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

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

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.

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。

2013电工杯数学建模A题论文(国家二等奖)

答卷编号：论文题目：A题：风电功率波动特性的分析姓名专业、班级有效联系电话参赛队员1参赛队员2参赛队员3指导教师：冯玉昌参赛学校：证书邮寄地址及收件人：答卷编号：阅卷专家1 阅卷专家2 阅卷专家3 专家签字风电功率波动特性的分析摘要本论文针对“风电功率波动特性的分析”问题，根据所给的风电机组功率数据建立风电功率波动特性的概率分布模型和灰色预测模型。

由此，进行相关的问题分析及解决。

对于问题一、二，借鉴分离min 级负荷的算法，采用滑动平均法分离s 级风电功率，且处理了丢失数据及错误数据，以此提高数据的准确性。

经过如此处理所给数据后，再采用Matlab 的概率密度拟合工具箱dfittool 得出五台风电机组的功率概率直方图及t location-scale 分布、正态分布、逻辑斯特分布的概率分布图。

发现t location-scale 分布比其他分布更适于拟合各风电场概率密度函数，并作相关分析及检验。

再用t location-scale 分布以每日为时间窗宽，对5个风电功率分别计算30个时段的概率分布参数并做出检验。

问题二与问题一方法一致，只需要从从上述5台机的风电功率数据中提取出间隔为1分钟的数据序列)(k m i t P ，重复问题一的方法即可。

对于问题三，由上可以分别获得采样间隔为5s 和1min 的相关波动特性参数，以此为依照进行波动特性与时间尺度关系的分析。

对于问题四，整合出20台机组的数据，再分别将采样间隔改为1min ，5min ，15min ，使用问题一中同样的方法分析处理即可。

对于问题五，预测未来四小时的风电场总功率，我们采用了灰色模型（Grey Model ，GM ），使用MATLAB 对灰色模型GM （1，1）编程得到预测值，残差，级比偏差等相关数据结果。

由于初步编程得出的预测模型为其累加后的方程，通过生成序列预测值及模型还原值之间的关系及之前所求的预测值模型易求的未来四小时风电场总功率预测模型。

2013高教社杯全国大学生数学建模竞赛A题论文

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。

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。

2013年美赛数模A题OUTSTADING_18439

Team # 18439
Page 1 of 21
Optimize the Shape For Brownie Pan
MCM 2013 Problem A
Contents
1 2 3 4 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Terms and Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . Model for Heat Distribution . . . . . . . . . . . . . . . . . . . . . 4.1 Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Initial Condition . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Between Air and Pans . . . . . . . . . . . . . . . . . . 4.3.2 The Top Face of the Brownie . . . . . . . . . . . . . . 4.3.3 Between Pan and Brownie . . . . . . . . . . . . . . . . 4.3.4 Inside the Brownie . . . . . . . . . . . . . . . . . . . . 4.4 Solve the Model . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 The Performance of the Model . . . . . . . . . . . . . . . . . 4.5.1 Four typical shapes . . . . . . . . . . . . . . . . . . . . 4.5.2 Heat Distribution Figure and a Preliminary Analysis Shape Evaluation Model . . . . . . . . . . . . . . . . . . . . . . 5.1 To Maximize the Number of Pans . . . . . . . . . . . . . . . 5.1.1 Arrange Rectangular Pans . . . . . . . . . . . . . . . 5.1.2 Arrange Round Pans . . . . . . . . . . . . . . . . . . 5.1.3 Arrange Rounded Rectangular Pans . . . . . . . . . 5.1.4 Evaluate the Performance . . . . . . . . . . . . . . . 5.2 To Maximize Even Distribution of Heat . . . . . . . . . . . 5.2.1 Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 The Variance of Four Typical Shapes . . . . . . . . . 5.3 Optimize a Shape Based on Two Conditions . . . . . . . . . 5.3.1 Score Algorithm . . . . . . . . . . . . . . . . . . . . 5.3.2 How to Get a Relatively Optimal Shape in Practice 5.4 The Performance of the Model . . . . . . . . . . . . . . . . 5.4.1 Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 3 3 4 4 4 4 5 5 6 6 6 6 7 8 9 9 10 11 12 14 14 15 15 15 16 17 17

2013国赛A题全国一等奖论文(英文版)

