2013年数学建模A题 优秀论文

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2013全国数模竞赛A题优秀论文祥解

2013全国数模竞赛A题优秀论文祥解

2013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):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):车道被占用对城市道路通行能力的影响摘要本文主要研究车道被占用对城市道路通行能力的影响并建立了相应的数学模型。

针对问题一,考虑到交通信号灯的周期,我们选择1分钟为周期,结合不同车辆的标准车当量的折算系数,求出每个采样点的交通量,通过MATLAB作图,从定性方面对道路通行能力进行分析,然后通过基本通行能力和4个修正系数建立动态通行能力的模型。

图像显示,事故发生后(采样点5附近),实际通行能力下降至一个较低水平,并且横断面处的实际能力变化过程呈先下后上的波形变化,在事故解决(第20个采样点)以后,由图像看出实际通行能力持续上升。

针对问题二,利用问题一建立的模型,结合视频二,比较交通事故所占不同车道时横断面的实际通行能力,可以发现二者实际通行能力变化趋势大致相同,但视频二实际通行能力大于视频一实际通行能力。

可见占用车流量大的车道使道路通行能力降低更多。

针对问题三,首先我们建立单车道排队车辆数目的积分模型,单个车道的滞留车辆为上游车流量和实际通行能力的差值。

我们以30s为一个时间段,对视频一中的车流量进行统计,得到横截面处每个监测段的实际通行能力。

本题要求考虑三车道,总体排队长度不容易通过积分模型确定,所以我们将队列长度问题转化为车辆数目问题,通过视频资料统计120米对应24辆车,据此关系转换,从而得到车辆排队长度与事故横断面实际通行能力、事故持续时间和上游车流量的关系。

针对问题四,在对问题3研究的基础上,根据问题3建立的数学模型,建立起某一段时间间隔车辆排队的长度,然后,通过求得的关系得到当排队长度为140m的时候所对应的时间段,由于每段时间间隔设为30s,因此,可以求得排队长度到达上游时用的时间为347.7273s。

关键词:交通事故车道占用通行能力排队论一、问题的重述车道被占用是指因交通事故、路边停车、占道施工等因素,导致车道或道路横断面通行能力在单位时间内降低的现象。

2013年五一数学建模联赛 A题论文

2013年五一数学建模联赛 A题论文

图 1 男女大学生体重指数一览表
人数
体重指数一览表
800
712
700
600 500
475
400
300
200 100
10264 38
237 183157 26
0
7267 5
低体重者
正常体重者
超重者
四类人群
肥胖者
学生数 男生数 女生数
2
4.2.3 结果与分析
对照表 1 及相关的频数表,反映出大学生 BMI 指数基本正常,正常体重人群相对集
关键词: 独立样本 t 检验 方差检验 K-S 检验 模糊综合评价 SPSS
一 问题重述
据数据显示,近年来中国大学生的体质健康水平呈下降趋势。学生或者过重或者过 瘦,对大学生体质健康的评价问题对于如何提高大学生的体质健康水平具有指导意义。
根据对某高校大一新生 36 个班级共 1000 多名学生进行的体质与健康测试,得到了 一组相关资料,由于测试过程中学生未能按照要求规范测试,导致测量结果中出现一些 偏差,进而影响了体质健康的测试,请结合各项测试评分标准,回答下列问题: 问题 1:影响大一新生的体质健康状况的因素很多,体重是体现体质健康状况的重要指 标,分析体重对体质健康的影响;在体质健康测试中,测试结果可能存在误差,在附表 1 中,有些测量资料不能反映同学的真实水平,根据附表 1 数据,请建立数学模型检验 测试结果的正确性和准确性,找出附表 1 中 1、2、3 班同学的可能偏差测试结果,并说 明理由。 问题 2:生源地是影响体质健康状况的因素,请在不同生源地选取适当的样本,试检验 不同地区学生的体能健康是否具有显著差别。 问题 3:目前,我国体能测试主要采用《国家体质健康标准》对学生体质进行评价,根 据附表 2 中(男生:sheet1;女生:sheet2)项目评价标准,试建立体质健康评价模型, 评价该校学生的体质健康状况,并对 1 班的 30 名同学进行体质健康评价。 问题 4:我国大多数高校学生体质健康合格率未达到国家要求,对于未达标的大一新生 来说,就如何让学生在在校期间提高自身的体质健康写一份建议报告书,其中包括提高 体质健康水平的措施和手段,如何量化提高体质健康指标等问题。

2013美国数学建模A题优秀论文

2013美国数学建模A题优秀论文

终极布朗尼烤烤盘一、摘要根据题意,我们把把要解决的分成三个问题;第一个就是建立一个模型来表示整个烤盘的外边缘热量的分布。

第二个就是优化组合题目中条件1和条件2,使得权重p和(1- p)能够描述随着W/L和p值的改变,最佳的烤烤盘形状和热量分布情况是如何改变的第三个问题就是为布朗尼美食家杂志准备一到两页的宣传广告,需要突出设计和结果。

对于第一个问题,我们结合傅里叶定律构建了二维热传导模型;然后通过模型中的S来限定范围得到六种不同形状烤盘对应的热传导偏微分方程。

然后对模型赋值和第二类边界条件(Neumann边界条件)下,应用comsol得出六种烤盘稳定热量分布图像和烤盘外边缘热量分布图像。

通过输出的图像,我们得出结论:矩形四角处温度较高,圆形外边缘热量分布比较均匀;随着烤盘边数的增加,烤盘外边缘热量分布愈加均匀,但在角处温度仍然会高一些对于问题二对于问题三关键词:二、问题重述当用一个长方形的平底烤盘(盘)烘烤时,热量被集中在4个角,在角落处,食物可能被烤焦了,而边缘处烤的不够熟。

