美赛论文

Camping along the Big Long RiverSummaryIn this paper, the problem that allows more parties entering recreation system is investigated. In order to let park managers have better arrangements on camping for parties, the problem is divided into four sections to consider.The first section is the description of the process for single-party's rafting. That is, formulating a Status Transfer Equation of a party based on the state of the arriving time at any campsite. Furthermore, we analyze the encounter situations between two parties.Next we build up a simulation model according to the analysis above. Setting that there are recreation sites though the river, count the encounter times when a new party enters this recreation system, and judge whether there exists campsites available for them to station. If the times of encounter between parties are small and the campsite is available, the managers give them a good schedule and permit their rafting, or else, putting off the small interval time t until the party satisfies the conditions.Then solve the problem by the method of computer simulation. We imitate the whole process of rafting for every party, and obtain different numbers of parties, every party's schedule arrangement, travelling time, numbers of every campsite's usage, ratio of these two kinds of rafting boats, and time intervals between two parties' starting time under various numbers of campsites after several times of simulation. Hence, explore the changing law between the numbers of parties (X) and the numbers of campsites (Y) that X ascends rapidly in the first period followed by Y's increasing and the curve tends to be steady and finally looks like a S curve.In the end of our paper, we make sensitive analysis by changing parameters of simulation and evaluate the strengths and weaknesses of our model, and write a memo to river managers on the arrangements of rafting.Key words: Camping；Computer Simulation; Status Transfer Equation1 IntroductionThe number of visits to outdoor recreation areas has increased dramatically in last three decades. Among all those outdoor activities, rafting is often chose as a family get-together during May to September. Rafting or white water rafting is a kind of interesting and challenging recreational outdoor activity, which uses an inflatable raft to navigate a river or sea [1]. It is very popular in the world, especially in occidental countries. This activity is commonly considered an extreme sport that usually done to thrill and excite the raft passengers on white water or different degrees of rough water. It can be dangerous.During the peak period, there are many tourists coming to experience rafting. In order to satisfy tourists to the maximum, we must make full use of our facilities in hand, which means we must do the utmost to utilize the campsites in the best way possible. What's more, to make more people feel the wildness life, we should minimize the encounters to the best extent; meanwhile no two sets of parties can occupy the same campsite at the same time. It is naturally coming into mind that we should consider where to stop, and when to stop of a party [2].In previous studies [3-5],many researchers have simulated the outdoor creation based on real-life data, because the approach is dynamic, stochastic, and discrete-event, and most recreation systems share these traits. But there exists little research aiming at describing the way that visitors travel and distribute themselves within a recreation system [6]. Hence, in our paper, we consider the whole process of parties in detail and simulate every party ’s behavior, including the location of their campsites, and how long it will last for them to stay in a campsite to finish their itineraries. Meanwhile minimize the numbers of encounters.Aiming at showing the whole process of rafting, we firstly focus on analyzing the situation s of a single-party's rafting by using status transfer equation, then consider the problems of two parties' encounters on the river. Finally, after several times of simulation on the whole process of rafting, we obtain the optimal value of X . 2 Symbols and DefinitionsIn this section, we will give some basic symbols and definitions in the following for the convenience.Table 1. Variable Definition SymbolsDefinition i vi pj i q , S dThe velocity of oar or motor 0-1 variables on choosing rafting transportation 0-1 variables on the occupation of campsitesLength of the riverAverage distance between two campsites3 General AssumptionsIn order to have a better study on this paper, we simplify our model by the following assumptions:1) 19 : 00 to 07: 00 is people's sleeping time, during this time, people are stationedin the campsite. The total time of sleeping is 12 hours, as rafting is an exiting sport game, after a day's entertainment, people have cost a lot of energy, and nearly tired out. So in order to have a better recreation for the next day, we set that people begin their trip at 07:00, and end at 19:00 for a day's schedule.2) Oar- powered rubber rafts and motorized parties can successfully raft from FirstLaunch to Final Exit, there exist no accident over the whole trips.3) All the rubber rafts and motorized boats have the same exterior except velocities;we regard a rubber raft or a motorized boat as a party and don't consider the tourists individuals on the parties.4) There is only one entrance for parties to enter the recreation system.5) Regardless of the effects that the physical features of the river brings to oar andmotorized parties, that is to say we ignore the stream ’s propulsion and resistance to both kinds of rafting boats. Oar and motorized parties can keep the average velocity of 4 mph and 8mph.6) Divide the whole river into N segments.4 Analysis of This Rafting ProblemRafting is a very popular spots game world-wide. In the peak period of rafting, there are more people choosing to raft, it often causes congestion that not all people can raft at any time they want. Hence, it is important for managers to set an optimal schedule for every party (from our assumptions, we regard a rafting boat as a party) in advance. Meanwhile, the parties need to experience wildness life, so the managers should arrange the schedules which minimize the encounters' time between parties to the best extent. What's more, no two sets of parties can occupy the same site at the same time.Our aim is to determine an optimal mix of trips over varying duration (measured YXNj i t ,j TtK Numbers of campsites Numbers of parties Numbers of attraction sites Time of the i th party finishing the whole trip ranges from6 days to 18 days Random staying time at each campsite Delay time of rafting from beginning Threshold value of encounterin nights on the river. That is to say,we must obtain an optimal value of X through lots of trails. This optimal value represents that the campsites have a high usage while more people are available to raft.The Long Big River is 225 miles long, if we discuss the river as a whole and consider all the parties together, it will be difficult for us to have a clear recognition on parties' behaviors. Hence, we divide the river into N attraction sites. Each of the attraction sites has Y/N campsites since the campsites are uniformly distributed throughout the river corridor. So build up a model based on single-party ’s behavior of rafting in small distance. At last, we can use computer simulation to imitate more complex situations with various rafting boats and large quantities of parties. 5 Mathematic Models5.1 Rafting of the Single-party Model （Status Transfer Equation [7]）From the previous analysis, in order to have a clear recognition of the whole rafting process, we must analyze every single-party's state at any time.In this model, we consider the situation that a single-party rafts from the First Launch to the Final Exit. So we formulate a model that focus on the behavior of one single-party.For a single-party, it must satisfy the following equation: status transfer equation. it represents the relationships between its former state and the latter state. State here means: when the i th party arrives at the j th campsites, the party may occupy the j th campsite or not.As a party can choose two kinds of transportation to raft: oar- powered rubber rafts(i v =4mph) and motorized rafts(i v =8mph). i v is the velocity of the rafting boats,and i p is the 0-1 variables of the selecting for boats. Therefore, we can obtainthe following equation:)1(84i i i p p v -+= (i=1,2,…,X ). (1) where i p =0 if the i th party uses motorized boat as their rafting tool, at thistime i v =8mph ; while i p =1 when , the i th party rafts with oar- powered rubber raft with i v =4mph. In fact, Eq.(1) denotes which kind of rafting boat a party can choose.A party not only has choice on rafting boats, but also can select where to camp based on whether the campsites are occupied or not. The following formulation shows the situation whether this party chooses this campsite or not:⎩⎨⎧=p a r t y p r e v i o u s a b y o c c u pi e d is campsite the 0,party previous a by occupied not is campsite the q ij ,1 (2) where i =1,2,…,X ; j =1,2,…,Y .Where the next one can’t set their camp at this place anymore, that is to say thelatter party’s behavior is determined by the former one.As campsites are fairly uniformly distributed throughout the river corridor, hence, we discrete the whole river into segments, and regard Y campsites as Y nodes which leaves out (Y +1) intervals. Finally we get the average distance between th e j th campsite and (j+1)th campsite:1+=Y S d (3) where is the length of the river, and its value is 225 miles.What’s more, the trip -days for a party is not infinite, it has fluctuating intervals: h t h j i 432144,≤≤ (4) where is the t i ，j itinerary time for a party ranges from 144 hours to 432 hours (6 to 18 nights).From Eq.(1), (2) and (3), the status transfer equation is given as follows: ),...2,1,,...2,1(11,1,,Y j X i T q v d t t j j i i j i j i ==⨯++=--- （5）The i th party’s arriving time at the j th campsite is determined by the time when the i th arrived at （j-1）campsite, the time interval i v d , and the time T j-1 random generated by computer shown in Eq.(5). It is a dynamic process and determined by its previous behavior.5.2 The Analysis of Two Parties’ Encounter on the RiverOur goal is to making full use of the campsites. Hence, the objective of all the formulation is to maximize the quantities of trips （parties ）X while consider getting rid of the congestion. If we reduce the numbers of the encounters among parties, there will be no congestion. In order to achieve this goal, we analysis the situations of when two parties’ to encounter, and where they will enco unter.In order to create a wildness environment for parties to experience wildness life, managers arrange a schedule that can make any two parties have minimal encounters with each other. Encounter is that parties meet at the same place and at the same time. Regarding the river as a whole is not convenient to study, hence, our discussion is based on a small distance where distance=d (Eq.3), between the j th and ( j+1)th campsites. Finally the encounter problem of the whole river is transferred into small fractions. On analyzing encounter problem in d and count numbers of each encounter in d together, we get a clear recognition of the whole process and the total numbers of encounter of two parties.The following Figure 1 represents random two parties rafting in d :Figure 1. Random two parties' encounter or not on the riverThe i th party arrives at j th campsite (t j k ,-t j i ,) time earlier than the k th party reaches the j th campsite. After t time, interval distance between the i th party and the k th party can be denoted by the following function:)()(t t t v t v t S ij kj i k j +-⨯-⨯=∆ (6) Where k,i =1,2,…,X , j =1,2,…,Y . k i ≠.Whether the two parties stationed on the j th campsite and(j +1)th campsite are based on the state of the campsites’occupation, yields we obtain:⎩⎨⎧=⨯01,,j k j i q q ( i,k =1,2,…,X ; j =1,2,…,Y ; k ≠i ) (7)Note that Eq.6 is constrained by Eq.7, for different value of )(t S J ∆ andj k j i q q ,,⨯we can obtain the different cases as follows:Case 1:⎩⎨⎧=⨯=∆10)(,,j k ji j q q t S (8) Which means both the i th and k th party don’t choose the j th campsite, they arerafting on the river. Hence, when the interval distance between the two parties is 0, that is )(t S J ∆=0,they encounter at a certain place in d on the river.Cases2:⎩⎨⎧=⨯=∆00)(,,j k ji j q q t S (9) Although the interval distance between the two parties is 0, the j th campsite is occupied by the i th party or the k th party. That is one of them stop to camp at a certain place throughout the river corridor. Hence, there is no possibility for them to encounter on the river.Cases 3:⎩⎨⎧=⨯≠∆10)(,,j k j i j q q t S⎩⎨⎧=⨯≠∆00)(,,j k j i j q q t S (10) No matter the j th campsite is occupied or not for )(t S J ∆≠0 , that is at the same time, they are not at the same place. Hence, they will not encounter at any place in d .5. 3 Overview of Computer Simulation Modeling to Rafting5.3.1 Computer SimulationSimulation modeling is a kind of method to imitate the real-word process or a system. This approach is especially suited to those tasks which are too complex for direct observation, manipulation, or even analytical mathematical analysis (Banks and Carson 1984, Law and Kelton 1991, Pidd 1992).The most appropriate approach for simulating out-door recreation is dynamic, stochastic, and discrete-event model, since most recreation systems share these traits. In all, simulation models can reflect the real-world accurately.5.3.2 Simulation for the Whole Process of Parties on Rafting [8]This simulation can approximate show a party’s behavior on the river under a wide rang of conditions. From the analysis of the previous study, we have known that the next party’s behavior is affected by the former one. Hence, when the first party enters the rafting system, there is no encounter, and it can choose every campsite. then the second party comes into the rafting system , at this time, we must consider the encounter between them, and the limit on choosing the campsite. As time goes by, more and more parties enter this system to raft which lead to a more complex situation. A party who satisfies the following two conditions will be removed from the current order to the next order. So he can’t “finish his trip” right away. The two conditions are as follows:(1) He chooses a campsite where has been occupied by other parties.(2) He has two many encounters with other parties.So in order to determine typical trip itineraries for various types of rafting boat ,campsite, and time intervals （See Trip Schedule Sheet 1），we need to perform a series of trails run that can represent the real-life process of rafting based on these considerations,. A main flowchart of the program is shown in Figure 2.Figure 2. Main simulation flowchartAfter several times of simulation, we obtain the optimal X (the numbers of campsites), minimal E(Encounter) and TP (Trip Time).Followed by Figure 2, we simulate the behavior of a party whether it can enterthe rafting system or not in Figure 3.Figure 3.Sub flowchart5.3.3 The Results of SimulationAfter simulating the whole process of parties rafting on the river, we get three figures (Figure 4, Figure 5 and Figure 6) to present the results.In order to simulate the rafting process more conveniently, we divide the whole river into 31 segments (31 attraction sites), and input an initial value of Y=155(numbers of campsites), where there are 5 campsites in every attraction sites.We represent the times of campsites occupied by various parties on Figure 2 by coordinates( x, y), where x is the order of the campsites from 0 to 155（these campsites are all uniformly distributed thorough the corridor）, and y is the numbers of each campsite occupied by different parties. For example, (140，1100)represents that at the campsite, there exists nearly 1100 times of occupation in total by parties over 180days. Hence, the following Figure 4 shows the times of campsites’ usage from March to September.Figure 4.Numbers of campsites' usage during six-month period from March to SeptemberFrom Figure 4, The numbers of campsites’ usage can be identified the efficiency of every campsites’ usage. The higher usage of the campsites, the higher efficiency they are. Based on these, we give a simple suggestion to managers (see in Memo to Managers).Figure 5. the ratio of usage on campsites with time going byFigure 5 shows the changes of the ratio on campsites. when t =0, the campsites are not used , but with time going by, the ratio of the usage of campsites becomes higher and higher.We can also obtain that when t >20, the ratio keeps on a steady level of 65% ; but when t >176 , the ratio comes down, that is, there are little parties entering the recreation system. In all, these changes are rational very much, and have high coincidence with real-world.Then we obtain 1599 parties arranged into recreation system after inputting the initial value Y=155, and set orders to every party from number 0 to number 1599. Plotting every party's travelling time of the whole process on a map by simulating, as follows:Figure 6. Every party’s travelling timeFigure 6 shows the itinerary of the travelling time, most of the travelling time is fluctuating between 13days and 15.3 days, and most of travelling time are concentrated around 14 days.In order to create an outdoor life for all parties, we should minimize the numbers of encounter among different parties based on equations (6) and (7)：So we get every party’s numbers of encounter by coordinates ( x, y), where x is the order of the parties from 0 to 1600, y is the numbers of encounters. Shown in Figure 7, as follows:Figure 7. Every party’s numbers of encounterFigure 7 shows every party’s numbers of encounter at each campsite. From this figure, we can know that the numbers of their encounter are relatively less, the highest one is 8 times, and most of the parties don’t encounter during their trips, which is coincident with the real-world data.Finally, according to the travelling time of a party from March to September, we set a plan for river managers to arrange the number of parties. Hence, by simulating the model, we obtain the results by coordinate ( x, y), where y is the days of travelling time, x is the numbers of parties on every day. The figure is shown as follows:Figure 8. Simulation on travelling days versus the numbers of parties From Figure 8.we set a suitable plan for river manager, which also provide reference on his managements.6 Sensitive AnalyzeSensitive analysis is very critical in mathematical modeling, it is a way to gauge the robustness of a model with respect to assumptions about the data and parameters. We try several times of simulation to get different numbers of parties on changing the numbers of campsites ceaselessly. Thus using the simulative data, we get the relationship between the numbers of campsites and parties by fitting. On the basis of this fitting, we revise the maximal encounter times (Threshold value) continually, and can also get the results of the relationships between the numbers of campsites and parties by fitting. Finally, we obtain a Figure 9denoting the relations of Y (numbers of campsites)and X(numbers of parties), as follows:Figure 9 .Sensitive analysis under different threshold values Given the permitted maximal numbers of encounters (threshold value= K), we obtain the relationships between Y(numbers of campsites) and X( numbers of trips). For example, when K=1 , it means no encounters are allowed on the river when rafting; when K=2, there is less than 2 chances for the boats to meet. So we can define the K=4,6,8 to describe the sensitivity of our model.From Figure 9, we get the information that with the increase of K , the numbers of boats available to rafts till increase. But when K>6 , the change of the numbers of boats is inconspicuous, which is not the main factor having appreciable impact on theIn all, when the numbers of campsites ( Y ) are less than 250, they would have a great effect on the numbers of boats. But in the diverse situations, like when Y>250 , the effect caused by adding the numbers of campsites to hold more boats is not notable.When K<6 , the numbers of boats available increases with the ascending of K , While K>6 , the numbers of boats don’t have g reat change .Take all these factors into consideration, it reflects that the numbers of the boats can’t exceed its upper limit. Increase the numbers of campsites and numbers of encounter blindly can’t bring back more profits.7 Strengths and WeaknessesStrengthsOur model has achieved all of the goals we set initially effectively. It is not only fast and could handle large quantities of data, but also has the flexibility we desire.Though we don’t test all possibilities, if we had chosen to inpu t the numbers of campsites data into our program, we could have produced high-quality results with virtually no added difficulty. Aswell, our method was robust.Based on general assumptions we have made in previous task, we consider a party’s state in the first place, then simulate the whole process of rafting. It is an exact reflection of the real-world. Hence, our main model's strength is its enormousedibility and stability and there are some key strengths:(1) The flowchart represents the whole process of rafting by given different initialvalues. It not only makes it possible to develop trip itineraries that are statistically more representatives of the total population of river trips, but also eliminates the tedious task of manual writing .(2)Our m odel focuses on parties’ behavior and interactions between each other, notthe managers on the arrangement of rafting, which can also get satisfactory and high-quality results.(3) Our model makes full use of campsites, while avoid too many encounters, whichleads to rational arrangements.WeaknessesOn the one hand, although we list the model's comprehensive simulation as a strength, it is paradoxically also the most notable weakness since we don’t take into account the carrying capacity of the water when simulates, and suppose that a river can bear as much weight as possible. But in reality, that is impossible. On the other hand, our results are not optimal, but relative optimal.8 ConclusionsAfter a serial of trials, we get different values of X based on the general assumptions we make. By comparing them, we choose a relative better one. From this problem, it verifies the important use of simulation especially in complex situations. Here we consider if we change some of the assumptions, it may lead to various results. For example,(a) Let the velocity of this two kinds of boats submit to normal distribution.In this paper, the average velocity of oar- powered rubber rafts and motorized boats are 4mphand 8mph, respectively. But in real-world, the speed of the boats can’t get rid of the impacts from external force like stream’s propulsion and resistance. Hence, they keep on changing all the time.(b) Add and reduce campsites to improve the ratio of usage on campsites. By analyzing and simulating, the usage of each campsite is different which may lead to waste or congestion at a campsite. Hence, we can adjust the distribution of campsites to arrive the best use.A Memo to River ManagersOur simulation model is with high edibility and stability in many occasions. It can imitate every party’s behavior when rafting so as to make a clear recognition of the process.Internal Workings of The ModelInputsOur model needs to input initial value of Y , as well as the numbers of attraction sites. Algorithm ( Figure 2, and Figure 3)Our algorithm represents the whole process of rafting, so we can use it to simulate the process of rafting by inputting various initial values.OutputsBased on the algorithm in our paper, our model will output the relative optimalnumbers of parties X. Furthermore, we can also get other information, such as the interval time between two parties at First Launch, a detailed schedule for each party of rafting , the relationship between X and Y and so on.Summary and RecommendationsAfter 100 times of simulating, we come to two conclusions:(a) The numbers of parties (X) have relations with the numbers of campsites(Y), that is to say, with the increasing of Y , the increasing speed of X goes fast at the first place and then goes down, finally it tends to be steady. Hence, we advice river managers to adjust the numbers of campsites properly to get the optimal numbers of parties.(b) Add campsites to the high usage of the former campsites and deduce campsites at the low usage of the former campsites. From Figure 4, we know that the ratios of every campsites are different, some campsites are frequently used, but some are not. Thus we can infer that the scenic views are attractive, and have attracted lots of parties camping at the campsite. so we can add campsites to this nodes. Else the campsites with low usage have lost attractions which we should reduce the numbers of campsites at those nodes.References[1] Karlo Šimović，Wikipedia，Rafting，/wiki/Rafting.[2] C. A. Roberts and R. Gimblett，Computer Simulation for Rafting Traffic on theColorado River，COMPUTER SIMULATION FOR RAFTING TRAFFIC, 2001, 19-30.[3] C.A. Roberts, D. Stallman，J. A. Bieri. Modeling complex human-environmentinteractions: the Grand Canyon river trip simulator，Ecological Modeling153(2002) 181-196.[4] J. A. Bieri and C. A. Roberts，Using the Grand Canyon River Trip Simulator toTest New Launch Scheduleson the Colorado River，Washington DC, AWISMagazine, V ol. 29, No. 3, 2000, 6-10.[5] A.H. Underhill and A. B. Xaba，The Wilderness Simulation Model as aManagement Tool for the Colorado River in Grand Canyon National Park，NATIONAL PARK SERVICE/UNIVERSITY OF ARIZONA Unit SupportProject CONTRIBUTION NO. 034/03.[6] B. Wang and R. E. Manning，Computer Simulation Modeling for RecreationManagement: A Study on Carriage Road Use in Acadia National Park, Maine, USA, USA, Vermont 05405, 1999.[7] M. M. Meerschaert，Mathematical Modeling(Third Edition). China MachinePress publishing,2009.[8] A.H. Underhill，The Wilderness Use Simulation Model Applied to Colorado RiverBoating in Grand Canyon National Park, USA，Environmental Management V ol.10, No. 3, 1986, 367-374.。

