2013MCM美国大学生数学建模竞赛获奖论文(一等奖)
美国大学生数学建模竞赛优秀论文
For office use onlyT1________________ T2________________ T3________________ T4________________Team Control Number7018Problem ChosencFor office use onlyF1________________F2________________F3________________F4________________ SummaryThe article is aimed to research the potential impact of the marine garbage debris on marine ecosystem and human beings，and how we can deal with the substantial problems caused by the aggregation of marine wastes.In task one，we give a definition of the potential long-term and short-term impact of marine plastic garbage. Regard the toxin concentration effect caused by marine garbage as long-term impact and to track and monitor it. We etablish the composite indicator model on density of plastic toxin，and the content of toxin absorbed by plastic fragment in the ocean to express the impact of marine garbage on ecosystem. Take Japan sea as example to examine our model.In ask two, we designe an algorithm, using the density value of marine plastic of each year in discrete measure point given by reference，and we plot plastic density of the whole area in varies locations. Based on the changes in marine plastic density in different years， we determine generally that the center of the plastic vortex is East—West140°W—150°W, South—North30°N—40°N. According to our algorithm, we can monitor a sea area reasonably only by regular observation of part of the specified measuring pointIn task three，we classify the plastic into three types，which is surface layer plastic，deep layer plastic and interlayer between the two. Then we analysis the the degradation mechanism of plastic in each layer. Finally，we get the reason why those plastic fragments come to a similar size.In task four， we classify the source of the marine plastic into three types，the land accounting for 80%，fishing gears accounting for 10%，boating accounting for 10%，and estimate the optimization model according to the duel-target principle of emissions reduction and management. Finally, we arrive at a more reasonable optimization strategy.In task five，we first analyze the mechanism of the formation of the Pacific ocean trash vortex， and thus conclude that the marine garbage swirl will also emerge in south Pacific，south Atlantic and the India ocean. According to the Concentration of diffusion theory, we establish the differential prediction model of the future marine garbage density，and predict the density of the garbage in south Atlantic ocean. Then we get the stable density in eight measuring point .In task six, we get the results by the data of the annual national consumption ofpolypropylene plastic packaging and the data fitting method, and predict the environmental benefit generated by the prohibition of polypropylene take-away food packaging in the next decade. By means of this model and our prediction，each nation will reduce releasing 1.31 million tons of plastic garbage in next decade.Finally, we submit a report to expediction leader，summarize our work and make some feasible suggestions to the policy- makers.Task 1:Definition:●Potential short-term effects of the plastic: the hazardeffects will be shown in the short term.●Potential long-term effects of the plastic: thepotential effects, of which hazards are great, willappear after a long time.The short- and long-term effects of the plastic on the ocean environment:In our definition, the short-term and long-term effects of the plastic on the ocean environment are as follows.Short-term effects:1）The plastic is eaten by marine animals or birds.2） Animals are wrapped by plastics, such as fishing nets, which hurt or even kill them.3）Deaden the way of the passing vessels.Long-term effects:1）Enrichment of toxins through the food chain: the waste plastic in the ocean has no natural degradation in theshort-term, which will first be broken down into tinyfragments through the role of light, waves,micro-organisms, while the molecular structure has notchanged. These "plastic sands", easy to be eaten byplankton, fish and other, are Seemingly very similar tomarine life’s food,causing the enrichment and delivery of toxins.2）Accelerate the greenhouse effect: after a long-term accumulation and pollution of plastics, the waterbecame turbid, which will seriously affect the marineplants (such as phytoplankton and algae) inphotosynthesis. A large number of plankton’s deathswould also lower the ability of the ocean to absorbcarbon dioxide, intensifying the greenhouse effect tosome extent.To monitor the impact of plastic rubbish on the marine ecosystem:According to the relevant literature, we know that plastic resin pellets accumulate toxic chemicals , such as PCBs、DDE , and nonylphenols , and may serve as a transport medium and soure of toxins to marine organisms that ingest them[]2. As it is difficult for the plastic garbage in the ocean to complete degradation in the short term, the plastic resin pellets in the water will increase over time and thus absorb more toxins, resulting in the enrichment of toxins and causing serious impact on the marine ecosystem.Therefore, we track the monitoring of the concentration of PCBs, DDE, and nonylphenols containing in the plastic resin pellets in the sea water, as an indicator to compare the extent of pollution in different regions of the sea, thus reflecting the impact of plastic rubbish on ecosystem.To establish pollution index evaluation model: For purposes of comparison, we unify the concentration indexes of PCBs, DDE, and nonylphenols in a comprehensive index.Preparations:1）Data Standardization2）Determination of the index weightBecause Japan has done researches on the contents of PCBs,DDE, and nonylphenols in the plastic resin pellets, we illustrate the survey conducted in Japanese waters by the University of Tokyo between 1997 and 1998.To standardize the concentration indexes of PCBs, DDE,and nonylphenols. We assume Kasai Sesside Park, KeihinCanal, Kugenuma Beach, Shioda Beach in the survey arethe first, second, third, fourth region; PCBs, DDE, andnonylphenols are the first, second, third indicators.Then to establish the standardized model:j j jij ij V V V V V min max min --= （1,2,3,4;1,2,3i j ==）wherej V max is the maximum of the measurement of j indicator in the four regions.j V min is the minimum of the measurement of j indicatorstandardized value of j indicator in i region.According to the literature [2], Japanese observationaldata is shown in Table 1.Table 1. PCBs, DDE, and, nonylphenols Contents in Marine PolypropyleneTable 1 Using the established standardized model to standardize, we have Table 2.In Table 2，the three indicators of Shioda Beach area are all 0, because the contents of PCBs, DDE, and nonylphenols in Polypropylene Plastic Resin Pellets in this area are the least, while 0 only relatively represents the smallest. Similarly, 1 indicates that in some area the value of a indicator is the largest.To determine the index weight of PCBs, DDE, and nonylphenolsWe use Analytic Hierarchy Process (AHP) to determine the weight of the three indicators in the general pollution indicator. AHP is an effective method which transforms semi-qualitative and semi-quantitative problems into quantitative calculation. It uses ideas of analysis and synthesis in decision-making, ideally suited for multi-index comprehensive evaluation.Hierarchy are shown in figure 1.Fig.1 Hierarchy of index factorsThen we determine the weight of each concentrationindicator in the generall pollution indicator, and the process are described as follows:To analyze the role of each concentration indicator, we haveestablished a matrix P to study the relative proportion.⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=111323123211312P P P P P P P Where mn P represents the relative importance of theconcentration indicators m B and n B . Usually we use 1，2，…，9 and their reciprocals to represent different importance. The greater the number is, the more important it is. Similarly, the relative importance of m B and n B is mn P /1（3,2,1,=n m ）.Suppose the maximum eigenvalue of P is m ax λ, then theconsistency index is1max --=n nCI λThe average consistency index is RI , then the consistencyratio isRICI CR = For the matrix P of 3≥n , if 1.0<CR the consistency isthougt to be better, of which eigenvector can be used as the weight vector.We get the comparison matrix accoding to the harmful levelsof PCBs, DDE, and nonylphenols and the requirments ofEPA on the maximum concentration of the three toxins inseawater as follows:⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=165416131431P We get the maximum eigenvalue of P by MATLAB calculation0012.3max =λand the corresponding eigenvector of it is()2393.02975.09243.0，，=W1.0042.012.1047.0<===RI CI CR Therefore,we determine the degree of inconsistency formatrix P within the permissible range. With the eigenvectors of p as weights vector, we get thefinal weight vector by normalization ()1638.02036.06326.0'，，=W . Defining the overall target of pollution for the No i oceanis i Q , among other things the standardized value of threeindicators for the No i ocean is ()321,,i i i i V V V V = and the weightvector is 'W ,Then we form the model for the overall target of marine pollution assessment, （3,2,1=i ）By the model above, we obtained the Value of the totalpollution index for four regions in Japanese ocean in Table 3T B W Q '=In Table3, the value of the total pollution index is the hightest that means the concentration of toxins in Polypropylene Plastic Resin Pellets is the hightest, whereas the value of the total pollution index in Shioda Beach is the lowest(we point up 0 is only a relative value that’s not in the name of free of plastics pollution)Getting through the assessment method above, we can monitor the concentration of PCBs, DDE and nonylphenols in the plastic debris for the sake of reflecting the influence to ocean ecosystem.The highter the the concentration of toxins,the bigger influence of the marine organism which lead to the inrichment of food chain is more and more dramatic.Above all, the variation of toxins’ concentration simultaneously reflects the distribution and time-varying of marine litter. We can predict the future development of marine litter by regularly monitoring the content of these substances, to provide data for the sea expedition of the detection of marine litter and reference for government departments to make the policies for ocean governance.Task 2:In the North Pacific, the clockwise flow formed a never-ending maelstrom which rotates the plastic garbage. Over the years, the subtropical eddy current in North Pacific gathered together the garbage from the coast or the fleet, entrapped them in the whirlpool, and brought them to the center under the action of the centripetal force, forming an area of 3.43 million square kilometers (more than one-third of Europe) .As time goes by, the garbage in the whirlpool has the trend of increasing year by year in terms of breadth, density, and distribution. In order to clearly describe the variability of the increases over time and space, according to “Count Densities of Plastic Debris from Ocean Surface Samples North Pacific Gyre 1999—2008”, we analyze the data, exclude them with a great dispersion, and retain them with concentrated distribution, while the longitude values of the garbage locations in sampled regions of years serve as the x-coordinate value of a three-dimensional coordinates, latitude values as the y-coordinate value, the Plastic Count per cubic Meter of water of the position as the z-coordinate value. Further, we establish an irregular grid in the yx plane according to obtained data, and draw a grid line through all the data points. Using the inverse distance squared method with a factor, which can not only estimate the Plastic Count per cubic Meter of water of any position, but also calculate the trends of the Plastic Counts per cubic Meter of water between two original data points, we can obtain the unknown grid points approximately. When the data of all the irregular grid points are known (or approximately known, or obtained from the original data), we can draw the three-dimensional image with the Matlab software, which can fully reflect the variability of the increases in the garbage density over time and space.Preparations:First, to determine the coordinates of each year’s sampled garbage.The distribution range of garbage is about the East - West 120W-170W, South - North 18N-41N shown in the “Count Densities of Plastic Debris from Ocean Surface Samples North Pacific Gyre 1999--2008”, we divide a square in the picture into 100 grids in Figure (1) as follows:According to the position of the grid where the measuring point’s center is, we can identify the latitude and longitude for each point, which respectively serve as the x- and y- coordinate value of the three-dimensional coordinates.To determine the Plastic Count per cubic Meter of water. As the “Plastic Count per cubic Meter of water” provided by “Count Densities of P lastic Debris from Ocean Surface Samples North Pacific Gyre 1999--2008”are 5 density interval, to identify the exact values of the garbage density of one year’s different measuring points, we assume that the density is a random variable which obeys uniform distribution in each interval.Uniform distribution can be described as below:()⎪⎩⎪⎨⎧-=01a b x f ()others b a x ,∈We use the uniform function in Matlab to generatecontinuous uniformly distributed random numbers in each interval, which approximately serve as the exact values of the garbage density andz-coordinate values of the three-dimensional coordinates of the year’s measuring points.Assumptions(1)The data we get is accurate and reasonable.(2)Plastic Count per cubic Meter of waterIn the oceanarea isa continuous change.