在一个圆形的平底烤盘(盘)热量被均匀地分布在整个外边缘,在边缘处食物不会被烤焦。

但是,大多数的烤箱的形状是矩形的,采用了圆形的烤盘(盘)相对于烤箱的使用空间而言效率不高。

为所有形状的烤盘(盘)----包括从矩形到圆形以及中间的形状,建立一个模型来表示整个烤盘(盘)的外边缘热量的分布。

假设:1. 形状是矩形的烤箱宽长比为W/L;2. 每个烤烤盘(盘)的面积为A;3. 每个烤箱最初只有两个均匀放置的烤架。

根据以下条件,建立一个能使用的最佳类型或形状的烤烤盘(盘):1.放入烤箱里的烤烤盘(盘)数量的最大值为(N);2.烤烤盘(盘)的平均分布热量最大值为(H);3.优化组合条件1和条件2,使得权重p和(1- p)能够描述随着W/L和p值的改变,最佳的烤烤盘形状和热量分布情况是如何改变的。

除了完成规定的解决方案,为布朗尼美食家杂志准备一到两页的宣传广告,需要突出你的设计和结果。

2013年全国数学建模竞赛A题

2013年全国数学建模竞赛A题

2013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《全国大学生数学建模竞赛章程》和《全国大学生数学建模竞赛参赛规则》(以下简称为“竞赛章程和参赛规则”,可从全国大学生数学建模竞赛网站下载)。

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章程和参赛规则的,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,严格遵守竞赛章程和参赛规则,以保证竞赛的公正、公平性。

如有违反竞赛章程和参赛规则的行为,我们将受到严肃处理。

我们授权全国大学生数学建模竞赛组委会,可将我们的论文以任何形式进行公开展示(包括进行网上公示,在书籍、期刊和其他媒体进行正式或非正式发表等)。

我们参赛选择的题号是(从A/B/C/D中选择一项填写): A我们的参赛报名号为(如果赛区设置报名号的话):所属学校(请填写完整的全名):参赛队员(打印并签名) :1.2.3.指导教师或指导教师组负责人(打印并签名):(论文纸质版与电子版中的以上信息必须一致,只是电子版中无需签名。

以上内容请仔细核对,提交后将不再允许做任何修改。

如填写错误,论文可能被取消评奖资格。

)日期: 2013 年月日赛区评阅编号(由赛区组委会评阅前进行编号):2013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):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)车道被占用对城市道路通行能力的影响摘要道路堵塞时车辆排队长度和排队持续时间时交通管理与控制部门制定和实施管理控制措施的重要依据,对道路堵塞时车辆排队和排队时间计算方法进行研究具有重要的实际意义和应用价值。

本文以交通事故为例讨论车道被占用对城市道路通行能力的影响,从而对交通管理部门正确引导车辆行驶、审批占道施工、设计道路渠化方案、设计路边停车位等问题提供理论依据。

2013高教社杯全国大学生数学建模竞赛A题论文.

2013高教社杯全国大学生数学建模竞赛A题论文.

2013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《全国大学生数学建模竞赛章程》和《全国大学生数学建模竞赛参赛规则》(以下简称为“竞赛章程和参赛规则”,可从全国大学生数学建模竞赛网站下载)。

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章程和参赛规则的,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,严格遵守竞赛章程和参赛规则,以保证竞赛的公正、公平性。

如有违反竞赛章程和参赛规则的行为,我们将受到严肃处理。

我们授权全国大学生数学建模竞赛组委会,可将我们的论文以任何形式进行公开展示(包括进行网上公示,在书籍、期刊和其他媒体进行正式或非正式发表等)。

我们参赛选择的题号是(从A/B/C/D中选择一项填写): A我们的参赛报名号为(如果赛区设置报名号的话):所属学校(请填写完整的全名):吉林医药学院参赛队员(打印并签名) :1. 于邦文2. 薛盈军3. 杨国庆指导教师或指导教师组负责人(打印并签名):霍俊爽(论文纸质版与电子版中的以上信息必须一致,只是电子版中无需签名。

以上内容请仔细核对,提交后将不再允许做任何修改。

如填写错误,论文可能被取消评奖资格。

)日期: 2013 年 9 月 16 日赛区评阅编号(由赛区组委会评阅前进行编号):2013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):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):车道被占用对城市道路通行能力的影响摘要本文通过对城市中车道因交通事故被占用问题的分析,探讨了事故所处道路横断面的实际通行能力的变化过程,并依据事故路段车辆排队长度与实际通行能力、事故持续时间、路段上游车辆流量之间的关系,最后针对各个问题建立模型并求解。

2013高教社杯全国大学生数学建模竞赛A题

2013高教社杯全国大学生数学建模竞赛A题

2013高教社杯全国大学生数学建模竞赛A题2013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《全国大学生数学建模竞赛章程》和《全国大学生数学建模竞赛参赛规则》。

我们完全明白,在竞赛开始后参赛队员不能以任何方式与队外的任何人研究、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章程和参赛规则的,如果引用别人的成果或其他公开的资料,必须按照规定的面的车辆数。

实际通行车流量的采集与处理视频1中出现车辆多种多样,要统计车流量数据,需先统一车流标准,把视频中出现的车辆进行折算,以小轿车做为标准,对各个型号车辆进行折算[2],折算系数如表1所示。

表1 车辆折算系数附件中出现汽车小轿车中型车大客车车辆折算系数在事故发生前,道路的通行能力足以应对上游车流量,当发生事故时,事故点上游共有10辆小轿车与5辆大客车,车流量为20pcu。

之后一分钟(16:42:32-16:43:32),上游又有车流量21pcu,但只通过了21pcu,说明造成了交通拥堵和排队情况。

“附件5”可知,相位时间为30s,红灯时间为30s,即60s为一个周期,进行统计时间周期也为60s,不会造成因交通灯引起的误差。

实际通行流量是指折算后通过事故横断面的车流,上游车流量是指折算后从各个路口驶入事故横断面的车流。

对附件1中事故横断面处的车流量进行统计,得出实际通行车流量情况,并统计横断面上游的车流量,在统计过程中发现视频并不是完全连续的,例如在16:49:40时出现了突变,直接到16:50:04,跳跃间隔为24s,但于堵车情况较重,可以根据车流量守恒原则和车辆追踪,统计出通过横断面处的车流量及上游车流量。