%% 本论文的排版主要参考了LaTeX2e插图指南(王磊), LaTeX2e用户手册, media的中文学位%% 论文宏包(CDT), happaytex的ORmain1.tex等文件以及ChinaTeX, CTeX论坛上的诸多贴子. %%% 本论文采用了Miktex2.2的方式在ChinaTeX.iso系统下得到了实现,其编译方式为%% latex(得到DVI文件)+dvips(得到PS文件)+ps2pdf(可得PDF文件).%%\documentclass[12pt]{article}%需要的一些宏包\usepackage{CJK} % 中文输入环境宏包\usepackage{titlesec,titletoc} % 配合命令在后面, 章节标题设置\usepackage{indentfirst} % 使首段首行缩进\usepackage{graphicx} % 插图宏包\usepackage{caption2} % 可以更改插图, 表格的标题样式\usepackage{subfigure} % 产生并列的子图或子表, 命令\subfigure, \subtable\usepackage{longtable} % 如果表格太长, 超过了一页时, 就可以试试longtable 宏包所定义的longtable 环境\usepackage{slashbox} % 在表格中绘制斜线\usepackage{fancyhdr} % 更改页眉的宏包, 并可在页眉插入图片\usepackage{times} % Times Roman + Helvetica + Courier\usepackage{amsmath} % 数学符号宏包AMS-LaTeX, 如下面的\overset需要此宏包%页面的设置\special{papersize=21cm,29.7cm} \setlength{\textwidth}{15cm}\setlength{\textheight}{23cm} \setlength{\evensidemargin}{0.46cm}\setlength{\oddsidemargin}{0.46cm} \setlength{\topmargin}{-1.84cm}\setlength{\headheight}{2.9cm} \setlength{\headsep}{0.4cm}%字号设置\newcommand{\chuhao}{\fontsize{42pt}{\baselineskip}\selectfont}\newcommand{\xiaochuhao}{\fontsize{36pt}{\baselineskip}\selectfont}\newcommand{\yihao}{\fontsize{26pt}{\baselineskip}\selectfont}\newcommand{\xiyihao}{\fontsize{24pt}{\baselineskip}\selectfont}\newcommand{\erhao}{\fontsize{22pt}{\baselineskip}\selectfont}\newcommand{\xiaoerhao}{\fontsize{18pt}{\baselineskip}\selectfont}\newcommand{\sanhao}{\fontsize{16pt}{\baselineskip}\selectfont}\newcommand{\xiaosanhao}{\fontsize{15pt}{\baselineskip}\selectfont}\newcommand{\sihao}{\fontsize{14pt}{\baselineskip}\selectfont}\newcommand{\xiaosihao}{\fontsize{12pt}{\baselineskip}\selectfont}\newcommand{\wuhao}{\fontsize{10.5pt}{\baselineskip}\selectfont}\newcommand{\xiaowuhao}{\fontsize{9pt}{\baselineskip}\selectfont}\newcommand{\liuhao}{\fontsize{7.5pt}{\baselineskip}\selectfont}\newcommand{\xiaoliuhao}{\fontsize{6.5pt}{\baselineskip}\selectfont}\newcommand{\qihao}{\fontsize{5.5pt}{\baselineskip}\selectfont}\newcommand{\bahao}{\fontsize{5pt}{\baselineskip}\selectfont}%页眉的设置, 要用到fancyhdr宏包\pagestyle{fancy} \fancyhead{} \fancyfoot{}\fancyhead[L]{\footnotesize Team \# 189}\fancyhead[R]{\footnotesize Page\ \thepage\ of\ 42}\fancypagestyle{plain}{%\fancyhead[L]{\footnotesize Team \# 189}\fancyhead[R]{\footnotesize Page\ \thepage\ of\ 42}}\setcounter{secnumdepth}{4}%更改\theparagraph的编号样式\makeatletter\renewcommand{\theparagraph}{\@arabic\c@paragraph}\makeatother%章节格式的设置\titleformat{\section}{\erhao\bf}{}{0em}{}[]\titleformat{\subsection}{\xiaoerhao\bf}{}{0em}{}[]\titleformat{\subsubsection}{\sanhao\bf}{}{0em}{}[]\titleformat{\paragraph}[hang]{\vspace*{0.5ex}\sihao\bf}{\hspace*{1em}\theparagraph)}{0.5em }{}[\vspace*{-0.5ex}]%更改插图的标题\renewcommand{\figurename}{\wuhao\bf\sf Figure}\renewcommand{\captionlabeldelim}{\ }%更改表格的标题\renewcommand{\tablename}{\wuhao\bf\sf Table}%更改图形或表格与其标题的间距\setlength{\abovecaptionskip}{10pt}\setlength{\belowcaptionskip}{10pt}%定义产生不浮动图形和表格的标题的命令\figcaption和\tabcaption\makeatletter\newcommand\figcaption{\def\@captype{figure}\caption}\newcommand\tabcaption{\def\@captype{table}\caption}\makeatother%自定义的可以调整粗细的水平线命令, 用于绘制表格, 调用格式\myhline{0.5mm}. \makeatletter\def\myhline#1{%\noalign{\ifnum0=`}\fi\hrule \@height #1 \futurelet\reserved@a\@xhline}\makeatother%第一层列表序号为带圈的阿拉伯数字\renewcommand{\labelenumi}{\textcircled{\arabic{enumi}}}%更改脚注设置\renewcommand{\thefootnote}{\fnsymbol{footnote}}\begin{document}\begin{CJK*}{GBK}{song}\CJKtilde\title{\bf\yihao Aviation Baggage Screening\\{\&} Flight Schedule}\author{}\date{}\maketitle\section{Introduction}Following the terrorist attacks on September 11, 2001, there isintense interest in improving the security screening process forairline passengers and their baggage. Airlines and airports areconsidered high-threat targets for terrorism, so aviation securityis crucial to the safety of the air-travelling public. Bombs andexplosives have been known to be introduced to aircraft by holdbaggage and cargo, carried on by passengers, and hidden withinaircraft supplies.At present To Screen or Not to Screen, that is a Hobson's choice.US Current laws mandate 100{\%} screening of all checked bags at the 429 passenger airports throughout the nation by explosive detection systems(EDS) by the end of the Dec 31 2003. However, because the manufacturers arenot able to produce the expected number of EDS required to meet the federal mandate, so it is significant to determine the correct number of devicesdeploy at each airport, and to take advantage of them effectively.The Transportation Security Administration (TSA) needs a complicatedanalysis on how to allocate limited device and how to best use them.Our paper contains the mathematical models to determine the number of EDSsand flight schedules for all airports in Midwest Region. We also discuss theETD devices as the additional security measures and the future developmentof the security systems.\section{Assumption and Hypothesis}\begin{itemize}\item The passengers who will get on the same airplane will arrive uniformly, namely the distribution is flat.\item The detection systems, both EDS and ETD, operate all the time during peak hour, except downtime.\item The airline checks the passengers randomly, according to its claim.\item The passengers, who are just landing and leave out, do not have to be checked through EDS or ETD.\item According to the literature, the aircraft loads approximately equal among the sets of departing flight during the peak hour.\item The landing flight did not affect the departure of the plane.\item Once a passenger arrives, he can go to EDS to be checked, except he has to wait in line.\item Once passengers finish screening, they can broad on the plane in no time.\item During peak hours, a set of flights departs at the same time every the same minutes.\item All the runways are used as much as possible during peak hours.\item The maximum number of the baggage is two, which a passenger can carry on plane. []\item The detection machine examines the bags at the same speed.\item EDS cannot make mistakes that it detect a normal object as an explosive.\end{itemize}\section{Variable and Definition}\begin{longtable}{p{100pt}p{280pt}}\caption{Variables}\\ %第一页表头的标题\endfirsthead %第一页的标题结束\caption{(continued)}\\ %第二页的标题\endhead %第二页的标题结束\hline\hline\textbf{Symbol}&\textbf{Description}\\\hline$n_{ij}$&The airplane number of the $i^{\mathrm{th}}$ type in the $j^{\mathrm{th}}$ flight set\\\hline${NP}_i\:(i=1,2,\ldots)$&The number of passengers on each airplanes of the same type.\\\hline$\xi_{ij}\:(i,j = 1,2,\cdots)$&The number of baggage on each airplane of the $j^{\mathrm{th}}$ flights\\\hline$a$&The maximal number of airplanes type\\\hline$B_j^{set}$&The total baggage number of each set of flight\\\hline${NF}_i$&Number of airplanes of each type\\\hline$\bar{\rho}$&The mean value of passengers' baggage coming per minute in every flight set\\ \hline$N_{set}$&The number of flight sets\\\hline$B_{total}$&The total number of checked baggage during the peak hour\\\hline$H_{peak}$&The length of the peak hour\\\hline$T_{set}$&The time length during which each flight set's passengers wait to be checked\\\hline$\Delta t$&The time interval between two consecutive flight set\\\hline$N_{EDS}$&The number of all the EDSs\\\hline$N_{shadow}$&The number of flight sets whose passengers will be mixed up before being checked\\\hline$v_{EDS}$&The number of baggage checking by one EDS per minute\\\hline$\rho_j$&The number of passengers' baggage coming per minute in one flight set\\\hline$N_{runway}$&The number of an airport's runway\\\hline\\*[-2.2ex]${\bar{B}}^{set}$&The mean value of checked baggage number of every flight set\\\hline$M$&The security cost\\\hline\hline\label{tab1}\end{longtable}\subsubsection{Definition:}\begin{description}\item[Flight set] A group of flights take off at the same time\item[The length of peak hour] The time between the first set of flight and the last set\end{description}\section{Basic Model}During a peak hour, many planes and many passengers would departfrom airports. Therefore, It is difficult to arrange for thepassengers to enter airports. If there are not enough EDSs forpassengers' baggage to check, it will take too long time for themto enter. That would result in the delay of airplanes. On thecontrary, if there are too many EDSs, it will be a waste. It isour task to find a suitable number of EDSs for airport. In orderto reach this objective, we use the linear programming method tosolve it.\subsection{Base analysis}The airplanes are occupied at least partly. The passengers'baggage would be checked by EDSs before they get on the airplanes.We have assumed that every passenger carry two baggages. Thisassumption would simplify the problem. According to the data fromthe problem sheet, we can obtain the useful information thatairlines claim 20{\%} of the passengers do not check any luggage,20{\%} check one bag, and the remaining passengers check two bags.Therefore, we can gain the total number of passengers' baggagethat should be carried on one plane: $\xi_{ij}$. Moreover, we canget the equation that calculate $\xi_{ij}$:\[\xi_{ij}={NP}_i\times 20\%+{NP}_i\times 60\%\times 2\]We define the matrix below as airplane baggage number matrix:\[\overset{\rightharpoonup}{\xi}_j=\left[\xi_{1j}\quad\xi_{2j}\quad\cdots\quad\xi_{ij}\quad\cdots\ right]\]We define the matrix below as flight schedule matrix:\[\left[\begin{array}{llcl}n_{11}&n_{12}&\cdots&n_{1N_{set}}\\n_{21}&n_{22}&\cdots&n_{2N_{set}}\\\multicolumn{4}{c}\dotfill\\n_{a1}&n_{a2}&\cdots&n_{aN_{set}}\end{array}\right]\]In this matrix, $n_{ij}$ is the airplane number of the$i^{\mathrm{th}}$ type in the $j^{\mathrm{th}}$ flight set whichwill take off. Apparently, this value is an integer.We define the matrix below as flight set baggage number matrix:\[\left[B_1^{set}\quad B_2^{set}\quad\cdots\quad B_j^{set}\quad\cdots\quad B_a^{set}\right] \]It is clear that they meet the relation below:\begin{equation}\begin{array}{cl}&\left[\xi_{1j}\quad\xi_{2j}\quad\cdots\quad\xi_{ij}\quad\cdots\right]\cdot\left[\begin{array}{llcl}n_{11}&n_{12}&\cdots&n_{1N_{set}}\\n_{21}&n_{22}&\cdots&n_{2N_{set}}\\\multicolumn{4}{c}\dotfill\\n_{a1}&n_{a2}&\cdots&n_{aN_{set}}\end{array}\right]\\=&\left[B_1^{set}\quad B_2^{set}\quad\cdots\quad B_j^{set}\quad\cdots\quad B_a^{set}\right]\end{array}\label{Flight:baggage}\end{equation}Then, we know:\[B_j^{set}=\sum\limits_{i=1}^a\xi_{ij}\times n_{ij}\]There are some constraints to the equation (\ref{Flight:baggage}).First, for each set of flight, the total number of airplanesshould be less than the number of runways. Second, the totalairplane number of the same type listed in the equation(\ref{Flight:baggage}) from every set of flight should be equal tothe actual airplane number of the same type during the peak hour.We can express them like these:\[\sum\limits_{i=1}^a n_{ij}\le N_{runway}\quad\quad\sum\limits_{j=1}^b n_{ij}={NF}_i \]We should resolve the number of flight sets. According to our assumptions,during the peak hour, the airlines should make the best use of the runways.Then get the number of flight sets approximately based on the number of allthe airplanes during the peak hour and that of the runways. We use anequation below to express this relation:\begin{equation}N_{set}=\left\lceil\frac{\sum\limits_{j=1}^{N_{set}}\sum\limits_{i=1}^an_{ij}}{N_{runway}}\right\rceil\label{sets:number}\end{equation}The checked baggage numbers of each flight set are equal to eachother according to our assumption. We make it based on literature.It can also simplify our model. We define $\bar{B}^{set}$ as themean value of checked baggage number of every flight set.Moreover, We define $\bar{\rho}$ as the mean value of checkedbaggage number of every flight set per minute:\[\bar{B}^{set}=\frac{B_{total}}{N_{set}}\]\[\bar{\rho}=\frac{\bar{B}^{set}}{T_{set}}=\frac{B_{total}}{T_{set}N_{set}}=\frac{B_{total}\ Delta t}{T_{set}H_{peak}}\]The course of passengers' arrival and entering airport isimportant for us to decide the number of EDSs and to make theflights schedule. Therefore, we should analyze this processcarefully. Passengers will arrive between forty-five minutes andtwo hours prior to the departure time, and the passengers who willget on the same airplane will arrive uniformly. Then we can getthe flow density of all checked baggage at any time duringpassengers' entering. This value is the sum of numbers ofpassengers' checked baggage coming per minute. To calculate thisvalue, firstly, we should obtain flow density of each flight set'schecked baggage. We define $\rho_j $, namely the number of checkedbaggage per minute of one flight set:\[\rho_j=\frac{B_j^{set}}{T_{set}}\]Secondly, we draw graphic to help us to understand. We userectangle to express the time length for all the passengers of oneflight set to come and check bags. In the graphic, the black partis the period for them to come. During the white part, nopassengers for this flight set come. According to the problemsheet, the former is 75 minute, and the latter is 45 minute. Thelength of rectangle is 120 minute. $T_{set}$ is the period duringwhich all passengers of one flight set wait to be checked. Sincewe have assumed that each time interval between two consecutiveflight set is same value, we define $\Delta t$ as it. Observe thesection that value we want to solve is $\sum\limits_j\rho_j$. Moreover, we can get another important equation from the graphic below:\begin{equation}N_{set}=\frac{H_{peak}}{\Delta t}\label{PeakHour}\end{equation}\begin{figure}[hbtp]\centering\includegraphics[width=298.2pt,totalheight=141.6pt]{fig01.eps}\caption{}\label{fig1}\end{figure}Each EDS has certain capacity. If the number of EDSs is $N_{EDS}$ and one EDS can check certain number of baggage per minute (Thatis checking velocity, marked by $v_{EDS}$), the total checking capacity is $N_{EDS}\cdot\frac{v_{EDS}}{60}$. $v_{EDS}$ is between 160 and 210.Now we can easily decide in what condition the passengers can be checked without delay:\[\sum\limits_j\rho_j\le v_{EDS}\]The passengers have to queue before being checked:$\sum\limits_j\rho_j>v_{EDS}$Well then, how can we decide how many $\rho_j$? It depends on how many flight sets whose passengers will be mixed up before being checked. We note it as $N_{shadow} $. Return to the Figure\ref{fig1}, we can know:\[N_{shadow}=\left\lfloor\frac{T_{set}}{\Delta t}\right\rfloor\]\begin{figure}%[htbp]\centering\includegraphics[width=240pt,totalheight=131.4pt]{fig02.eps}\caption{}\label{fig2}\end{figure}From Figure \ref{fig1} and Figure \ref{fig2}, we can get theresult as follows:\begin{enumerate}\item If $N_{shadow}\le N_{set}$, namely $H_{peak}>T_{set}$, then $\sum\limits_{j=1}^{N_{shadow}}\rho _j\le N_{EDS}\frac{v_{EDS}}{60}$\renewcommand{\theequation}{\arabic{equation}a}That is:\begin{equation}N_{EDS}\ge\frac{60}{v_{EDS}}\sum\limits_{j=1}^{N_{shadow}}\rho_j\approx\frac{60}{v_{ EDS}}N_{shadow}\bar{\rho}=\frac{60B_{total}\Deltat}{v_{EDS}T_{set}H_{peak}}N_{shadow}\label{EDS:number:a}\end{equation}\item If $N_{shadow}>N_{set}$, namely $H_{peak}\le T_{set}$, then $\sum\limits_{j=1}^{N_{set}}\rho_j\le N_{EDS}\frac{v_{EDS}}{60}$\setcounter{equation}{3}\renewcommand{\theequation}{\arabic{equation}b}That is:\begin{equation}N_{EDS}\ge\frac{60}{v_{EDS}}\sum\limits_{j=1}^{N_{set}}\rho_j\approx\frac{60}{v_{EDS} }N_{set}\bar{\rho}=\frac{60B_{total}\Delta t}{v_{EDS}T_{set}H_{peak}}N_{set}\label{EDS:number:b}\end{equation}\end{enumerate}\subsection{The number of EDSs}Then we begin to resolve the number of EDSs assisted by the linearprogramming method.EDS is operational about 92{\%} of the time. That is to say, whenever it isduring a peak hour, there are some EDSs stopping working. Then the workingefficiency of all the EDSs is less than the level we have expected.Therefore, the airline has to add more EDSs to do the work, which can bedone with less EDSs without downtime.We use binomial distribution to solve this problem. $N$ is the number ofactual EDSs with downtime and $k$ is the number of EDSs without downtime. Ifprobability is $P$, we can get the equation below:\[\left(\begin{array}{c}N\\k\end{array}\right)\cdot98\%^k\cdot(1-98\%)^{N-k}=P\]We can obtain $N$ when we give $P$ a certain value. In this paper,$P$ is 95{\%}. The $N_{EDS}$ is the actual number we obtainthrough the equation above.Now we have assumed that passengers can be checked unless be delayed by the people before him once he arrives at airport. Apparently, if the time length between two sets of flight is short, the density of passengers will begreat. It will bring great stress to security check and may even make some passengers miss their flight. To resolve this question, the airline has toinstall more EDSs to meet the demand. However, this measure will cost much more money. Consequently, we have to set a suitable time interval between two set of flight.Based on the base analysis above. We can use the equation(\ref{sets:number}) to decide the number of flight sets $N_{set}$assuming we know the number of runways of a certain airport. Thenbased on the equation (\ref{PeakHour}), we can decide the peakhour length $H_{peak}$ when we assume a time interval between two consecutive flight sets. Then we use \textcircled{1} and\textcircled{2} to decide which to choose between equation(\ref{EDS:number:a}) and equation (\ref{EDS:number:b}). In consequence, we can obtain the minimum of EDSs number.If we choose different numbers of runways and the time intervalsbetween two flight sets, we can get different EDSs numbers. Inthis paper that followed, we gain a table of some value of$N_{runway}$ and $\Delta t$ with the corresponding EDSs numbers. Moreover, we draw some figure to reflect their relation.For a certain airport, its number of runway is known. Givencertain time interval ($\Delta t$), we can get the length of thepeak hour ($H_{peak}$). When the $N_{runway}$ is few enough,perhaps $H_{peak}$ is too long to be adopted. However, for acertain airline, they can decide the time interval of their ownpeak hour. In this given time interval, they could find theminimum of $N_{runway}$ through the Figure \ref{fig3}. We draw asketch map to describe our steps.\begin{figure}[hbtp]\centering\includegraphics[width=352.8pt,totalheight=214.2pt]{fig03.eps}\caption{}\label{fig3}\end{figure}\subsection{The Flight Schedule }According to the base analysis, we can know that the flightschedule matrix and $\Delta t$ is one form of flight timetable. In``The number of EDSs'', we can get suitable $\Delta t$. Then weshould resolve the flight schedule matrix.Because we have assumed that the checked baggage numbers of each flight setare equal to each other. It can be described as follows:\[\left\{\begin{array}{l}\rho_j\approx\bar{\rho}\\B_j^{set}\approx\bar{B}^{set}\end{array}\right.\begin{array}{*{20}c}\hfill&{j=1,2,\cdots,N_{set}}\hfill\end{array}\]The flight schedule matrix subject to this group:\[\left\{\begin{array}{ll}\sum\limits_{j=1}^{N_{set}}n_{ij}={NF}_i&i=1,2,\cdots\\\sum\limits_{i=1}^a n_{ij}\le N_{runway}&j=1,2,\cdots,N_{set}\\n_{ij}\ge0,&\mathrm{and}\:n_{ij}\:\mathrm{is}\:\mathrm{a}\:\mathrm{Integer} \end{array}\right.\]In order to make the best use of runway, we should make$\sum\limits_{i=1}^a n_{ij}$ as great as we can unless it exceed$N_{runway}$.Then we can see that how to resolve the flight schedule matrix is a problemof divide among a group of integers. This group is all the numbers of eachflight passengers' baggage in one flight set. We program for this problemusing MA TLAB and we get at least one solution in the end. However, thematrix elements we have obtained are not integer, we have to adjust them tobe integers manually.\subsection{Results and Interpretation for Airport A and B}The number of passengers in a certain flight (${NP}_i$), the timelength of security checking ($T_{set}$), the checking velocity ofEDS ($v_{EDS}$), and the number of baggage carried by onepassenger are random.\subsubsection{Data Assumption:}\begin{itemize}\item $T_{set}$ is 110 minutes, which is reasonable for airline.\item To simplify the problem, we assume that every passenger carry 2 baggage. If some of thepassengers carry one baggage, the solution based on 2 baggages per passenger meets therequirement.\item The number of runways in airport A and airport B is 5.\end{itemize}\subsubsection{Airport A:}Once the number of runway and the number of the flights aredecided, the flight schedule matrix is decided, too. We producethis matrix using MATLAB. This matrix companied by $\Delta t$ isthe flight schedule for airport A. $\Delta t$ will be calculatedin (\ref{Flight:baggage}), (\ref{sets:number}) and(\ref{PeakHour}).We calculate $N_{EDS}$ and make the flight timetable in threeconditions. The three conditions and the solution are listed asfollowed:\paragraph{Every flight are fully occupied}The checking speed of EDS is 160 bags/hour.\begin{table}[htbp]\centering\caption{}\begin{tabular}{*{11}c}\myhline{0.4mm}$\mathbf{\Deltat(\min)}$&\textbf{2}&\textbf{4}&\textbf{6}&\textbf{8}&\textbf{10}&\textbf{12}&\textbf{14} &\textbf{16}&\textbf{18}&\textbf{20}\\\myhline{0.4mm}$N_{EDS}(\ge)$&31&31&31&31&31&29&24&22&20&17\\\hline$H_{peak}(\min)$&20&40&60&80&100&120&140&160&180&200\\\myhline{0.4mm}\end{tabular}\label{tab2}\end{table}We assume that the suitable value of $H_{peak}$ is 120 minutes.Then the suitable value of $\Delta t$ is about 12 minutes, and$N_{EDS}$ is 29 judged from Figure \ref{fig4}. Certainly, we canwork $\Delta t$ and $N_{EDS}$ out through equation.\begin{figure}[htbp]\centering\includegraphics[width=294.6pt,totalheight=253.2pt]{fig04.eps}\caption{}\label{fig4}\end{figure}\paragraph{Every flight is occupied by the minimal number of passengers onstatistics in the long run.}The checking speed of EDS is 210 bags/hour.\begin{table}[htbp]\centering\caption{}\begin{tabular}{*{11}c}\myhline{0.4mm}$\mathbf{\Deltat(\min)}$&\textbf{2}&\textbf{4}&\textbf{6}&\textbf{8}&\textbf{10}&\textbf{12}&\textbf{14} &\textbf{16}&\textbf{18}&\textbf{20}\\\myhline{0.4mm}$N_{EDS}(\ge)$&15&15&15&15&15&14&13&12&10&7\\\hline$H_{peak}(\min)$&20&40&60&80&100&120&140&160&180&200\\\myhline{0.4mm}\end{tabular}\label{tab3}\end{table}We assume that the suitable value of $H_{peak}$ is 120 minutes.Then the suitable value of $\Delta t$ is about 12 minutes, and$N_{EDS}$ is 14 judging from Figure \ref{fig5}. Certainly, we canwork $\Delta t$ and $N_{EDS}$ out through equation.\begin{figure}[htbp]\centering\includegraphics[width=294.6pt,totalheight=253.2pt]{fig05.eps}\caption{}\label{fig5}\end{figure}\paragraph{${NP}_i$ and $v_{EDS}$ are random value produced by MATLAB.}\begin{table}[htbp]\centering\caption{}\begin{tabular}{*{11}c}\myhline{0.4mm}$\mathbf{\Deltat(\min)}$&\textbf{2}&\textbf{4}&\textbf{6}&\textbf{8}&\textbf{10}&\textbf{12}&\textbf{14} &\textbf{16}&\textbf{18}&\textbf{20}\\\myhline{0.4mm}$N_{EDS}(\ge)$&15&22&21&21&15&17&21&16&13&14\\\hline$H_{peak}(\min)$&20&40&60&80&100&120&140&160&180&200\\\myhline{0.4mm}\end{tabular}\label{tab4}\end{table}We assume that the suitable value of $H_{peak}$ is 120 minutes.Then the suitable value of $\Delta t$ is about 12 minutes, and$N_{EDS}$ is 17 judging from Figure \ref{fig6}. Certainly, we canwork $\Delta t$ and $N_{EDS}$ out through equation.\begin{figure}[htbp]\centering\includegraphics[width=294.6pt,totalheight=249.6pt]{fig06.eps}\caption{}\label{fig6}\end{figure}\subsubsection{Interpretation:}By analyzing the results above, we can conclude that when$N_{EDS}$ is 29, and $\Delta t$ is 12, the flight schedule willmeet requirement at any time. The flight schedule is:\\[\intextsep]\begin{minipage}{\textwidth}\centering\tabcaption{}\begin{tabular}{c|*{8}c|c|c}\myhline{0.4mm}\backslashbox{\textbf{Set}}{\textbf{Type}}&\textbf{1}&\textbf{2}&\textbf{3}&\textbf{4}&\te xtbf{5}&\textbf{6}&\textbf{7}&\textbf{8}&\textbf{Numbers of Bags}&\textbf{Numbers of Flights}\\\myhline{0.4mm}1&2&0&0&0&2&1&0&0&766&5\\\hline2&2&0&2&0&2&0&0&0&732&4\\\hline3&0&1&1&1&2&0&0&0&762&4\\\hline4&0&1&0&0&2&1&0&0&735&4\\\hline5&0&1&0&0&2&1&0&0&735&5\\\hline6&2&0&0&0&1&0&0&1&785&5\\\hline7&2&0&0&0&2&0&1&0&795&5\\\hline8&0&1&0&0&2&1&0&0&735&4\\\hline9&2&0&0&0&2&1&0&0&766&5\\\hline10&0&0&0&2&2&0&0&0&758&5\\\hlineTotal&10&4&3&3&19&5&1&1&7569&46\\\myhline{0.4mm}\end{tabular}\label{tab5}\end{minipage}\\[\intextsep]We have produced random value for ${NP}_i$ and $v_{EDS}$. On thiscondition, the number of EDSs is 17, which is less than 29 that wedecide for the airport A. That is to say our solution can meet thereal requirement.\subsubsection{Airport B:}\paragraph{The passenger load is 100{\%}}The checking speed of EDS is 160 bags/hour.\begin{table}[htbp]\centering。