(3)Density of the plastic in the gyre is a variable by region.Density of the plastic in the gyre and its surrounding area is interdependent , However, this dependence decreases with increasing distance . For our discussion issue, Each data point influences the point of each unknown around and the point of each unknown around is influenced by a given data point. The nearer a given data point from the unknown point, the larger the role.Establishing the modelFor the method described by the previous,we serve the distributions of garbage density in the “Count Pensities of Plastic Debris from Ocean Surface Samples North Pacific Gyre 1999--2008”as coordinates ()z y,, As Table 1:x,Through analysis and comparison, We excluded a number of data which has very large dispersion and retained the data that is under the more concentrated the distribution which, can be seen on Table 2.In this way, this is conducive for us to get more accurate density distribution map.Then we have a segmentation that is according to the arrangement of the composition of X direction and Y direction from small to large by using x co-ordinate value and y co-ordinate value of known data points n, in order to form a non-equidistant Segmentation which has n nodes. For the Segmentation we get above,we only know the density of the plastic known n nodes, therefore, we must find other density of the plastic garbage of n nodes.We only do the sampling survey of garbage density of the north pacificvortex,so only understand logically each known data point has a certain extent effect on the unknown node and the close-known points of density of the plastic garbage has high-impact than distant known point.In this respect,we use the weighted average format, that means using the adverse which with distance squared to express more important effects in close known points. There're two known points Q1 and Q2 in a line ,that is to say we have already known the plastic litter density in Q1 and Q2, then speculate the plastic litter density's affects between Q1、Q2 and the point G which in the connection of Q1 and Q2. It can be shown by a weighted average algorithm22212221111121GQ GQ GQ Z GQ Z Z Q Q G +*+*=in this formula GQ expresses the distance between the pointG and Q.We know that only use a weighted average close to the unknown point can not reflect the trend of the known points, we assume that any two given point of plastic garbage between the changes in the density of plastic impact the plastic garbage density of the unknown point and reflecting the density of plastic garbage changes in linear trend. So in the weighted average formula what is in order to presume an unknown point of plastic garbage density, we introduce the trend items. And because the greater impact at close range point, and thus the density of plastic wastes trends close points stronger. For the one-dimensional case, the calculation formula G Z in the previous example modify in the following format:2212122212212122211111112121Q Q GQ GQ GQ Q Q GQ Z GQ Z GQ Z Z Q Q Q Q G ++++*+*+*=Among them, 21Q Q known as the separation distance of the known point, 21Q Q Z is the density of plastic garbage which is the plastic waste density of 1Q and 2Q for the linear trend of point G . For the two-dimensional area, point G is not on the line 21Q Q , so we make a vertical from the point G and cross the line connect the point 1Q and 2Q , and get point P , the impact of point P to 1Q and 2Q just like one-dimensional, and the one-dimensional closer of G to P , the distant of G to P become farther, the smaller of the impact, so the weighting factor should also reflect the GP in inversely proportional to a certain way, then we adopt following format:221212222122121222211111112121Q Q GQ GP GQ GQ Q Q GQ GP Z GQ Z GQ Z Z P Q Q Q Q G ++++++*+*+*=Taken together, we speculated following roles:(1) Each known point data are influence the density of plastic garbage of each unknown point in the inversely proportional to the square of the distance;(2) the change of density of plastic garbage between any two known points data, for each unknown point are affected, and the influence to each particular point of their plastic garbage diffuse the straight line along the two known particular point; (3) the change of the density of plastic garbage between any two known data points impact a specific unknown points of the density of plastic litter depends on the three distances: a. the vertical distance to a straight line which is a specific point link to a known point;b. the distance between the latest known point to a specific unknown point;c. the separation distance between two known data points.If we mark 1Q ，2Q ，…，N Q as the location of known data points,G as an unknown node, ijG P is the intersection of the connection of i Q ，j Q and the vertical line from G to i Q ，j Q()G Q Q Z j i ,,is the density trend of i Q ，j Q in the of plasticgarbage points and prescribe ()G Q Q Z j i ,,is the testing point i Q ’ s density of plastic garbage ,so there are calculation formula:()()∑∑∑∑==-==++++*=Ni N ij ji i ijGji i ijG N i Nj j i G Q Q GQ GPQ Q GQ GP G Q Q Z Z 11222222111,,Here we plug each year’s observational data in schedule 1 into our model, and draw the three-dimensional images of the spatial distribution of the marine garbage ’s density with Matlab in Figure (2) as follows:199920002002200520062007－2008(1)It’s observed and analyzed that, from 1999 to 2008, the density of plastic garbage is increasing year by year and significantly in the region of East – West 140W-150W, south - north 30N-40N. Therefore, we can make sure that this region is probably the center of the marine litter whirlpool. Gathering process should be such that the dispersed garbage floating in the ocean move with the ocean currents and gradually close to the whirlpool region. At the beginning, the area close to the vortex will have obviously increasable about plastic litter density, because of this centripetal they keeping move to the center of the vortex ,then with the time accumulates ,the garbage density in the center of the vortex become much bigger and bigger , at last it becomes the Pacific rubbish island we have seen today.It can be seen that through our algorithm, as long as the reference to be able to detect the density in an area which has a number of discrete measuring points,Through tracking these density changes ,we Will be able to value out all the waters of the density measurement through our models to determine,This will reduce the workload of the marine expedition team monitoring marine pollution significantly, and also saving costs .Task 3:The degradation mechanism of marine plasticsWe know that light, mechanical force, heat, oxygen, water, microbes, chemicals, etc. can result in the degradation of plastics . In mechanism ,Factors result in the degradation can be summarized as optical ，biological，and chemical。
2013年全国大学生数学建模竞赛B题全国一等奖论文.
碎纸片的拼接复原【摘要】破碎文件的拼接在司法物证复原、历史文献修复以及军事情报获取等领域都有着重要的应用。
本文主要解决碎纸机切割后的碎纸片拼接复原问题。
针对第一问，附件1、2分别为沿纵向切割后的19张中英文碎纸片，本文在考虑破碎纸片携带信息量较大的基础上，利用MATLAB对附件1、2的碎纸片图像分别读入，以数字矩阵的方式进行存储。
利用数字矩阵中包含图像边缘灰度这一特征，本文采用贪心算法的思想，在首先确定原文件左右边界的基础上，以Manhattan距离来度量两两碎纸片边界差异度，利用计算机搜索依次从左往右搜寻最匹配的碎纸片进行横向配对并达成排序目的。
最终，本文在没有进行人工干预，成功地将附件1、2碎纸片分别拼接复原，得到复原图片见附录2.1、2.2，纵切中文及英文结果表分别如下:思想仍为贪心算法，整体思路为先对209张碎纸片进行聚类还原成11行，再对分好的每行进行横向排序，最后对排序好的各行进行纵向排序。
本文在充分考虑汉字与拉丁字母结构特征差异以及每块碎纸片携带信息减少的基础上，创新地提出一种特征线模型来分别描述汉字及拉丁文字母的特征用于行聚类。
对于行聚类后碎片的横向排序，本文综合了广义Jaccard系数、一阶差分法、二阶差分法、Spearman系数等来构建扩展的边界差异度模型，刻画碎片间的差异度。
对于计算机横向排序存在些许错误的情况，本文给出了人工干预的位置节点和方式。
对于横向排序后的各行，由于在一页纸上，文字的各行是均匀分布的，本文基于各行文字的特征线，在确定首行的位置后，估计出其他行的基准线位置，得到一页的基准线网格，并通过各行基准线在基准线网格上的适配实现纵向的排序。
最终，本文成功的将附件3、4碎纸片分别拼接复原得到复原图片及结果表见附录1.3、1.4、2.3、2.4，同时本文给出了横向排序中人工干预的位置节点和方式。
针对第三问，附件5为双面文件既横切又纵切后的209张碎片（包含正反面），即包含418张图像。
2013美国大学生数学建模竞赛论文
