美赛论文LaTeX模板
美赛论文LaTeX模板
%% 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。
用latex软件撰写科技议论文模板.
Latex 实用例子通过实验本例子可以基本掌握科技排版的方法：\documentclass[twocolumn]{article}\usepackage{amsmath}\renewcommand{\rmdefault}{ptm}\begin{document}\title{Physical Model Order Reduction}\author{Qiang Wang and Guo-Hua Li\\Department of Electronic Engineering,the East University of China}\date{December 17, 2009}\maketitle\begin{abstract}This paper presents a novelapproach for model order reduction for multilayer lossy RFembedded passives.\end{abstract}\section{Introduction}As the Radiofrequency modules having been designed more compactthan ever before, the parasitic effects due to the tightly coupledinter-connections on the circuit layout are inevitable. Therefore,an efficient method that can derive a circuit model of such circuitlayout is highly desirable. A few techniques such as PEEC were developed to extract equivalent circuits from an electromagnetic model.Classic PEEC solver converts the layout into lumped RLCinterconnection networks, including mutual couplings. Once a circuit model is generated, any circuit solver, such as SPICE, can manage the rest of the job. Unfortunately, the numbers of nodes and elements in the circuits are excessive. Therefore, researchers have been searching for an effective measure that can reduce the model order for accelerating the analysis of the circuit model.Although exploited Krylovsubspace methods and provided ways to speed up the simulation; they all lack the physical insight. In fact, they are mathematics-based MOR. Several realizable model order reduction approaches are also proposed. However, can only handle RC networks. Besides, they both concern with matching the first two or three moments of the system via Taylor's expansion. Thus, these methods can not provide a clear physical explanation to the reduced circuit either. It is worth mentioning that has showed some insight fordealing with coupled inductances, even though the complexity of the scheme itself might have already limited its practical use.The work presented in this paper is an extension to,in which a derived physically realizable lossless expressive circuit model reduction method is introduced. In this paper, a lossy modelis the major concern. The passivity of the resultant circuit modelby the new reduction scheme is guaranteed.\section{Theory}The circuit model generated by traditional quasi-static PEEC model for a multi-layer circuit layout with very thin conducting stripscan easily incorporate the conducting loss, which is a major origin of the circuit loss.Since the meshes used in solving MPIE (Mixed Potential Integrated Equations of the PEEC algorithm are all in regular shapes, thus we could first evaluate their losses piecewisely and then superimpose this pre-calculated loss model to the generated circuit model to represent the conductor loss of the circuit. Therefore reasonable and time-saving approaches to calculate the loss for different meshing geometries are investigated.Since the conductor loss is generally determined by the skin depth effect at RF frequency, a coarse but rapid approximation to this type of loss is to find out the skin depth and other shape factorsof the mesh. Then the equivalent surface impedance can be easily calculated by $R_{L} = l / 2\delta S$, where $l$ is the mesh length, along which the current flows, $S$ is the area of the equivalent crossing section where the current goes through, $\delta$ is the skin depth:\begin{align}\delta=\frac{1}{\sqrt{\smash[b]{\pi f \mu \sigma}}},\label{Equ1}\end{align}in which $f$ is the frequency, $\mu$ is the magnetic permeability, and $\sigma$ is the conductivity of the metal. In addition to this, various other empirical formulas, such as those incould also be used todetermine the conducting loss.Capacitor's loss is mainly due to bypassing leak-age current. Considering the basic relationship be-tween the current and the charge stored in the capacitor, its losscontribution could be computed by $G_{C} = \sigma C / \varepsilon$, where $G_{C}$ is the bypassing conductance, $C$ is the capacitance, $\varepsilon$ and $\sigma$ are the dielectric constant and the conductivity of the substrate, respectively....\end{document}形成的PDF效果图：。