Effect of Lane Occupied on Traffic CapacityAbstract:The article aims at the problem how the lane occupation affects the traffic capacity. By the means of wavelet analysis and time series analysis method, we study on the periodicity and stability of the actual capacity which varies with time going. Meanwhile, we get the expression of the vehicle equivalent queue length based on the Two Fluid Theory.For the question 1: Firstly, considering that the number of vehicles which arrive at the cross-section in the accident obeying periodical distribution, we choose the beginning of green signal time as a starting point and calculate the traffic density every 15 seconds at the cross-section. (Average traffic density within the range from the accident site to the point 120 meters away instead). Secondly, according to the relationship among the three parameters of the traffic flow, we get saturated traffic flow at each time, which represents the actual traffic capacity at the moment that the accident happened. Finally, we analyze the periodicity and stability of the actual traffic capacity by utilizing wavelet analysis and SAS software tools, which result in the conclusion that the actual capacity decays periodically and transits from a stable state to another one.For the question 2: We obtain the actual capacity at the same cross-section when different kinds of accidents happen in video 2. Through the analysis, it is easy to find that both variation laws are basically the same, and the stable value is also basically the same, but the latter decays faster than the former. Therefore, we establish explicit equation about the real traffic capacity and draw the following conclusion: When the road capacity is not saturated, the outer one of lanes blocked by the accident has the greater influence on the actual traffic capacity than the inner one but the influence will disappear when the road cannot bear more cars.For the question 3: we can introduce the concept of the equivalent queue length to deal with the real line length which may be quite complicated. And then we classify all the vehicle running states into two equivalently based on the Two Fluid Theory. This processing method helps us achieve the expression of the equivalent queue length. Here we list the expression '00()()()()t t o o m j m N q t dt W t dt k LM L t M k k +--=-⎰⎰. At last we examine our model by the means of thethat the model we established can preliminarily explain the relationship amongst the four parameters.For the question 4: We put the data into the expression in question 3. And then we get that the vehicle queue length will reach the upstream intersection after 3.34 minutes.Keywords:Wavelet Analysis, Time Series Analysis, Computer Simulation, the Two Fluid Theory, the Equivalent Queue Length.ContentEffect of Lane Occupied on Traffic Capacity (1)Abstract: (1)1. Description of the Problem (3)2. Assumptions (3)3. Definition (3)3.1 Description of symbols (3)3.2 Definition of terms (4)4. Primary Analysis (4)5. Preparations for the model (5)5.1 Determining the cross section of the actual traffic capacity (5)5.2 Basic relationships of three traffic flow parameters (6)6. Foundation and solution of model (6)6.2 Establishing and solving of the model for problem two (10)6.3 Establishment and solution to the model of Question 3 (13)6.4Establishment and Solution of the Model for the Forth Question .. 187. Model Test (19)8 Advantages and Disadvantages of the model (19)9. Reference (20)Appendix (21)1. Description of the ProblemThe phenomenon of lane occupied will lead to lower lane or cross section traffic capacity. A city’s traf fic flow could have large road density and strong continuity, which will reduce the road capacity even if only one lane is occupied. Even though the time is short, it may also cause a queue of vehicles even a traffic jam. And if handled inappropriately, it will lead to emergent regional congestion.The variety of lane occupied is complicated though, it will properly offer the theory basis for the traffic management if we are able to estimate the impact of lane occupied on the road traffic capacity our city.According to the different situation where one or two lanes are blocked shown in the video 1 and video 2, you need to solve the questions below.1: Describe the changing process of the actual capacity at the cross-section during period from traffic accident happens to the end.2: Explain the difference when one or two lane is occupied based on the conclusion drawn from the question 1and the video 2.3: Construct a mathematical model to analyze the relationship among vehicle queue length, lane traffic capacity, the ongoing time of the accident and the upstream traffic flow.4: Try to estimate the duration from the beginning of the accident to the moment the queue line reaches the upstream junction. Assuming that the distance between the traffic accident site and the junction becomes 140 meters, the downstream flow demand is unchanged, road traffic upstream flow is 1500pcu / h, the initial queue length is zero when accident happens and the accident lasts for enough long time.2. AssumptionsThe event that vehicles arrived at the accident site is independent.The standard car equivalent number through the upstream cross-sectional obeys the Poisson Distribution.Ignoring the influence on the actual traffic flow statistics which arises from non-standard vehicle on the road;The vehicles reach into the section where the accident happens at a constant speed.3. Definition3.1 Description of symbols3.2 Definition of