summaryOur solution paper mainly deals with the following problems:·How to measure the distribution of heat across the outer edge of pans in differentshapes and maximize even distribution of heat for the pan·How to design the shape of pans in order to make the best of space in an oven·How to optimize a combination of the former two conditions.When building the mathematic models, we make some assumptions to get themto be more reasonable. One of the major assumptions is that heat is evenly distributedwithin the oven. We also introduce some new variables to help describe the problem.To solve all of the problems, we design three models. Based on the equation ofheat conduction, we simulate the distribution of heat across the outer edge with thehelp of some mathematical softwares. In addition, taking the same area of all the pansinto consideration, we analyze the rate of space utilization ratio instead of thinkingabout maximal number of pans contained in the oven. What’s more, we optimize acombination of conditions (1) and (2) to find out the best shape and build a function toshow the relation between the weightiness of both conditions and the width to lengthratio, and to illustrate how the results vary with different values of W/L and p.To test our models, we compare the results obtained by stimulation and our models, tofind that our models fit the truth well. Yet, there are still small errors. For instance, inModel One, the error is within 1.2% .In our models, we introduce the rate of satisfaction to show how even thedistribution of heat across the outer edge of a pan is clearly. And with the help ofmathematical softwares such as Matlab, we add many pictures into our models,making them more intuitively clear. But our models are not perfect and there are someshortcomings such as lacking specific analysis of the distribution of heat across theouter edge of a pan of irregular shapes. In spite of these, our models can mainlypredict the actual conditions, within reasonable range of error.For office use onlyT1 ________________T2 ________________T3 ________________T4 ________________ Team Control Number18674 Problem Chosen AFor office use only F1 ________________ F2 ________________ F3 ________________ F4 ________________2013 Mathematical Contest in Modeling (MCM) Summary Sheet(Attach a copy of this page to your solution paper.)Type a summary of your results on this page. Do not includethe name of your school, advisor, or team members on this page.The Ultimate Brownie PanAbstractWe introduce three models in the paper in order to find out the best shape for the Brownie Pan, which is beneficial to both heat conduction and space utility.The major assumption is that heat is evenly distributed within the oven. On the basis of this, we introduce three models to solve the problem.The first model deals with heat distribution. After simulative experiments and data processing, we achieve the connection between the outer shape of pans and heat distribution.The second model is mainly on the maximal number of pans contained in an oven. During the course, we use utility rate of space to describe the number. Finally, we find out the functional relation.Having combined both of the conditions, we find an equation relation. Through mathematical operation, we attain the final conclusion.IntroductionHeat usage has always been one of the most challenging issues in modern world. Not only does it has physic significance, but also it can influence each bit of our daily life. Likewise,space utilization, beyond any doubt, also contains its own strategic importance. We build three mathematic models based on underlying theory of thermal conduction and tip thermal effects.The first model describes the process and consequence of heat conduction, thus representing the temperature distribution. Given the condition that regular polygons gets overcooked at the corners, we introduced the concept of tip thermal effects into our prediction scheme. Besides, simulation technique is applied to both models for error correction to predict the final heat distribution.Assumption• Heat is distributed evenly in the oven.Obviously, an oven has its normal operating temperature, which is gradually reached actually. We neglect the distinction of temperature in the oven and the heating process, only to focus on the heat distribution of pans on the basis of their construction.Furthermore, this assumption guarantees the equivalency of the two racks.• Thermal conductivity is temperature-invariant.Thermal conductivity is a physical quantity, symbolizing the capacity of materials. Always, the thermal conductivity of metal material usually varies with different temperatures, in spite of tiny change in value. Simply, we suppose the value to be a constant.• Heat flux of boundaries keeps steady.Heat flux is among the important indexes of heat dispersion. In this transference, we give it a constant value.• Heat conduction dom inates the variation of temperature, while the effects ofheat radiation and heat convection can be neglected.Actually, the course of heat conduction, heat radiation and heat convectiondecide the variation of temperature collectively. Due to the tiny influence of other twofactors, we pay closer attention to heat conduction.• The area of ovens is a constant.I ntroduction of mathematic modelsModel 1: Heat conduction• Introduction of physical quantities:q: heat fluxλ: Thermal conductivityρ: densityc: specific heat capacityt: temperature τ: timeV q : inner heat sourceW q : thermal fluxn: the number of edges of the original polygonsM t : maximum temperaturem t : minimum temperatureΔt: change quantity of temperatureL: side length of regular polygon• Analysis:Firstly, we start with The Fourier Law:2(/)q gradt W m λ=- . (1) According to The Fourier Law, along the direction of heat conduction, positionsof a larger cross-sectional area are lower in temperature. Therefore, corners of panshave higher temperatures.Secondly, let’s analyze the course of heat conduction quantitatively.To achieve this, we need to figure out exact temperatures of each point across theouter edge of a pan and the variation law.Based on the two-dimension differential equation of heat conduction:()()V t t t c q x x y yρλλτ∂∂∂∂∂=++∂∂∂∂∂. (2) Under the assumption that heat distribution is time-independent, we get0t τ∂=∂. (3)And then the heat conduction equation (with no inner heat source)comes to:20t ∇=. (4)under the Neumann boundary condition: |W s q t n λ∂-=∂. (5)Then we get the heat conduction status of regular polygons and circles as follows:Fig 1In consideration of the actual circumstances that temperature is higher at cornersthan on edges, we simulate the temperature distribution in an oven and get resultsabove. Apparently, there is always higher temperature at corners than on edges.Comparatively speaking, temperature is quite more evenly distributed around circles.This can prove the validity of our model rudimentarily.From the figure above, we can get extreme values along edges, which we callM t and m t . Here, we introduce a new physical quantity k , describing the unevennessof heat distribution. For all the figures are the same in area, we suppose the area to be1. Obviously, we have22sin 2sin L n n n ππ= (6) Then we figure out the following results.n t M t m t ∆ L ksquare 4 214.6 203.3 11.3 1.0000 11.30pentagon 5 202.1 195.7 6.4 0.7624 8.395hexagon 6 195.7 191.3 4.4 0.6204 7.092heptagon 7 193.1 190.1 3.0 0.5246 5.719octagon 8 191.1 188.9 2.2 0.4551 4.834nonagon 9 188.9 187.1 1.8 0.4022 4.475decagon 10 189.0 187.4 1.6 0.3605 4.438Table 1It ’s obvious that there is negative correlation between the value of k and thenumber of edges of the original polygons. Therefore, we can use k to describe theunevenness of temperature distribution along the outer edge of a pan. That is to say, thesmaller k is, the more homogeneous the temperature distribution is.