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。