terms1. The basic capacity W: the maximum number of standard vehicles which pass through cross-sectional lane during the unit time at the situation where road, traffic and environment are ideal.2. The saturation capacity Q: the maximum number of standard vehicles which pass through cross-sectional lane during the unit time at the situation where road, traffic and environment are ideal and traffic flow is continuous.3. Chain block: referring to the situation where the next cyclical vehicles arrive before the former vehicles entirely pass the accident site.4. Primary AnalysisThis problem is a comprehensive problem, including the changes of the actual capacity at the cross-section, the influence of different lanes occupied, the queue length on the road and the relationship among them. It may involve the Poisson Distribution, the Computer Simulations and other knowledge.Firstly, we need to extract and process the data. After researching the video 1 and video 2, we record the actual number of vehicles through the accident site before, during and after car the accident. And then we convert it into the standard vehicle equivalent number.Question 1 is to describe the changing process of the actual traffic capacity in the accident cross-section during the car accident. The key to solve the question is how todetermine the dynamic process of the vehicle flow. After analysis, we find that the actual road traffic capacity can be valued by referring to the correlation about the traffic flow, vehicle speed and the traffic density. In such way, we can draw a diagram to vividly illustrate the dynamic changing process of the flow.Question 2 is a problem that aims at comparing the difference. Firstly, we need to draw the graphs respectively about the flow density at the two different situations. And then we can tell the similarities and differences between the two cases. Secondly, We need to find the reason causing this phenomenon. Through the analysis, we should divide the blockage into two steps from the moment accident happens to evacuates, which are chain blockage and partial blockage. During the chain blockage, in video 1W o is a constant value the same in video 2. However, it’s complicated in partial blockage. Finally, we build a function model using the vehicle flow proportional coefficient as variable. Obviously we can obtain the different influence on actual traffic capacity as a result of different occupied lanes.For the question 4: It’s vital for us to build an expression about the equivalent queue length and time. According to the analysis of question 3, what we need is just make some appreciate changes for model of question 3. Then we can get the answer of question 4.For the modified model, we analyze the value of each parameter and ultimately solve the equation based on the assumption of the title. The final solution is the unknown quantity t. Finally, we tested the rationality and validity of the model by comparing the actual recorded data with the data of theoretical analysis.5. Preparations for the model5.1 Determining the cross section of the actual traffic capacityConsidering the actual traffic flow, the road is controlled by the traffic light at the upstream intersection. If vehicles are released by the first phase stopped by the second phase, we can see the road’s traffic appears as cyclical trends. Besides both the time of first and second phase are 30 seconds, therefore 30 seconds can be taken as a time interval. Through the analysis of video 1 and 2, we recorded every 30s of actual traffic volume to obtain the relevant data traffic from two minutes before accident happened to two minutes after accident withdrawing. Due to different vehicles having different effects on the ability of W and Q,we introduce standard car equivalent number C. namely the vehicles which are the four-wheeled vehicles and more vehicles or electric vehicles amount to equivalent vehicles. Specific data tables are in Appendix 1 1-1. Conversation coefficients [ 2 ] is shown in Table 1.Table 1 the traffic investigation vehicle type and vehicle conversion coefficient5.2 Basic relationships of three traffic flow parametersAccording to Green Hilts flow density model, the relation between the parameters of the model provisions comply with the following formula, we can see:Q VK =... (1)According to Green Hilts speed density model , we can see:(1)f fk V V k =- (2) According to Green Hilts flow density relationship model, we can see:(1)f fk Q kV k =-…………(3) The expression (3) shows the average flow rate and an average density of a quadratic function.6. Foundation and solution of modelIn view of question 1, we are asked to build a model to describe the changing process of actual capacity from the time that the accident happens to evacuation. Apparently, this question is a dynamic analysis problem. First we need draw a picture to describe the changing process visually. Then we can use Matlab and SAS software to analyze qualitatively the periodicity of the actual capacity change process and the stability of the stability of time series.