Camping along the Big Long RiverSummaryIn this paper,the problem that allows more parties entering recreation system is investigated.In order to let park managers have better arrangements on camping for parties,the problem is divided into four sections to consider.The first section is the description of the process for single-party's rafting.That is, formulating a Status Transfer Equation of a party based on the state of the arriving time at any campsite.Furthermore,we analyze the encounter situations between two parties.Next we build up a simulation model according to the analysis above.Setting that there are recreation sites though the river,count the encounter times when a new party enters this recreation system,and judge whether there exists campsites available for them to station.If the times of encounter between parties are small and the campsite is available,the managers give them a good schedule and permit their rafting,or else, putting off the small interval time t∆until the party satisfies the conditions.Then solve the problem by the method of computer simulation.We imitate the whole process of rafting for every party,and obtain different numbers of parties,every party's schedule arrangement,travelling time,numbers of every campsite's usage, ratio of these two kinds of rafting boats,and time intervals between two parties' starting time under various numbers of campsites after several times of simulation. Hence,explore the changing law between the numbers of parties(X)and the numbers of campsites(Y)that X ascends rapidly in the first period followed by Y's increasing and the curve tends to be steady and finally looks like a S curve.In the end of our paper,we make sensitive analysis by changing parameters of simulation and evaluate the strengths and weaknesses of our model,and write a memo to river managers on the arrangements of rafting.Key words:Camping；Computer Simulation;Status Transfer Equation1IntroductionThe number of visits to outdoor recreation areas has increased dramatically in last three decades.Among all those outdoor activities,rafting is often chose as a family get-together during May to September.Rafting or white water rafting is a kind of interesting and challenging recreational outdoor activity,which uses an inflatable raft to navigate a river or sea [1].It is very popular in the world,especially in occidental countries.This activity is commonly considered an extreme sport that usually done to thrill and excite the raft passengers on white water or different degrees of rough water.It can be dangerous.During the peak period,there are many tourists coming to experience rafting.In order to satisfy tourists to the maximum,we must make full use of our facilities in hand,which means we must do the utmost to utilize the campsites in the best way possible.What's more,to make more people feel the wildness life,we should minimize the encounters to the best extent;meanwhile no two sets of parties can occupy the same campsite at the same time.It is naturally coming into mind that we should consider where to stop,and when to stop of a party [2].In previous studies [3-5],many researchers have simulated the outdoor creation based on real-life data,because the approach is dynamic,stochastic,and discrete-event,and most recreation systems share these traits.But there exists little research aiming at describing the way that visitors travel and distribute themselves within a recreation system [6].Hence,in our paper,we consider the whole process of parties in detail and simulate every party ’s behavior,including the location of their campsites,and how long it will last for them to stay in a campsite to finish their itineraries.Meanwhile minimize the numbers of encounters.Aiming at showing the whole process of rafting,we firstly focus on analyzing the situation s of a single-party's rafting by using status transfer equation,then consider the problems of two parties'encounters on the river.Finally,after several times of simulation on the whole process of rafting,we obtain the optimal value of X .2Symbols and DefinitionsIn this section,we will give some basic symbols and definitions in the following for the convenience.Table 1.Variable Definition Symbols Definitioni v i p j i q ,S dThe velocity of oar or motor0-1variables on choosing rafting transportation0-1variables on the occupation of campsitesLength of the riverAverage distance between two campsites3General AssumptionsIn order to have a better study on this paper,we simplify our model by thefollowing assumptions:1)19:00to 07:00is people's sleeping time,during this time,people are stationedin the campsite.The total time of sleeping is 12hours,as rafting is an exiting sport game,after a day's entertainment,people have cost a lot of energy,and nearly tired out.So in order to have a better recreation for the next day,we set that people begin their trip at 07:00,and end at 19:00for a day's schedule.2)Oar-powered rubber rafts and motorized parties can successfully raft from FirstLaunch to Final Exit,there exist no accident over the whole trips.3)All the rubber rafts and motorized boats have the same exterior except velocities;we regard a rubber raft or a motorized boat as a party and don't consider the tourists individuals on the parties.4)There is only one entrance for parties to enter the recreation system.5)Regardless of the effects that the physical features of the river brings to oar andmotorized parties,that is to say we ignore the stream ’s propulsion and resistance to both kinds of rafting boats.Oar and motorized parties can keep the average velocity of 4mph and 8mph.6)Divide the whole river into N segments.4Analysis of This Rafting ProblemRafting is a very popular spots game world-wide.In the peak period of rafting,there are more people choosing to raft,it often causes congestion that not all people can raft at any time they want.Hence,it is important for managers to set an optimal schedule for every party (from our assumptions,we regard a rafting boat as a party)in advance.Meanwhile,the parties need to experience wildness life,so the managers should arrange the schedules which minimize the encounters'time between parties to the best extent.What's more,no two sets of parties can occupy the same site at the same time.Our aim is to determine an optimal mix of trips over varying duration (measured YXNj i t ,jT t∆KNumbers of campsites Numbers of parties Numbers of attraction sites Time of the i th party finishing the whole trip ranges from6days to 18daysRandom staying time at each campsiteDelay time of rafting from beginning Threshold value of encounterin nights on the river.That is to say,we must obtain an optimal value of X through lots of trails.This optimal value represents that the campsites have a high usage while more people are available to raft.The Long Big River is 225miles long,if we discuss the river as a whole and consider all the parties together,it will be difficult for us to have a clear recognition on parties'behaviors.Hence,we divide the river into N attraction sites.Each of the attraction sites has Y/N campsites since the campsites are uniformly distributed throughout the river corridor.So build up a model based on single-party ’s behavior of rafting in small distance.At last,we can use computer simulation to imitate more complex situations with various rafting boats and large quantities of parties.5Mathematic Models5.1Rafting of the Single-party Model （Status Transfer Equation [7]）From the previous analysis,in order to have a clear recognition of the whole rafting process,we must analyze every single-party's state at any time.In this model,we consider the situation that a single-party rafts from the First Launch to the Final Exit.So we formulate a model that focus on the behavior of one single-party.For a single-party,it must satisfy the following equation:status transfer equation.it represents the relationships between its former state and the latter state.State here means:when the i th party arrives at the j th campsites,the party may occupy the j th campsite or not.As a party can choose two kinds of transportation to raft:oar-powered rubber rafts(i v =4mph)and motorized rafts(i v =8mph).i v is the velocity of the rafting boats,and i p is the 0-1variables of the selecting for boats.Therefore,we can obtainthe following equation:)1(84i i i p p v −+=(i=1,2,…,X ).(1)where i p =0if the i th party uses motorized boat as their rafting tool,at thistime i v =8mph ;while i p =1when ,the i th party rafts with oar-powered rubber raft with i v =4mph.In fact,Eq.(1)denotes which kind of rafting boat a party can choose.A party not only has choice on rafting boats,but also can select where to camp based on whether the campsites are occupied or not.The following formulation shows the situation whether this party chooses this campsite or not:⎩⎨⎧=party previous a by occupied is campsite the 0,party previous a by occupied not is campsite the q ij ,1(2)where i =1,2,…,X ;j =1,2,…,Y .Where the next one can’t set their camp at this place anymore,that is to say thelatter party’s behavior is determined by the former one.As campsites are fairly uniformly distributed throughout the river corridor,hence,we discrete the whole river into segments,and regard Y campsites as Y nodes which leaves out (Y +1)intervals.Finally we get the average distance between th e j th campsite and (j+1)th campsite:1+=Y Sd (3)where is the length of the river,and its value is 225miles.What’s more,the trip-days for a party is not infinite,it has fluctuating intervals:h t h j i 432144,≤≤(4)where is the t i ，j itinerary time for a party ranges from 144hours to 432hours (6to 18nights).From Eq.(1),(2)and (3),the status transfer equation is given as follows:),...2,1,,...2,1(11,1,,Y j X i T q v d t t j j i i j i j i ==×++=−−−（5）The i th party’s arriving time at the j th campsite is determined by the time when the i th arrived at （j-1）campsite,the time interval i v d ,and the time T j-1random generated by computer shown in Eq.(5).It is a dynamic process and determined by its previous behavior.5.2The Analysis of Two Parties Parties’’Encounter on the River Our goal is to making full use of the campsites.Hence,the objective of all the formulation is to maximize the quantities of trips （parties ）X while consider getting rid of the congestion.If we reduce the numbers of the encounters among parties,there will be no congestion.In order to achieve this goal,we analysis the situations of when two parties’to encounter,and where they will encounter.In order to create a wildness environment for parties to experience wildness life,managers arrange a schedule that can make any two parties have minimal encounters with each other.Encounter is that parties meet at the same place and at the same time.Regarding the river as a whole is not convenient to study,hence,our discussion is based on a small distance where distance=d (Eq.3),between the j th and (j+1)th campsites.Finally the encounter problem of the whole river is transferred into small fractions.On analyzing encounter problem in d and count numbers of each encounter in d together,we get a clear recognition of the whole process and the total numbers of encounter of two parties.The following Figure 1represents random two parties rafting in d :Figure 1.Random two parties'encounter or not on the riverThe i th party arrives at j th campsite (t j k ,-t j i ,)time earlier than the k th party reaches the j th campsite.After t time,interval distance between the i th party and the k th party can be denoted by the following function:)()(t t t v t v t S ij kj i k j +−×−×=∆(6)Where k,i =1,2,…,X ,j =1,2,…,Y .k i ≠.Whether the two parties stationed on the j th campsite and(j +1)th campsite are based on the state of the campsites’occupation,yields we obtain:⎩⎨⎧=×01,,j k j i q q (i,k =1,2,…,X ;j =1,2,…,Y ;k ≠i )(7)Note that Eq.6is constrained by Eq.7,for different value of )(t S J ∆andj k j i q q ,,×we can obtain the different cases as follows:Case 1:⎩⎨⎧=×=∆10)(,,j k j i j q q t S (8)Which means both the i th and k th party don’t choose the j th campsite,they are rafting on the river.Hence,when the interval distance between the two parties is 0,that is )(t S J ∆=0,they encounter at a certain place in d on the river.Cases2:⎩⎨⎧=×=∆00)(,,j k j i j q q t S (9)Although the interval distance between the two parties is 0,the j th campsite is occupied by the i th party or the k th party.That is one of them stop to camp at a certain place throughout the river corridor.Hence,there is no possibility for them to encounter on the river.Cases 3:⎩⎨⎧=×≠∆10)(,,j k j i j q q t S ⎩⎨⎧=×≠∆00)(,,j k j i j q q t S (10)No matter the j th campsite is occupied or not for )(t S J ∆≠0,that is at the same time,they are not at the same place.Hence,they will not encounter at any place in d .5.3Overview of Computer Simulation Modeling to Rafting5.3.1Computer SimulationSimulation modeling is a kind of method to imitate the real-word process or a system.This approach is especially suited to those tasks which are too complex for direct observation,manipulation,or even analytical mathematical analysis (Banks and Carson 1984,Law and Kelton 1991,Pidd 1992).The most appropriate approach for simulating out-door recreation is dynamic,stochastic,and discrete-event model,since most recreation systems share these traits.In all,simulation models can reflect the real-world accurately.5.3.2Simulation for the Whole Process of Parties on Rafting [8]This simulation can approximate show a party’s behavior on the river under a wide rang of conditions.From the analysis of the previous study,we have known that the next party’s behavior is affected by the former one.Hence,when the first party enters the rafting system,there is no encounter,and it can choose every campsite.then the second party comes into the rafting system ,at this time,we must consider the encounter between them,and the limit on choosing the campsite.As time goes by,more and more parties enter this system to raft which lead to a more complex situation.A party who satisfies the following two conditions will be removed from the current order to the next order.So he can’t “finish his trip”right away.The two conditions are as follows:(1)He chooses a campsite where has been occupied by other parties.(2)He has two many encounters with other parties.So in order to determine typical trip itineraries for various types of rafting boat ,campsite,and time intervals （See Trip Schedule Sheet 1），we need to perform a series of trails run that can represent the real-life process of rafting based on these considerations,.A main flowchart of the program is shown in Figure 2.Figure2.Main simulation flowchartAfter several times of simulation,we obtain the optimal X(the numbers of campsites),minimal E(Encounter)and TP(Trip Time).Followed by Figure2,we simulate the behavior of a party whether it can enterthe rafting system or not in Figure3.Figure3.Sub flowchart5.3.3The Results of SimulationAfter simulating the whole process of parties rafting on the river,we get three figures(Figure4,Figure5and Figure6)to present the results.In order to simulate the rafting process more conveniently,we divide the whole river into31segments(31attraction sites),and input an initial value of Y=155(numbers of campsites),where there are5campsites in every attraction sites.We represent the times of campsites occupied by various parties on Figure2by coordinates(x,y),where x is the order of the campsites from0to155（these campsites are all uniformly distributed thorough the corridor）,and y is the numbers of each campsite occupied by different parties.For example,(140，1100)represents that at the campsite,there exists nearly1100times of occupation in total by parties over180days. Hence,the following Figure4shows the times of campsites’usage from March to September.Figure4.Numbers of campsites'usage during six-month period from March to SeptemberFrom Figure4,The numbers of campsites’usage can be identified the efficiency of every campsites’usage.The higher usage of the campsites,the higher efficiency they are.Based on these,we give a simple suggestion to managers(see in Memo to Managers).Figure5.the ratio of usage on campsites with time going byFigure5shows the changes of the ratio on campsites.when t=0,the campsites are not used,but with time going by,the ratio of the usage of campsites becomes higher and higher.We can also obtain that when t>20,the ratio keeps on a steady level of65%;but when t >176,the ratio comes down,that is,there are little parties entering the recreation system.In all,these changes are rational very much,and have high coincidence with real-world.Then we obtain1599parties arranged into recreation system after inputting the initial value Y=155,and set orders to every party from number0to number1599.Plotting every party's travelling time of the whole process on a map by simulating,as follows:Figure6.Every party’s travelling timeFigure6shows the itinerary of the travelling time,most of the travelling time is fluctuating between13days and15.3days,and most of travelling time are concentrated around14days.In order to create an outdoor life for all parties,we should minimize the numbers of encounter among different parties based on equations(6)and(7)：So we get every party’s numbers of encounter by coordinates(x,y),where x is the order of the parties from0to1600,y is the numbers of encounters.Shown in Figure7,as follows:Figure7.Every party’s numbers of encounterFigure7shows every party’s numbers of encounter at each campsite.From this figure,we can know that the numbers of their encounter are relatively less,the highest one is8times,and most of the parties don’t encounter during their trips,which is coincident with the real-world data.Finally,according to the travelling time of a party from March to September,we set a plan for river managers to arrange the number of parties.Hence,by simulating the model,we obtain the results by coordinate(x,y),where y is the days of travelling time,x is the numbers of parties on every day.The figure is shown as follows:Figure8.Simulation on travelling days versus the numbers of parties From Figure8.we set a suitable plan for river manager,which also provide reference on his managements.6Sensitive AnalyzeSensitive analysis is very critical in mathematical modeling,it is a way to gauge the robustness of a model with respect to assumptions about the data and parameters. We try several times of simulation to get different numbers of parties on changing the numbers of campsites ceaselessly.Thus using the simulative data,we get the relationship between the numbers of campsites and parties by fitting.On the basis of this fitting,we revise the maximal encounter times(Threshold value)continually,and can also get the results of the relationships between the numbers of campsites and parties by fitting.Finally,we obtain a Figure9denoting the relations of Y(numbers of campsites)and X(numbers of parties),as follows:Figure9.Sensitive analysis under different threshold values Given the permitted maximal numbers of encounters(threshold value=K),we obtain the relationships between Y(numbers of campsites)and X(numbers of trips). For example,when K=1,it means no encounters are allowed on the river when rafting;when K=2,there is less than2chances for the boats to meet.So we can define the K=4,6,8to describe the sensitivity of our model.From Figure9,we get the information that with the increase of K,the numbers of boats available to rafts till increase.But when K>6,the change of the numbers of boats is inconspicuous,which is not the main factor having appreciable impact on the numbers of boats.>In all,when the numbers of campsites(Y)are less than250,they would have a great effect on the numbers of boats.But in the diverse situations,like when Y>250, the effect caused by adding the numbers of campsites to hold more boats is not notable.When K<6,the numbers of boats available increases with the ascending of K, While K>6,the numbers of boats don’t have great change.Take all these factors into consideration,it reflects that the numbers of the boats can’t exceed its upper limit.Increase the numbers of campsites and numbers of encounter blindly can’t bring back more profits.7Strengths and WeaknessesStrengthsOur model has achieved all of the goals we set initially effectively.It is not only fast and could handle large quantities of data,but also has the flexibility we desire.Though we don’t test all possibilities,if we had chosen to input the numbers of campsites data into our program,we could have produced high-quality results with virtually no added difficulty.Aswell,our method was robust.Based on general assumptions we have made in previous task,we consider a party’s state in the first place,then simulate the whole process of rafting.It is an exact reflection of the real-world.Hence,our main model's strength is its enormousedibility and stability and there are some key strengths:(1)The flowchart represents the whole process of rafting by given different initialvalues.It not only makes it possible to develop trip itineraries that are statistically more representatives of the total population of river trips,but also eliminates the tedious task of manual writing.(2)Our model focuses on parties’behavior and interactions between each other,notthe managers on the arrangement of rafting,which can also get satisfactory and high-quality results.(3)Our model makes full use of campsites,while avoid too many encounters,whichleads to rational arrangements.WeaknessesOn the one hand,although we list the model's comprehensive simulation as a strength,it is paradoxically also the most notable weakness since we don’t take into account the carrying capacity of the water when simulates,and suppose that a river can bear as much weight as possible.But in reality,that is impossible.On the other hand,our results are not optimal,but relative optimal.8ConclusionsAfter a serial of trials,we get different values of X based on the general assumptions we make.By comparing them,we choose a relative better one.From this problem,it verifies the important use of simulation especially in complex situations. Here we consider if we change some of the assumptions,it may lead to various results. For example,(a)Let the velocity of this two kinds of boats submit to normal distribution.In this paper,the average velocity of oar-powered rubber rafts and motorized boats are 4mphand8mph,respectively.But in real-world,the speed of the boats can’t get rid of the impacts from external force like stream’s propulsion and resistance.Hence,they keep on changing all the time.(b)Add and reduce campsites to improve the ratio of usage on campsites.By analyzing and simulating,the usage of each campsite is different which may lead to waste or congestion at a campsite.Hence,we can adjust the distribution of campsites to arrive the best use.A Memo to River ManagersOur simulation model is with high edibility and stability in many occasions.It can imitate every party’s behavior when rafting so as to make a clear recognition of the process.Internal Workings of The ModelInputsOur model needs to input initial value of Y,as well as the numbers of attraction sites. Algorithm(Figure2,and Figure3)Our algorithm represents the whole process of rafting,so we can use it to simulate the process of rafting by inputting various initial values.OutputsBased on the algorithm in our paper,our model will output the relative optimalnumbers of parties X.Furthermore,we can also get other information,such as the interval time between two parties at First Launch,a detailed schedule for each party of rafting,the relationship between X and Y and so on.Summary and RecommendationsAfter100times of simulating,we come to two conclusions:(a)The numbers of parties(X)have relations with the numbers of campsites(Y), that is to say,with the increasing of Y,the increasing speed of X goes fast at the first place and then goes down,finally it tends to be steady.Hence,we advice river managers to adjust the numbers of campsites properly to get the optimal numbers of parties.(b)Add campsites to the high usage of the former campsites and deduce campsites at the low usage of the former campsites.From Figure4,we know that the ratios of every campsites are different,some campsites are frequently used,but some are not.Thus we can infer that the scenic views are attractive,and have attracted lots of parties camping at the campsite.so we can add campsites to this nodes.Else the campsites with low usage have lost attractions which we should reduce the numbers of campsites at those nodes.References[1]KarloŠimović，Wikipedia，Rafting，/wiki/Rafting.[2] C.A.Roberts and R.Gimblett，Computer Simulation for Rafting Traffic on theColorado River，COMPUTER SIMULATION FOR RAFTING TRAFFIC,2001, 19-30.[3] C.A.Roberts,D.Stallman，J.A.Bieri.Modeling complex human-environmentinteractions:the Grand Canyon river trip simulator，Ecological Modeling153(2002)181-196.[4]J.A.Bieri and C.A.Roberts，Using the Grand Canyon River Trip Simulator toTest New Launch Scheduleson the Colorado River，Washington DC,AWISMagazine,Vol.29,No.3,2000,6-10.[5] A.H.Underhill and A.B.Xaba，The Wilderness Simulation Model as aManagement Tool for the Colorado River in Grand Canyon National Park，NATIONAL PARK SERVICE/UNIVERSITY OF ARIZONA Unit SupportProject CONTRIBUTION NO.034/03.[6] B.Wang and R.E.Manning，Computer Simulation Modeling for RecreationManagement:A Study on Carriage Road Use in Acadia National Park,Maine,USA,USA,Vermont05405,1999.[7]M.M.Meerschaert，Mathematical Modeling(Third Edition).China MachinePress publishing,2009.[8]A.H.Underhill，The Wilderness Use Simulation Model Applied to Colorado RiverBoating in Grand Canyon National Park,USA，Environmental Management Vol.10,No.3,1986,367-374.。

