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2021年国赛数学建模A题优秀论文

2021年国赛数学建模A题优秀论文本文基于FAST的工作原理，通过机理分析、坐标変换、非线性最小二乘优化等方法，建立了反射面板谟节优化模型・并利用BFGS 算法、蒙特卡洛积分算法等算法，对不同条件下反射光线吸收比率进行了研究。

问题一中，首先基于固定的仰角观测目标S、圆心C和焦点P・利用旋转抛物面的中心对祢性，选取焦距作为自由度控制变量，构建在极坐标系下开口竖直向上的二维抛物线方程.得到不同偏转角度下原点到抛物线的距离.进而导出三维下的旋转掀物面方程。

其次，以焦距为决策变量，将口径300米的拋物面作为积分域•将理想抛物面到原点的距离与基准球面半径差值平方作为被积函数进行积分作为最小化目标函数.建立了确定理想抛物面的优化模型。

最后，使用二分法求得目标函数导函数在定义区间上的零点.得到理想抛物面焦距的精确值为280.854,误差平方积分的最小偵为10.112o此时对应理想抛物面的解析式为z=Q+#)2/561.708300.841,问题二中，首先利用球坐标下不同轴线方向抛物面的旋转不变性.在原坐标系和问题一的坐标系之间建立了双向可逆的变换关系.得到了不同方位角下理想抛物面到原点的距离。

其次.以主索节点的工作坐标和促动器的伸缩长度为决策变量:.以积分域覆盖的主索节点到原点的距离与理想抛物面到原点的距厲之差的平方和为最小化目标函数.分别考虑下拉索长度固定、相邻节点的距离变化幅度不超过0.07%.促动器的伸缩范围在±0.6m为约束条件.建立反射面板调节优化模型。

最后，使用拉格朗日乗子法和BFGS算法进行求解.得到误差平方在抛物面口径上的积分的最小值为5.1353X109.理想抛物线的顶点坐标为(-49.392,-36.943,-294.450).调节后反射面300米口径内的主索节点编号、位置坐标、各促动器的伸缩量等结果见文件result.xlsxe问题三中，首先通过旋转变换.将反肘问题的倾斜入射光线转化为垂直入射光线。

2012年数学建模A题优秀论文

基于数理分析的葡萄评价体系摘要葡萄酒质量的好坏主要依赖于评酒员的感观评价，由于人为主观因素的影响，对于酒质量的评价总会存在随机差异，为此找到一种简单有效的客观方法来评酒，就显得尤为重要了。

本文通过研究酿酒葡萄的好坏与所酿葡萄酒的质量的关系，以及葡萄酒和酿酒葡萄检测的理化指标的关系，以及葡萄酒理化指标与葡萄酒质量的关系，旨在通过客观数据建立数学模型，用客观有效的方法来评价葡萄酒质量。

对于问题一，我们首先用配对样品t 检验方法研究两组评酒员评价差异的显著性，将红葡萄酒与白葡萄酒进行分类处理，用SPSS 软件对两组评酒员的评分的各个指标以及总评分进行了配对样本t 检验。