但16:56:04等时间,跳跃时间较长,近2分钟,无法精确统计,如表2处“空缺”所示。

在17:00:07到17:01:20时视频发生跳变,在此期间事故车辆驶离道路,之后为事故恢复时间。

为了描述事故发生开始到车辆离开车道全程的实际通行能力变化情况,将视频中空缺数据通过灰色预测(程序见附录)进行填补,结果如表2所示。

2013高教杯数学建模A题


从视频一获取数据,把 16:42 分事故发生作为起点,每隔 30s 设置一个时间点, 时间起始点分别记为 t0 , t1 , t2
ti
面 的 车 辆 以 及 上 游 进 入 的 小 客 车 数 目 , 记 为 n1 0, n 1 1 ni 1 , 大 客 车 数 目 记 为
n20 , n21 n2i
我们参赛选择的题号是(从 A/B/C/D 中选择一项填写) : 我们的参赛报名号为(如果赛区设置报名号的话) :136A06 所属学校(请填写完整的全名) : 参赛队员 (打印并签名) :1. 2. 3. 浙江大学 赵东伟 张世豪 陆韶琦 数模组
A
指导教师或指导教师组负责人 (打印并签名):
(论文纸质版与电子版中的以上信息必须一致,只是电子版中无需签名。以上内容 请仔细核对,提交后将不再允许做任何修改。如填写错误,论文可能被取消评奖资格。) 日期: 2013 年 9 月 16 日
关键词: 道路通行能力 排队长度 层次分析法 回归模型 统计检验 流体力学模型 神经 网络模型
1
1
问题重述
车道被占用是指因交通事故等因素, 导致车道或道路横断面通行能力在单位时间内 降低的现象。如果车道被占用,可能降低路段所有车道的通行能力,引起车辆排队,造 成交通阻塞。正确估算车道被占用对城市道路通行能力的影响程度,可以为道路交通部 门的交通管理提供决策依据。 现根据附件提供的资料研究以下问题: 问题一:分析描述交通事故发生至撤离期间,事故所处横断面实际通行能力的变化 过程。 问题二: 分析说明同一横断面交通事故所占车道不同对该横断面实际通行能力影响 的差异。 问题三:构建数学模型,研究交通事故所影响的路段车辆排队长度与事故横断面实 际通行能力、事故持续时间、路段上游车流量间的关系。 问题四:如果交通事故所处横断面距离上游路口变为 140 米,路段下游方向需求不 变, 路段上游车流量为 1500pcu/h,事故发生时车辆初始排队长度为零, 且事故持续不撤 离。估算从事故发生开始到车辆排队长度将到达上游路口所用的时间。

2013全国数学建模A题优秀作品

第六届“认证杯”数学中国数学建模网络挑战赛承诺书我们仔细阅读了第六届“认证杯”数学中国数学建模网络挑战赛的竞赛规则。

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。

我们允许数学中国网站()公布论文,以供网友之间学习交流,数学中国网站以非商业目的的论文交流不需要提前取得我们的同意。

我们的参赛队号为:2261参赛队员(签名) :队员1:张述平队员2:魏方征队员3:乔赛参赛队教练员(签名):参赛队伍组别:2261第六届“认证杯”数学中国数学建模网络挑战赛编号专用页参赛队伍的参赛队号:2261竞赛统一编号(由竞赛组委会送至评委团前编号):竞赛评阅编号(由竞赛评委团评阅前进行编号):2013年第六届“认证杯”数学中国数学建模网络挑战赛题目护岸框架减速效果的优化方案关键词护岸框架减速效果单因素方差分析最小二乘拟合回归分析摘要:四面六边透水框架是一种新型江河护岸工程技术,对于降低岸边流速、稳定河道、保护堤岸有显著的作用。

本文针对所引用参考文献中的图像、数据,从四面六边透水框架群框架尺寸、架空率和长度三方面出发,对框架群的水力特性及其影响因素进行分析,探讨三要素对减速效果的影响,建立三个模型,为这种“亲水”式生态防护技术在工程中推广运用提供参考依据。

模型一:架空率对减速效果影响的分析。

首先,利用单因素方差分析,推断出架空率对减速效果的影响较显著;其次,采取最小二乘法拟合曲线具体演示二者的发展趋势,并得到关系模型,依照此模型得出架空率对减速效果影响显著。