2013建模美赛B题思路数学建模美赛B题论文摘要水资源是极为重要生活资料，同时与政治经济文化的发展密切相关，北京市是世界上水资源严重缺乏的大都市之一。

本文以北京为例，针对影响水资源短缺的因素，通过查找权威数据建立数学模型揭示相关因素与水资源短缺的关系，评价水资源短缺风险并运用模型对水资源短缺问题进行有效调控。

首先，分析水资源量的组成得出影响因素。

主要从水资源总量（供水量）和总用水量（需水量）两方面进行讨论。

影响水资源总量的因素从地表水量，地下水量和污水处理量入手。

影响总用水量的因素从农业用水，工业用水，第三产业及生活用水量入手进行具体分析。

其次，利用查得得北京市2001-2008年水量数据，采用多元线性回归，建立水资源总量与地表水量，地下水量和污水处理量的线性回归方程yˆ=-4.732+2.138x1+0.498x2+0.274x3根据各个因数前的系数的大小，得到风险因子的显著性为rx1>rx2>rx3(x1, x2，x3分别为地表水、地下水、污水处理量)。

再次，利用灰色关联确定农业用水、工业用水、第三产业及生活用水量与总用水量的关联程度ra =0.369852，rb= 0.369167，rc=0.260981。

从而确定其风险显著性为r a>r b>r c。

再再次，由数据利用曲线拟合得到农业、工业及第三产业及生活用水量与年份之间的函数关系，a=0.0019（t-1994）3-0.0383（t-1994）2-0.4332（t-1994）+20.2598;b=0.014（t-1994）2-0.8261t+14.1337;c=0.0383（t-1994）2-0.097（t-1994）+11.2116;D=a+b+c；预测出2009-2012年用水总量。