得到的部分结果显示：红葡萄酒外观色调、香气质量的评价存在显著性差异，其他单指标的评价不存在显著差异，白葡萄、红葡萄以及整体的评价存在显著性差异。

接着我们建立了数据可信度评价模型比较两组数据的可信性，将数据的可信度评价转化成对两组评酒员评分的稳定性评价。

首先我们对单个评酒员评分与该组所有评酒员评分的均值的偏差进行了分析，偏差不稳定的点就成为噪声点，表明此次评分不稳定。

然后我们用两组评酒员评分的偏差的方差衡量评酒员的稳定性。

得到第 2 组的方差明显小于第1 组的，从而得出了第2 组评价数据的可信度更高的结论。

对于问题二，我们根据酿酒葡萄的理化指标和葡萄酒质量对葡萄进行了分级。

一方面，我们对酿酒葡萄的一级理化指标的数据进行标准化，基于主成分分析法对其进行了因子分析，并且得到了27 种葡萄理化指标的综合得分及其排序。

另一方面，我们又对附录给出的各单指标百分制评分的权重进行评价，并用信息熵法重新确定了权重，用新的权重计算出27 种葡萄酒质量的综合得分并排序。

最后我们对两个排名次序用基于模糊数学评价方法将葡萄的等级划分为1-5 级。

对于问题三，首先我们将众多的葡萄理化指标用主成分分析法综合成 6 个主因子，并将葡萄等级也列为主因子之一。

对葡萄的 6 个主因子，以及葡萄酒的10 个指标用SPSS 软件进行偏相关分析，得到酒黄酮与葡萄的等级正相关性较强等结论。

第十一届五一数学建模联赛A优秀论文

第十一届五一数学建模联赛A优秀论文摘要：本文对第十一届五一数学建模联赛A题进行了深入的研究，通过对问题的分析与解决方案的设计，提出了一种可行的数学模型，并进行了相应的实验验证和结果分析。

该模型通过考虑影响因素，有效地解决了题目中的实际问题，并得出了一系列实用的结论。

本文的研究结果对于实际工作和实际问题的解决具有重要的理论和应用价值。

1. 引言第十一届五一数学建模联赛A题要求我们通过建立数学模型，研究某市公交车调度问题。

该问题涉及到公交车乘客数量、站点的服务能力、乘客等待时间等多个因素。

解决该问题对于公众出行的便利性和公交系统的高效性具有重要意义。

2. 问题分析在分析题目要求和实际问题背景的基础上，我们发现该问题主要存在以下几个方面的挑战：2.1 公交车乘客数量的变化公交车乘客数量的变化是该问题中的一个重要因素。

乘客数量的变化影响到了公交站点的服务能力和公交车的调度策略。

我们需要考虑如何根据乘客数量的变化来制定相应的调度方案，以满足乘客的需求。

2.2 站点的服务能力每个公交车站点的服务能力直接影响乘客的等待时间和公交车的运行效率。

我们需要考虑如何确定每个站点的服务能力，以便在给定乘客数量的情况下，有效地分配公交车资源，减少乘客的等待时间。

2.3 公交车调度策略公交车调度策略是解决该问题的核心。

我们需要考虑如何根据乘客数量的变化和站点的服务能力，合理地安排公交车的发车时间和车辆数量，以最大程度地满足乘客的需求。

同时，我们还需要考虑如何高效地调度公交车，以减少乘客的等待时间。

3. 模型建立基于对问题的分析，我们建立了以下数学模型来解决该问题：3.1 乘客数量模型我们通过统计历史乘客数量数据和相关因素的变化趋势，建立了乘客数量的数学模型。

该模型可以预测未来乘客数量的变化规律，并为公交车调度提供参考依据。

3.2 站点服务能力模型我们通过考虑站点的运行时间、人员分配和设备设施等因素，建立了站点的服务能力模型。

全国大学生数学建模优秀论文(A题) 国家一等奖

地下储油罐的变位分析与罐容表标定摘要加油站地下储油罐在使用一段时间后，由于地基变形等原因会发生纵向倾斜及横向偏转，导致与之配套的“油位计量管理系统”受到影响，必须重新标定罐容表。

本文即针对储油罐的变位时罐容表标定的问题建立了相应的数学模型。

首先从简单的小椭圆型储油罐入手，研究变位对罐容表的影响。

在无变位、纵向变位的情况下分别建立空间直角坐标系，在忽略罐壁厚度等细微影响下，运用积分的方法求出储油量和测量油位高度的关系。

将计算结果与实际测量数据在同一个坐标系中作图，经计算得误差均保持在3.5%以内。

纵向变位中，要分三种情况来进行求解，然后将三段的结果综合在一起与变位前作比较，可以得到变位对罐容表的影响。

通过计算，具体列表给出了罐体变位后油位高度间隔为1cm 的罐容表标定值。

进一步考虑实际储油罐，两端为球冠体顶。

把储油罐分成中间的圆柱体和两边的球冠体分别求解。

中间的圆柱体求解类似于第一问，要分为三种情况。

在计算球冠内储油量时为简化计算，将其内油面看做垂直于圆柱底面。

根据几何关系，可以得到如下几个变量之间的关系：测量的油位高度0h 实际的油位高度h 计算体积所需的高度H于是得到罐内储油量与油位高度及变位参数（纵向倾斜角度α和横向偏转角度β ）之间的一般关系。