得出框架率ε=4.2~4.8 之间时,框架群的减速率比较高,能够使得框架群的阻水消能作用和“亲水”功能较好的结合起来。

2013年数学建模美赛A题二等奖作品

The perfect pan for ovenThe heat transfer in the oven includes heat conduction, heat radiation and heatconvection. We use two-dimensional Fourier heat conduction equation ∂u∂t −α(ð2u∂x2+ð2u∂y2)=f(x, y, t) to make a research on distribution of heat for the pan. Heat source heats the pan by heat radiation. The pan interacts with air in the oven in the way of natural convection, so the pan realizes heat dissipation.We calculate heat radiation based on radiation ability of heat source and heating tube area. We use heat dissipation function to show the pan's different parts' loss of heat caused by natural convection. Both of them consist in heat source function f.The area of the pan is fixed at 0.085m2in this paper. When comparing temperatures at the edges of rectangular pans with different length to width ratios ξ, we can get that the smaller ξ is, the lower the temperature of the edges is. But as long as it is still a rectangle, the amplitude of its drop won't be very big. When we make the pans with fixed area vary from square to round square to round, we find that the bigger the fillet radius is, the lower the temperature of its corners is and the extent of temperature's reducing is large.We fix the bottom area of the oven and area of the pan. Through study, we find that round square's capacity for uniform distribution of heat is far higher than other shape's (except round). The larger the fillet radius of the round square is, the larger the pan’s waste of space is. But heat distribution is more uniform. We work out the optimal solution of pan’s size under different weights p through optimizing the relationship between two conditions. Then we get several oven's width to length ratios of W/L by arranging the pans with the optimal size.I. IntroductionThe temperature of each point in the pan is different. For a rectangular pan, the corners have the highest temperature, so the food is easily overcooked. While the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges in the round pan.To illustrate the model further, the following information is worth mentioning1.1 Floor space of the panThe floor space of each pan is not the square itself necessarily. In this paper, there are 3 kinds of pans with different shapes, as rectangular pans, round pans and round rectangle pans.For rectangular pan, the floor space is the square itself, and the pans can connect closely without space.For round pans, the diagrammatic sketch of the floor space is as follows:shade stands for the round pan; square stands for the floor spaceFigure 1Round pans have the largest floor space for a certain area. The space between each pan is larger than other two kinds of pans. The coefficient of utilization for the round pans is the lowest.For round rectangle pans, the diagrammatic sketch of the floor space is as follows:shade stands for the round rectangle pan; square stands for the floor spaceFigure 2If the area of the round rectangle is the same as the other two, its floor space is between them. The coefficient of utilization of oven decreases with the radius of expansion.1.2 Introduction of ovenThe oven is usually a cube, no matter it is used in home or for business. A width to length ratio for the oven is not a certain number. There are always two racks in the oven, evenly spaced. There are one or more pans on each rack. To preserve heat for the oven, food is heated by radiation. Heating tube can be made of quartz or metal. The temperature of the tube can reach 800℃ high when the material is quartz. The heating tube is often in the top and bottom of the oven. Heating mode can be heating from top or heating from bottom, and maybe both[1].1.3 Two dimensional equation of conductionTo research the heat distribution of pan, we draw into two dimensional equation[2]of conduction:∂u ∂t −α(ð2u∂x2+ð2u∂y2)=f(x, y, t)In this equation:u- temperature of the pant- time from starting to heatx- the abscissay-ordinateα- thermal diffusivityf- heat source functionThe heat equation is a parabolic partial differential equation which describes the distribution of heat (or variation in temperature) in a given region over time. The heat equation is of fundamental importance in diverse scientific fields. In mathematics, it is the prototypical parabolic partial differential equation. In probability theory, the heat equation is connected with the study of Brownian motion via the Fokker–Planck equation. The diffusion equation, a more general version of the heat equation, arises in connection with the study of chemical diffusion and other related processes.II. The Description of the Problem2.1 The original problemWhen baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient withrespect to using the space in an oven.Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes -rectangular to circular and other shapes in between.Assume1. A width to length ratio of W/L for the oven which is rectangular in shape.2. Each pan must have an area of A.3. Initially two racks in the oven, evenly spaced.Develop a model that can be used to select the best type of pan (shape) under the following conditions:1. Maximize number of pans that can fit in the oven (N)2. Maximize even distribution of heat (H) for the pan3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p.In addition to your MCM formatted solution, prepare a one to two page advertising sheet for the new Brownie Gourmet Magazine highlighting your design and results.2.2 Problem analysisWe analyze this problem from 3 aspects, showing as follows:2.2.1 Why the edge of the pan has the highest temperature?The form of heat transfer includes heat radiation, heat conduction and heat convection. Energy of heat radiation comes from heat resource. The further from heat resource, the less energy it gets. Heat conduction happens in the interior of the pan, and heat transfers from part of high temperature to the part of low with the temperature contrast as its motivation. With the two forms of heat transfer above, we find the result is that pan center has the high temperature and the boundary has the low. D epending on that, we can’t explain why the product gets overcooked at the corners while at the edges not. We think that there is natural convection between pan and gas, because the temperature of pan is much higher than that of gas. The convection is connected with the contact area. The point in pan center has a larger contact range with air, so the energy loss from convection is more. While the point in the corners of a rectangular pan has a narrow contact area, the energy loss is less than that in the inner part. Because that above, the energy in the pan center is more than that in corner, and the corners have higher temperature.2.2.2 Analysis of heat distribution in pans in different shapesThe shape of pan includes rectangle, round rectangle and round. When these pans' area is fixed, the rectangles with different shapes can be shown with different length to width ratios. Firstly, we study temperature (maximum temperature) in the corners of rectangles with different length to width ratios. Then we study how temperature in the corners changes when the pans vary from square to round square to round. After that, we select some rectangular panand make it change from rectangle to round rectangle to study changes of temperature in the corners. To calculate distribution of heat for the pan, we would use three main equations. The first one is the Fourier equation, namely heat conduction equation, the second one is the radiation transfer equation of heat source and the third one is the equation of heat dissipation through convection. In the radiation transfer equation of heat source, we take heat source as a point. We get its radiating capacity through its absolute temperature, blackening and Stefan-Boltzmann law. We combine radiating capacity with surface area of quartz heating tube to get quantity of heat emitted by heat source per second, then we can get heat flux at each point of the pan. In the equation of heat loss through convection, Heat dissipating capacity is proportional to area of heat dissipation, its proportional coefficient can be found from the related material. Through the establishment of above three main equations, we can use pdetool in matlab to draw the figure about distribution of heat for the pan.2.2.3 How to determine the shape of the pan?To make heat distribution of the pan uniform, we must make it approach round. But under the circumstances of the pan's fixed area, the closer the pan approaches round, the larger its floor space is. In other words, the closer the pan approaches round, the lower the utilization rate of the oven is.More uniform distribution of heat for the pan is, the lower temperature in the corners of the pan is. Assuming the bottom area of the oven is fixed, the number of most pans which the oven can bear is equal to the quotient of the bottom area of the oven divided by floor space per pan. So in a certain weight P, we can get the best type of pan (shape) by optimizing the