(1) Method to find the actual capacity:Step 1: The Extraction of raw data In the Video 1, we record a value of the traffic density every 15 minutes from the time that green signal just bright to the evacuation of the accident. Generally speaking, it’s thought that vehicles which are below 120 meters away from the accident site will be influenced and meanwhile the average velocity of vehicles with this range approximately is equal to the speed of vehicles in the accident site. For simplifying the model, it’s reasonable to think the average traffic flow density as the traffic flow density in the accident site. Step 2: Formula for Solving the Actual CapacityWhen the upstream intersection can provide unlimited traffic , the value of Q in the formula ( 3) can be considered as the actual capacity under different speed.In accordance with the relevant literature [1],when the actual traffic flow density is greater than half of the blocking traffic density, it can also be thought that road is congested, otherwise road is in a smooth state. This can be described byk (1),k 2k ,k 42j j f j j f j k k V k W k V ⎧->⎪⎪=⎨⎪<⎪⎩ (4) Step 3: Estimation of values tor j k and f VIn order to facilitate the calculation below, we estimate the best traffic flow density at the same.Calculations of values for j k and f V :According to the traffic flow theory [6], the relationship between traffic density and flow can be described with many models. We use the classical inverted U-shaped model to go on a qualitative description of the position where j kappears.Figure 1 is an inverted U-shaped flow-density diagram:Figure1 Inverted U-shaped flow —Schematic density relationshipFigure 1 shows that j k is the maximum density when the velocity of the traffic obstruction is 0;m k is traffic density when road traffic flow reaches the maximum density. According to the related literature and requirements of road vehicle safetydistance, j k and m k are related to road conditions, the driver’s literacy and other factors, so in practice the two values should be calibrated according to survey data at the road section. Here, we take the data in Video 1 as standard value. Through observation, we can find that before the traffic accident, the road traffic flow density corresponding to the traffic flow is the optical traffic flow density with its value being 48.45,that is to say, m k =48.45 .After the traffic accident, the traffic jam will appear at the cross-sectional area of the accident. We identify several time points of the most serious blockage and calculate the traffic density corresponding to the time point, of which the average value is 287.5, that is to say,j k =287.5.m k and j k here are the sum of the traffic flow density of three lanes,j k and m k are determined by a weighted average of traffic flow density ratio of the downstream direction. Calculating the value of f V : During the time when accident has not happened, weselect multiple free vehicles and calculate the time of driving by using the following formula:1n i f i i S V t ==∑（5）Where:i S is the journey that the first I vehicle travelled. i t is the time that the first I vehicle spent.At this time, the actual capacity is obtained. The specific values have been givenin Schedule 1 to 3of Appendix 1. The waveform is shown in Figure 2.Figure 2 accident cross-sectional actual capacity changes over time graph in Video 1To the next, we made up a Matlab program in order to go on the periodical analysis of wavelet data. Taking the characteristic of the actual data into account, we choose db4 wavelet function to decompose the data signal into 10 layers, then we reconstruct 1 to 10 layers detail signal.Therefore, we get the second layer detail signal d2, as shown in Figure 3 (The program is shown in program2-1 of Appendix 2):Figure 3 Trend of the detail signal of the actual capacityFrom Figure 3, we can find that the actual capacity shows a trend of cyclical fluctuation.Next, we get the autocorrelation function diagram of the actual capacity by using SAS software to analyze the stability of the data, which is shown in Figure 4:Figure 4 The autocorrelation function diagram of the actual capacity According to the knowledge of time series theory, it is clear that this time series has the stability [5] if the autocorrelation coefficient of a time series whose expectation is zero shows the tendency of rapid attenuation to the range within the double standard deviation. Analyzing Figure 4, we can find that the autocorrelation coefficients are within the range of two times of the standard deviation. Therefore the trends of road capacity can be considered as stable.With comprehensive analysis of Figure 2, Figure 3 and Figure 4, the following conclusions can be drawn:Before the accident, namely it is 16:40:15—16:41:45, the actualcapacityW shows periodic fluctuation. By further analysis, we find that the oroad is open before the accident. So the actual road capacity is mainlycontrolled by traffic lights. In the second phase, vehicles from the upstreamare prohibited. It can be seen that these vehicles which pass by the pointwhere an accident happens mainly comes from vehicles of the first phase inthe previous stage, so the value of o W is smaller. However in the secondphase, vehicles from the upstream part start to move forward. Due to themore vehicle sources, capacity gets even bigger here. From what has beendiscussed above, the changing trend of actual traffic is periodic.During the accident, it is f16:42:30—16:55:30 (corresponding to [4,25] of xaxis in Figure 1),we can find that the actual capacity decreases over time andtends to be a stable value in the end except the abnormal part of time missingin the video.With deep analysis of the underlying causes, it can be found that when the traffic accident happens, two lanes are occupied, so all vehicles can only choose the only lane to travel through. In the sphere of influence at accident section, the speed of the car need to decrease, and then it makes the actual capacity to be reduced to a constant