再利用附表2中的数据列方程组寻找α与β最准确的取值。

αβ一、问题重述通常加油站都有若干个储存燃油的地下储油罐，并且一般都有与之配套的“油位计量管理系统”，采用流量计和油位计来测量进/出油量与罐内油位高度等数据，通过预先标定的罐容表（即罐内油位高度与储油量的对应关系）进行实时计算，以得到罐内油位高度和储油量的变化情况。

许多储油罐在使用一段时间后，由于地基变形等原因，使罐体的位置会发生纵向倾斜和横向偏转等变化（以下称为变位），从而导致罐容表发生改变。

按照有关规定，需要定期对罐容表进行重新标定。

题目给出了一种典型的储油罐尺寸及形状示意图，其主体为圆柱体，两端为球冠体。

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年美国大学生数学建模大赛A题一等奖

最终的布朗尼蛋糕盘Team #23686 February 5, 2013摘要Summary/Abstract为了解决布朗尼蛋糕最佳烤盘形状的选择问题，本文首先建立了烤盘热量分布模型，解决了烤盘形态转变过程中所有烤盘形状热量分布的问题。

又建立了数量最优模型，解决了烤箱所能容纳最大烤盘数的问题。

然后建立了热量分布最优模型，解决了烤盘平均热量分布最大问题。

最后，我们建立了数量与热量最优模型，解决了选择最佳烤盘形状的问题。

模型一：为了解决烤盘形态转变过程中所有烤盘形状热量分布的问题，我们假设烤盘的任意一条边为半无限大平板，结合第三边界条件下非稳态导热公式，建立了不同形状烤盘的热量分布模型，模拟出不同形状烤盘热量分布图。

最后得到结论：在烤盘由多边形趋于圆的过程中，烤焦的程度会越来越小。

模型二：为了解决烤箱所能容纳最大烤盘数的问题，本文建立了随烤箱长宽比变化下的数量最优模型。

求解得到烤盘数目N 随着烤箱长宽比和烤盘边数n 变化的函数如下：AL W L W cont cont cont N 4n2nsin 1222⎪⎭⎫ ⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+⋅--=π模型三：本文定义平均热量分布H 为未超过某一温度时的非烤焦区域占烤盘边缘总区域的百分比。

为了解决烤盘平均热量分布最大问题，本文建立了热量分布最优模型，求解得到平均热量分布随着烤箱长宽比和形状变化的函数如下：n sin n cos -n 2nsin 22ntan1H ππδπδπ⎪⎪⎪⎪⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛⋅-=A结论是：当烤箱长宽比为定值时，正方形烤盘在烤箱中被容纳的最多，圆形烤盘的平均热量分布最大。

当烤盘边数为定值时，在长宽比为1:1的烤箱中被容纳的烤盘数量最多，平均热量分布H 最大。

模型四：通过对函数⎪⎭⎫ ⎝⎛n ,L W N 和函数⎪⎭⎫⎝⎛n ,L W H 作无量纲化处理，结合各自的权重p 和()p -1，本文建立了数量和热量混合最优模型，得到烤盘边数n 随p值和LW的函数。

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。

2016年美国大学生数学建模大赛A题获奖论文A Hot Bath

Part II Finding the best strategy ...................................................................12 VI. Results ........................................................................................................ 12
The first part has five sections: air’s heat radiation, bathtub wall’s heat radiation, person in, hot water in, bubble existed. We discuss some factors that affect water temperature, such as the shape and the volume of the bathtub and person, especially the motions made by the person in the bathtub because the temperature in the bathtub has a great connection with person. Finally, we get the water temperature variation and distribution model.
In this article, we establish two models. One is water temperature variation and distr one is finding best strategy model. We put forward some acceptable hypothesis to simplify the model. What’s more, we clear the meaning of the word “noticeably”.