relationship between temperature in the corners and the number of most pans which the oven can bear. Then we get the oven's width to length ratio of W/L by arranging the pans with the optimal size.2.3 Practical problem parameterizationu: temperature of each point in the oven;t: heating time;x: the abscissa values;y: the ordinate value;α:thermal diffusivity;ε: degree of blackness of heat resource;E: radiating capacity of heat resource;T: absolute temperature of heat resource;T max: Highest temperature of pans’ edge;ξ: the length to width ratio for a rectangular pan;R: radius for a round pan;L: length for the oven;W: width for the oven;q: heat flux;k: coefficient of the convective heat transfer;Q: heat transfer rate;P: minimum distance from pan to the heat resource;Other definitions will be given in the specific models below2.4 Assumption of all models1. We assume the heat resource as a mass point, and it has the same radiation energy in all directions.2. The absorbtivity of pan on the radiation energy is 100%.3. The rate of heat dissipation is proportional to area of heat dissipation.4. The area of heat dissipation changes in a linear fashion from the centre of the pan to its border.5. Each pan is just a two-dimensional surface and we do not care about its thickness.6. Room temperature is 25 degrees Celsius.7.The area of the pan is a certain number.III. ModelsConsider the pan center as origin, establishing a coordinate system for pan as follows:Figure 33.1 Basic ModelIn order to explain main model better, the process of building following branch models needs to be explained specially, the explanation is as follows:3.1.1 Heat radiation modelFigure 4The proportional of energy received by B accounting for energy from A is4π(x2+y2+P2) The absolute temperature of heat resource A (the heating tube made of quartz) T= 773K when it works. The degree of blackness for quartz ε=0.94.The area of quartz heating tube is 0.0088m2.Depending on Stefan-Boltzmann law[3]E=σεT4 (σ=5.67*10-8)we can get E=18827w/m2.The radiation energy per second is 0.0088E.At last, we can get the heat flux of any point of pan. =0.0088E4π(x2+y2+P2)Synthesizing the formulas above, we can get:=13.24π(x2+y2+P2)(1) 3.1.2 Heat convection model3.1.2.1 Round panheat dissipation area of round panFigure 4When the pan is round, the coordinate of any point in the pan is (x,y). When the point is in the centre of a circle, its area of heat dissipation is dxdy. When the point is in theboundaries of round, its area of heat dissipation is 12dxdy. According to equation of heatPdissipation through convection dQ=kdxdy and assumption that the area of heat dissipation changes in a linear fashion along the radius[4], we can get the pan's equation of heat dissipation:q =k[1-0.5(x2+y2)0.5/R] (2) 3.1.2.2 Rectangular panHeat dissipation area of rectangular pan(length: M width: N)Figure 5When the pan is rectangular, the coordinate of any point in the pan is (x,y). When the point is in the centre of a rectangle, its area of heat dissipation is largest, namely dxdy. Whenthe point is in the center of the rectangular edges, its area of heat dissipation is 1dxdy. When2the point is in rectangular vertices, its area of heat dissipation is minimum, namely According to equation of heat dissipation dQ=kdxdy and assumption that the area of heat dissipation changes in a linear fashion from the centre of a rectangle to the center of the rectangular edges.The area of any other point in the pan can be regarded as the result of superposing two corresponding points' area in two lines.Heat dissipating capacity of any point is:=1−y/N(3)1−x/M3.2 Pan heat distribution Model3.2.1 Heat distribution of rectangular pansAssume that the rectangle's length is M and its width is N,the material of the pan is iron.For rectangular pans, we change its length to width ratio, establishing a model to get thetemperature of the corners (namely the highest temperature of the pan).We assume the area of pan is a certain number 0.085m2,The distance from the pan above to the top of the oven P=0.23m,By checking the data, we can know that the coefficient of the convective heat transfer k is approximately 25 if the temperature contrast between pan and oven is 100~200℃.then get a several kinds rectangular pans following:In the two dimensional equation of conduction,we get the thermal diffusivity of iron is 0.000013m2/s through checking data.Heat source function equals received thermal radiation minus loss of heat caused by heat dissipation through ly equation (1) minus equation (3).In this way, we get a more complicated partial differential equation. For example, through analyzing the NO.1 pan, we can get the following partial differential equation.∂u ∂t −α(ð2u∂x2+ð2u∂y2)=13.24π(x2+y2+0.529)−1−y/0.291551−x/0.29155It is hardly to get the analytic solutions of the partial differential equation. We utilize the method of finite element partition to analyze its numerical solution, and show it in the form of figures directly.Pdetool in matlab can solve the numerical solution to differential equation in the regular form quickly and show distribution of heat by the three-dimensional image[5]. We enter partial differential equation of two-dimensional heat conduction into it, then we get the figures about distribution of heat in different pans.When we use pdetool, we take Neumann condition as boundary condition, and we suppose that the boundary is insulated, In fact, it is not insulated, and the heat dissipation will show in the heat source function.We heat the pan for 8 minutes no matter what kind of shape the pan is. By Pdetool, the heat distribution of each pan shows as follows:Figure 6(heat distribution for NO.1 pan) Figure 7(heat distribution for NO.2 pan) Figure 8(heat distribution for NO.3 pan)This pan is just a square pan, with the lowest length to width ratio. From the figure, we can know that corners have the highest temperature which is 297.7℃.The corners have the highest temperature for the pan, which is 296℃The corners have the highest temperature for the pan, which is 294.8℃The temperature ofcorners for the pan isslightly lower than thehighest temperature, andthe highest temperature is293.7 ℃.Figure 9(heat distribution for NO.4 pan)To get a more accurate relationship between the length to width ratio(ξ) and highest temperature(T max), we make several more figures of heat distribution based on the different length to width ratio. At last, we can get its highest temperature. The specific result is as follows:ξ is the argument and T max is the dependent variable. The points in the chart are scaled out in the coordinate system by mathematical software Origin. Connecting the points by smooth curve, we can get the figure following:Figure 10(the relationship between ξ and T max )From the figure we can find that, with the length to width ratio increases, the temperature of corners will sharply fall at first, and it is namely that the heat distributes evenly . When the length to width ratio increases further, the temperature of corners drops obscurely . When the ratio reaches about 2.125, the length to width ratio of rectangular pans has little effect on the heat distribution. In contrast, the ratio is too big, it is difficult for practical application.In addition, we can also find that as long as the shape of the pan is a rectangle. The highest temperature in the corners of the pan won't change a lot whether its length to width ratio changes, From the figure 10, we can find that maximum range is about 6 degrees Celsius. So in general, it is very difficult to change high temperature in the corners of the rectangular pan.3.2.2 Heat distribution of square pans to round pans 3.2.2.1 Size definition of round squareWe need some sizes to define round square, our definition isas follows:T maxThe size of round square isdecided by l and r (l stands forthe length of the straight flange;r stands for the radius of thefillet).Figure 11We assume that the area of the pan is 0.085m2. The r of round square has its range, which is 【0,0.164488】,When the r reaches the two extremums, round square becomes square and circle.3.2.2.2 ModelWhen we make this kind of pans’ heat distribution figures, we take the heat source function as (1) and (3). When the round square becomes circle, the heat source function is (1) and (2).We get a several round squares with different r and l, and take a circle as an example. The specific examples show in the table below:The pan is still made of iron. Here we also heat the pan for 8 minutes.By Pdetool, we can obtain their heat distribution figures as follows:Figure 11(heat distribution for NO.1 pan) Figure 12(heat distribution for NO.2 pan) Figure 13(heat distribution for NO.3pan)The temperature of the corners about the pan is relatively low, and the highest temperature is 264℃. On the contrary, the heat distributes quite evenly.The highest temperature of the pan is 271.1℃The highest temperature of the pan is 274℃The highest temperatureof the pan is 279.8℃Figure 14(heat distribution for NO.4pan)The highest temperatureof the pan is 283.7℃Figure 15(heat distribution for