的数据）可以忽略不考虑，可是并不是所有的数据都是可以去掉的，对于这些数据就要进行数据的拟合近似进行数据总体的平滑性处理。
2．对数据应该仔细分析，一点一点认真看，把所有的噪点进行处理3．这其中可以包含对数据的假设性的近似（在你的模型需要的基础上），这就是数据的预处理这个paper可能对数据的预处理做得比较好。
是亮点。
Problem analysis and assumption这个部分写的就比较凌乱了，如果分条陈述可能更加直观一些，实话说，这个问题的假设比较困难，也就是说看起来没有什么好假设的，这个时候就应该把公理摆出来，就是公认为正确的不用用语言说明的基本假设。
假设所有的同谋者肯定至少和一个其他的同谋者进行交流，假设发出信息和受到信息具有相同的作用，没有什么区别（对于做出是否是同谋者的结论）。
2012数学建模大赛C题论文
封一答卷编号（参赛学校填写）：答卷编号（竞赛组委会填写）：论文题目：C组别：本科生参赛队员信息(必填)：姓名参赛队员1 沈倩参赛队员2 王青原参赛队员3 付新新参赛学校：黑龙江工程学院封二答卷编号（参赛学校填写）：答卷编号（竞赛组委会填写）：评阅情况（学校评阅专家填写）：学校评阅1.学校评阅2.学校评阅3.评阅情况（联赛评阅专家填写）：联赛评阅1.联赛评阅2.联赛评阅3.2012年“深圳杯”全国大学生数学建模夏令营C题：3Ｄ仿真机房建模问题分析摘要随着经济的发展、计算机的普及，人们对数据的处理越来越多。
机房的设计问题也越来越受到人们的关注，如何在满足工作的前提条件下，做到最低的消耗，成了很多公司发现商机的、创造价值的有利方向。
通过对机房设计，得到相应的实验数据，建立确定的数学模型，找到最佳的设计方案成了人们关注的焦点。
建立模型的出发点，影响因素有距空调的位置，高度，机柜摆放方式，任务量，空调送风速度。
对于第一问，根据分析附件1的数据，用MATLAB软件进行插值，绘出冷、热通道的热分布及流场分布图（共四幅），并且确定出室内最高温度位置。
对于第二问，利用附表2提供的数据，经过分析发现当固定其中某一个物理量时，其他的未知量之间会成现出特定的曲线关系。
通过MATLAB软件拟合出各个影响因素与温度之间的图像发现特定关系，通过多元非线性回归解得函数关系。
建立热分布的数学模型及算法，同时与测试案例进行比较。
对于第三问，结合前两问的结论，通过分析在不同任务量时绘制出的热分布图确定最优任务的分配方案，并且找到室内最高温度。
分析附表2中改变任务量对通道3的温度影响，从而假设实际任务量为0.8和0.5的分配方案，再通过问题二中得到论证。
对于第四问，按照《电子信息系统机房设计规范》C级要求，在任务量一定的情况下，热点温度超过规范要求时，通过调节出风槽风速或出风槽温度从而降低温度，保证服务器的健康工作。
通过多元线性回归找到热点温度与出风槽温度之间的间接关系，从而进行调节，实现任务量的合理使用和降低机房内热点温度的节能目的。
2012年数学建模国家优秀论文
承诺书我们仔细阅读了中国大学生数学建模竞赛的竞赛规则.我们完全明白，在竞赛开始后参赛队员不能以任何方式（包括电话、电子邮件、网上咨询等）与队外的任何人（包括指导教师）研究、讨论与赛题有关的问题。
我们知道，抄袭别人的成果是违反竞赛规则的, 如果引用别人的成果或其他公开的资料（包括网上查到的资料），必须按照规定的参考文献的表述方式在正文引用处和参考文献中明确列出。
我们郑重承诺，严格遵守竞赛规则，以保证竞赛的公正、公平性。
如有违反竞赛规则的行为，我们将受到严肃处理。
我们授权全国大学生数学建模竞赛组委会，可将我们的论文以任何形式进行公开展示（包括进行网上公示，在书籍、期刊和其他媒体进行正式或非正式发表等）。
我们参赛选择的题号是（从A/B/C/D中选择一项填写）：我们的参赛报名号为（如果赛区设置报名号的话）：所属学校（请填写完整的全名）：参赛队员(打印并签名) ：1.2.3.指导教师或指导教师组负责人(打印并签名)：日期：年月日赛区评阅编号（由赛区组委会评阅前进行编号）：编号专用页赛区评阅编号（由赛区组委会评阅前进行编号）：全国统一编号（由赛区组委会送交全国前编号）：全国评阅编号（由全国组委会评阅前进行编号）：太阳能小屋的设计摘要本文针对光伏建筑设计时对外表面光伏电池板优化铺设及逆变器选用优化问题，建立太阳辐射模型、多目标优化模型，并引入运筹学中松弛约束、动态规划、启发式算法、等步长探索思想求解优化模型，解决不同安装方式下（贴附、架空）光伏电池阵列最优排布并合理选择逆变器的问题，达到优化目标。
继而，在计算求得电池板最佳倾角的基础上，提出了一套合理化太阳能小屋建设方案。
光伏电池发电原理为光电效应，能量来源为太阳能。
模型I对经典太阳辐射模型进行适当改进，以求不同方位角γ和水平倾角β下倾斜平面接收的太阳辐射能量。
借助Matlab软件编程求解，得到位于大同地区的小屋朝南倾斜屋顶和东、南、西、北立面接收的年太阳辐射量分别为1564.49、594.21、1050.16、881.23、261.47（单位：kw·h/㎡）。
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For office use onlyT1________________ T2________________ T3________________ T4________________Team Control Number7013Problem ChosenCFor office use onlyF1________________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 includethe name of your school, advisor, or team members on this page.Modeling for Crime BustingYour organization, the Intergalactic Crime Modelers (ICM), is investigating a conspiracy to commit a criminal act. The investigators are highly confident they know several members of the conspiracy, but hope to identify the other members and the leaders before they make arrests. The conspirators and the possible suspected conspirators all work for the same company in a large office complex. The company has been growing fast and making a name for itself in developing and marketing computer software for banks and credit card companies. ICM has recently found a small set of messages from a group of 82 workers in the company that they believe will help them find the most likely candidates for the unidentifiedco‐conspirators and unknown leaders. Since the message traffic is for all the office workers in the company, it is very likely that some (maybe many) of the identified communicators in the message traffic are not involved in the conspiracy.