最后，通过定义缺水程度S=（D-y）/D=1-y/D，计算出1994-2008的缺水程度，绘制出柱状图，划分风险等级。

我们取多年数据进行比较，推测未来四年地表水量和地下水量维持在前八年的平均水平，污水处理量为近三年的平均水平，得出2009-2012年的预测值，并利用回归方程yˆ=-4.732+2.138x1+0.4982x2+0.274x3计算出对应的水资源总量。

Winter is approaching, may the dragon’s wings grow moreabundantSummaryIn the game of thrones, Daenerys Targaryen depicts the image of a dragon. In eastern and western cultures, the phenomenon of dragons is not uncommon. If dragons live in modern society, how can we raise these war monsters? Research, and applied the cross disciplines of biology, physics, and chemistry to build a mathematical model and solve it to achieve the maximum growth of the dragon. Of course, dragons do not exist in real life, so we likened pterosaurs, modern Aircraft and chemical burner to derive the specific physiological characteristics of the dragon to ensure the rationality and scientificity of the research.First, we studied the flight and fire-spitting models of dragons. Through analogical reasoning, our hypothetical dragon's fire-spitting principle is similar to modern alcohol flamethrowers. For dragon flight, we used fluid mechanics to get the dragon's flight speed. And glucose energy loss. Combining the two to get the energy loss model of the dragon. Second, we studied the basic physical characteristics of the dragon. For the relationship between the body length and body age of the dragon, we established an elastic model of growth. Because the weight and body length of dragons have upper and lower limits, in order to comply with basic ecology, we have defined the dragon's bone saturation value as the cut-off value, and conducted a segmented study. When studying the relationship between weight and body length, We know that the weight of the dragon is proportional to the cube of the body length. Then, because the dragon needs resources to replenish like other animals, we built a dragon's food supply model. Suppose that the three dragons have the same competitiveness and the daily sheep Resources are the same. According to ecology, when the number of sheep in a certain area reaches k / 2, we need to migrate the dragon. Finally, the temperature will affect the living environment of the dragon, so the dragon needs to followMigration was selected for changes in temperature, and we selected three areas of drought, cold, and warmth to study the dragon, and integrated the model of the regional area of the dragon by the appealing model.In addition, we wrote a letter to the author of the Song of Ice and Fire, giving some suggestions on the actual ecological foundation of the dragon, hoping to be adopted. Although the dragon does not exist in our real life, the dragon can be broken down into Part of our modern society. For the dragon's flying spitfire energy loss model, we can further study the aircraft's fluid mechanics and modern flamethrowers. The study of non-existent organisms also prepares us for the arrival of new species .table of ContentsWinter is approaching, may the dragon’s wings grow more abundant (1)Summary (1)table of Contents (2)1 Introduction (3)1.1 restatement (3)1.2 Problem Analysis (3)2 Assumptions and reasons (4)3 Symbol Definition (4)4. Mathematical modeling (5)4.1 About Dragon Flight and Spitfire Consumption (5)4.2 About the relationship between dragon's body length and weight and age (7)4.3 About Dragon's Food Supply (8)4.4 Regulating the area of dragons by region (9)5 Sensitivity analysis (10)6 Model evaluation and outlook (11)6.1 Model evaluation (11)6.2 Further discussion (12)7 to a letter from George RR Martin (12)8.Appendix: (13)8.1 References (13)8.2 Matlab code (13)1 Introduction1.1 restatementIn the magical TV series "Game of Thrones", Daenerys Targaryen, known as the Mother of Dragons, raised three dragons as an aggressive army. Dragons have always been the most mysterious monsters in Eastern and Western cultures, but if Dragons live in the present era, how should we feed the three dragons in pursuit of maximum growth? In this article, we assume that the growth rules of dragons are in line with basic biology. To study them, we build mathematical models to solve problem.a. Analyze the change of the dragon's weight length with age, and estimate the value of the dragon's weight length corresponding to the age group.b. Investigate the loss of self energy during dragon fire, flight, and breathing, so as to estimate the minimum supply value of dragon for external activitiesc. Dragons need food and survival areas like other animals in the real world. Through certain assumptions and calculations, we can determine the total amount of food that dragons need daily and the size of living areas in three areas.d. Sensitivity analysis: As temperature and climate change, dragons will also migrate to different regions. Therefore, we need to analyze the differences in the impact of dragons on the survival of arid regions, temperate regions, and cold regions.1.2 Problem AnalysisBecause dragons do not exist in real life, we need to use some things in the real world to compare dragons in order to achieve the purpose of studying dragons. In analyzing the biological morphological characteristics of dragons, we use the knowledge of ecology and basic elements of biology Let's conceive the basic biological characteristics of the dragon such as weight and body length. For the energy loss model of the dragon, we have studied three aspects to describe its loss. Here we compare the modern flamethrower and establish related chemical equations to achieve the research of the dragon. Spitfire loss. In addition, in TV series such as "Game of Thrones" we will find that dragons can fly in common sense, so we have derived the dragon's flight loss. Of course, all aerobic organisms can breathe. Dragons are no exception, so there is a loss of breathing to maintain body temperature. At the same time, in order to make up for the loss of dragons in daily activities, we have established a material reserve model, in which materials are cattle and sheep in real life, etc. Finally, during the cyclical changes in climate and food, the dragons we feed will also migrate to some extent, so we analyzed the impact of different regions on the growth of dragons.Into account various factors that we can more scientific training of dragons, have achieved our purpose.2 Assumptions and reasonsAfter a comprehensive analysis of the problem, in order to increase the enforceability, we make the following assumptions to ensure the rationality of our model establishment.2.1 Assumptions: The basic biological characteristics of dragons are in line with the law of biological growth. In modern life, the growth and development of dragons should also be similar to other animals and conform to basic biology.2.2 Assumption: The dragon will spit fire and fly, and its flight conforms to the physical environment of fluid mechanicsReason: In Game of Thrones, the image of the dragon was once able to fly and spit fire.2.3 Assumption: In the single field we are studying, the environment of a certain area will not change abruptly and maintain a dynamic stability.2.4 Hypothesis: Dragons are top predators in the food chain, but dragons do not cause devastating harm to the biosphere.2.5 Assumption: The weight distribution of the dragon is uniform, and the body length reaches 30 to 40 cm at the time of birth.Reason 2.6: We refer to ancient biology and some dinosaur fossils.2.7 Hypothesis: Except for the skull, heart, liver, lungs, kidneys, bones, etc., the sum of other body masses is proportional to the cube of height.Reason: The hypothesis is obtained by counting the relationship between body length and weight of modern organisms.2.8 Hypothesis: The dragon is a constant temperature animal whose body temperature is not affected by external factors.Reason: A few pterosaur fossils have traces of "hair" on the surface, while the dragons in Game of Thrones are similar to pterosaurs.2.9 Hypothesis: The dragon is fully aerobic during the flight to provide energy2.10 Hypothesis: A certain fixed ratio of the amount of energy that is not assimilated by the growth and metabolism of the dragon's breathing and other organisms2.11 Hypothesis: Dragon's Flight Similar to Modern Fighter3 Symbol Definition4. Mathematical modeling4.1 About Dragon Flight and Spitfire Consumption4.1.1 Proposed modelConsidering that dragons fly and spit fire during activities, we have established an energy loss model. Comparing the principle of dragon's spitfire with modern flamethrowers, modern flamethrowers consume hydrocarbons or alcohols. It does not cause any impact, so the dragon's fire-breathing principle is in line with the alcohol flame-thrower principle. Considering that the formaldehyde produced by the metabolism of methanol in the animal body is harmful to the body, we stipulate that ethanol is the fuel used by the dragon's flame. In the process, the relationship between the dragon's flight speed and glucose energy consumption is obtained according to fluid mechanics. In this process, we assume that the aerobic respiration is completely performed, and the energy consumed by the dragon due to flight is obtained according to the glucose consumption. In summary, the dragon energy loss model is obtained. .4.1.2 Establishment and Solution of Dragon's Spitfire ModelThe thermochemical equation for ethanol combustion is: C2H5OH (l) + 3O2 (g) = 2CO2 (g) + 2H2O (l) △H = -12KJ / gSpecify the energy released per unit mass of ethanol combustion x1When the dragon spit fire in unit time t, the unit mass of ethanol consumption is a fixed valueThe energy consumed by the fire time t1 is w1The mass consumed by the fire time T1 is m4Let the energy emitted by the combustion of unit mass of ethanol be w1 'Then W1 = x1 * tm4=W1/W1’Solve m4 = x1 * t / W1 '4.1.3 Establishment and Solution of Dragon Flight ModelDuring the flight of the dragon, it will be affected by the air resistance. In the ideal situation, the dragon's flight can be considered as a uniform acceleration and then a uniform speed, and it will decelerate when it is about to reach its destination.When Long uniform acceleration is specified, the acceleration is aSince the flight of the dragon is similar to that of a fighter, a = 30m / s ^ 2The speed of the dragon during uniform motion is v0The total flight length of the dragon during flight is sBecause air resistance is proportional to the speed of movement, that is, F1 = k * v (where k is a constant)Since the dragon's flight is similar to an airplane, we can get k = 3.2325Available according to the relevant kinematic formulaThe flying distance of the dragon during uniform acceleration is s1 = (v0) ^ 2 / 2aThe flying distance of the dragon during uniform deceleration is s3 = (v0) ^ 2 / 2aThe flying distance of the dragon during uniform motion is s2 = s-s1-s3Average air resistance during uniform acceleration F1 '= k * (0 + v0) / 2The average air resistance during uniform motion is F1 '' = k * v0Average air resistance during uniform deceleration f1 '' '= k * (v0 + 0) / 2According to the law of conservation of energyThe energy w2 consumed by the dragon during flight is all used for air resistance workW2=F1’*s1+F1’’*s2+F1’’’*s3Solve W2 = 3.2325 * v0 * s-3.2325 * (v0) ^ 3 / (2 * 30)During the flight of the dragon, the principle of energy provided by aerobic respiration isC6H12O6+6O2=6CO2+6H2OAmong them, the energy produced when 1g of glucose is completely consumed is 16KJThen the weight consumed in this process is m6 = W2 / 16[v,s]=meshgrid(0:0.1:100;0:0.1:100);m=3.2325*v*s-3.2325*v^3/60mesh(v,s,m)4.2 About the relationship between dragon's body length and weight and age4.2.1 Proposed ModelFirst, in order to study the relationship between the weight, length, and age of the dragon, that is, morphological characteristics, we established a model of elasticity during growth. The above-mentioned change curve is continuous, so we use the weight of the dragon at birth, and consider the weight and length of the dragon. The relationship between age changes can be used to derive the normal weight and body length of dragons in all ages. When analyzing the weight changes of dragons, biological knowledge shows that the amount of assimilation of the dragon is equal to the intake amount minus the amount of unassimilated amount Considering that the growth rate of the dragon in adulthood is a watershed, we use the saturation value of the dragon's head, heart, and liver as a cutoff value to estimate the relationship between the dragon's weight and age, respectively. When studying the body length of the dragon, according to the existing morphological knowledge, the head to hip of the dragon is used as the length standard. Because the weight of the dragon is proportional to the cube of the dragon's length, we get the weight and length Functional relationship. Of course, the daily weight gain of the dragon must be less than the daily energy consumption. In summary, we have a dragon intake model.4.2.2 Model establishmentSpecify the weight of the dragon as mDragon was born with a weight of m0 (known m0 = 10kg)Assume that the mass of cattle and sheep fed by a train every day is m2The assimilation amount of the dragon is fixed at a%A certain fixed ratio of the amount of unabsorbed energy due to growth and metabolism of organisms such as dragon's respiration, recorded as b%The weight gain of the dragon is m 'The sum of the weight of the dragon's head, heart, liver, lungs, kidneys, bones, etc. m1 increases with age y until adulthoodDragon is y1 when he is an adultThe growth rate of m1 is v1The mass of m1 at birth is m0Before the dragon reaches y1m1=m0+v1*yAfter the dragon reaches y1m1’=m0+v1*y14.2.3 Model Solvingm’=m2*(1-a%)*(1-b%)-m4-m6So the weight of the dragon m = m '+ m0Except for the dragon, except for the head, heart, liver, lungs, kidneys, bones, etc., the sum of other body masses is proportional to the cube of height, and the body length is recorded as l When the age of the dragon does not reach y1, l = (m-m1) ^ (1/3)When the age of the dragon reaches y1, l '= (m-m1') ^ (1/3)M2 =y=0:0.1:20function[y]= (m2*(1-a%)*(1-b%)-m4-m6-v1*y)y=20:0.1:100function[y]= (m2*(1-a%)*(1-b%)-m4-m6-v1*20)power(y,1/3)4.3 About Dragon's Food Supply4.3.1 Proposed modelBased on the above analysis, we studied the living area of the three dragons in the region andtheir impact on the ecological community in the region. For the sake of research, we assume that the other creatures in the region are cattle and sheep, and the competitiveness of the three dragons is comparable, Being a top predator in the food chain.4.3.2 Model establishmentThe local food chain can be approximated as: grass → cow or sheep → dragonAssume that the weight of the grass in the arid region, the warm temperate region, and the Arctic region is the same as m8.Remember that the mass of each cow and sheep is the same as m7We provide the same initial number of cattle and sheep in all three regionsAssume that the daily growth rate of cattle and sheep is c%The initial number of cattle and sheep is n1And n1 is the number of populations reaching k in the regionDragons live in this area. When the number of cattle and sheep reaches k / 2, in order to ensure the balance of the ecological environment, the dragons need to be moved to other regions.4.3.3 Model SolvingThe initial amount of cattle and sheep on day 1 is: n1The initial amount of cattle and sheep on the second day is: N2 = (N1-3 * m2 / m7) * (1 + c%) The initial amount of cattle and sheep on the third day is: N3 = ((N1-3 * m2 / m7) * (1 + c%)-3 * m2 / m7) * (1 + c%)……From this we can get the initial amount of Ni of cattle and sheep on day iI can be solved by the equation Ni = K / 2That is, the dragon needs to change a living area after living in the area for i days.4.4 Regulating the area of dragons by region4.4.1 Proposed modelIn order to ensure the normal growth of the dragon, we provide fixed-quality cattle and sheep as the supply of resources for the survival of the dragon region, and assume that the number of cattle and sheep is proportional to the size of the regional living area. Considering the growth rate of cattle and sheep, we have established a differential The equation draws the relationship between the growth rate of cattle and sheep and the age of the dragon. However, cattle and sheep will reach a growth saturation value at a certain moment, we will consider it in segments to ensure that the data is more scientific. In order to comply with ecology, cattle The supply of sheep should also have a lower limit. In summary, we have established a dragon-cow-sheep-living area function model.4.4.2 Model establishmentRemember that the assimilation rate of cattle and sheep grazing in this area is d%Because the solar energy received by the surface area of the three areas is different, the total area required for the grass under the same quality conditions is different. The utilization rate of the solar energy is required to be e% (0.5 <e <1 under the natural conditions of the search data)The solar energy per unit area in the arid area is q1Unit area solar energy in warm zone is q2Solar energy per unit area in the Arctic is q34.4.3 Model SolvingAccording to the utilization of solar energy, we can find:Area required to support the arid areas where the three dragons live: S1 = m8 / (q1 * e%)Warm zone: S2 = m8 / (q2 * e%)Arctic region: S3 = m8 / (q3 * e%)5 Sensitivity analysisImpact of climatic conditions on dragon lifeThe effect of climatic conditions on dragon growth can be obtained from the logistic growth model dm/dt=r*m*(1-m/k)That is m = 15 / (4 * t + 20);(Where m is the mass that the dragon can eventually grow into)Where m0 = 10 (k is the maximum carrying capacity of the ecosystem and r is a parameter of the environmental carrying capacity)k is 0.75r is 0.8dm/dt=0.8*m*(1-m/0.75)t=0:0.1:100;m=15./(4*t+20);plot(t,m)6 Model evaluation and outlook6.1 Model evaluationFor the idealized model of Yanglong, we have performed various aspects of modeling and solving, and the scope is relatively broad. Of course, the content has been streamlined to facilitate understanding and application. We have used physical and biological models based on The mathematical formulas are also encountered in the middle school stage. In these more basic models, we have solved efficiently, and at the same time, for the interdisciplinary problems of question a, we have considered the field that the ideal biology of dragons may involve and solve The process is relatively complete. In addition, the four models are closely related and logical. First, we consider the consumption of dragons in daily life, and use the results of consumption to calculate the weight and length of the dragon at various ages. In order to meet the requirements of all ages, we have established the ecological supply model of dragons, and discussed the problem of periodic alternating fields. Second, the fields are also scoped. Therefore, we calculated the scope of three areas with different climates. Interval problems. However, the models we build are idealized, the data is also streamlined, and the assumptions set are also fallible. In reality,The data is diverse and complex, and our considerations are obviously lacking, and further optimization is needed in the later stage. In summary, the model we built is very consistent with the solution of the problem. Although there are some flaws, it does not affect the specific Specific analysis of the problem.6.2 Further discussionCombining the models and evaluations described above, we will improve in the later stages. If this model is used in a specific environment, by statistic large amounts of real data, we can optimize the model. At the same time research also It will be more scientific and rigorous, and it will be more efficient for raising a fictional creature.7 to a letter from George RR MartinDear George RR MartinHope you are wellAfter reading the Song of Ice and Fire, we watched the "Game of Thrones". We became very curious about the mysterious giant that appeared in it-the dragon. Dragons are not uncommon in Eastern and Western cultures. In previous impressions However, there are few studies on dragons. So if we imagine that dragons live in modern times, what would it look like?According to the description of the dragon in the novel, we discussed the following questions. What are the ecological impacts and requirements of the dragon? What is the energy consumption of the dragon, what are their calorie intake requirements? How much area is needed to support the three dragons? Energy loss during fire? In response to these problems, we constructed a multivariate non-linear objective programming model of dragon's growth index and function, size, diet, growth changes, and other animal-related features. Considering the physical characteristics of dragons, we will Its fire-spitting ability is analogized to modern flame-throwers to ensure scientific and rational research.Based on these, we have established a mathematical model. The weight and length of the dragon also grows with the age of the dragon. When the dragon grows slowly at the initial 10 kilograms, the mass of sheep it needs each year also varies The growth of the supply chain of resources and the size of the ecological community should also change. The fire and flight of the dragon will also have a certain impact on the ecological environment. As the dragon and other creatures will migrate with changes in temperature, we choose The three regions of the cold zone, temperate zone and arid zone were taken as key research objects to find out the impact of climate change on Long.Therefore, we make the following suggestions, hoping that the survival of the dragon in the realm of science is more reasonable and scientific.When the herd resource is saturated, the dragon needs to expand the area living area.Dragons like warm, hydrated areas, and migrate to warm areas in the cold winter.A dragon has a certain weight and length when it is just born, and it will grow over time, but it also has an upper limit. It cannot grow endlessly.The daily energy intake of the dragon is limited, and the dragon spitfire flight consumes energy, which requires that the dragon's flight distance and spitfire time are limited, and it is related to the age of the dragon body.Because the living conditions of the three areas are different, the unit area will also receive solar energy differently, resulting in different resource distributions in each area, which means thatthe speed of dragon growth should also be different in different areas.The environmental carrying capacity of each area is limited, and the dragon does not stay in one place for long.The above content is the result of our research on the Queen of Dragons. We sincerely hope that you can adopt it, and we have been looking forward to your new book.Your fans: 27 groupsJanuary 7, 20208.Appendix:8.1 References1) Chen Yun.Research on Environmental Carrying Capacity of Yuhuan County [j] .Energy and Energy Conservation, 2014 (4): 31-33.2) Zhu Ziqiang.Aerodynamic design of modern aircraft [m] .Beijing: National Defense Industry Press, 2011-10-13) Jin Lan.Environmental Ecology [m] .Higher Education Press, 19928.2 Matlab codeModeling the flight of a dragon[v,s]=meshgrid(0:0.1:100;0:0.1:100);m=3.2325*v*s-3.2325*v^3/60mesh(v,s,m)Sensitivity Analysis of the Impact of Climate Conditions on Lifet=0:0.1:100;m=15./(4*t+20);plot(t,m)。