value. Along with the subsequent vehicles to arrive, the actual demand also increases. It is consistent with of conclusions of the stability analysis .Therefore , the actual demand is close to even greater than saturated traffic Q. According to the Theorem 1, traffic jams occur in line with the actual situation in the Video 1.At the same time, considering the impact of traffic lights alone, the traffic volume is also periodical and it’s period is about 1 minute in line with the periodic of the wavelet analysis and it is close to the period of the traffic light signal.To sum up: After a traffic accident, actual capacity at the cross-section where an accident happened cyclically shocks and decays over time besides transiting from a stable state to another one.6.2 Establishing and solving of the model for problem two(1) Analysis of problem twoFor problem two, it asks us to analyze and illustrate different influences in the actual capacity, with the same section of traffic accident and different occupied lane.Firstly, we need get original data of the actual capacity of various time points, so we can select data obtained in appendix1. Secondly, we build a model to describethe influences on the actual capacity at the same section of traffic accident and different occupied lane through a comprehensive comparison of video 1 and 2.(2)Establishing and solving of the modelBefore the accident , comparing to the value named o W of video 1(specific datacan be seen in the Appendix 1 ), we find that the actual capacity o W of Video 1 and Video 2 are all within the range from 1100 to 1400. Because the study is the actual capacity where are the same lanes before these lanes are occupied, the actual capacity in Video 1 and Video 2 is equal before an accident apparently.In order to research on changes of the data in video 2, during the accident occurring till evacuating, we described it by drawing it into graphs as the following graph 4 shows:Graph 4The diagram showing the capability of transportation during the accident 2 occurring till evacuating by comparing graph 1 with graph 4, we might notice visibly that the actual capability of transportation display.The oscillation-decaying tendency and after several cycles of the signal, the decay of video 1 and 2 are almost keeping on the line about 10,but the rate and amplitude of them have some major differences. Considering the situation above, we will divide the whole process into two parts.First part: before turning to stability, we call it partial jam stage.Based on the conclusion which is qualitatively analyzed above, we build the mathematics model and quantize the index to further clarify the difference of influence on the actual capacity of transportation by traffic accidents' occupation on different roads.Through the following ideas we build the model:For accident 2, the lane 1 and lane 2 are occupied, so we analyze the partial jam situation of the road first. To simplify the model, we think cars on road 3 directly drive across the intersecting surface to the accident site along the road and vehicles on road 2 drive to the intersecting surface of the accident earlier and then turn to road 3. Of course vehicles on road 1 also drive to the intersecting surface of the accident, stride road 2 and turn to road 3 to drive out.We define a set{}123,,V v v v =,iv is a0-1variable which indicates whether theroad is occupied. ‘1’ means road has been occupied and ‘0’ means contrarily. So the Set V is a collection of situations to road's occupation.We can know the accident 2 by the definition above{}1,1,0V =We can know t car flux of the road by the analysis above'i i Q Q α= (6)The time that the flow of traffics going straight needs at Lane 3 is '3(b d )Q μ+.The time that the traffic flow turns at Lane 2 is2Q t∆.The time that the flow of traffics turning and changing the lane at Lane 1 spendsis 13Q d μ .Consequentially, the total time of traffics passing is'312(b d )Q 3Q dt Q t μμ+=+∆+(7)In view of the simplified model that have been discussed earlier, we make a correction and unification according to the practical problems, then the total time that all of the traffics require to pass the place where accident happens is'312(b d )Q 3i i ii i i i Q dv i t t Q v t μμ=-+==+∆+∑. (8)According to this model, traffic capacity of a lane in unit time is'O Q W t=(9)So we can put Formula (1) and Formula (2) to the Formula (3), then we will get aresult thatis'123(b d )3k t (3t )k O W d μμμ=++∆+∆+ (10), andandare respectively the coefficient of traffic flow at the nearestoccupied lane and the farthest occupied lane.1) In Formula (9), traffic flow of a lane in unit time decreases with the increaseofandare respectively replaced with 21% and 35% in Video 1 and Video 2.A conclusion can be drawn that Lane 3 has more impact on traffic capability than Lane 1, which makes the amplitudeofchange more largely and quickly,according to the situation Figure 2 has more quick attenuation of the amplitude than Figure 1.2)According to 2o W k ∂∂＜1OW k ∂∂＜0,we know that the farthest occupied lane hasgreater effecton ,in other word ,the occupied Lane 1 or Lane 3 makes thedecreasing of amplitude of larger than what the Lane 2 does.From what has been discussed above, the degree of effect that different occupied lane have on practical traffic capacity can be ordered from big to small. Occupied Lane3 an Lane2 have the biggest effect, while occupied Lane1 and Lane2 have the least.So the result above indicates that during the accident occurring till evacuating in the road partial jam stage, if