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 addition to your one page summary sheet, prepare a one page memo to the managers of the river describing your key findings. Through the careful studying, we abstract the collision system into a simple harmonic motion system which is constituted by 4 parts of objects andelastic force between them. By analyzing the Simple harmonic system, we build differential equations model for the collision, and gain the ball leave speed at all locations of the hitting point. Then, we can get the location of sweet point by finding the maximum return speed of the ball. For problem 2 and 3, we change some coefficient of the model used in problem1 according to the situation, and concluded that corking a bat will slow the speed of ball, while using a bat made of aluminum will increase the ball speed.Keywords: simple harmonic motion system , differential equations model , collision systemThe Probability Model for Predicting the Area of A Serial Crime Key words: the Bayes Theorem equation; "center of mass"; crime; the Distance Decay Function; reliability; model; the geographical profile.Contents1 Introduction 31.1 Comprehension to the Problem (3)1.2 The Goal of the Model (4)1.3 The Analysis of the Problem (4)2 Notation and Definitions 43 Simplifying Assumptions 44 The Model4.1 Mark the Locations of Crime (5)4.2 Predict the Locations (7)4.3 Two Methods to find the "center of mass" (8)4.3.1 The First Method (8)4.3.2 The Result of the First Method (8)4.3.3 The Second Method (10)4.3.4 The Result of the Second Method (12)4.3.5 The Combination of the Two Results (13)5 Case Study: The Test of the Model 146 Model Evaluation 167 Executive Summary 178 Literature References 199 Additional Lists 201 IntroductionAfter getting the problem, we search the following sites for relevant information. /wiki/Peter_Sutcliffe/hiper22/sutcliffe_cf.htm/group/topic/6142810/Peter William Sutcliffe (born 2 June 1946 in Bingley, West Riding of Yorkshire) is an English serial killer who was dubbed The Yorkshire Ripper. Sutcliffe committed his first documented assault on the night of 5 July 1975 in Keighley.On 2 January 1981, Sutcliffe was stopped by the police with 24 year old prostitute Olivia Reivers in the driveway of Light Trades House, Melbourne Avenue, Broomhill, Sheffield, South Yorkshire.A police check revealed the car was fitted with false number plates and Sutcliffe was arrested for this offence and transferred to Dewsbury Police Station, Sutcliffe was convicted in 1981 of murdering 13 women and attacking several others.Details are as follows:Murder Type/Practices - Serial Killer / Sadism.Method/Weapons Used - Stabbing, Strangulation, Bashing / Ball-Pein Hammer, Knives, Claw Hammer, Hacksaw, Screwdrivers, Rope.Organization - DisorganizedMobility - Stable..Victim Vicinity - Yorkshire, Northern England.(Leeds,Bradford,Huddersfield, Manchester)Murder Time Span - July 1975 - January1981.Victim Type - Prostitutes.Victims - Wilma McCann (Died 30 Oct 1975), Emily Jackson (Died 20 Jan 1976), Irene Richardson (Died 5 Feb 1977), Patricia Atkinson (Died 23 Apr 1977), JayneMacDonald (Died 26 Jun 1977), Jean Jordan (Died 1 Oct 1977), Yvonne Pearson (Died 21 Jan 1978), Helen Rytka (Died 31 Jan 1978), Vera Millward (Died 16 May 1978), Josephine Whitaker (Died 4 Apr 1979), Barbara Leach (Died 2 Sep 1979), Marguerite Walls (Died 18 Aug 1980), Jacqueline Hill (Died 17 Nov 1980) .There are a lot of murders like this, which lead to plenty of victims everyday. We have developed a number of ways to predict the murder based on these problems in order to make the innocent people away from the suffering.1.1 Comprehension to the problemThe given problem requires us to solve serial crimes. On one hand, we have to predict the possible locations in the next stages based on the time and locations of the past crime scenes. On the other hand, estimating the reliability of our prediction is what we have to do. In addition, the given problem also asks us to accomplish additional two-page executive summary which will be read by the chief of police. After the above analysis of the problem's requirements, we understand that how to determine the locations and environment of the past crime scenes is the key factor to solve the problem. The model we have established should find a way to solve this problem. Only in this way can we make the best prediction.1.2 The goal of the modelThe goal of our model is to predict the possible locations of the crime. In order to achieve this goal, we take the following steps.Step 1: Analyzing the locations and environments of the past crime scenes.Step 2: Making use of two different