Your Paper's Title Starts Here: Please Centeruse Helvetica (Arial) 14论文的题目从这里开始：用Helvetica (Arial)14号FULL First Author1, a, FULL Second Author2,b and Last Author3,c第一第二第三作者的全名1Full address of first author, including country第一作者的地址全名，包括国家2Full address of second author, including country第二作者的地址全名，包括国家3List all distinct addresses in the same way第三作者同上a email,b email,c email第一第二第三作者的邮箱地址Keywords:List the keywords covered in your paper. These keywords will also be used by the publisher to produce a keyword index.关键字：列出你论文中的关键词。

这些关键词将会被出版者用作制作一个关键词索引。

For the rest of the paper, please use Times Roman (Times New Roman) 12论文的其他部分请用Times Roman (Times New Roman) 12号字Abstract. This template explains and demonstrates how to prepare your camera-ready paper for Trans Tech Publications. The best is to read these instructions and follow the outline of this text.Please make the page settings of your word processor to A4 format (21 x 29,7 cm or 8 x 11 inches); with the margins: bottom 1.5 cm (0.59 in) and top 2.5 cm (0.98 in), right/left margins must be 2 cm (0.78 in).摘要：这个模板解释和示范供稿技术刊物有限公司时，如何准备你的供相机使用文件。

ContentsⅠIntroduction (1)1.1Problem Background (1)1.2Previous Research (2)1.3Our Work (2)ⅡGeneral Assumptions (3)ⅢNotations and Symbol Description (3)3.1 Notations (4)3.2 Symbol Description (4)ⅣSpread of Ebola (5)4.1 Traditional Epidemic Model (5)4.1.1.The SEIR Model (5)4.1.2 (6)4.1.3 (6)4.2 Improved Model (7)4.2.1.The SEIHCR Model (8)4.2.2 (9)ⅤPharmaceutical Intervention (9)5.1 Total Quantity of the Medicine (10)5.1.1.Results from WHO Statistics (10)5.1.2.Results from SEIHCR Model (11)5.2 Delivery System (12)5.2.1.Locations of Delivery (13)5.2.2 (14)5.3 Speed of Manufacturing (15)ⅥOther Important Interventions (16)6.1 Safer Treatment of Corpses (17)6.2 Conclusion (18)ⅦControl and Eradication of Ebola (19)7.1 How Ebola Can Be Controlled (20)7.2 When Ebola Will Be Eradicated (21)ⅧSensitivity Analysis (22)8.1 Impact of Transmission Rate (23)8.2 Impact of the Incubation Priod (24)ⅨStrengths and Weaknesses (25)9.1 Strengths (26)9.2 Weaknesses (27)9.3 Future Work (28)Letter to the World Medical Association (30)References (31)ⅠIntroduction1.1.Promblem Background1.2.Previous Research1.3.Our WorkⅡGeneral Assumptions●●ⅢNotations and Symbol Description3.1. Notataions3.2. Symbol DescriptionSymbol DescriptionⅣSpread of Ebola4.1. Traditional Epidemic Model4.1.1. The SEIR Model4.1.2. Outbreak Data4.1.3. Reslts of the SEIR Model4.2. Improved Model4.2.1. The SEIHCR Model4.2.2. Choosing paametersⅤPharmaceutical Intervention 5.1. Total Quantity of the Medicine 5.1.1. Results from WHO Statistics5.2. Delivery System5.2.1. Locations of Delivery5.2.2. Amount of Delivery5.3. Speed of Manufacturong5.4. Medicine EfficacyⅥOther Important Interventions 6.1. Safer Treatment of Corpses6.2. ConclusionⅦControl and Eradication of Ebola 7.1. How Ebola Can Be Controlled7.2. When Ebola Will Be EradicatedⅧSensitivity Analysis8.1. Impact of Transmission Rate8.2. Impact of Incubation PeriodⅨStrengths and Weaknesses 9.1. Strengths●●●9.2. Weaknesses●●●9.3.Future WorkLetter to the World Medical AssociationTo whom it may concern,Best regards,Team #32150References ［1］［2］［3］［4］。

2023年美国大学生社会学建模E题中文版论文引言本文旨在探讨2023年美国大学生社会学建模竞赛中的E题。

本竞赛是一个模拟实践社会学的机会，参赛者需利用相关数据和模型建议解决一个社会问题。

E题作为其中的一道题目，关注的是.................................（简要介绍E题背景）。

方法为了解决E题，我们采用了如下的方法：1. 收集数据：我们首先收集了关于..............................................（提供数据收集方法和来源）。

2. 数据清洗：我们对收集到的数据进行了清洗和整理，去除了异常值并确保数据的准确性和一致性。

3. 模型选择：在分析数据之前，我们比较了多种社会学建模方法，并选择了最适用的模型来解决E题。

4. 模型建立：我们根据所选模型的要求和理论基础，建立了相应的数学模型，并运用相关软件工具进行计算和模拟。

5. 结果分析：通过对模型的运行结果进行分析和解读，我们得出了结论和建议。

结果经过分析和模拟实验，我们得出了以下结论：1. .......................................（列出结论1）2. .......................................（列出结论2）3. .......................................（列出结论3）讨论我们认识到本研究有一些局限性，主要包括：1. .......................................（列出局限性1）2. .......................................（列出局限性2）3. .......................................（列出局限性3）尽管如此，我们的研究结果仍具有一定的实际应用意义，并为解决E题提供了一种可能的途径和思路。

分析溃坝：针对南卡罗来纳州大坝坍塌建立模型 摘要萨鲁达大坝建立在卡罗莱纳州的墨累湖与萨鲁达河之间，如果发生地震大坝就会坍塌。

本文通过建立模型来分析以下四种大坝决口时水的流量以及洪水泛滥时水的流量：● 大坝的绝大部分被瞬间侵蚀看成是大坝瞬间彻底坍塌；● 大坝的绝大部分被缓慢侵蚀看成是大坝延期彻底坍塌；● 管涌就是先形成一个小孔，最终形成一个裂口；● 溢出就是大坝被侵蚀后，形成一个梯形的裂口。

本文建立了两个模型来描述下游洪水的泛滥情况。

两个模型都采用离散网格的方法，将一个地区看成是一个网格，每个网格都包含洪水的深度和体积。

复力模型运用了网格的速度、重力以及邻近网格的压力来模拟水流。

下坡模型假定水流速度与邻近网格间水位高度的成正比例。

下坡模型是高效率的、直观的、灵活的，可以适用于已知海拔的任何地区。

它的两个参数稳定并限制了水流，但该模型的预测很少依赖于它们的静态值。

对于萨鲁达溃坝，洪水总面积为25.106km ；它还没有到达国会大厦。

罗威克里克的洪水向上游延伸了km 4.4，覆盖面积达24.26.1km -变量及假设表1说明了用来描述和模拟模型的变量，表2列出了模拟程序中的参数。

表 1模型中的变量.变量 定义溃坝时的水流量速率1TF Q 瞬间彻底坍塌2TF Q 延期彻底坍塌PIPE Q 管涌OT Q 溢出peak Q 最大流速溃坝时水流出到停止所用时间1TF t 瞬间彻底坍塌2TF t 延期彻底坍塌PIPE t 管涌OT t 溢出V ∆ 溃坝后从墨累湖里流出的水的总体积Lm Vol 墨累湖的原来体积LM Area 墨累湖的原来面积breach d 从裂口到坝顶距离breach t 从裂口开始到溃坝形成的时间 近似圆锥的墨累湖的侧面一般假设● 正常水位是在溃坝前的湖水位置。

● 河道中的水流不随季节变化而变动。

● 墨累湖里的水的容积可以看作为一个正圆锥（图1 ）。

表2 模拟程序中的参数 参数 所取值 意义BREACH_TYPE 变量 瞬间彻底坍塌，延期彻底坍，管涌，溢出模型中的一种 T ∆ 0.10 时间不长的长度(s)MIN_DEPTH 0001.0 网格空时的水的深度(m) FINAT T 100000 大坝彻底决口所用时间 b T 3600 溃坝达最大值的时间(s) peak Q 25000 溃坝的最大流速(m 3/s) breach d 30 蓄水池的最初深度(m) LM Volume 910714.2⨯ 墨累湖的总体积(m 3) LM Area 610202⨯ 墨累湖的总面积(m 2)k 504.0 扩散因素 (控制两网格间交换的水的数量) MAX_LOSS_FRAC 25.0 单位网格中水的最大流失量图 1. 水库近似一个正圆锥.大坝假设● 萨鲁达大坝在以下四种方式之一坍塌：-瞬间彻底坍塌，-延期彻底坍塌，-管涌，-溢出。

Team Control NumberFor office use only27688For office use onlyT1 ________________F1 ________________T2 ________________F2 ________________T3 ________________Problem Chosen F3 ________________ T4 ________________C F4 ________________2014Mathematical Contest in Modeling (MCM/ICM) Summary SheetThe research of influence based on the characteristic of a network To find the influential nodes in the network, the key is the definition of “influential”and how to measure the influence. In this paper, we use two kinds of metrics to measure the influence of coauthor network and citation network. In coauthor network, both the Authority and Importance of the researchers are proposed to measure the influential of researcher. And the second one in citation network take the citation times, publication time and the position in the network into account.For the evaluation of coauthor, we first construct a coauthor network with 511 vertices and 18000 edges and it is an undirected graph. Next, we use software UCInet to analyze the degree centrality, eigenvector centrality, closeness centrality and betweenness centrality of the network. Since there is no evident transfer relationship in the coauthor network, we using Authority and Importance to measure the influence of a research. In detail, the Authority is correlated with the coauthoring times with Paul Erdös and the Importance is measured by eigenvector centrality. Finally, we rank the researchers whose authority is larger than 2 according to their importance. And the top 5 most influential researchers are: RODL, VOJTECH; LOVASZ, LASZLO; GRAHAM, RONALD LEWIS; PACH, JANOS; BOLLOBAS, BELA. Finally, we search for some data through websites and verify these people are really influential.For the evaluation of papers, we first compare the difference between the citation network and coauthor network. According to the characteristic of Directed Acyclic Graph(DAG), we define a contribution coefficient and self-contribution coefficient by making an analogy with the energy transfer in the food chain. Considering the less-effectiveness of PageRank Algorithm and Hits Algorithm, we design an algorithm, which is effective in solving the DAG problem, to calculate the contribution coefficient. We find 3 most influential papers: Paper 14, Paper 4 and Paper 2 in the NetSciFoundation.pdf.In the third part, we implement our model to analyze a corporation ownership network. We use the value of the company’s cash, stock, real estate, technical personnel, patent and relationships to define its value. And we use the proportion of stock to measure the control ability of parent company. Applying the model and algorithm of citation network, we find 15 influential companies. Then we find that 9 of them are in the top 20 of authoritative ranking, which verifies the rationality of our result.Finally, we describe how we can utilize these influential models to do some socialized service, to aid in making decision on company acquisition and to carry out strategic attack.Team #27688Page 1 of 18 1. IntroductionNowadays, coauthor network and citation network are built to determine influence of academic research. Paul Erdös, one of the most influential researchers who had over 500 coauthors and published over 1400 technical research papers. There exists a coauthor network among those who had coauthored with Erdös and those who had coauthored with Erdös’s directed coauthors.In this paper, we first analyze this coauthor network and find some researchers who have significant influence. Then, we analyze the citation network of some set of foundational papers in the emerging field of network science. Furthermore, we determine some measures to find some most influential papers. After that, we use the data of US Corporate Ownership to construct a new network and test the applicability of our model and algorithm. Finally, we describe some applications of using the analysis of different networks.In section 3, the coauthor network is an undirected graph. We first analyze four kinds of centrality: Degree Centrality, Eigenvector Centrality, Closeness Centrality and Betweenness Centrality. Additional, the Degree distribution and Clustering coefficient are also the important properties of the network. Then, we define Authority and Importance to measure the influence of a researcher. Authority can be measured by the coauthoring times with Erdös. It is clearly that the researcher who coauthors with more people is more important. Since this is not a problem about “information flow”, we only cons ider the influence of those directed coauthor and neglect the transitivity of influence. That is to say, Importance can be measure by Eigenvector Centrality. Finally, we choose some people with higher authority and rank them according to their Eigenvector Centrality.In section 4, the citation network is different from the coauthor network. As the citation relation is related to publication time, the citation network is a Directed Acyclic Graph(DAG). Traditionally, we calculate the nodes’ importance of a n etwork by using PageRankAlgorithm[17] and HITS Algorithm[18]. However, both of them involve matrix multiplicationand repeated iterative process, which is less-effective. Since the network satisfies theproperty of Directed Acyclic Graph(DAG), we draw on the thought of topological sorting to design a more effective algorithm. In this citation network, there exists transitive relation that does not exist in the coauthor network. We first use software UCInet to calculate thecentrality of each paper. And then we take these metrics, publication time and times cited count into account to develop a new model. In this model, we learn from the energy transfersin the food chain and define an initial contribution coefficient to measure its authority. In addition, we define a self-contribution coefficient to measure the influence from other papers. Finally, we design an algorithm to calculate each paper’s final contribution coefficientto measure the paper’s influence.In section 5, we use nearly 500 US Media Companies to construct an ownership network. Then we set the initial value of each company according to their case, stock, real estate, technical personnel, patent and relationships. And we set a control coefficient to measure the ownership between two companies. Then we can use the algorithm in citation network to find someTeam #27688Page 2 of 18influential companies.In the fourth part, we utilize these influential models to do some social service, aid in making decision on company acquisition and carry out strategic attack.In general, the article is written follows:(1)Build a coauthor network for question 1.(2)Build the evaluation index of the most influential coauthor to estimate the influenceof coauthors in the coauthor network.(3)Build citation network and define the influence criterion of papers to estimate themost influential paper.(4)Implement our model to the US Corporate Ownership network to analyze theimportance and the value of the company.(5)Finally, we discuss about the basic theory, the use and effectiveness of the science ofnetwork.2. Assumptions and Justification(1)We use number 1..16 to represent the paper given in the NetSciFoundation.pdf according totheir sequence. It is worth mentioning that the information of paper 7 given in the file seems to be wrong. Hence, we regard it as an isolated vertex in the network.(2)The researchers’ authority, it is correlated with the coauthoring times with Paul Erdös. Inthe coauthor network, we know that all of them have coauthored with Erdös and Erdös is such an excellent mathematician. So it is suitable for us to assume that more times coauthored with Erdös, more authority the researcher is.(3)We do not consider the influence of the paper’s content and field because the cited times indifferent fields have no comparability. In question 3 we know that 16 papers are in the emerging field of network science, so it is reasonable for us to simplify this problem.(4)When constructing the citation network, we only take those papers citing more than twopapers in 16 given papers and also having been cited by other papers. Absolutely, the citation network is infinite. In this paper, we aim to find influential papers. Hence, we give up those less important papers and restrict the scale of our network.(5)We assume that the citation relation is effective. If a paper cited other papers, we considerthat the author admitted the positively effect of the cited paper. Since the influence of a paper is related to the citation times, our assumption can improve the validity of the result.(6)The data in our paper is effective. Our dataset is searched in Web of Science and GoogleScholar, which are equipped with high authority.Team #27688Page 3 of 18 3. Coauthor Network3.1 Building the modelA coauthor network can be built to help analyze the influence of the researchers whose Erdös Number are 1. Obviously, this is a social network. In the network, each node represents a researcher who has coauthored with Paul Erdös and each link could represent the coauthoring relationship between two researchers. Since the coauthor matrix is symmetrical, we know that there is no different between A coauthors withB and B coauthors with A. Therefore, the coauthor network is an undirected network which has 511 vertices. We use software Gephi to draw the graph and the network diagram is shown in Figure 1.Figure 1: the co-author networkIn this graph, the vertex represents a researcher and the edge represents the coauthoring relation. The size of the vertex represents its coauthoring times with Erdös and the darker the color is, the more people he coauthored with. There are 511 vertices and 18000 edges.In this network, there are many basic measures and metrics, such as Degree, Centrality, Clustering coefficient, Density, Betweenness and so on. In this paper, we first choose several important measures for analyzing this network and show them as follows. [1]Of course, the common property is CENTRALITY. Centrality is a crucial metric to evaluate the influence of a vertex. In the following, we discuss several classic Centralities and analyze theirdifference.⏹DEGREE CENTRALITYThe degree of a vertex in a graph is the number of edges connected to it. We will denote thedegree of vertex i by d i. And the simplest centrality measure, which is called degree centrality ( C d ), is just the degree of a vertex. That means:C d(i ) d iTeam #27688 Page 4 of 18In a social network, for instance, it seems reasonable to suppose that individuals whohave connections to many others might have more influence, more access to information, or more prestige than those who have fewer connections.⏹ EIGENVECTOR CENTRALITYSometimes, all neighbors of a vertex are not equivalent. Hence, Bonacich [2] puts forwardEigenvector centrality to cope with this situation. It assigns relative scores to all nodes in the network based on the concept that connections to high-scoring nodes contribute more to the score of the node in question than equal connections to low-scoring nodes.λd i = ∑r ij d j jWhere:r ij represents the elements in the adjacency matrix; d irepresents the degree centrality of vertex iUsually, we choose the eigenvector corresponding to the maximal eigenvalue to be the eihenvector centrality( C e )[3]. ⏹ CLOSENESS CENTRALITYCloseness centrality measures the mean distance from a vertex to other vertices, which canused to analyze the position of a vertex in the network [1]. ∑ D ijC c (i ) = j ( ≠i )-n 1Where: D ij is the distance between vertex i and vertex j ;C c (i ) is the closeness centrality of vertex i ;n is the number of vertices. ⏹ BETWEENNESS CENTRALITYBetweenness centrality measures the extent to which a vertex lies on paths between othervertices [1]. That is to say, a vertex with a higher betweenness centrality plays a moreimportant role in the connection of the network.n nC b ( k ) = ∑∑[ g ij ( k ) / g ij ] i jWhere: g ij ( k ) represents the number of shortest path between i andj through k ; g ij represents the number of shortest path between i andjThen, we use the UCINET to calculate some basic metrics and show them in table 1.Table 1: the basic data of centralitytype Degree Closeness Betweenness EigenvectorAverage 1.292 2.115 0.461 3.055Minimum 0.000 0.196 0.0000.000 Maximum 10.392 2.201 7.508 36.515According to the above table and Figure 1, we can know that about 30 vertices have 3 times more than the average degree. That is to say, these researchers have many coauthors. In addition, since the average value of closeness is close to its maximum, we know that there are few vertices。