2015年全国数学建模竞赛A题全国一等奖论文14

阳光线的影响； 6、假设春分日为每年的 3 月 21 日，夏至日为每年的 6 月 22 日，秋分日为每年
的 9 月 23 日，冬至日为每年的 12 月 22 日。
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到各个点的空间坐标：A R cos,0, Rsin ，BR cos cos, R cos sin, Rsin ， C R cos, Rsin,0 ， D R,0,0 。
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通过对包含点 A,B 的最大圆进行几何学分析，我们得到长度为 AE 的物体在太阳光的照射下，投影长度为 AF，则：
子与 Y 轴夹角 arctan(xi / yi )，进一步求出 20 组相邻时刻的影子之间的夹角 i arctan(xi / yi ) arctan(xi1 / yi1) 作为实际值。接着再引入影子与正北方向的夹角作为参数。我们运用几何学知识可以求出与各参数, , 之间的函数关系。并且与上一模型类似，我们对直杆所在地点的经度，纬度，测量时间 t 进行穷举法遍历，通过建立的模型对于每一组 ( , ) 求解出 20 组 i i i1 作
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一、问题重述
确定视频的拍摄地点和拍摄日期是视频数据分析的重要方面，太阳影子定位技术就是通过分析视频中物体的太阳影子变化，确定视频拍摄的地点和日期的一种方法。
1、建立影子长度变化的数学模型，分析影子长度关于各个参数的变化规律，并用建立的模型画出 2015 年 10 月 22 日北京时间 9:00-15:00 之间天安门广场（北纬 39 度 54 分 26 秒,东经 116 度 23 分 29 秒）3 米高的直杆的太阳影子长度的变化曲线。

大学生数学建模竞赛A题优秀论文A题葡萄酒定稿版

大学生数学建模竞赛A 题优秀论文A题葡萄酒 HUA system office room 【HUA16H-TTMS2A-HUAS8Q8-HUAH1688】葡萄酒质量的评价摘要葡萄酒质量的好坏主要依赖于评酒员的感观评价，由于人为主观因素的影响，对于酒质量的评价总会存在随机差异，为此找到一种简单有效的客观方法来评酒，就显得尤为重要了。

首先，采用双因子可重复方差分析方法，对红、白葡萄酒评分结果分别进行检验，利用Matlab软件得到样品酒各个分析结果，结合01-数据分析，发现对于红葡酒有70.3%的评价结果存在显着性差异，对于白葡萄酒只有53%的评价结果存在显着性差异。

通过比较可知，两组评酒员对红葡萄酒的评分结果更具有显着性差异，而对于白葡萄酒的评分，评价差异性较为不明显。

为了评价两组结果的可信度，借助Alpha模型用克伦巴赫α系数衡量，并结合F检验，得出红葡萄酒第一组评酒员的评价结果可信度更高，而对白葡萄酒的品尝评分，第二组评酒员的评价结果可信度更高。

综合来看，主观因素对葡萄酒质量的评价具有不确定性。

结合已分析出的两组品酒师可靠性结果，对葡萄酒的理化指标进行加权平均，最终得出十位品酒师对样品酒的综合评价得分。

将每一样品酒的综合得分与其所对应酿酒葡萄的理化指标（一级指标）共同构成一个数据矩阵，采用聚类分析法，利用SPSS软件对葡萄酒样进行分类，根据分类的结果以及各葡萄样品酒综合得分最终将酿酒葡萄分为A（优质）、B（良好）、C（中等）、D（差）四个等级，客观地反映了酿酒葡萄的理化指标与葡萄酒质量之间的联系。

为了分析酿酒葡萄与葡萄酒理化指标之间的联系，采用相关分析法，能有效地反映出两者间的联系，取与葡萄各成分相关性显着的葡萄酒理化指标，与葡萄成分做多元线性回归得出葡萄酒理化指标与酿酒葡萄的拟合方程，从而反映酿酒葡萄与葡萄酒理化指标之间的联系。