the traffic accident occupies the road with more car flux, the rate of oscillation decaying of the road's intersecting surface is higher and altitude is greater.After decaying to the stable entire jam stage, occupation of different roads of the same intersecting surface doesn't influence more on the actual capability of the intersecting surface.6.3 Establishment and solution to the model of Question 3（1）establishment of the model of question 3For question 3, we are required to establish mathematical model between the vehicle-queue length of the accident-affecting road, the actual transportation capability of intersecting surface of the accident, duration time of the accident and upstream car flux of the road. Firstly, we should define the queue length on concept in order to quantify it. so we bring in the equivalent L--queue length. Then we can establish a function relationship equation among these variables on the Two Fluid Theory.1）Introduction of the Equivalent Queue LengthIt is necessary to determine a new definition on the queue length that may differ from the additional one, which is also what we prefer to take advantage of to describe the situation showed in the video. Thus we introduce the concept of the equivalent queue length in our model. The new concept, in other words, is the ideal length in the Two Fluid theory in the traffic flow which not only reflects the stationary vehicles’influence on the queue length, but also takes the effect that the length has on the moving vehicles into consideration. And it will definitely produce more accurate result by combining the two dimensions together. The actual traffic flow operation state and the two fluid operation state diagrams are shown in the Figure 5 and Figure 6 below.Fig. 5 the actual running state of traffic flow on the road during the accidentFig. 6 accidents to evacuate Two Fluid operation state during the road traffic flowThe traffic flow in the actual running state can be divided into three conditions-the stagnation traffic flow Z, the uniform traffic flow Y and the transition traffic flow G , while in the two fluid theory, it only needs to be divided into two categories-the moving ones 'Z and the stationary ones 'Y . That is to say, in consideration of the gradual change and indeterminacy of the transition flow, in the theory, we can simply our model by putting the transition flow into two parts, each of which can be reckoned as the condition where the running state is the most familiar .Therefore, the theoretical situation itself includes the real three conditions, and it indicates that the theory is more than accuracy. We can draw a conclusion from the above analysis that the expression for the equivalent queue length is 'L Z Z =+.2) Establishment of the modelIn this part, our goal is to establish an useful expression that can best describe the variation of the equivalent queue length L when the real road traffic capacity 0W , accident duration t or the upstream flow changes (t)N based on the two fluid theory. First of all, let us figure out the comparatively easier situation where only the single lane is taken into account. And then we make some adjustments to apply it to the multiple lanes.i ） Single LaneAccording to the flow conservation principle, we get the equation of traffic flow on the road:N (t)N (t)(t)o u d N N +=+∆(11)Where :o N is the number of the vehicles between the accident site and the upstream intersection when the accident just happens.(t)u N is the accumulation of the upstream flow on the road at moment of t.(t)d N is the sum of vehicles which pass by the accident site at the moment of t.(t)N ∆is the number of vehicles between the accident site and the upstreamintersection at the moment of t.Combining the two fluid theory and the figure 6，we get'()()()j m N t k L t k L L t ⎡⎤∆=+-⎣⎦(12)Where :()L t is the equivalent queue length between the accident site and the upstreamintersection at the moment of t.'L is the extent of the road, here it is 240m.m k and j k respectively are the optical density of traffic flow and the traffic jam density between the accident site and the upstream intersection. They are constants.Simultaneous equations (11) and (12) can solve the model of equivalent length on the single lane:'()()()o u d m j mN N t N t k L L t k k +--=-(13) ii ）Adjustments applying for multiple lanesFor the situation of multiple lanes, the equivalent queue length on a lane is totally different when flow rate changes or the cars switch the lane. To describe the extent of vehicle line with a variable, we could introduce a method to measure the whole situation. And the approach is to get the average of the three lanes. After that,we get the model of the equivalent length on multiple lanes:'()()()()tto o m j m N q t dt W t dt k LML t M k k +--=-⎰⎰(14)Where:01()(,)Mtu i q t dt N i t ==∑⎰,1()(,)Mtodi W t dt Ni t ==∑⎰and the number of the lanesM=3.（2）Solution of the model for question 3 1）Deterioration of parameters in the average equivalent queue length on multiple lanesStep 1：o N is the number of the vehicles between the accident site and the upstream intersection when the accident just happens. We assume the happening time is 16:42:30, and we can count the number of vehicles in the video. The result is016N =.Step 2：Determine the value of1(,)Mui Ni t =∑, the expression means theaccumulation of the upstream cars flow at the moment of t;Through the analysis, we know that the upstream flowq is unpredictable, thuswe are capable of adopting the Poisson Distribution to describe it. The next is the。