2022美赛A题相关论文美国大学生数学建模竞赛(MCM/ICM)是世界范围内最具影响力的大学生数学建模竞赛（含金量仅次于数模国赛）。

MCM/ICM着重强调研究和解决方案的原创性、团队合作、交流及结果的合理性。

竞赛题目内容涉及经济、管理、环境、资源、生态、医学、安全、未来科技等众多领域。

每年题目为6个，分别是：有同学会好奇ICM和MCM的区别大么？他们之间难度的区别大么？等等，这个因人而异，看你对哪个题目更熟悉，能够做的出彩，他们其实都需要数学建模，然后采用一些算法解决，最终以论文的形式呈现，究竟怎么选择还是看你对哪个题目更有感觉。

我个人感觉，MCM的题目可能更看重数学模型，ICM的更强调算法，比如ICM会用到一些复杂的算法，MCM一般不出现很复杂的算法。

2021年的C题比往年数据量大很多，是关于图像和视频处理的，所以需要用到一些新的知识和方法，想做C题的朋友要注意这个变化。

具体如何准备美赛我会在后文细说。

美赛的奖项大家应该都有所了解，奖项的英文首字母作为简称，和我们通常奖项对应如下：可以看出MCM比ICM的获奖比例高一些，这是因为2021年MCM题目的B和C比较难，大家普遍选ICM问题尤其是F题的的队伍比较多，所以获奖难度就大了一些。

这点说明，大家不要忙着往“简单”的选，反而可能获奖难度会增大。

还是切合队伍三人的能力，选合适的问题，而不能一味追求容易。

补充一点，如果你可以拿到O奖，那么你还可能会遇到冠名奖，甚至有极个别的F奖也拿到了冠名奖，冠名奖大概有六七种吧，官网都有介绍。

我拿到的是美国工业与应用数学学会奖，这个当然就是美国工业与应用数学学会设置的，每题1个，一共六个。

还有一个是弗兰克乔丹诺奖，这是为了表达对美赛执教20年的教授而设置的，颁发给MCM的唯一一支队伍，当然还有一些其他的比如COMAP等等，其实这些冠名奖是我们看到美赛证书的时候才去了解的hhh。

经常会有学弟学妹们问我，有没有好的队友推荐啊之类的，但其实这个还需要自己去努力寻找，随意组队是会出现问题的，下面我通过几个问题来向大家展示三个人应该有怎样的配合。