schemes to generate a geographical profile. Step 3: Combining the results of the schemes and generate a useful prediction. Step 4: Estimate the reliability of our prediction and accomplish the summary.1.3 The analysis of the problemFirst, we mark the locations of murders and attacks on the map and find the murders and their attack time. Then, using longitude and latitude to represent these points, and regard them as the coordinates of these points. What's more, finding out the “center of mass” is what we have to do. We also have to use two different methods to form a geographical profile. Finally, the probability of next murdering in each region can be predicted. In addition to that, we have to test and verify our model. Thus, the problem analysis is completed.2 Notation and DefinitionsNotation meaningXi The past crime pointsYi The predicted pointsd(x,y) The distance between x and yf(d) The distance decay functionS(y) The hit score functionLon LongitudeLat LatitudeWon The latitude's actual distanceWat The longitude's actual distanceΔon The longitude differenceΔat The latitude differenced The expected valueZi The divided aream The "center of mass" point3 Simplifying Assumptions1.During a short period, the criminal's mental feature doesn't alter.2.The criminal doesn't suffer from the unexpected things, such as the accident, the illness and so on.3.Nature disasters don't appear in this region.4.The criminal has a stable residence.4 The model4.1 Mark the Locations of crimeIn order to easily make a forecast, we find out the detailed locations and time of the murders and the attacks. Then we mark them out on the map.Sutcliffe is convicted of murdering the following 13 victims:Table.1 locations and time of the crimeDate Name ofvictimAgeatdeathBody foundLocationonmap30 October 1975 WilmaMcCann28 Prince Phillip Playing Fields, Leeds 120 January 1976 EmilyJackson42 Manor Street, Sheepscar, Leeds at 25 February 1977 IreneRichardson28 Roundhay Park, Leeds at 323 April 1977 PatriciaAtkinson32 Flat 3, 9 Oak Avenue, Bradford 426 June 1977 JayneMacDonald16 Adventure playground, Reginald Street, Leeds 51 October 1977 Jean Jordan 20Allotments next to Southern Cemetery,Manchester621 January 1978 YvonnePearson21Under a discarded sofa on waste ground offArthington Street, Bradford731 January 1978 Helen Rytka 18Timber yard in Great Northern Street,Huddersfield816 May 1978 VeraMillward40 Grounds of Manchester Royal Infirmary 94 April 1979 JosephineWhitaker19 Savile Park, Halifax 102 September 1979 BarbaraLeach20 Back of 13 Ashgrove, Bradford 1120 August 1980 MargueriteWalls47Garden of a house called "Claremont", NewStreet, Farsley, Leeds1217 November 1980 JacquelineHill20Waste ground off Alma Road, Headingley，Leeds1313 locations are marked on the following map:Fig.1 The locations marked on the mapTaking into account the complexity of the locations on the map, we can use the longitude and latitude to represent these points. Taking the locations of attacks into account, we receive a total of 21 points. Then the coordinates of these 21 points can be obtained.Table.2 The coordinates of 21 locationspoint latitude longitude1 53.7996374 -1.549112 53.807214836 -1.5220355983 53.837335272 -1.495385174 53.810002329 -1.76583766945 53.81818635 -1.53370857246 53.429531854 -2.25863456737 53.800403705 -1.77133083348 53.653413662 -1.77972078329 53.460702343 -2.225418090810 53.710348595 -1.872653961211 53.780800484 -1.847376823412 53.808051104 -1.672024726913 53.822955402 -1.575529575314 53.8656925 -1.909745215 53.914662 -1.937945816 53.8375092 -1.496272617 53.793853 -1.752442218 53.7996374 -1.5491119 53.7996374 -1.5491120 53.6451179 -1.78487621 53.72438 -1.86157664.2 Predict the locationsAfter we have marked out these points on the map, we can find out the environmental conditions near the points.Table.3 The environmental conditionspoint geographical profile1 park2 business center3 near the park,residential area4 residential area5 small playground6 near the cemetery7 near the Recycle bin8 lumber yard9 hospital10 park11 residential