2017年全国大学生数学建模竞赛A题一等奖

针对问题四，由新项目任务的位置分布图可得，新项目任务主要分布在已结束项目中未完成任务所在区域。为使 APP 开发商可以用最少的成本（任务总定价）得到最多的商检和信息数据（最高的任务完成率），建立双优化定价模型（DOP 模型）。用 Matlab 软件编程求解得任务完成率为 80.1%，APP 开发商应给出的最低总定价为 75827 元，同时得到 2066 个任务的具体定价。最后对该模型进行 10 次模拟仿真，每次模拟仿真所得值与实际值的误差都小于 5%，说明该定价计划的实施效果很好。
针对问题二，需制定新的任务定价计划，属于优化问题，解决该问题需找到每个任务的最优定价。由附件一可知任务标价与位置和执行情况都有关，故可按距离关系和任务完成率分别制定一组定价计划。按距离制定的计划，关键在于距离会员近的任务定价低些，而距离会员远的任务定价则按一定比例高些。由问题一得到定价按位置大致分为几类，将任务按这几类划分区域，分别算出各区域内每个任务的最优定价。按任务完成率制定的计划，关键在于新计划定价与任务执行情况之间的关系，可用 0-1 整数线性规划建立相应模型求解。最后，分别计算
2
在原计划和两组新的定价计划下，该平台一组项目需支付的总定价，比较其值大小。三者中总定价少且任务完成率高的计划为最优定价计划。
针对问题三，只需考虑任务位置与定价之间的关系，故在问题二按距离关系所建立的定价计划的基础上做出改进即可。先将原来的 835 个任务按距离进行聚类分析，利用可打包任务间的距离范围确定聚类个数。对于仍是未打包的任务（单个任务）而言，定价不变；对于打包在一起的多个任务，可整体看成一个任务，聚类的中心即这个新任务的位置，即从问题二中的点与点之间距离变成点与集合之间的距离。题目中提到会员之间对任务有竞争关系，则此时的距离不再是任务与最近会员间的距离，此距离还与时间有关，可以基于序贯算法（优先级的先后次序）来改进定价模型。首先，对打包后的任务集合以会员能接受的最远距离为半径画圆，得到可能会竞争这个任务集合的会员集合。其次，每个会员都有其任务预定限额，若该任务集中任务个数超过某个会员的限额数，则该会员失去竞争力，从而缩小会员集合。每个会员的任务开始预定时间也不相同，挑选预定时间最早的会员得到进一步缩小的会员集。最后，在上述会员集中按问题二的定价模型，找到与该任务集距离最短的会员，则这个任务集就被这个会员所领取了。按上述算法思想来修改问题三种按距离关系建立的定价计划。另一方面，打包后会员可选择的总任务数减少，之前由于距离太远未被选择的任务可能会因此被没有抢到任务的会员选择，导致任务完成率增大；而打包被领取的任务的完成情况不受打包的影响。因此，整体任务完成率增大。

2019年全国大学生数学建模竞赛A题题目及论文精选

2019全国大学生数学建模竞赛A题目及优秀论文精选A题高压油管的压力控制燃油进入和喷出高压油管是许多燃油发动机工作的基础，图1给出了某高压燃油系统的工作原理，燃油经过高压油泵从A处进入高压油管，再由喷口B喷出。

燃油进入和喷出的间歇性工作过程会导致高压油管内压力的变化，使得所喷出的燃油量出现偏差，从而影响发动机的工作效率。

图1高压油管示意图问题1.某型号高压油管的内腔长度为500mm，内直径为10mm，供油入口A处小孔的直径为1.4mm，通过单向阀开关控制供油时间的长短，单向阀每打开一次后就要关闭10ms。

喷油器每秒工作10次，每次工作时喷油时间为2.4ms，喷油器工作时从喷油嘴B处向外喷油的速率如图2所示。

高压油泵在入口A处提供的压力恒为160MPa，高压油管内的初始压力为100MPa。

如果要将高压油管内的压力尽可能稳定在100MPa左右，如何设置单向阀每次开启的时长？如果要将高压油管内的压力从100MPa增加到150MPa，且分别经过约2s、5s和10s的调整过程后稳定在150MPa，单向阀开启的时长应如何调整？图2喷油速率示意图问题2.在实际工作过程中，高压油管A处的燃油来自高压油泵的柱塞腔出口，喷油由喷油嘴的针阀控制。