%% 本论文的排版主要参考了LaTeX2e插图指南(王磊), LaTeX2e用户手册, media的中文学位%% 论文宏包(CDT), happaytex的ORmain1.tex等文件以及ChinaTeX, CTeX论坛上的诸多贴子. %%% 本论文采用了Miktex2.2的方式在ChinaTeX.iso系统下得到了实现,其编译方式为%% latex(得到DVI文件)+dvips(得到PS文件)+ps2pdf(可得PDF文件).%%\documentclass[12pt]{article}%需要的一些宏包\usepackage{CJK} % 中文输入环境宏包\usepackage{titlesec,titletoc} % 配合命令在后面, 章节标题设置\usepackage{indentfirst} % 使首段首行缩进\usepackage{graphicx} % 插图宏包\usepackage{caption2} % 可以更改插图, 表格的标题样式\usepackage{subfigure} % 产生并列的子图或子表, 命令\subfigure, \subtable\usepackage{longtable} % 如果表格太长, 超过了一页时, 就可以试试longtable 宏包所定义的longtable 环境\usepackage{slashbox} % 在表格中绘制斜线\usepackage{fancyhdr} % 更改页眉的宏包, 并可在页眉插入图片\usepackage{times} % Times Roman + Helvetica + Courier\usepackage{amsmath} % 数学符号宏包AMS-LaTeX, 如下面的\overset需要此宏包%页面的设置\special{papersize=21cm,29.7cm} \setlength{\textwidth}{15cm}\setlength{\textheight}{23cm} \setlength{\evensidemargin}{0.46cm}\setlength{\oddsidemargin}{0.46cm} \setlength{\topmargin}{-1.84cm}\setlength{\headheight}{2.9cm} \setlength{\headsep}{0.4cm}%字号设置\newcommand{\chuhao}{\fontsize{42pt}{\baselineskip}\selectfont}\newcommand{\xiaochuhao}{\fontsize{36pt}{\baselineskip}\selectfont}\newcommand{\yihao}{\fontsize{26pt}{\baselineskip}\selectfont}\newcommand{\xiyihao}{\fontsize{24pt}{\baselineskip}\selectfont}\newcommand{\erhao}{\fontsize{22pt}{\baselineskip}\selectfont}\newcommand{\xiaoerhao}{\fontsize{18pt}{\baselineskip}\selectfont}\newcommand{\sanhao}{\fontsize{16pt}{\baselineskip}\selectfont}\newcommand{\xiaosanhao}{\fontsize{15pt}{\baselineskip}\selectfont}\newcommand{\sihao}{\fontsize{14pt}{\baselineskip}\selectfont}\newcommand{\xiaosihao}{\fontsize{12pt}{\baselineskip}\selectfont}\newcommand{\wuhao}{\fontsize{10.5pt}{\baselineskip}\selectfont}\newcommand{\xiaowuhao}{\fontsize{9pt}{\baselineskip}\selectfont}\newcommand{\liuhao}{\fontsize{7.5pt}{\baselineskip}\selectfont}\newcommand{\xiaoliuhao}{\fontsize{6.5pt}{\baselineskip}\selectfont}\newcommand{\qihao}{\fontsize{5.5pt}{\baselineskip}\selectfont}\newcommand{\bahao}{\fontsize{5pt}{\baselineskip}\selectfont}%页眉的设置, 要用到fancyhdr宏包\pagestyle{fancy} \fancyhead{} \fancyfoot{}\fancyhead[L]{\footnotesize Team \# 189}\fancyhead[R]{\footnotesize Page\ \thepage\ of\ 42}\fancypagestyle{plain}{%\fancyhead[L]{\footnotesize Team \# 189}\fancyhead[R]{\footnotesize Page\ \thepage\ of\ 42}}\setcounter{secnumdepth}{4}%更改\theparagraph的编号样式\makeatletter\renewcommand{\theparagraph}{\@arabic\c@paragraph}\makeatother%章节格式的设置\titleformat{\section}{\erhao\bf}{}{0em}{}[]\titleformat{\subsection}{\xiaoerhao\bf}{}{0em}{}[]\titleformat{\subsubsection}{\sanhao\bf}{}{0em}{}[]\titleformat{\paragraph}[hang]{\vspace*{0.5ex}\sihao\bf}{\hspace*{1em}\theparagraph)}{0.5em }{}[\vspace*{-0.5ex}]%更改插图的标题\renewcommand{\figurename}{\wuhao\bf\sf Figure}\renewcommand{\captionlabeldelim}{\ }%更改表格的标题\renewcommand{\tablename}{\wuhao\bf\sf Table}%更改图形或表格与其标题的间距\setlength{\abovecaptionskip}{10pt}\setlength{\belowcaptionskip}{10pt}%定义产生不浮动图形和表格的标题的命令\figcaption和\tabcaption\makeatletter\newcommand\figcaption{\def\@captype{figure}\caption}\newcommand\tabcaption{\def\@captype{table}\caption}\makeatother%自定义的可以调整粗细的水平线命令, 用于绘制表格, 调用格式\myhline{0.5mm}. \makeatletter\def\myhline#1{%\noalign{\ifnum0=`}\fi\hrule \@height #1 \futurelet\reserved@a\@xhline}\makeatother%第一层列表序号为带圈的阿拉伯数字\renewcommand{\labelenumi}{\textcircled{\arabic{enumi}}}%更改脚注设置\renewcommand{\thefootnote}{\fnsymbol{footnote}}\begin{document}\begin{CJK*}{GBK}{song}\CJKtilde\title{\bf\yihao Aviation Baggage Screening\\{\&} Flight Schedule}\author{}\date{}\maketitle\section{Introduction}Following the terrorist attacks on September 11, 2001, there isintense interest in improving the security screening process forairline passengers and their baggage. Airlines and airports areconsidered high-threat targets for terrorism, so aviation securityis crucial to the safety of the air-travelling public. Bombs andexplosives have been known to be introduced to aircraft by holdbaggage and cargo, carried on by passengers, and hidden withinaircraft supplies.At present To Screen or Not to Screen, that is a Hobson's choice.US Current laws mandate 100{\%} screening of all checked bags at the 429 passenger airports throughout the nation by explosive detection systems(EDS) by the end of the Dec 31 2003. However, because the manufacturers arenot able to produce the expected number of EDS required to meet the federal mandate, so it is significant to determine the correct number of devicesdeploy at each airport, and to take advantage of them effectively.The Transportation Security Administration (TSA) needs a complicatedanalysis on how to allocate limited device and how to best use them.Our paper contains the mathematical models to determine the number of EDSsand flight schedules for all airports in Midwest Region. We also discuss theETD devices as the additional security measures and the future developmentof the security systems.\section{Assumption and Hypothesis}\begin{itemize}\item The passengers who will get on the same airplane will arrive uniformly, namely the distribution is flat.\item The detection systems, both EDS and ETD, operate all the time during peak hour, except downtime.\item The airline checks the passengers randomly, according to its claim.\item The passengers, who are just landing and leave out, do not have to be checked through EDS or ETD.\item According to the literature, the aircraft loads approximately equal among the sets of departing flight during the peak hour.\item The landing flight did not affect the departure of the plane.\item Once a passenger arrives, he can go to EDS to be checked, except he has to wait in line.\item Once passengers finish screening, they can broad on the plane in no time.\item During peak hours, a set of flights departs at the same time every the same minutes.\item All the runways are used as much as possible during peak hours.\item The maximum number of the baggage is two, which a passenger can carry on plane. []\item The detection machine examines the bags at the same speed.\item EDS cannot make mistakes that it detect a normal object as an explosive.\end{itemize}\section{Variable and Definition}\begin{longtable}{p{100pt}p{280pt}}\caption{Variables}\\ %第一页表头的标题\endfirsthead %第一页的标题结束\caption{(continued)}\\ %第二页的标题\endhead %第二页的标题结束\hline\hline\textbf{Symbol}&\textbf{Description}\\\hline$n_{ij}$&The airplane number of the $i^{\mathrm{th}}$ type in the $j^{\mathrm{th}}$ flight set\\\hline${NP}_i\:(i=1,2,\ldots)$&The number of passengers on each airplanes of the same type.\\\hline$\xi_{ij}\:(i,j = 1,2,\cdots)$&The number of baggage on each airplane of the $j^{\mathrm{th}}$ flights\\\hline$a$&The maximal number of airplanes type\\\hline$B_j^{set}$&The total baggage number of each set of flight\\\hline${NF}_i$&Number of airplanes of each type\\\hline$\bar{\rho}$&The mean value of passengers' baggage coming per minute in every flight set\\ \hline$N_{set}$&The number of flight sets\\\hline$B_{total}$&The total number of checked baggage during the peak hour\\\hline$H_{peak}$&The length of the peak hour\\\hline$T_{set}$&The time length during which each flight set's passengers wait to be checked\\\hline$\Delta t$&The time interval between two consecutive flight set\\\hline$N_{EDS}$&The number of all the EDSs\\\hline$N_{shadow}$&The number of flight sets whose passengers will be mixed up before being checked\\\hline$v_{EDS}$&The number of baggage checking by one EDS per minute\\\hline$\rho_j$&The number of passengers' baggage coming per minute in one flight set\\\hline$N_{runway}$&The number of an airport's runway\\\hline\\*[-2.2ex]${\bar{B}}^{set}$&The mean value of checked baggage number of every flight set\\\hline$M$&The security cost\\\hline\hline\label{tab1}\end{longtable}\subsubsection{Definition:}\begin{description}\item[Flight set] A group of flights take off at the same time\item[The length of peak hour] The time between the first set of flight and the last set\end{description}\section{Basic Model}During a peak hour, many planes and many passengers would departfrom airports. Therefore, It is difficult to arrange for thepassengers to enter airports. If there are not enough EDSs forpassengers' baggage to check, it will take too long time for themto enter. That would result in the delay of airplanes. On thecontrary, if there are too many EDSs, it will be a waste. It isour task to find a suitable number of EDSs for airport. In orderto reach this objective, we use the linear programming method tosolve it.\subsection{Base analysis}The airplanes are occupied at least partly. The passengers'baggage would be checked by EDSs before they get on the airplanes.We have assumed that every passenger carry two baggages. Thisassumption would simplify the problem. According to the data fromthe problem sheet, we can obtain the useful information thatairlines claim 20{\%} of the passengers do not check any luggage,20{\%} check one bag, and the remaining passengers check two bags.Therefore, we can gain the total number of passengers' baggagethat should be carried on one plane: $\xi_{ij}$. Moreover, we canget the equation that calculate $\xi_{ij}$:\[\xi_{ij}={NP}_i\times 20\%+{NP}_i\times 60\%\times 2\]We define the matrix below as airplane baggage number matrix:\[\overset{\rightharpoonup}{\xi}_j=\left[\xi_{1j}\quad\xi_{2j}\quad\cdots\quad\xi_{ij}\quad\cdots\ right]\]We define the matrix below as flight schedule matrix:\[\left[\begin{array}{llcl}n_{11}&n_{12}&\cdots&n_{1N_{set}}\\n_{21}&n_{22}&\cdots&n_{2N_{set}}\\\multicolumn{4}{c}\dotfill\\n_{a1}&n_{a2}&\cdots&n_{aN_{set}}\end{array}\right]\]In this matrix, $n_{ij}$ is the airplane number of the$i^{\mathrm{th}}$ type in the $j^{\mathrm{th}}$ flight set whichwill take off. Apparently, this value is an integer.We define the matrix below as flight set baggage number matrix:\[\left[B_1^{set}\quad B_2^{set}\quad\cdots\quad B_j^{set}\quad\cdots\quad B_a^{set}\right] \]It is clear that they meet the relation below:\begin{equation}\begin{array}{cl}&\left[\xi_{1j}\quad\xi_{2j}\quad\cdots\quad\xi_{ij}\quad\cdots\right]\cdot\left[\begin{array}{llcl}n_{11}&n_{12}&\cdots&n_{1N_{set}}\\n_{21}&n_{22}&\cdots&n_{2N_{set}}\\\multicolumn{4}{c}\dotfill\\n_{a1}&n_{a2}&\cdots&n_{aN_{set}}\end{array}\right]\\=&\left[B_1^{set}\quad B_2^{set}\quad\cdots\quad B_j^{set}\quad\cdots\quad B_a^{set}\right]\end{array}\label{Flight:baggage}\end{equation}Then, we know:\[B_j^{set}=\sum\limits_{i=1}^a\xi_{ij}\times n_{ij}\]There are some constraints to the equation (\ref{Flight:baggage}).First, for each set of flight, the total number of airplanesshould be less than the number of runways. Second, the totalairplane number of the same type listed in the equation(\ref{Flight:baggage}) from every set of flight should be equal tothe actual airplane number of the same type during the peak hour.We can express them like these:\[\sum\limits_{i=1}^a n_{ij}\le N_{runway}\quad\quad\sum\limits_{j=1}^b n_{ij}={NF}_i \]We should resolve the number of flight sets. According to our assumptions,during the peak hour, the airlines should make the best use of the runways.Then get the number of flight sets approximately based on the number of allthe airplanes during the peak hour and that of the runways. We use anequation below to express this relation:\begin{equation}N_{set}=\left\lceil\frac{\sum\limits_{j=1}^{N_{set}}\sum\limits_{i=1}^an_{ij}}{N_{runway}}\right\rceil\label{sets:number}\end{equation}The checked baggage numbers of each flight set are equal to eachother according to our assumption. We make it based on literature.It can also simplify our model. We define $\bar{B}^{set}$ as themean value of checked baggage number of every flight set.Moreover, We define $\bar{\rho}$ as the mean value of checkedbaggage number of every flight set per minute:\[\bar{B}^{set}=\frac{B_{total}}{N_{set}}\]\[\bar{\rho}=\frac{\bar{B}^{set}}{T_{set}}=\frac{B_{total}}{T_{set}N_{set}}=\frac{B_{total}\ Delta t}{T_{set}H_{peak}}\]The course of passengers' arrival and entering airport isimportant for us to decide the number of EDSs and to make theflights schedule. Therefore, we should analyze this processcarefully. Passengers will arrive between forty-five minutes andtwo hours prior to the departure time, and the passengers who willget on the same airplane will arrive uniformly. Then we can getthe flow density of all checked baggage at any time duringpassengers' entering. This value is the sum of numbers ofpassengers' checked baggage coming per minute. To calculate thisvalue, firstly, we should obtain flow density of each flight set'schecked baggage. We define $\rho_j $, namely the number of checkedbaggage per minute of one flight set:\[\rho_j=\frac{B_j^{set}}{T_{set}}\]Secondly, we draw graphic to help us to understand. We userectangle to express the time length for all the passengers of oneflight set to come and check bags. In the graphic, the black partis the period for them to come. During the white part, nopassengers for this flight set come. According to the problemsheet, the former is 75 minute, and the latter is 45 minute. Thelength of rectangle is 120 minute. $T_{set}$ is the period duringwhich all passengers of one flight set wait to be checked. Sincewe have assumed that each time interval between two consecutiveflight set is same value, we define $\Delta t$ as it. Observe thesection that value we want to solve is $\sum\limits_j\rho_j$. Moreover, we can get another important equation from the graphic below:\begin{equation}N_{set}=\frac{H_{peak}}{\Delta t}\label{PeakHour}\end{equation}\begin{figure}[hbtp]\centering\includegraphics[width=298.2pt,totalheight=141.6pt]{fig01.eps}\caption{}\label{fig1}\end{figure}Each EDS has certain capacity. If the number of EDSs is $N_{EDS}$ and one EDS can check certain number of baggage per minute (Thatis checking velocity, marked by $v_{EDS}$), the total checking capacity is $N_{EDS}\cdot\frac{v_{EDS}}{60}$. $v_{EDS}$ is between 160 and 210.Now we can easily decide in what condition the passengers can be checked without delay:\[\sum\limits_j\rho_j\le v_{EDS}\]The passengers have to queue before being checked:$\sum\limits_j\rho_j>v_{EDS}$Well then, how can we decide how many $\rho_j$? It depends on how many flight sets whose passengers will be mixed up before being checked. We note it as $N_{shadow} $. Return to the Figure\ref{fig1}, we can know:\[N_{shadow}=\left\lfloor\frac{T_{set}}{\Delta t}\right\rfloor\]\begin{figure}%[htbp]\centering\includegraphics[width=240pt,totalheight=131.4pt]{fig02.eps}\caption{}\label{fig2}\end{figure}From Figure \ref{fig1} and Figure \ref{fig2}, we can get theresult as follows:\begin{enumerate}\item If $N_{shadow}\le N_{set}$, namely $H_{peak}>T_{set}$, then $\sum\limits_{j=1}^{N_{shadow}}\rho _j\le N_{EDS}\frac{v_{EDS}}{60}$\renewcommand{\theequation}{\arabic{equation}a}That is:\begin{equation}N_{EDS}\ge\frac{60}{v_{EDS}}\sum\limits_{j=1}^{N_{shadow}}\rho_j\approx\frac{60}{v_{ EDS}}N_{shadow}\bar{\rho}=\frac{60B_{total}\Deltat}{v_{EDS}T_{set}H_{peak}}N_{shadow}\label{EDS:number:a}\end{equation}\item If $N_{shadow}>N_{set}$, namely $H_{peak}\le T_{set}$, then $\sum\limits_{j=1}^{N_{set}}\rho_j\le N_{EDS}\frac{v_{EDS}}{60}$\setcounter{equation}{3}\renewcommand{\theequation}{\arabic{equation}b}That is:\begin{equation}N_{EDS}\ge\frac{60}{v_{EDS}}\sum\limits_{j=1}^{N_{set}}\rho_j\approx\frac{60}{v_{EDS} }N_{set}\bar{\rho}=\frac{60B_{total}\Delta t}{v_{EDS}T_{set}H_{peak}}N_{set}\label{EDS:number:b}\end{equation}\end{enumerate}\subsection{The number of EDSs}Then we begin to resolve the number of EDSs assisted by the linearprogramming method.EDS is operational about 92{\%} of the time. That is to say, whenever it isduring a peak hour, there are some EDSs stopping working. Then the workingefficiency of all the EDSs is less than the level we have expected.Therefore, the airline has to add more EDSs to do the work, which can bedone with less EDSs without downtime.We use binomial distribution to solve this problem. $N$ is the number ofactual EDSs with downtime and $k$ is the number of EDSs without downtime. Ifprobability is $P$, we can get the equation below:\[\left(\begin{array}{c}N\\k\end{array}\right)\cdot98\%^k\cdot(1-98\%)^{N-k}=P\]We can obtain $N$ when we give $P$ a certain value. In this paper,$P$ is 95{\%}. The $N_{EDS}$ is the actual number we obtainthrough the equation above.Now we have assumed that passengers can be checked unless be delayed by the people before him once he arrives at airport. Apparently, if the time length between two sets of flight is short, the density of passengers will begreat. It will bring great stress to security check and may even make some passengers miss their flight. To resolve this question, the airline has toinstall more EDSs to meet the demand. However, this measure will cost much more money. Consequently, we have to set a suitable time interval between two set of flight.Based on the base analysis above. We can use the equation(\ref{sets:number}) to decide the number of flight sets $N_{set}$assuming we know the number of runways of a certain airport. Thenbased on the equation (\ref{PeakHour}), we can decide the peakhour length $H_{peak}$ when we assume a time interval between two consecutive flight sets. Then we use \textcircled{1} and\textcircled{2} to decide which to choose between equation(\ref{EDS:number:a}) and equation (\ref{EDS:number:b}). In consequence, we can obtain the minimum of EDSs number.If we choose different numbers of runways and the time intervalsbetween two flight sets, we can get different EDSs numbers. Inthis paper that followed, we gain a table of some value of$N_{runway}$ and $\Delta t$ with the corresponding EDSs numbers. Moreover, we draw some figure to reflect their relation.For a certain airport, its number of runway is known. Givencertain time interval ($\Delta t$), we can get the length of thepeak hour ($H_{peak}$). When the $N_{runway}$ is few enough,perhaps $H_{peak}$ is too long to be adopted. However, for acertain airline, they can decide the time interval of their ownpeak hour. In this given time interval, they could find theminimum of $N_{runway}$ through the Figure \ref{fig3}. We draw asketch map to describe our steps.\begin{figure}[hbtp]\centering\includegraphics[width=352.8pt,totalheight=214.2pt]{fig03.eps}\caption{}\label{fig3}\end{figure}\subsection{The Flight Schedule }According to the base analysis, we can know that the flightschedule matrix and $\Delta t$ is one form of flight timetable. In``The number of EDSs'', we can get suitable $\Delta t$. Then weshould resolve the flight schedule matrix.Because we have assumed that the checked baggage numbers of each flight setare equal to each other. It can be described as follows:\[\left\{\begin{array}{l}\rho_j\approx\bar{\rho}\\B_j^{set}\approx\bar{B}^{set}\end{array}\right.\begin{array}{*{20}c}\hfill&{j=1,2,\cdots,N_{set}}\hfill\end{array}\]The flight schedule matrix subject to this group:\[\left\{\begin{array}{ll}\sum\limits_{j=1}^{N_{set}}n_{ij}={NF}_i&i=1,2,\cdots\\\sum\limits_{i=1}^a n_{ij}\le N_{runway}&j=1,2,\cdots,N_{set}\\n_{ij}\ge0,&\mathrm{and}\:n_{ij}\:\mathrm{is}\:\mathrm{a}\:\mathrm{Integer} \end{array}\right.\]In order to make the best use of runway, we should make$\sum\limits_{i=1}^a n_{ij}$ as great as we can unless it exceed$N_{runway}$.Then we can see that how to resolve the flight schedule matrix is a problemof divide among a group of integers. This group is all the numbers of eachflight passengers' baggage in one flight set. We program for this problemusing MA TLAB and we get at least one solution in the end. However, thematrix elements we have obtained are not integer, we have to adjust them tobe integers manually.\subsection{Results and Interpretation for Airport A and B}The number of passengers in a certain flight (${NP}_i$), the timelength of security checking ($T_{set}$), the checking velocity ofEDS ($v_{EDS}$), and the number of baggage carried by onepassenger are random.\subsubsection{Data Assumption:}\begin{itemize}\item $T_{set}$ is 110 minutes, which is reasonable for airline.\item To simplify the problem, we assume that every passenger carry 2 baggage. If some of thepassengers carry one baggage, the solution based on 2 baggages per passenger meets therequirement.\item The number of runways in airport A and airport B is 5.\end{itemize}\subsubsection{Airport A:}Once the number of runway and the number of the flights aredecided, the flight schedule matrix is decided, too. We producethis matrix using MATLAB. This matrix companied by $\Delta t$ isthe flight schedule for airport A. $\Delta t$ will be calculatedin (\ref{Flight:baggage}), (\ref{sets:number}) and(\ref{PeakHour}).We calculate $N_{EDS}$ and make the flight timetable in threeconditions. The three conditions and the solution are listed asfollowed:\paragraph{Every flight are fully occupied}The checking speed of EDS is 160 bags/hour.\begin{table}[htbp]\centering\caption{}\begin{tabular}{*{11}c}\myhline{0.4mm}$\mathbf{\Deltat(\min)}$&\textbf{2}&\textbf{4}&\textbf{6}&\textbf{8}&\textbf{10}&\textbf{12}&\textbf{14} &\textbf{16}&\textbf{18}&\textbf{20}\\\myhline{0.4mm}$N_{EDS}(\ge)$&31&31&31&31&31&29&24&22&20&17\\\hline$H_{peak}(\min)$&20&40&60&80&100&120&140&160&180&200\\\myhline{0.4mm}\end{tabular}\label{tab2}\end{table}We assume that the suitable value of $H_{peak}$ is 120 minutes.Then the suitable value of $\Delta t$ is about 12 minutes, and$N_{EDS}$ is 29 judged from Figure \ref{fig4}. Certainly, we canwork $\Delta t$ and $N_{EDS}$ out through equation.\begin{figure}[htbp]\centering\includegraphics[width=294.6pt,totalheight=253.2pt]{fig04.eps}\caption{}\label{fig4}\end{figure}\paragraph{Every flight is occupied by the minimal number of passengers onstatistics in the long run.}The checking speed of EDS is 210 bags/hour.\begin{table}[htbp]\centering\caption{}\begin{tabular}{*{11}c}\myhline{0.4mm}$\mathbf{\Deltat(\min)}$&\textbf{2}&\textbf{4}&\textbf{6}&\textbf{8}&\textbf{10}&\textbf{12}&\textbf{14} &\textbf{16}&\textbf{18}&\textbf{20}\\\myhline{0.4mm}$N_{EDS}(\ge)$&15&15&15&15&15&14&13&12&10&7\\\hline$H_{peak}(\min)$&20&40&60&80&100&120&140&160&180&200\\\myhline{0.4mm}\end{tabular}\label{tab3}\end{table}We assume that the suitable value of $H_{peak}$ is 120 minutes.Then the suitable value of $\Delta t$ is about 12 minutes, and$N_{EDS}$ is 14 judging from Figure \ref{fig5}. Certainly, we canwork $\Delta t$ and $N_{EDS}$ out through equation.\begin{figure}[htbp]\centering\includegraphics[width=294.6pt,totalheight=253.2pt]{fig05.eps}\caption{}\label{fig5}\end{figure}\paragraph{${NP}_i$ and $v_{EDS}$ are random value produced by MATLAB.}\begin{table}[htbp]\centering\caption{}\begin{tabular}{*{11}c}\myhline{0.4mm}$\mathbf{\Deltat(\min)}$&\textbf{2}&\textbf{4}&\textbf{6}&\textbf{8}&\textbf{10}&\textbf{12}&\textbf{14} &\textbf{16}&\textbf{18}&\textbf{20}\\\myhline{0.4mm}$N_{EDS}(\ge)$&15&22&21&21&15&17&21&16&13&14\\\hline$H_{peak}(\min)$&20&40&60&80&100&120&140&160&180&200\\\myhline{0.4mm}\end{tabular}\label{tab4}\end{table}We assume that the suitable value of $H_{peak}$ is 120 minutes.Then the suitable value of $\Delta t$ is about 12 minutes, and$N_{EDS}$ is 17 judging from Figure \ref{fig6}. Certainly, we canwork $\Delta t$ and $N_{EDS}$ out through equation.\begin{figure}[htbp]\centering\includegraphics[width=294.6pt,totalheight=249.6pt]{fig06.eps}\caption{}\label{fig6}\end{figure}\subsubsection{Interpretation:}By analyzing the results above, we can conclude that when$N_{EDS}$ is 29, and $\Delta t$ is 12, the flight schedule willmeet requirement at any time. The flight schedule is:\\[\intextsep]\begin{minipage}{\textwidth}\centering\tabcaption{}\begin{tabular}{c|*{8}c|c|c}\myhline{0.4mm}\backslashbox{\textbf{Set}}{\textbf{Type}}&\textbf{1}&\textbf{2}&\textbf{3}&\textbf{4}&\te xtbf{5}&\textbf{6}&\textbf{7}&\textbf{8}&\textbf{Numbers of Bags}&\textbf{Numbers of Flights}\\\myhline{0.4mm}1&2&0&0&0&2&1&0&0&766&5\\\hline2&2&0&2&0&2&0&0&0&732&4\\\hline3&0&1&1&1&2&0&0&0&762&4\\\hline4&0&1&0&0&2&1&0&0&735&4\\\hline5&0&1&0&0&2&1&0&0&735&5\\\hline6&2&0&0&0&1&0&0&1&785&5\\\hline7&2&0&0&0&2&0&1&0&795&5\\\hline8&0&1&0&0&2&1&0&0&735&4\\\hline9&2&0&0&0&2&1&0&0&766&5\\\hline10&0&0&0&2&2&0&0&0&758&5\\\hlineTotal&10&4&3&3&19&5&1&1&7569&46\\\myhline{0.4mm}\end{tabular}\label{tab5}\end{minipage}\\[\intextsep]We have produced random value for ${NP}_i$ and $v_{EDS}$. On thiscondition, the number of EDSs is 17, which is less than 29 that wedecide for the airport A. That is to say our solution can meet thereal requirement.\subsubsection{Airport B:}\paragraph{The passenger load is 100{\%}}The checking speed of EDS is 160 bags/hour.\begin{table}[htbp]\centering。

美赛（MCM/ICM）评阅流程可分为三个阶段：
第一阶段，也称为鉴别阶段，所有论文按质量被评判为三类：1.可以进入下一阶段评审的论文（这类论文通常具有较高
的质量，有机会获得较高的奖项）。

2.满足竞赛要求，但不足以进入下一评审阶段的论文（这
类论文为合格论文，将获得成功参赛奖）。

3.不符合竞赛要求的论文（这类论文为不合格论文，将不
会获得任何奖项）。

在这一阶段，每篇论文由两名评委独立评审并打分。

如果意见无法统一，则增加第三名评委进行评审。

主审和竞赛主席共同商定进入第二阶段评审的分数线，使只有略少于一半的参赛论文能进入第二阶段的评审。

第二阶段，论文按连续数学和离散数学进行题型分类，分别进入第二阶段的评审。

第二阶段的评审又分成若干轮，通过的评审轮数越多，论文评定的级别将越高。

在进入下一轮评审之前，每篇论文都将经过多名评委的评阅。

第三阶段，评委主要关注参赛小组的解题思路和建模过程，以及是否给出了清晰的描述。

他们会检查论文是否对赛题给出了满意的解读方式，并对赛题中可能出现的模糊概念给予了必要的澄清；是否明确列出了建模用到的所有前提条
件及假设，并对其合理性给出了满意的解释或论证；是否通过对赛题的分析给出了建模的动机或论证了建模的合理性；是否设计出了能有效地解答赛题的模型；是否对模型给出了稳定性测试；是否讨论了模型的优缺点，并给出了清晰的结论；是否给出了圆满的摘要。

美赛e题优秀论文翻译E题中文翻译：问题E：需要可持续城市！背景：许多社区正在实施智能增长计划，以考虑长期，可持续的规划目标。

“聪明的成长是关于帮助每个城镇和城市变得更加经济繁荣，社会公平和环境可持续的生活地方。

”[2]智能增长的重点是建设拥抱可持续发展的城市 - 经济繁荣，社会公平，环境可持续。

这个任务比以往任何时候都重要，因为世界正在迅速城市化。

预计到2050年，世界人口的66％将是城市人口 - 这将导致25亿人口被纳入城市人口。

[3]因此，城市规划变得越来越重要和必要，以确保人们获得公平和可持续的家园，资源和就业机会。

智能增长是一种城市规划理论，起源于1990年代，作为遏制城市持续蔓延和减少城市中心周围农田损失的手段。

智能增长的十大原则是[4]1混合土地利用2利用紧凑的建筑设计3创造一系列住房机会和选择4创建可步行的社区5培养独特的，有吸引力的社区，具有强烈的地方感6保留开放空间，农田，自然美景和关键环境区域7加强和指导现有社区的发展8提供多种交通选择9使开发决策具有可预测性，公平性和成本效益10鼓励社区和利益相关者在发展决策中进行合作这些广泛的原则必须适应社区的独特需求，才能有效。

因此，任何成功的衡量都必须包括一个城市的人口统计，增长需求和地理条件，以及坚持三个E的目标。

任务：国际城市管理集团（ICM）需要您帮助实施智能增长理论到世界各地的城市设计。

在两个不同的大陆选择两个中型城市（人口在10万和50万之间的任何城市）。

1.定义衡量城市智能增长成功率的指标。

它应该考虑可持续性的三个E和/或智能增长的十个原则。

2.研究选定城市的当前增长计划。

衡量和讨论每个城市目前的增长计划是否符合智能增长原则。

根据您的指标，当前的计划是否成功？3.使用智能增长原则在未来几十年内为两个城市制定增长计划。

支持您为什么根据您的城市的地理位置，预期增长率和经济机会选择您的计划的组件和计划。

使用您的指标评估您的智能增长计划的成功。