• Usability testing:We use regular hendecagon to test the availability of the model.Based on the existing figures, we get a fitting function to analyze the trend of thevalue of k. Again, we introduce a parameter to measure the value of k.Simply, we assume203v k =, (7) so that100v ≤. (8)n k v square 4 11.30 75.33pentagon 5 8.39 55.96hexagon 6 7.09 47.28heptagon 7 5.72 38.12octagon 8 4.83 32.23nonagon9 4.47 29.84 decagon 10 4.44 29.59Table 2Then, we get the functional image with two independent variables v and n.Fig 2According to the functional image above, we get the fitting function0.4631289.024.46n v e -=+.(9) When it comes to hendecagons, n=11. Then, v=26.85.As shown in the figure below, the heat conduction is within our easy access.Fig 3So, we can figure out the following result.vnActually,2026.523tvL∆==.n ∆t L k vhendecagons 11 187.1 185.8 1.3 0.3268 3.978 26.52Table 3Easily , the relative error is 1.24％.So, our model is quite well.• ConclusionHeat distribution varies with the shape of pans. To put it succinctly, heat is more evenly distributed along more edges of a single pan. That is to say, pans with more number of peripheries or more smooth peripheries are beneficial to even distribution of heat. And the difference in temperature contributes to overcooking. Through calculation, the value of k decreases with the increase of edges. With the help of the value of k, we can have a precise prediction of heat contribution.Model 2: The maximum number• Introduction of physical quantities:n: the number of edges of the original polygonsα: utility rate of space• Analysis:Due to the fact that the area of ovens and pans are constant, we can use the area occupied by pans to describe the number of pans. Further, the utility rate of space can be used to describe the number. In the following analysis, we will make use of the utility rate of space to pick out the best shape of pans. We begin with the best permutation devise of regular polygon. Having calculated each utility rate of space, we get the variation tendency.• Model Design:W e begin with the scheme which makes the best of space. Based on this knowledge, we get the following inlay scheme.Fig 4Fig 5According to the schemes, we get each utility rate of space which is showed below.n=4 n=5 n=6 n=7 n=8 n=9 n=10 n=11 shape square pentagon hexagon heptagon octagon nonagon decagon hendecagon utility rate(％)100.00 85.41 100.00 84.22 82.84 80.11 84.25 86.21Table 4Using the ratio above, we get the variation tendency.Fig 6 nutility rate of space• I nstructions:·The interior angle degrees of triangles, squares, and regular hexagon can be divided by 360, so that they all can completely fill a plane. Here, we exclude them in the graph of function.·When n is no more than 9, there is obvious negative correlation between utility rate of space and the value of n. Otherwise, there is positive correlation.·The extremum value of utility rate of space is 90.69％,which is the value for circles.• Usability testing:We pick regular dodecagon for usability testing. Below is the inlay scheme.Fig 7The space utility for dodecagon is 89.88％, which is around the predicted value. So, we’ve got a rather ideal model.• Conclusion:n≥), the When the number of edges of the original polygons is more than 9(9 space utility is gradually increasing. Circles have the extreme value of the space utility. In other words, circles waste the least area. Besides, the rate of increase is in decrease. The situation of regular polygon with many sides tends to be that of circles. In a word, circles have the highest space utility.Model 3: Rounded rectangle• Introduction of physical quantities:A: the area of the rounded rectanglel: the length of the rounded rectangleα: space utilityβ: the width to length ratio• Analysis:Based on the combination of consideration on the highest space utility of quadrangle and the even heat distribution of circles, we invent a model using rounded rectangle device for pans. It can both optimize the cooking effect and minimize the waste of space.However, rounded rectangles are exactly not the same. Firstly, we give our rounded rectangle the same width to length ratio (W/L) as that of the oven, so that least area will be wasted. Secondly, the corner radius can not be neglected as well. It’ll give the distribution of heat across the outer edge a vital influence. In order to get the best pan in shape, we must balance how much the two of the conditions weigh in the scheme.• Model Design:To begin with, we investigate regular rounded rectangle.The area224r ar a A π++= (10) S imilarly , we suppose the value of A to be 1. Then we have a function between a and r :21(4)2a r r π=+--(11) Then, the space utility is()212a r α=+ (12) And, we obtain()2114rαπ=+- (13)N ext, we investigate the relation between k and r, referring to the method in the first model. Such are the simulative result.Fig 8Specific experimental results arer a ∆t L k 0.05 0.90 209.2 199.9 9.3 0.98 9.49 0.10 0.80 203.8 196.4 7.4 0.96 7.70 0.15 0.71 199.6 193.4 6.2 0.95 6.56 0.20 0.62 195.8 190.5 5.3 0.93 5.69 0.25 0.53 193.2 189.1 4.1 0.92 4.46Table 5According to the table above, we get the relation between k and r.Fig 9So, we get the function relation3.66511.190.1013r k e -=+. (14) After this, we continue with the connection between the width to length ratioW Lβ=and heat distribution. We get the following results.krFig 10From the condition of heat distribution, we get the relation between k and βFig 11And the function relation is4.248 2.463k β=+ (15)Now we have to combine the two patterns together:3.6654.248 2.463(11.190.1013)4.248 2.463r k e β-+=++ (16)Finally, we need to take the weightiness (p) into account,(,,)()(,)(1)f r p r p k r p βαβ=⋅+⋅- (17)To standard the assessment level, we take squares as criterion.()(,)(1)(,,)111.30r p k r p f r p αββ⋅⋅-=+ (18) Then, we get the final function3.6652(,,)(1)(0.37590.2180)(1.6670.0151)1(4)r p f r p p e rββπ-=+-⋅+⋅++- (19) So we get()()3.6652224(p 1)(2.259β 1.310)14r p f e r r ππ--∂=-+-+∂⎡⎤+-⎣⎦ (20) Let 0f r∂=∂,we can get the function (,)r p β. Easily,0r p∂<∂ and 0r β∂>∂ (21) So we can come to the conclusion that the value of r decreases with the increase of p. Similarly, the value of r increases with the increase of β.• Conclusion:Model 3 combines all of our former analysis, and gives the final result. According to the weightiness of either of the two conditions, we can confirm the final best shape for a pan.