NO.5pan)To get the different relationship between l/r and T max, we take l/r as argument, and T max as dependent variable. We change l and r of round square, and make several heat distribution figures. In this way, we can get the highest temperature T max, and the results show in the table below:Here, we alsouse the mathematical drawing software origin. We use l/r and T max to express coordinates. We use smooth curve to connect points, then we can get the trend line.Figure 16(the relationship between l/r and T max )From the figures we can know that, the value of l/r is smaller, the temperature of the corners about pan is higher. It is namely that the pan is more closely to circle, and heat distribution is more evenly. With the value of l/r increases, the temperature of corners rise very quickly at first, then the amplitude is getting smaller. When the value of l/r is infinitely great, the highest temperature of the pan go to a certain number.Analyzing in a theoretical way , the shape of pan goes to square when the value of l/r is infinitely great. At the same time, the temperature of the round square’s corners approach to that of square’s. From the figures, we can know that this function has a upper boundary ,T maxwhose value is close to the corner temperature of square. Through this, we can verify the correctness of our models.3.2.3 Heat distribution of round rectangle (except round square)From the model about heat distribution of rectangular pan, we can learn that drop of temperature in the corners of rectangular pans will be very little when its length to width ratio is bigger than 2.125. So we select the rectangle with length to width ratio of 2.125. We let the pan vary from the rectangle to round rectangle. So we can study changes of temperature in the corners of the pan.3.2.3.1 Size definition of round rectangleThe specification of the roundsquare is decided by l and r. Weset its width the same as therectangular pan before, namely0.2m.(l stands for the length ofstraight long side, and r stands forthe radius of the fillet.)Figure 17We assume that the area of the pan is 0.085m2, For round rectangle pans, with r decreases constantly, the pan finally approaches the rectangular pan before .If the r increases constantly, its shape will become that of playground. The range of r is 【0,0.1】3.2.3.2 ModelWhen we draw the figure about heat distribution of this kind of pan, heat source function equals equation (1) minus equation (2).We can get some different round rectangles by changing the value of r and l. Their detailed specifications are shown in the following table:The pan is still made of iron. Here we also heat the pan for 8 minutes.By Pdetool, we can obtain their heat distribution figures as follows:The corner temperaturewhich is 291.7℃andslightly lower than thehighest temperature ofthe pan.Figure 18(heat distribution for NO.1pan)The corner temperaturewhich is 292.54℃andslightly lower than thehighest temperature ofthe pan.Figure 19(heat distribution for NO.2pan)The corner temperaturewhich is 293.5℃andslightly lower than thehighest temperature ofthe pan.Figure 20(heat distribution for NO.3pan)To get the different relationship between l/r and T max, we take l/r as argument, and T max as dependent variable. We change l and r of round square, and make several heat distribution figures. In this way, we can get the highest temperature T max, and the results show in the table below:Here, we also use the mathematical drawing software origin. We use l/r and T max to express coordinates. We use smooth curve to connect points, then we can get the trend line.Figure 21(T max)From the figures we can get that, with l/r increases, the highest temperature of pan goes down. That is to say, the bigger radius of the fillet is, the more evenly heat distributes. In addition, with l/r increases, T max rises quickly at first, then the extent is smaller. When l/r is infinitely great, the round rectangle pans become rectangle pans, and the temperature approaches to that of the rectangle pan before.Secondly, from the figure, we can learn that change of the largest temperature is very little and its largest temperature is all very high when the pan varies from rectangle to round rectangle. compared with round square pan, round rectangle pan has much worse capacity of distributing heat.3.3 Best type of pan selection ModelFrom the model of heat distribution, we can know that the extent of heat distribution for round square pan is more than the extent of other pans with other kinds of shape( except T maxcircle). It is difficult to accept it for people because the food is easily overcook, no matter what the number of pan is. Depending on that, people will choose the round square pan.In the following models, we mainly discuss the advantages and disadvantages of round square pans with different specifications.3.3.1 Local parametersFor study's convenience, we take commercial oven of bottom area 1.21m2as an example. The distance between heating tube on the top and the nearest rack is P=0.23m.The number of pans which the oven can contain is n;The fillet radius of round square is r;The floor space of pan is S;The weight of the number of pans in the oven is P;The area of pan is still 0.085m2, the material is still iron.3.3.2 The relationship between T max and rThrough the models before, we know the relationship between l/r and T max. We transform it as the function of r and T max, and shows in the form of table below:We take r as the argument, and T max is the dependent variable. Making the dots in the coordinate system by the mathematical software Origin[6].The figure is as follows:Figurebetween r and T max )To our surprise, we can find that there is nearlylinear relation between r and T max fromthe figure. We may fit the relation with a linear function, so we can get the function of r and T max .FigureThe function we are getting is: T max =-174.5r+291.5 (4)3.3.3 The relationship between n and rWe have introduced that the floor space is not the area of the pan itself in theT maxT maxIntroduction. So we can get the formula below of the area of round square pan and r:S=r2(4−π)+0.085We have already known the floor space of the oven. So we can know the maximum number of pans that the oven can hold , in which condition the shape of pan is sure. Above all, we can get the function of n and r.n= 1.21r2(4−π)+0.085(5) 3.3.3 The optimum solutionThe weight of the number of pans which the oven can hold is P, while the weight of heat distribution is (1-P). The dimensions of the two are different. The effects are also different with the unit change of r. Depending on the message above, in order to induce the weight P. we need to eliminate their dimensions[7].Through observing the figure 23, we can know that the range of T max is 27℃with the domain of r.Then we change the value of r in its domain of definition, then we can get n's approximate range: 3.05mThen we eliminate their dimension, so they are transformed into value which could be compared.They are T max/27 and n/3.5 respectively.We hope that we can get a smaller T max and a larger n. We let T max/27 multiply by -1, then add them (T max/27 and n/3.5) together, the final result is K. K has no practical significance, and we just want to know its relative value.K=-(1-P)T max/27+P n/3.5After simplification, we can obtain:K=-(1-P)(-6.463r+10.796)+0.345Pr2(4−π)+0.085(6)How to get the pan we want with the idealized shape and its corresponding width to length ratio of the oven by using this formula?We explain it by an exampleIf some one’s ideal weight P is 0.6, the function(6) of K becomes:K=-0.4(-6.463r+10.796)+0.207r2(4−π)+0.085We can get the figure of K within the domain of r, by using the matlab. The figure is presented as follow:Figure 24(the relationship between K and r)We plugged the value of r into the equation (5), then we can get n=13.76Because the number of the pan should be an integer, we round up n, namely n=13.The largest number of pans which the oven can contain is prime number, the oven has only a width to length ratio of 1/13.So at last, we get the following conclusion:When P=0.6, n=13, W/L=1/13, r=0.058m is the best solution.To make the coefficient of oven reaches the top (namely without space), we take the radius of round square into equation (5). We should notice that n must be an integer. We adopt the method of exhaustion, and the result shows in the table below:This table will be used in the following advertizing.3.3.4 Model verificationWe can see the equation (6) , when the P tends to 0, which means the largest number of pans the oven can contain make no sense, the equation becomes: K=- (-6.463r+10.796) ; the optimum solution is r=0.164488m. which means maximize even distribution of heat for the pan is most important.On the contrary, when the P tends to 1, which means the maximize even distribution of heat for the pan make no sense, the equation becomes: K =0.345r 2(4−π)+0.085; the optimum solution is r=0m. which means the largest number of pans the oven can contain is most The value of r for thecorresponding peak valuein the figure is 0.058m。