area,private courtyard12 residential areas, gardens13 residential areas, wastelandThrough the analysis of the 13 points, we find that the majority of offenders commit crimes are in the places where there are few people and a relatively open space, such as parks, gardens, wasteland, pile wood field, the cemeteries and so on. Therefore,very few people and open place is the common point of the locations of the crime. Then we can take the park for an example. We can find the six parks that have the similar environmental conditions.Then we can find out the Longitudes and latitudes of the six places. Table.4 Six locations predictedplace longitude latitudeLister Park -1.7781543731689453 53.81083854068482Pollard Park -1.7348957061767578 53.803438755384576Victoria Park -2.3321914672851562 53.505558395542735Middleton park -1.549673080444336 53.75424262648326Norman Park -1.785020828247070 53.665391639441296Lister park -1.7774677276611328 53.815855460240394.3 Two methods to find the "center of mass"4.3.1 The first methodHere is a total of 21 points of the crime locations. We use x 1, x 2, x 3, x 4, x 5......x 21 to represent them.Then the six predicted points can be represented by y 1, y 2, y 3, y 4, y 5, y 6.The distance between the point x and point y will be d(x,y).We found that the distances are linked to the probability of occurrence of these crimes. We can regard one location as the center. If one point is farther away from the center, the probability of occurrence is smaller. So we can use the distance decay function to represent these relationships.distance decay function [1].,)(2da d f = (1) Next we need to find the reliability of the six predicted points. Through the distance decay function, we find the predicted point to these 21 points' f value. Then we add all the values together and we get the construct a hit score function S(y) by computing. This function can be the probability of the predicted point.hit score function:∑=++==211211)).,((...)),(()),(()(i i y x d f y x d f y x d f y S (2))(y S reflects possibility of the murder in y point. If )(y S is greater, the possibility of the murder is greater. If smaller, the possibility is smaller.There will exist a maximum )(y S of these six predicted points. Then, the point that has the maximum )(y S is the "center of mass". Then we get the "center of mass".4.3.2 The result of the first methodTo solve it through the distance decay function, we have to compute the distances between points. Our coordinates is represented by latitude and longitude. So we have to convert the latitude and longitude into the actual distance. The following is the way to the conversion.We use Lon to represent the longitude and Lat to represent the latitude.The latitude's actual distance can be represented by Won and the longitude's actual distance can be represented by Wat.Δon and Δat represent the longitude difference and the latitude difference.Now we can equate the Wat and Won by means of Lon, Lat, Δon and Δat.)cos(111Lat on Won ⨯⨯=Δ. (3)111⨯=at Wat Δ. (4)Through the above equations, we can equate the Wat and Won by means of Lon, Lat, Δon and Δat. Now take the data of the Table.2 the Table.4 into the equations. We can get the f and )(y S .Taking the complexity of the data into account, now we take the y 1 point for an example. We put the other points' data into the additional list.(list.1) Table.5 The data of the y 1 pointpoint Δat Δon Won Wat d(x i ,y 1) f(d(x i ,y 1)) X 1 -0.011201 0.229044 -23.48998 -1.243327 23.522856 0.0018072 X 2 -0.003623 0.2561187 -26.18346 -0.402231 26.186554 0.0014582 X 3 0.0264967 0.2827692 -28.52664 2.9411372 28.677866 0.0012159 X 4 -0.000836 0.012316 -1.257668 -0.092819 1.2610890 0.6287941 X 5 0.0073478 0.2444458 -24.87264 0.8156068 24.886017 0.0016146 X 6 -0.381306 -0.480480 53.319853 -42.32504 68.076545 0.0002157 X 7 -0.010434 0.0068235 -0.699575 -1.158266 1.3531400 0.5461532 X 8 -0.157424 -0.001566 0.1686226 -17.47416 17.474975 0.0032746 X 9 -0.350136 -0.447263 49.574901 -38.86511 62.993398 0.0002520 X 10 -0.100489 -0.094499 10.010757 -11.15438 