高压油泵柱塞的压油过程如图3所示，凸轮驱动柱塞上下运动，凸轮边缘曲线与角度的关系见附件1。

柱塞向上运动时压缩柱塞腔内的燃油，当柱塞腔内的压力大于高压油管内的压力时，柱塞腔与高压油管连接的单向阀开启，燃油进入高压油管内。

柱塞腔内直径为5mm，柱塞运动到上止点位置时，柱塞腔残余容积为20mm3。

柱塞运动到下止点时，低压燃油会充满柱塞腔（包括残余容积），低压燃油的压力为0.5MPa。

喷油器喷嘴结构如图4所示，针阀直径为2.5mm、密封座是半角为9°的圆锥，最下端喷孔的直径为1.4mm。

针阀升程为0时，针阀关闭；针阀升程大于0时，针阀开启，燃油向喷孔流动，通过喷孔喷出。

在一个喷油周期内针阀升程与时间的关系由附件2给出。

2018年全国大学生数学建模大赛A题高温作业专用服装设计及三篇优秀论文精选

2018年高教社杯全国大学生数学建模竞赛A题目及论文精选（请先阅读“全国大学生数学建模竞赛论文格式规范”）A题高温作业专用服装设计在高温环境下工作时，人们需要穿着专用服装以避免灼伤。

专用服装通常由三层织物材料构成，记为I、II、III层，其中I层与外界环境接触，III层与皮肤之间还存在空隙，将此空隙记为IV层。

为设计专用服装，将体内温度控制在37ºC的假人放置在实验室的高温环境中，测量假人皮肤外侧的温度。

为了降低研发成本、缩短研发周期，请你们利用数学模型来确定假人皮肤外侧的温度变化情况，并解决以下问题：(1)专用服装材料的某些参数值由附件1给出，对环境温度为75ºC、II层厚度为6mm、IV层厚度为5mm、工作时间为90分钟的情形开展实验，测量得到假人皮肤外侧的温度（见附件2）。

建立数学模型，计算温度分布，并生成温度分布的Excel文件（文件名为problem1.xlsx）。

(2)当环境温度为65ºC、IV层的厚度为5.5mm时，确定II层的最优厚度，确保工作60分钟时，假人皮肤外侧温度不超过47ºC，且超过44ºC的时间不超过5分钟。

(3)当环境温度为80C 时，确定II层和IV层的最优厚度，确保工作30分钟时，假人皮肤外侧温度不超过47ºC，且超过44ºC的时间不超过5分钟。

附件1.专用服装材料的参数值附件2.假人皮肤外侧的测量温度优秀论文第一篇高温作业服设计第二篇高温作业专用服装设计第三篇基于非稳态导热的高温作业专用服装设计高温作业服设计【摘要】高温作业服可以避免高温灼伤，在实际作业中有广泛应用。

本文对高温作业服的优化设计进行研究，分析作业服的传热过程，综合考虑各种传热方式、边界和初始条件，建立非稳态一维传热模型，并应用于作业服厚度的优化设计。

对于问题一：通过分析传热模型特点，将三维问题简化为一维问题，研究非稳态传热过程。

2017年数学建模国赛A题全国优秀论文32.pdf

基于规划模型的太阳影子定位策略摘要本文研究的是由太阳影子变化确定地点和日期的问题，根据太阳高度角及相关参数的算法建立太阳影子定位和定时的模型，在实际生活中有较强的实用性。

对于问题一，首先根据太阳高度角、时角、均时差和太阳赤纬的算法建立影子长度与经度、纬度、时间、日期、杆高的数学关系，表明影子长度与这5个参数有关，为进一步分析影子长度与各个参数的关系，通过控制变量的方法，建立5个模型分别观察影子长度随5个影响参数的变化情况。

然后根据影子长度随时间变化的模型，绘制出3米高直杆影子长度的变化曲线，发现在北京时间11点58分时影长最短。

基于问题一可知影长与5个参数有关，问题二中纬度、经度和杆高为未知量，问题三中纬度、经度、杆高和日期为未知量，均可建立杆高关于其他未知量的数学关系式，采用遍历算法求解：对于问题二，对经纬度进行遍历运算，对于问题三，对经纬度和日期进行遍历运算。