西工大2013年数学建模竞赛A题二等奖论文

一、问题重述：全球化竞争的加剧促使越来越多的企业开始采用供应链管理策略，以实现企业的一体化管理。

供应链是一个复杂的网状结构系统，每一部分都面临着各种潜在的风险，任何一部分出现问题都可能给整个供应链带来严重的影响，因此如何分析、评价和提高供应链系统的可靠性变得日益迫切。

设施系统是供应链的核心，在供应链研究中有着极其重要的地位。

在一个设施系统中，某些个设施由于自然灾害或者其他因素的影响可能失效，例如911恐怖袭击事件、2004年的印度洋海啸、2008年的汶川地震等都对诸多行业的设施系统造成了严重的破坏。

现有某物流公司要在全国各城市之间建立供应链网络。

需要选定部分城市作为供应点，将货物运输到各城市。

通常每个供应点的货物是充足的，可以充分满足相应城市的需求。

设该公司考虑共考虑49个城市的网络，城市的坐标见表1。

城市之间的道路连接关系见表2。

在每个城市建立配送中心的固定费用和需求量表3，并假定作为供应点的城市其供应量可以满足有需要的城市的需求。

现将要建立一个供应网络，为各城市提供货物供应。

货物运输利用汽车进行公路运输。

设每吨每公里运输费用为0.5元。

现提出如下问题：1.现在要从49个城市中选取部分城市做为供给点供应本城市及其它城市。

建立供给点会花费固定费用，从供应点运输到需求点会产生运输费用，要使总费用最小，问建立多少个供应点最好。

给出选中作为供应点的城市，并给出每个供应点供应的城市。

同时根据坐标作出每一个供应点到需求点的连接图。

2.假定有某组织对该供应网络的道路进行破坏。

并非所有的道路都可以被破坏，可破坏的道路见表4。

当某条道路被破坏后，该条道路就不能再被使用，以前运输经过该道路的只有改道，但总是沿最短路运输。

如果破坏方选取的策略是使对方总费用增加25%，而每破坏一条道路都需要成本和代价，因此需要破坏最少的道路。

问破坏方选取哪几条线路进行破坏。

给出具体的破坏道路和总费用。

3.假定各道路能否被破坏具有随机性，当某条道路被破坏后，该条道路就不能再被使用，以前运输经过该道路的只有改道，但总是沿最短路运输。

2013年研究生数学建模竞赛A题全国三等奖论文

参赛密码（由组委会填写）第十届华为杯全国研究生数学建模竞赛学校参赛队号队员姓名1.2.3.参赛密码（由组委会填写）第十届华为杯全国研究生数学建模竞赛题目变循环发动机部件法建模及优化摘要变循环发动机可以实现超音速时的高推力与亚音速时的低油耗，其内在性能的优势引起了各航空强国的高度重视，因此是目前航空发动机的重要研究方向。