14.987846 0.0044518 X 11 -0.030038 -0.069222 7.1533152 -3.334224 7.8922095 0.0160547 X 12 -0.002787 0.1061296 -10.84597 -0.309405 10.850388 0.0084939 X 13 0.0121168 0.2026247 -20.57421 1.3449716 20.618127 0.0023523 X 14 0.0548539 -0.131590 13.097272 6.0887894 14.443353 0.0047936 X 15 0.1038234 -0.159791 15.500586 11.524403 19.315280 0.0026803 X 16 0.0266706 0.2818817 -28.43485 2.9604431 28.588547 0.0012235 X 17 -0.016985 0.0257121 -2.643220 -1.885395 3.2467411 0.0948647 X 18 -0.011201 0.229044 -23.48997 -1.243326 23.522856 0.0018072 X 19 -0.011201 0.2290443 -23.48997 -1.243326 23.522856 0.0018072 X 20 -0.165720 -0.006721 0.7250615 -18.39499 18.409275 0.0029507 X 21 -0.086458 -0.083422 8.7976123 -9.596898 13.01915 0.0058997We add the f values together and then we get the 1)(y S =0.111932.Using this method, we can equate the other )(y S value.Table.6 )(y S value point)(y S Y 10.111932 Y 20.725086 Y 30.001451 Y 4 0.017765 Y 5 0.060981 Y 60.082784From the table.6, we can see that the 2)(y S value of the y 2 point is the maximum. So we can conclude that the possibility of the murder in y 2 point is the greatest. The y 2 point is the "center of mass".4.3.3 The second methodSuppose only that the offender chooses to offend at the location x with probability density P(x).The probability density P(x) depends on the following three variables.1. The "center of mass" point m of the murder.2. The "center of mass" point must have a well-defined meaning- e.g. the murder ’s place of residence.3. The "center of mass" point needs to be stable during the crime series.The average distance d is the expected value of the locations of murder.∑=⨯=211),(211)(i i y x d y d (5) We are left with the assumption that a murder with "center of mass" point mand mean murder distance d commits an offense at the location x with probability density ),(d m x P .Our method then can be summarized as follows.Given a sample x 1, x 2, . . . , x 21 (the crime sites) from a probability distribution ),(d m x P , and estimate the parameter m (the "center of mass").This is a well-studied mathematical problem.Our approach is the theory of the maximum likelihood.Construct the likelihood function:),()...,(),(),(211211d y x P d y x P d y x P d y F i i ==∏= (6)This is a Joint Conditional Probability function. Another likelihood function is:),(ln ...),(ln ),(ln ),(211211d y x P d y x P d y x P d y i i ++==∑=λ (7)Then the best choice of m is the choice of y that makes the likelihood as large as possible. Then this is the "center of mass".We use Δi to represent the subtraction between the )(y d and ),(y x d i . Δi reflects the difference between the locations of crime and the expected value.)(),(y d y x d i i -=Δ (8)1. Δi is greater than zero.),()(),(y x d y d d y x P i i =(9)2. Δi is smaller than zero.)(),(),(y d y x d d y x P i i =(10) Put the Eq.9 and Eq.10 into the Eq.6 and Eq.7, we can get the maximal F value and the λ value. Then this point is the "center of mass".After we have found the "center of mass" point, we should calculate the probability of the points. Here we can use the Bayes Theorem equation [2].),...,()(),,...,(),...,(211211211x x P m H d m x x P x x d m P ⨯=(11))(m H is the prior distribution of "center of mass" points.It represents our knowledge of the probability density for the "center of mass"point m and average murder distance d before we incorporate information about the serial crime.We will assume that the offender’s choice s of crime sites are mutually independent, so that:),()...,(),,...,(211211d m x P d m x P d m x x P = (12)To equate the probability ),...,(211x x P , we can divide the crime areas into several small areas. The longitude and latitude is divided into 16 equal portions. By means of the environmental conditions near the conditions of crime and the time of crime, we narrowed the region to four main areas A, B, C and D.Fig.2 Locations of crime in the areasThen we equate each point's probability of crime in these four areas. We use )(i x P to represent each area's probability of crime. Then we multiply the four area's probability. So the ),...,(211x x P can be equated.)