latex的论文格式模板
latex的论文格式模板LaTeX是一种基于ΤΕΧ的排版系统即使使用者没有排版和程序设计的知识也可以充分发挥由TeX所提供的强大功能，那它的论文格式是怎么样的呢?下面是小编精心推荐的一些latex的论文格式模板，希望你能有所感触!latex的论文格式模板1、题目：应简洁、明确、有概括性，字数不宜超过20个字。
2、摘要：要有高度的概括力，语言精练、明确，中文摘要约100—200字;3、关键词：从论文标题或正文中挑选3～5个最能表达主要内容的词作为关键词。
4、目录：写出目录，标明页码。
5、正文：论文正文字数一般应在3000字以上。
论文正文：包括前言、本论、结论三个部分。
前言(引言)是论文的开头部分，主要说明论文写作的目的、现实意义、对所研究问题的认识，并提出论文的中心论点等。
前言要写得简明扼要，篇幅不要太长。
本论是论文的主体，包括研究内容与方法、实验材料、实验结果与分析(讨论)等。
在本部分要运用各方面的研究方法和实验结果，分析问题，论证观点，尽量反映出自己的科研能力和学术水平。
结论是论文的收尾部分，是围绕本论所作的结束语。
其基本的要点就是总结全文，加深题意。
6、谢辞：简述自己通过做论文的体会，并应对指导教师和协助完成论文的有关人员表示谢意。
7、参考文献：在论文末尾要列出在论文中参考过的专著、论文及其他资料，所列参考文献应按文中参考或引证的先后顺序排列。
8、注释：在论文写作过程中，有些问题需要在正文之外加以阐述和说明。
9、附录：对于一些不宜放在正文中，但有参考价值的内容，可编入附录中。
关于养生的论文范文浅析中医养生智慧摘要：《黄帝内经》讲道："法于阴阳，和于术数，食饮有节，起居有常，不妄作劳，故能形与神俱，而尽终其天年，度百岁乃去。
"只有在平时养成良好的生活习惯，并以"经典"养生方法，点滴积累，持之以恒，才能令体质强健，年过百而动作不衰也，真正达到益寿延年的目的。
latex英文作业模板
latex英文作业模板英文回答：Introduction.In this essay, I will discuss the various factors that influence the popularity of online games, including game design, marketing, and social aspects.Game Design.One of the most important factors that influence the popularity of online games is the game design itself. Games that are well-designed are more likely to be enjoyed by players and to keep them coming back for more. There are a number of elements that go into good game design, including:Gameplay: The gameplay is the core of any game, and it is what keeps players engaged. Games that are challenging and rewarding are more likely to be popular than games thatare too easy or too difficult.Graphics: The graphics of a game can also play a role in its popularity. Games with high-quality graphics are more likely to attract players than games with poor graphics.Story: The story of a game can also be a factor in its popularity. Games with engaging stories are more likely to keep players invested in the game world.Marketing.Marketing is another important factor that can influence the popularity of online games. Games that are marketed well are more likely to reach a wider audience and to generate interest in the game. There are a number of different marketing strategies that can be used to promote online games, including:Advertising: Advertising is one of the most common ways to market online games. Games can be advertised on avariety of platforms, including television, radio, and the internet.Social media: Social media can also be a powerful tool for marketing online games. Games can be promoted on social media platforms by creating engaging content and by interacting with potential players.Influencer marketing: Influencer marketing is another effective way to market online games. Influencers are people who have a large following on social media and who can help to promote games to their followers.Social Aspects.The social aspects of online games can also play a role in their popularity. Games that allow players to interact with each other are more likely to be popular than games that are played solo. There are a number of different ways that players can interact in online games, including:Chat: Chat is one of the most common ways for playersto interact in online games. Players can chat with each other in real-time, which can help to build relationships and create a sense of community.Clans and guilds: Clans and guilds are groups of players who band together to play together and achieve common goals. Clans and guilds can provide players with a sense of belonging and support.Player-versus-player (PvP) PvP is a type of online game in which players compete against each other. PvP can be a lot of fun, and it can also help to build rivalries and create a sense of competition.Conclusion.The popularity of online games is influenced by a number of factors, including game design, marketing, and social aspects. By understanding these factors, game developers can create games that are more likely to be popular with players.中文回答：简介。