• References:[1] Xingming Qi. Matlab 7.0. Beijing: Posts & Telecom Press, 2009: 27-32[2] Jiancheng Chen, Xinsheng Pang. Statistical data analysis theory and method. Beijing: China's Forestry Press, 2006: 34-67[3] Zhengshen Fan. Mathematical modeling technology. Beijing: China Water Conservancy Press, 2003: 44-54Own It NowYahoo! Ladies and gentlemen, please just have a look at what a pan we have created-the Ultimate Brownie Pan.Can you imagine that just by means of this small invention, you can get away of annoying overcookedchocolate Brownie Cake? Pardon me, I don’t want to surprise you, but I must tell you , our potential customers, that we’ve made it! Believing that it’s nothing more than a common pan, some people may think that it’s not so difficult to create such a pan. To be honest, it’s not just a simple pan as usual, and it takes a lot of work. Now let me show you how great it is. Here we go!Believing that it’s nothing more than a common pan, some people may think that it’s not so difficult to create such a pan. To be honest, it’s not just a simple pan as usual, and it takes a lot of work. Now let me show you how great it is. Here we go!Maybe nobody will deny this: when baked in arectangular pan, cakes get easily overcooked at thecorners (and to a lesser extent at the edges).But neverwill this happen in a round pan. However, round pansare not the best in respects of saving finite space in anoven. How to solve this problem? This is the key pointthat our work focuses on.Up to now, as you know, there have been two factors determining the quality of apan -- the distribution of heat across the outer edge of and thespace occupied in an oven. Unfortunately, they cannot beachieved at the same time. Time calls for a perfect pan, andthen our Ultimate Brownie Pan comes into existence. TheUltimate Brownie Pan has an outstandingadvantage--optimizing a combination of the two conditions. As you can see, it’s so cute. And when you really begin to use it, you’ll find yourself really enjoy being with it. By using this kind of pan, you can use four pans in the meanwhile. That is to say you can bake more cakes at one time.So you can see that our Ultimate Brownie Pan will certainly be able to solve the two big problems disturbing so many people. And so it will! Feel good? So what are you waiting for? Own it now!。
2013年美国大学生数学建模大赛B题获奖论文
For office use onlyT1________________ T2________________ T3________________ T4________________Team Control Number22599Problem ChosenAFor office use onlyF1________________F2________________F3________________F4________________ 2013Mathematical Contest in Modeling(MCM/ICM)Summary Sheet(Attach a copy of this page to your solution paper.)Heat Radiation in The OvenHeat distribution of pans in the oven is quite different from each other,which depends on their shapes.Thus,our model aims at two goals.One is to analyze the heat distri-bution in different ovens based on the locations of electrical heating cubes.Further-more,a series of heat distribution which varies from circular pans to rectangular pans could be got easily.The other is to optimize the pans placing,in order to choose a best way to maximize the even heat and the number of pans at the same time.Mathematically speaking,our solution consists of two models,analyzing and optimi-zing.In part one,our whole-local approach shows the heat distribution of every pan.Firstly,we use the Stefan-Boltzmann law and Fourier theorem to describe the heat distribution in the air around the electrical heating tube.And then, based on plane in-tercept method and simplified Monte Carlo method,the heat distribution of different shapes of pans is obtained.Finally,we explain the phenomenon that the corners of a pan always get over heated with water waves stirring by analogy.In part two,our discretize-convert approach optimizes the shape and number of the pans.Above all,we discre-tize the side length of the oven, so that the number and the average heat of the pans vary linearly.In the end,the abstract weight P is converted into a specific length,in order to reach a compromise between the two factors.Specially,we create a unique method to convert the variables from the whole space to the local section.The special method allows us to draw the heat distribution of every single section in the oven.The algorithm we create does a great job in flexibility,which can be applied to all shapes of pans.Type a summary of your results on this page.Do not includethe name of your school,advisor,or team members on this page.Heat Radiation in The OvenSummaryHeat distribution of pans in the oven is quite different from each other,which depends on their shapes.Thus,our model aims at two goals.One is to analyze the heat distri-bution in different ovens based on the locations of electrical heating cubes.Further-more,a series of heat distribution which varies from circular pans to rectangular pans could be got easily.The other is to optimize the pans placing,in order to choose a best way to maximize the even heat and the number of pans at the same time.Mathematically speaking,our solution consists of two models,analyzing and optimi-zing.In part one,our whole-local approach shows the heat distribution of every pan. Firstly,we use the Stefan-Boltzmann law and Fourier theorem to describe the heat distribution in the air around the electrical heating tube.And then,based on plane in-tercept method and simplified Monte Carlo method,the heat distribution of different shapes of pans is obtained.Finally,we explain the phenomenon that the corners of a pan always get over heated with water waves stirring by analogy.In part two,our discretize-convert approach optimizes the shape and number of the pans.Above all, we discre-tize the side length of the oven,so that the number and the average heat of the pans vary linearly.In the end,the abstract weight P is converted into a specific length,in order to reach a compromise between the two factors.Specially,we create a unique method to convert the variables from the whole space to the local section.The special method allows us to draw the heat distribution of every single section in the oven.The algorithm we create does a great job in flexibility, which can be applied to all shapes of pans.Keywords:Monte Carlo thermal radiation section heat distribution discretizationIntroductionMany studies on heat conduction wasted plenty of time in solving the partial differential equations,since it’s difficult to solve even for computers.We turn to another way to work it out.Firstly,we study the heat radiation instead of heat conduction to keep away from the sophisticated partial differential equations.Then, we create a unique method to convert every variable from the whole space to section. In other words,we work everything out in heat radiation and convert them into heat contradiction.AssumptionsWe make the following assumptions about the distribution of heat in this paper.·Initially two racks in the oven,evenly spaced.·When heating the electrical heating tubes,the temperature of which changes from room temperature to the desired temperature.It takes such a short time that we can ignore it.·Different pans are made in same material,so they have the same rate of heat conduction.