()()()()(),...,(4321211211z P z P z P z P z P x x P i i ⨯⨯⨯==∏= (13)By means of these above equations, we can equate the ),...,(211x x d m P . Then we equate the probability of the "center of mass" point.4.3.4 The result of the second methodIn the beginning, we have to equate the expected value )(i y d . ),(y x d i can be acquired from the Table.5 and List.1. So we can equate the )(i y d by means of Eq.5. Data are as following:Table.7 The expected value Y 1Y 2 Y 3 Y 4Y 5 Y 6 )(i y d 23.4726 22.8237 67.376825.78125.391121.7063Then by means of Eq.8, Eq.9 and Eq.10, we can equate the ),(d y x P i . The data are put into the additional list.(list 2)By means of Eq.6 and Eq.7, we can equate the F and λ as followings.Table.8 F and λY 1 Y 2 Y 3 Y 4 Y 5 Y 6F 1.81E-6 1.96E-8 1.19E-3 1.72E-7 7.7E-6 1.02E-6 λ -13.22 -17.75 -6.73 -15.58 -11.77 -13.8From the Table.8, we can see that y 3 point's F value is the greatest. So the y 3 point is the "center of mass".Now we have to equate the probability of the "center of mass" point. Since the y 3 point is the "center of mass", we can use m point to replace y 3 point. By means of Eq.12 and the additional list.2, we can equate the probability001193.0),,...,(211=d m x x P .Then 2857.021/6),(==d m H .The following is what we can conclude from the Fig.2:Table.9 P(z i )area The number of crimes P(z i ) A 9 0.42857142857 B 5 0.2380952381 C 5 0.2380952381 D 2 0.0952********Then 000624372.0)()()()()(),...,(4321211211=⨯⨯⨯==∏=z P z P z P z P z P x x P i iSo that,54594834.0000624372.028*******.0001193062.0),...,()(),,...,(),...,(211211211=⨯=⨯=x x P m H d m x x P x x d m P Then we get the probability of the "center of mass" point is 0.546.4.3.5 The combination of the two resultsBy means of the first method, we know that y 2 point is the "center of mass". But through the other method, we equate that the y3 point is the "center of mass". So we have to combine the two results of the methods.Taking advantage of the longitude and latitude, we can calculate the distance between the y2 point and the y3 point. It is about 995m.Now we use the distance between the two points as the diameter to generate a circle. The area inside this circle can be regarded as the locations of the next possible crime. At this point our model building is completed.5 Case Study: The Test of ModelTaking the following case for example.Nickname:The Night StalkerReign of terror:17/3/85 - 31/8/85Motive:He enjoyed it.Crimes:Ramirez was charged with 63 crimes: including 13 murders, and 30 other offences including attempted murder, rape and sexual assaults.Table.10 The offencesDate Victim Crime Method Place June 1984 Jennie Vincow Rape andmurder- -17/3/85 MariaHernandez AttemptedmurderShot Rosemead, LosAngeles17/3/85 Dayle Okasaki Murder Shot Rosemead, LosAngeles 17/3/85 Tsai Lian Yu Murder Shot MontereyPark, LA27/3/85 VincentZazzara Murder Shot Near the SanGabrielfreeway27/3/85 MaxineZazzara Murder Shot, stabbedand her eyeswere gougedoutNear the SanGabrielfreeway14/4/85 William Doi Murder Shot in thehead Monterey Park, LA14/4/85 Lillie Doi Attemptedmurder SavagelybeatenMontereyPark, LA30/5/85 Carol Kyle Rape - Burbank, LA 27/6/85 Patty Elaine Murder Slashed throat Arcadia, LAHiggins2/7/85 Mary LouiseCannonMurder Slashed throat Arcadia, LA5/7/85 WhitneyBennet AttemptedmurderBeaten with acrowbarArcadia, LA7/7/85 Joyce LucilleNelson Murder Beaten todeath andstrangledMontereyPark, LA7/7/85 SophieDickman Rape andRobbery- Monterey Park20/7/85 Max and LelaKneiding Murder Shot. Lela wasalso stabbedrepeatedly.Glendale, LA20/7/85 ChainarongKhovananthMurder Shot Sun Valley, LA20/7/85 SomkidKhovananthGBH and rape - Sun Valley, LA5/8/85 Christopherand VirginiaPeterson Attemptedmurder- -9/8/85 Elyas Abowath Murder Shot San GabrielValley9/8/85 SakinaAbowath GBH and rape Beaten San GabrielValley17/8/85 Barbara andPeter Pan Murder Shot in theheadLake Merced24/8/85 William Carns Attemptedmurder Shot 3 times inthe headMission Viego24/8/85 Inez Erickson Rape - Mission