并依据测量时刻和影子变化方向对经纬度范围加以约束，对得到的每一组经纬度值，回代入关系式求出各北京时刻的杆高h，基于实际杆高一定，根据杆高方差最小原则筛选经纬度和日期，最后求得附件1的直杆可能位于海南、云南和越南，对附件2的直杆存在8个可能的日期和一个可能的地点新疆，对附件3的直杆存在8个可能的日期和5个可能的地点，分别为湖北、陕西、甘肃、重庆、河南。

对于问题四，首先对所给视频进行压缩，然后读取视频并按照一定的时间间隔提取画面，利用Matlab软件自带的像素坐标系，测得直杆底端和影子顶端的坐标，从而求得影长。

采用问题二的遍历算法模型，拍摄日期已知时，建立影长与经纬度的数学关系式，对经纬度进行遍历运算；拍摄日期未知时，建立影长关于经纬度和日期的关系式，对经纬度和日期进行遍历运算。

并依据测量时刻和影子变化方向对经纬度范围加以约束，对得到的每一组经纬度值，回代入关系式求出各北京时刻的影长，根据回代影长与测量影长的平均相对误差最小原则筛选经纬度和日期。

国赛数学建模A题优秀论文之欧阳引擎创编

葡萄酒的评价模型欧阳引擎（2021.01.01）海军航空工程学院(烟台) 史成巍许志鹏王鑫指导教师司守奎专家点评：本文格式基本规范，表达较清晰。

解决问题一方法适当，结论正确；问题二以相关系数筛选出与葡萄酒质量相关性较大的理化指标与葡萄酒质量一起作为评估葡萄质量的评价指标，进行聚类分析，思路简明，结论较合理。

问题三进行理化指标的相关性分析，切入准确，但对结果的说明不够充分。

不足之处是在问题二到问题四中没有充分考虑芳香类物质的使用，问题四中对如何判定“葡萄和葡萄酒的理化指标是否能用来评价葡萄酒”时方法略有不妥，导致结论不当。

点评人：济南大学数学科学学院许振宇副教授摘要：本文主要针对葡萄酒的评价问题建立了相关数学模型。

在对两组评酒员的评价是否存在显著性差异的问题中，首先验证了两组评酒员的评价结果服从正态分布，并通过方差分析法对两组评酒员的评价结果进行了分析，发现两组评酒员对于红葡萄酒和白葡萄酒的评价结果均存在显著性差异，由于第二组评酒员的评分方差更小，故评价结果均衡度更好,其结果可信度更大。

在对酿酒葡萄进行分级的问题中，首先以相关系数衡量葡萄理化指标与葡萄酒质量的相似性程度，然后筛选出与葡萄酒质量相关性较大的理化指标与葡萄酒质量一起作为评估葡萄质量的评价指标，利用筛选出的评价指标对酿酒葡萄进行聚类分析，将红葡萄和白葡萄均分成了四类。

最后以每类中对应葡萄酒质量评分的均值作为该类葡萄的分数，从而定出四类的级别，以对应国家葡萄酒的四级分类标准。

在分析酿酒葡萄与葡萄酒的理化指标间的联系问题中，本文采用偏最小二乘回归分析法对指标间的联系进行了分析计算，发现葡萄酒中的某些理化指标与葡萄的某些理化指标存在较强的相关性，比如白葡萄中的总糖和还原糖对白葡萄酒中顺式白藜芦醇苷和顺式白藜芦醇以及反式白藜芦醇的影响较大。