2013年全国大学生数学建模竞赛A题

车道被占用对城市道路通行能力的影响摘要在城市道路常会发生交通异常事件,导致车道被占用,事发地段的通行能力也会因此受到影响。

当交通需求大于事发断剩余通行能力时,车辆排队,产生延误,行程时间增加,交通流量发生变化。

根据这些特点,我们以城市道路基本路段发生交通事故为例,主要分析了交通事故发生后道路的通行能力的变化,以及不同时间段事故点及其上下游路段交通流量的变化,用于以后进一步突发事件下交通流的预测。

针对问题一,根据道路通行能力的定义,考虑到车身大小不同,我们把所有车辆进行标准化。

运用统计估算模型对视频一的车辆进行分段统计,得出未发生事故前道路通行能力2555(辆/h )。

因为车辆所占车道未达到数学理论计算要求,所以我们利用修正过后城市干道通行能力的数学计算模型,计算出交通事故发生至撤离期间的理论通行能力为1356(辆/h ),进而与实际数据对比,得出相对误差。

针对问题二,我们基于问题一的模型,以及附件三数据分析所得,不同车道的通行流量比例不同,对视频二的车辆各项数据的分段统计分析,得到道路实际通行能力。

再根据修正的理论数学计算模型,得出理论通行能力。

得到的结果与问题一的结果相比较,得出结论:在同一横断面上的实际通行能力与交通事故所占车道的车流量呈负相关性。

针对问题三,我们运用了两种模型,一种结合层次分析与线性回归模型,得到理想化的函数关系式。

基于层次分析模型,我们将进行问题分解,把车辆长度作为目标层,其他三个量作为准则层。

通过查阅资料对各因素进行打分,计算出事故持续时间、车道通行能力、上游车流量对车辆排队长度的权重。

层次分析模型得到各个指标对目标层的影响关系的大小,然后我们用线性回归模型求出各指标与目标层的具体的函数关系式为130.0430.09263.623y x x =-+-。

第二,我们运用车流波动相关理论,得到理论模型,继而得出它们之间的关系。

针对问题四,我们首先考虑的是上游来车在红绿灯下的时间间断问题,所以把来车的情况作周期性分析,假设来车是间隔相同的时间连续的到来,求出一个周期能通过的最大车流量数。