研究的重中之重是燃气涡轮发动机的特性，本文在了解变循环发动机的构造及工作原理的基础上，运用线性插值、遗传算法、多目标规划等方法给出各部件的出口总温、总压、流量和功率的算法、程序，进一步给出了平衡方程组成的非线性方程组的算法及其有效性分析，得到了发动机性能最优的相关约束条件，进而解决了发动机性能最优时的参数取值。

针对问题一，运用Matlab软件进行线性插值得到相应换算转速下增压比、流量、效率。

1）对附录4中风扇特性数据表中增压比进行标准化处理得到对应压比值，根据附录1中2.2.2得到相应换算转速下的流量，通过Matlab软件中Spline插值方法画出了相应换算转速下流量随压比函数值变化的图形。

2）根据附录1中2.1.2得到了进气道出口总温、总压，再由压气机（风扇、CDFS）进口总温、总压和物理转速得到的换算转速，利用线性插值得到对应增压比、流量、效率，再由附录1中2.2.2求出了压气机（风扇、CDFS）的出口总温、总压和流量。

针对问题二，给出了求解由发动机7个平衡方程组成的非线性方程组的遗传算法，进一步运用Matlab软件编写了子程序并进行有效性分析。

对于发动机7个平衡方程涉及的变量进行分类处理，结合附录1中变循环发动机各部件的计算公式给出对应类的算法及程序。

再将变量代入相应平衡方程，得到关于发动机参数的非线性方程组。

针对问题三，依据附录1给出的发动机性能参数：推力、单位推力和耗油率，建立了多目标规划模型。

根据所建模型，找出了发动机CDFS导叶角度、低压涡轮导叶角度和喷管喉道面积3个量与推力、单位推力和耗油率之间的关系，进而通过Lingo求解出3个量为多少时，发动机的性能最优。

2013年美国数学建模A题论文中文版

4.2.2 热量的均匀分布
4.2.3(二) 如何得到最佳烤盘从上边两种情况可以分别得到不同形状的烤盘排列在烤炉中时的空间利用率，以及它们各自在达到平衡时的热量分布，同时我们由温度的方差得出温度的分布均匀性。考虑到实际情况中，我们常常既想追求温度的最均匀分布，因为此时得到的蛋糕品味最佳，又想使空间得到最有效利用，不致于资源浪费，显然这两者不能同时满足。这时我们应该考虑针对不同的需要，使用不同形状的烤盘，下面我们来解决这个问题。就像在招聘员工考虑不同因素来为求职者打分一样，我们从温度的均匀分布和空间利用率两方面考察一个烤盘的性能时，可以给以不同的权重 p 与 1-p，我们将在不同的 p 值下考察不同形状烤盘的性能。
占用率 Q=
NA 84 % . WL
WL 2(2 3 ) X 2
]=2[
L2
2(2 3 ) A
],
由此，我们比较四种不同图形在烤箱平面中数目和占用率，可以容易的得知：矩形（以正方形为例）烤盘在烤箱平面中占用率最高，为 100%，而圆形盘占用率最小仅为 84%。矩形盘和圆盘分为两个极端，中间分布着其他图形的数目和占用率。当选用烤箱平面中烤盘的最大数量时，显然矩形(正方形)最大。
为了减少由于方差 S 过大而引起的影响，我们引入了参数 U，以正方形的温度分布为基准，表示温度分布的相对不均匀度 Ui =
Si S1
由此得到不同情况下温度分布的相对不均匀度从而， U1 = 1， U2 = 0.526， U3 = 0.427， U4 = 0.397 U1 , U2 , U3 , U4 分别对应于正方形，正六边形，正八边形，圆形。我们再引入一个参数 R 来反映烤盘的相对综合性能 R=p∗Q− 1−p ∗U 这里的 R 相当于烤盘的性能得分，R 的值越大，则说明其性能越好。注意，这里对 U 的处理比较特殊，因为其值越小反映的烤盘的温度分布越均匀，故这里要用减号。（1） p 一定时，不同烤盘的性能比较

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