MCM美赛论文集
高教社杯全国大学生数学建模竞赛承诺书我们仔细阅读了中国大学生数学建模竞赛的竞赛规则。
我们完全明白，在竞赛开始后参赛队员不能以任何方式（包括电话、电子邮件、网上咨询等）与队外的任何人（包括指导教师）研究、讨论与赛题有关的问题。
我们知道，抄袭别人的成果是违反竞赛规则的,如果引用别人的成果或其他公开的资料（包括网上查到的资料），必须按照规定的参考文献的表述方式在正文引用处和参考文献中明确列出。
我们郑重承诺，严格遵守竞赛规则，以保证竞赛的公正、公平性。
如有违反竞赛规则的行为，我们将受到严肃处理。
我们参赛选择的题号是（从A/B/C/D中选择一项填写）：A我们的参赛报名号为（如果赛区设置报名号的话）：99999所属学校（请填写完整的全名）：西安交通大学参赛队员(打印并签名）：1.一作者2.二作者3.三作者指导教师或指导教师组负责人(打印并签名)：导师日期：2011年8月1日赛区评阅编号（由赛区组委会评阅前进行编号）：2011高教社杯全国大学生数学建模竞赛编号专用页赛区评阅编号（由赛区组委会评阅前进行编号）：赛区评阅记录（可供赛区评阅时使用）：评阅人评分备注全国统一编号（由赛区组委会送交全国前编号）：全国评阅编号（由全国组委会评阅前进行编号）:全国大学生数学建模竞赛L A T E X2ε模板摘要这是数学建模论文模板mcmthesis的示例文件。
特别地，这篇文档是“全国大学生数学建模竞赛（CUMCM）”模板的示例文件。
这个模板使用于参加高教社杯全国大学生数学竞赛的同学准备他们的建模论文，帮助他们更多的关注于论文内容而非论文的排版。
这个模板的设计是根据2010年修订的《全国大学生数学建模竞赛论文格式规范》[1]制作，完全符合该论文格式规范，但是该模板未得到官方认可，请使用者自己斟酌使用。
这个示例文档逐条展示其对[1]的实现效果，并对所有自定义命令进行说明。
这个示例文件还包含了一些对公示、插图、表格、交叉引用、参考文献、代码等的测试部分，以展示其效果，并作简要的使用说明。
latex数学模板样例
当您在撰写数学论文或报告时，使用LaTeX数学模板可以帮助您更轻松地组织和格式化数学内容。
以下是一个简单的LaTeX数学模板样例，用于演示如何使用LaTeX编写数学公式和符号。
首先，确保您已经安装了LaTeX软件包和相应的数学包。
然后，按照以下步骤创建数学模板文档：1. 创建一个新的LaTeX文档，并使用适合您的模板或新建一个空白文档。
2. 在文档中添加一个环境来包含您的数学公式和符号。
常用的环境是`equation`、`align`或`gather`等。
3. 在环境中，使用LaTeX数学命令来创建公式和符号。
以下是一些常用的命令示例：* `\frac{分子}{分母}`：用于分数。
* `\sqrt{x}`：用于平方根。
* `\lim_{x \rightarrow ∞}`：用于极限符号。
* `\oint_{x=a到b}`：用于积分符号。
* `\pm`、` \times`、` \div`等：基本的运算符。
4. 添加必要的上下文和说明。
例如，如果您要写一个数学定理的证明过程，请提供相关的定义、前提和结论。
5. 使用LaTeX文档类和数学包来正确显示公式和符号。
通常，LaTeX文档类提供了一些预设的样式和符号，而数学包提供了更多的符号和格式选项。
6. 在编译LaTeX文档时，确保正确设置了排版选项和格式化参数。
下面是一个简单的LaTeX数学模板样例，其中包含一些常见的数学公式和符号：```latex\documentclass{article} % 选择适合您的文档类，例如report或book\usepackage{amsmath} % 添加数学包以提供额外的符号和格式选项\begin{document}\begin{equation}\lim_{x \rightarrow \infty} \frac{x}{e^x} = 1\end{equation}\begin{align}a +b &=c \\x^2 + y^2 &= z^2 \\\frac{d}{dx} f(x) &= g(x)\end{align}\end{document}```这个样例包含了一个极限公式、一个加法公式和一个微分公式。
latex论文模板
latex论文模板
在LaTeX中，有许多可以用于撰写论文的模板。
以下是一些
常用的LaTeX论文模板：
1. `\documentclass{article}`：这是LaTeX的基本模板类别之一，可用于撰写一般论文。
你可以根据需要随意添加其他包和自定义设置。
2. `elsevier`：这是适用于撰写科技领域论文（如工程、计算机
科学等）的模板。
这个模板提供了相应的论文格式和引用风格。
3. `IEEEtran`：这是适用于撰写电气工程和计算机科学领域论
文的模板。
它符合IEEE期刊论文的格式要求。
4. `acmart`：这是适用于计算机科学和信息技术领域的模板，
符合ACM期刊和会议论文的格式要求。
5. `memoir`：这是一个全功能的文档类，可以用于撰写各种类
型的论文，包括书籍、学位论文等。
它提供了丰富的布局和样式选项。
这只是一小部分可以用于LaTeX论文撰写的模板示例。
你可
以根据自己的需求选择相应的模板，并按照模板提供的文档结构和命令进行撰写。
latex中文文稿模板
latex中文文稿模板以下是一个基本的 LaTeX 中文文稿模板，你可以根据自己的需求进行修改和扩展。
```latex\documentclass[UTF8]{ctexart}\usepackage{ctex}\usepackage{graphicx}\usepackage{geometry}\usepackage{hyperref}\usepackage{caption}\usepackage{subcaption}\usepackage{float}\usepackage{enumitem}\usepackage{makeidx}\makeindex\geometry{a4paper, left=3.18cm, right=3.18cm, top=2.54cm, bottom=2.54cm} \title{中文文稿标题}\author{作者姓名}\date{\today}\begin{document}\maketitle\section{引言}在此处插入引言。
\section{正文}在此处插入正文内容。
\section{结论}在此处插入结论。
\appendix\section{附录}在此处插入附录内容。
\end{document}```此模板使用 `ctexart` 文档类，并采用了 `UTF8` 编码以支持中文字符。
它包含了常用的 LaTeX 包，如 `ctex` 用于中文排版，`graphicx` 用于插入图片，`geometry` 用于页面布局，`hyperref` 用于生成超链接，`caption` 和 `subcaption` 用于图片标题和子标题，`float` 用于浮动图形和表格，`enumitem` 用于定制列表环境，以及 `makeidx` 用于生成索引。
你可以根据自己的需求修改和扩展这个模板，例如添加章节、图表、引用等内容。
如果你有特定的要求或需要进一步的帮助，请提供更多细节。
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- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
%% 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。