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2013 高教社杯全国大学生数学建模竞赛
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车道被占用是指因交通事故、 路边停车、 占道施工等因素, 导致车道或道路横断面通 行能力在单位时间内降低的现象。车道被占用的情况种类繁多、复杂,正确估算车道被 占用对城市道路通行能力的影响程度,将为交通管理部门正确引导车辆行驶、审批占道 施工、 设计道路渠化方案、 设置路边停车位和设置非港湾式公交车站等提供理论依据。 针对问题一,通过对视频1中交通事故发生至撤离期间,各种数据的采集,确定了事 故横断面实际通行能力,运用Gibbs 抽样仿真方法,通过excel软件解决数据缺失问题,并 用eviews进行ARMA模型拟合, 发现实际通行能力的变化过程为yt = −2.0232+ϵt −0.7243ϵt−1 − 0.2757ϵt−2 。 针对问题二,我们结合视频1,采用与问题一同样的方法对视频2进行数据采集和分 析, 得到了横断面的实际通行能力, 并将视频1和视频2仔细对比, 通过视频1和视频2中对 交通事故发生至撤离期间,事故所处横断面实际通行能力, 运用SPSS软件进行两独立样本 的曼―惠特尼U检验,再根据路段附近交通设置,车辆流向比例,司机心理,周围地形等 因素, 分析出产生差异的原因是主要是在不同车道车辆流量的比例。 针对问题三, 通过分析视频1塞车情况, 分别建立非稳态排队论模型和分段差分方程 模型,并运用matlab软件编程绘制图像,解释出视频1中路段车辆排队长度与事故横断面 实际通行能力、 事故持续时间、 路段上游车流量间的关系为: 实际通行能力越大交通事故 所影响的路段车辆排队长度越小,或者说路段车辆排队长度增长越慢;路段上游车流量 越大交通事故所影响的路段车辆排队长度越大,或者说路段车辆排队长度增长越快;事 故持续时间越长, 交通事故所影响的路段车辆排队长度越大。 针对问题四,利用问题三所建立的模型和题目所给数据,如果采用非稳态排队论模 型,运用matlab软件求解,得出从事故发生开始,经过6.656分钟,车辆排队长度将到达上 游路口; 如果通过分段差分方程模型, 事故发生时上游路口刚好是绿灯时需要经过6.42分 钟,事故发生时上游路口刚好是红灯时需要经过7.97分钟,与上一模型所得结果相吻合。 说明两模型有很高的合理性和实用性。 【关键词】通行能力 Gibbs抽样仿真 ARMA模型拟合 曼―惠特尼U检验 非稳态 排队论 分段差分方程
2. 3. A A0114
指导教师或指导教师组负责人(打印并签名) : (论文纸质版与电子版中的以上信息必须一致, 只是电子版中无需签名。 以上内容请 仔细核对,提交后将不再允许做任何修改。如填写错误,论文可能被取消评奖资格。 ) 日期: 2013 年 9 月 15 日
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2013 高教社杯全国大学生数学建模竞赛
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我们仔细阅读了 《全国大学生数学建模竞赛章程》 和 《全国大学生数学建模竞赛参赛 规则》 (以下简称为”竞赛章程和参赛规则”, 可从全国大学生数学建模竞赛网站下载) 。 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章程和参赛规则的,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, 严格遵守竞赛章程和参赛规则, 以保证竞赛的公正、 公平性。 如有违 反竞赛章程和参赛规则的行为, 我们将受到严肃处理。 我们授权全国大学生数学建模竞赛组委会,可将我们的论文以任何形式进行公开展 示(包括进行网上公示, 在书籍、 期刊和其他媒体进行正式或非正式发表等) 。 我们参赛选择的题号是(从A/B/C/D中选择一项填写) : 我们的参赛报名号为(如果赛区设置报名号的话) : 所属学校(请填写完整的全名) : 参赛队员(打印并签名) : 1.
1
1 问题重述
车道被占用是指因交通事故、路边停车、占道施工等因素,导致车道或道路横断面 通行能力在单位时间内降低的现象。一条车道被占用,也可能降低路段所有车道的通行 能力。附件中视频1和视频2的两个交通事故处于同一路段的同一横断面,且完全占用两 条车道。 问题1:根据视频1(附件1) ,描述视频中交通事故发生至撤离期间,事故所处横断面 实际通行能力的变化过程。 问题2:根据问题1所得结论,结合视频2(附件2) ,分析说明同一横断面交通事故所 占车道不同对该横断面实际通行能力影响的差异。 问题3:构建数学模型,分析视频1(附件1)中交通事故所影响的路段车辆排队长度 与事故横断面实际通行能力、 事故持续时间、 路段上游车流量间的关系。 问题4:假如视频1(附件1)中的交通事故所处横断面距离上游路口变为140米, 路段 下游方向需求不变, 路段上游车流量为1500pcu/h, 事故发生时车辆初始排队长度为零, 且 事故持续不撤离。 请估算, 从事故发生开始, 经过多长时间, 车辆排队长度将到达上游路 口。
3 符号说明
• yt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t时刻视频1实际通行能力 • ϵt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t时刻的误差 • I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 单位矩阵 • i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 虚数i2 = −1
2 模型假设
1. 视频中所统计数据真实可靠。 2. 排队所占车道车辆数与对应车道行驶方向车辆数成正比, 即: 车道一车辆数: 车道二
车辆数: 车道三车辆数=0.21:0.44:0.35。
3. 除事故车辆外的其他车辆严格遵守交通规则, 红灯停, 绿灯行。 4. 车辆到达率与正在排队车辆数量无关, 无论有多少车在排队, 车辆到达率不变。 5. 车辆来源是无限的。 6. 堵车期间该路段没有其他交通事故发生。 7. 在堵车状况下相邻两辆车车头之间间距为7米。
2
• t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 堵车持续时间 • λj . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . j状态下的车辆到达率 • µj . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . j状态下的车辆离开率 • Pj (t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t时刻为状态j的概率 • A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 常微分方程组系数矩阵 • xk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 矩阵A的第k个特征值 • e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 自然指数e ≈ 2.718 • ck . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 常微分方程通解系数 • m(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t时刻的平均车辆数 • s(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t时刻路段车辆排队长度 • a1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 小区1进出的车净到达率 • a2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 小区2进出的车净到达率 • λ′ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 红灯时的上游路口到达率 • λ′ 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 绿灯时的上游路口到达率 • Pj′ (t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . t时刻状态j下的概率变化率 • l0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 平均车间距与平均车长之和 • rk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 第k个特征值对应的特征向量 • Pi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 车道i所含排队车辆数占总排队车辆数比例 • P {N (t + ∆t) | N (t)} . . . . . . t时刻状态为N(t)时, t + ∆t时刻状态为N (t + ∆t)的条件概率
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