Viego The areas of crime on the map are as follows.Fig.3 The mapThe steps of establishing the model are as follows.Step.1 We mark the points on the map.Step.2 We analyze the environmental conditions of these points and find out the common features of these points.Step.3 Then we find out the other points which have the similar environmental conditions.Step.4 By means of the model, we calculate the "center of mass".Step.5 We calculate the probability of the "center of mass" point.Through the comparison of the data, the "center of mass" point is suited with the practical conditions. So the reliability of our model is relevantly high.6 Model EvaluationThe model we established has advantages and disadvantages. The advantages are as follows:1.Our model has a very clear structure and a good level. By means of two different schemes to generate a geographic profile, we predict the locations of the crime. So a simple structure is a major feature of this model.2.Starting from the issues, we find a great deal of information and analyse various situations that we should consider into our model. With model, we also make a case study, which can make our model more practical.3.We use Excel software to deal with the data, which make our model more objective. Moreover, the calculation of the probability enhances the explanatory power of our model power of the persuasion. So our model is also relatively objective.4.In order to establish the model, we use many tables and figures, which could makeour model more simple and clear.Meanwhile, there are disadvantage of our model:1.During our prediction of the locations of the crime based on the time and locations in the past, we only choose six points to make the prediction. Lack of the predicting points is one of shortcomings of our model.2.In order to simplify the model, we make some idealized assumptions which might make the results less reliable.7 Executive SummaryOur model is established based on the background of serial murders. The purpose of this model is to assist the police in forecasting the locations of the possible serial murders.Firstly, according to the time and locations of the past crime scenes, we analyze the environmental conditions near the locations of murder and investigate the characteristics of the crime time. Secondly, with these characteristics, we find out the locations of the similar environment in the areas where there are frequent murders. These locations of the similar environment are the places where he might make a murder. Thirdly, we use two different methods to predict the locations of the crime. The first method takes advantage of the Distance Decay Function, and the other method is a probability model. Using the model, policeman can predict the next locations of committing the crime as long as the relevant crime data are given. However, it is necessary to pay attention to the following several points when using the model:a. The model can be used to make forecasts under some conditions.1. The crime is a serial attacks or murders.2. The locations of the crime have the common features and have the similar environmental conditions of the crime locations. This is the basement of the model. If the feature has not existed, the model will not work effectively.3. The locations of the crime must be located between several cities or small areas because the model is based on the small areas between several cities and large areas can't be suit with the model.4. The model can be used in the conditions where single people's crime as well as the group's crime.5. The model can be used in the conditions where terrain is not complicated. Remote places suit with the model best.b. On some conditions, the model can't be used to make forecasts1.The crime is not a serial attack or murder because single crime can't be satisfied with the conditions the model requirements.2.The locations of the crime have not the common features or the similar environmental conditions. If the environmental conditions have great differences, the model can't be used to predict.。