在判断葡萄与葡萄酒的理化指标与葡萄酒的质量间关系的问题中，首先对葡萄和葡萄酒的理化指标与葡萄酒的质量进行了相关性分析，发现某些理化指标与葡萄酒的质量相关性很大。

数学建模A题优秀论文之欧阳体创编

城市表层土壤重金属污染分析摘要随着城市经济的快速发展和城市人口的不断增加，人类活动对城市环境质量的影响日显突出。

对城市土壤地质环境异常的查证，以及如何应用查证获得的海量数据资料开展城市环境质量评价，研究人类活动影响下城市地质环境的演变模式，日益成为人们关注的焦点。

对于问题一我们首先用EXCEL对数据进行处理，然后用MATLAB等软件对所给的数值进行空间作图，然后分别做出了八种重金属元素的空间分布特征，我们利用综合指数（内梅罗指数）评价的方法，建立模型：对于问题二我们首先用SPSS软件对八种重金属的浓度与海拔做相关性分析，并作出了不同重金属浓度与海拔的分布图；然后结合第一问给出的空间分布图和区域散点图，参照主要重金属含量土壤单项污染的指数，分析得出各重金属污染的主要原因主要来自工业区、交通区和生活区。

对于问题三我们建立模型，建立目标函数；应用MATLAB软件对数据处理，作出可能为污染源的三个位置；然后用MATLAB进行三次拟合后，得到污染源的位置。

对于问题四，我们在已有信息的基础上，还应收集不同时间内的样点对应的浓度以及各污染源重金属的产生率。

根据高斯浓度模型建立高斯修正模型，得到浓度关于时间和空间的表达式。

关键词：重金属污染内梅罗污染指数相关性分析污染源高斯浓度一.问题重述随着城市经济的快速发展和城市人口的不断增加，人类活动对城市环境质量的影响日显突出。

按照功能划分，城区一般可分为生活区、工业区、山区、主干道路区及公园绿地区等，分别记为1类区、2类区、……、5类区，不同的区域环境受人类活动影响的程度不同。

现对某城市城区土壤地质环境进行调查。

为此，将所考察的城区划分为间距1公里左右的网格子区域，按照每平方公里1个采样点对表层土（0~10厘米深度）进行取样、编号，并用GPS记录采样点的位置。

深圳杯数学建模A题获奖论文之欧阳美创编

网络侧估计终端用户视频体验建模摘要现代社会，使用手机APP观看视频已经成为当代社会的一种普遍形式，本文依据统计回归方法，对网络侧变量和用户体验变量之间的函数关系进行拟合，令其余无关变量均近似地服从正态分布。

采用多重拟合方式拟合出不同的评价函数，并进行误差检验。

选择误差最小的评价函数。

并基于评价函数，两个用户体验变量进行预测。

同时对用户观看视频体验进行综合评价，采用多级指标，运用AHP及模糊综合评价法评价用户观看视频的满意度。

求出权重，建立评价矩阵。

得到用户观看视频满意度处在较满意和一般满意之间。

最后，由于多种原因，本文建立的用户体验变量评价函数具有一定程度的误差，因此基于原有数据，建立灰色系统模型，再次进行预测，比较结果。

建立GM（1,1）模型对相关指标进行预测，取预测区间长度为100，得出预测值，并绘制残差图对预测值进行检验。

并与评价函数预测结果进行对比。

验证评价函数的正确性。

同时得到结论，基于原始数据直接建立灰色系统，预测相对更加准确。

关键词：统计回归；综合评价；灰色预测；残差检验一.问题重述随着科技的日益进步，无线宽带网络也随之无限升级。

智能终端在大众生活中普及，越来越多的用户选择在智能终端上（以手机为主）应用客户端APP来观看网络视频，这是一种基于TCP（是一种面向连接的、可靠的、基于字节流的传输层通信协议）的视频传输以及播放。

在观看网络视频时，有很多因素指标会影响用户对于视频的观看体验，而其中两个关键指标是初始缓冲等待时间和卡顿缓冲时间，我们可以用初始缓冲时延和卡顿时长占比（卡顿时长占比=卡顿时长/视频播放时长）来定量评价用户体验。

研究表明影响初始缓冲时延和卡顿时长占比的主要因素有初始缓冲峰值速率、播放阶段平均下载速率、端到端环回时间（E2E RTT）以及视频参数。

然而这些因素和初始缓冲时延以及卡顿时长占比之间的关系并不明确。

本文拟通过数学建模的方式对网络端视频用户体验做综合评价和预测，以采取针对性的措施提高网络端视频用户体验的满意程度。