·The inner walls of the oven are blackbodies.The pan is a gray body.The inner walls of the oven absorb heat only and reflect no heat.·The heat can only be reflected once when rebounded from the pan.Heat Distribution ModelOur approach involves four steps:·Use the Fourier theorem to calculate the loss energy when energy beams are spread in the medium.So we can get the heat distribution around each electrical heating tube.The heat distribution of the entire space could be go where the heat of two electrical heating tubes cross together.·When different shapes of the pans are inserted into the oven,the heat map of the entire space is crossed by the section of the pan.Thus,the heat map of every single pan is obtained.·Establish a suitable model to get the reflectivity of every single point on the pan with the simplified Monte Carlo method.And then,a final heat distribution map of the pan without reflection loss is obtained.·A realistic conclusion is drawn due to the results of our model compared with water wave propagation phenomena.First of all,the paper will give a description of the initial energy of the electrical hea-ting tube.We see it as a blackbody who reflects no heat at all.Electromagnetic know-ledge shows that wavelength of the heat rays ranges from um 110−to um 210as shown below[1]:Figure 1.Figure 2.We apply the Stefan-Boltzmann’s law[2]whose solution is ()1/512−=−T c b e c E λλλ(1)()λλλλλd e c d E E T c b b ∫∫∞−∞−==0/51012(2)Where b E means the ability of blackbody to radiate. 1c and 2c are constants.Obviously,,the initial energy of a black body is )(0122398.320m w e E b ×+=.Combine Figure 1with Figure 2,we integrate (1)from 1λto 2λto get the equation as follow:λλλλλλd E E b b ∫=−2121)((3)Figure 3.From Figure 3,it can be seen how the power of radiation varies with wavelength.Secondly,based on the Fourier theorem,the relation between heat and the distance from the electrical heating tubes is:dxdt S Q λ−=(4)Where Q is the power of heat (W s J =/),S is the area where the energy beamradiates (2m ),dxdt represents the temperature gradient along the direction of energy beam.[3]It is known that the energy becomes weaker as the distance becomes larger.According to the fact we know:dxdQ =ρ(5)Where ρis the rate of energy changing.We assume that the desired temperature of electrical heating tube is 500k.With the two equations,the distribution of heat is shown as follow:Figure 4.(a)Figure 4.(b)In order to draw the map of heat distribution in the oven,we use MATLAB to work on the complicated algorithm.The relation between the power of heat and the distance is shown in Figure4(a).The relation between temperature and distance is presented in Figure4(b).The spreading direction of energy beam is presented in Figure5.Figure5.The shape of electrical heating tube is irregular.The heat distribution of a single electrical heating tube can be draw in3D space with MATLAB.The picture is shown in Figure6.After superimposing,the total heat distribution of two tubes is shown below in Figure7and Figure8.Figure6.Figure7.Figure8.The pictures above show the energy in an oven with no pan.We put in a rectangular pan whose area is A,and intercept the maps with MATLAB.The result is show in Figure9.Figure9.Figure10.Put in a circular pan to intercept the maps,whose area is A,also.The distribution of heat is shown in Figure11.Figure11.When put in a pan in transition shape,which is neither rectangular nor circular.The area of it is A,also.The heat distribution on such a pan is shown as follow:Figure12(a)Figure12(b).Figure13.Next,learning from the Monte Carlo simulation[4],a model is established to get obtain the reflectivity.We generate a random number between0and1to determine if the energy beam on certain point is reflected.•Firstly,to demonstrate the question better,we construct a simple model:Figure 14.Where θis the viewing angle from electrical heating tube to the pan.360θ=R is the proportion of the beams radiated to the pan.•What is more,we assume the total beam is 1M .Ideally,the number of absorption is3601θ×M .Then,each element of the pan is seen as a grid point.Each grid point can generate a-3601θ×M -random-number vector between 0and 1in MATLAB.•After MATLAB simulating,the number of beams decreased by 2M ,due to thereflection.So we define a probability θρ12360M M ×=to describe the number of beams reflected.The conclusion is :•If R ≤ρ,the energy beam is absorbed.•If R >ρ,the energy beam is reflected.[5]Based on the analysis above,our model get a final result of heat-distribution on the pan as shown below:Figure 15(a)Figure15(b).The conclusion is known that the closer the shape of pans is to circle,the more evenly the heat is distributed.Moreover,the phenomenon that the corners always get over heated can be explained by water wave propagation in different containers.When there is a fluctuation in the center of the water,the ripples will fluctuate and spread in concentric circles,as shown in Figure16.The fluctuation stirs waves up when contacting the pared with the waves with one boundary,the waves in corner make a higher amplitude.The thermal conduction on the pan is exactly the inverse process of the waves propagation.The range of thermal motion is much smaller than it on the side.That’s why the corners is easy to get over heated.In order to make the heat evenly distributed on the pan,the sides of the pan should be as few as possible.Therefore,if nothing is considered about the utilization of space,a circle pan is the best choice.Figure16,the water waves propagation[6]According to the analysis above and Figure7,the phenomenon shows that the heat conduction is similar to water waves propagation.So it is proved that heatconcentrates in the four corners of the rectangular pan.The Super Pan ModelAssumptions•The width of the oven(W)is mm100,the length is L.•There are three pans at most in vertical direction.•Each pan’s area is A.The first part.Calculate the maximum number of pans in the oven.Different shapes of pan have different heat distribution which affects the number of pans,judging from the previous solution.According to the conclusion in the first model,the heat is distributed the most evenly on a circular pan rather than a rectangular one.However, the rectangular pans make fuller use of the space the space than circular ones.Both factors considered,a polygonal pan is chosen.A circle can be regard as a polygon whose number of boundaries tends to infinity. Except for rectangle,only regular hexagon and equilateral triangle can be closely placed.Because of the edges of equilateral triangle,heat dissipation is worse than rectangle.So,hexagonal pans are adopted after all the discussion.Considering the gaps near boundaries,we place the hexagonal pans closely attached each other on the long side L.There are two kinds of programs as shown below.Program1.Program2.Obviously,Program2is better than Program1when considering space utilization.So scheme 1is adopted.Then,design a size of each hexagonal pan to make the highest space utilization.With the aim of utilization,hexagonal pans has to be placed contact closely with each other on both sides.It is necessary to assume a aspect ratio of the oven to work out the number of pans(N ).Assume that the side length of a regular hexagon is a ,the length-width ratio of the oven is λand L ∆is the increment in discretization.Because the number of pans can not change continuously when ⋅⋅⋅=+∈3,2,1),1,(m m m n ,the equations would be as follows.⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎧⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⋅−====∆⋅+<⎥⎦⎤⎢⎣⎡∆⋅+−∆⋅+≤⋅=∆∆⋅+=<<=a W L n W L N Lk L W W L k L W L k L a L L k L L L W aW 23,810;23105000λ(6)Result:⎪⎪⎩⎪⎪⎨⎧⋅⋅⋅==⋅+=⋅⋅⋅=−=+−⋅+=3,2,1,2233,2,1,1212130201k k n n N N k k n n N N Where 1N represents the number of pans when n is odd,2N represents the number of pans when n is even.The specific number of pans is depended on the width-length ratio of oven.The second part.Maximum the heat distribution of the pans.We define the average heat(H )as the ratio of total heat and total area of the pans.Aiming to get the most average heat,we set the width-length ratio of the oven λ.Space utilization is not considered here.A conclusion is easy to draw from Figure 8that a square area in the oven from 150mm to 350mm in length shares the most heat evenly.So the pans should beplaced mainly in this area.From model1we know that the corners of the oven are apt to gather heat.Besides,four more pans are added in the corners to absorb more heat. Because heat absorbing is the only aim,there is no need to consider space utilization. Circular pans can distribute heat more evenly than any other shape due to model1.So circular pans are used in Figure17.Figure8.Figure17.We set the heat of the pans in the most heated area(the middle row)as Q.Pans in the corners receive more heat but uneven theoretically.And the square of the four pans in the corners is so small compared with the total square that we set the heat of the four as Q too.When the length of oven(L)increases,the number of pans increases too. It makes the square of the gaps between pans bigger,meanwhile.If each pan has a same radius(r)and square(A),the equation about average heat,length-width ratio and number of pans would be(7).⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎧=+=⎥⎦⎤⎢⎣⎡⋅−====∆⋅+<⎦⎤⎢⎣⎡∆⋅+−∆⋅+≤⋅=∆∆⋅+=<<⋅=⋅⋅=...3,2,12;71021053410002k nN N r W L n W L N L k L W W L k L W L k L r L L k L L L W r A W r λπ(7)Here we get the most average heat (H ):29400WQ H ⋅=πThe third part.We discussed two different plans in the previous parts of the paper.One is aimed to get the most average heat,while the other aimed to place the most pans.The two plans are contradictory with each other,and can not be achieved together.Firstly,the weight of plan 1is P and the weight of plan 2is P −1.Obviously,this kind optimization has difficulty in solving and understanding.So we turn to another way to make it a easier and linear question.It has been set that the width of the oven is a constant W and there should be three pans at most in vertical direction.We make the weight P a proportion of the two plans.Thus the two plans could be achieved together due to proportion P and P −1,as shown in Figure18.Figure 18.As been told in model 1,the corners have a higher temperature than other parts of the oven.So plan 1is used in district 1(in Figure 10)and plan 2is used in district 2(in Figure 10).A better compromise could be reached in this way,as shown in Figure 19.Figure 19.Every pan has a square of A .Radius of circular ones is r .Side length of regular hexagon is a .1.1:23322=⇒⋅=⋅r a a r π(8)Based on the equation (8),if the pans are placed as shown in Figure 19,regular hexagons are placed full of district 1,the circular ones will be placed beyond the border line.If the circular ones are placed full of district 2,there will be more gaps in district 1,which will be wasted.So we change our plan of placing pans as Figure 20.Figure 20.The number of circular ones decreases by two,but the space in district 1is fully used,and no pan will be placed beyond the borderline.We assume that P is bigger than P −1,so that,the heat in district 1will be fully used.By simple calculating,we know that the ratio of the heat absorbed in circular pan (1H )and in regular hexagon (2H )is 1.2:1.Figure21.So,based on the pans placing plan,a equation on heat can be got as follow:⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎧=⎥⎦⎤⎢⎣⎡−⋅−=⎥⎦⎤⎢⎣⎡−⋅====∆⋅+<⎥⎦⎤⎢⎣⎡∆⋅+−∆⋅+≤=∆∆⋅+=<<⋅=⋅==≈...3,2,1)1(,911233;23212kxWLPnxWLPnWLNLkLWWLkLWLkLxLLkLLLWraAxraλπ(9)Resolution:⎪⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎪⎨⎧=⋅⋅+⋅+⋅⋅+=⋅⋅+⋅+⋅⋅−+==⋅+⋅+=−=+−⋅+⋅+=...3,2,1)24(2.1)325()24(2.1)3215()2(232)12(12132221212111121211201k A N n Q Q n H A N n Q Q n H k n n n N N k n n n N N (10)1N and 1H means the number of pans and average heat absorbed when n is odd.2N and 2H means the number of pans and average heat absorbed when n is even.For example:(1)When 37.0=λ,6.0=P :16=N ,AQ H 075.1=.The best placing plan is:(2)When 37.0=λ,7.0=P :18=N ,AQ H ⋅=044.1.The best placing planis:(3)When 58.0=λ,6.0=P ：12=N ,AQ H ⋅=067.1.The best placing plan is:A conclusion is easy to draw that when the ratio of width and length of the oven (λ)is a constant,the number of pans increases with an increasing P,but the average heat decreases (example (1)and (2)).When the weight P is a constant,the number of pans decreases with an increasing λ,and the average heat decreases also.So,the actual plan should be base on your specific needs.ConclusionIn conclusion,our team is very certain that the method we came up with is effective in heat distribution analysis.Based on our model,the more edges the pan has,the more evenly the heat distribute on.With the discretize-convert approach,we know that when the ratio of width and length of the oven (γ)is a constant,the number of pans increases with an increasing P ,but the average heat decreases.When the weight P is a constant,the number of pans decreases with an increasing γ,and the average heat decreases also.So,the actual plan should be base on your specific needs.Strengths &WeaknessesStrengths•Difficulties Avoided Avoided..In model 1,we turn to another way to work simulate the heat distribution instead of work on heat conduction directly.Firstly,we simulate heat radiation not heat conduction to keep away from the sophisticated partial differential equations.Then,we create a unique method to convert every variable from the whole space to section.In other words,we work everything out in heat radiation and convert them into heat contradiction.•Close to Reality.Our model considers both the thermal radiation and surface reflection,which is relatively close to the actual situation.•Flexibility Provided.Our algorithm does a great job in flexibility.The heat distribution map on sections are intercepted from the heat distribution maps of the entire space.All shapes of sections can be used in the algorithm.The heat distribution in the whole space is generated based on the location of the electrical heating tubes and the decay curve of the heat, which can be modified at any time.•Innovation.Based on our model,the space of an oven can be divided into six parts with different hear distribution.In order to make full use of the inner space,we invent a new pan which allows users to cook six different kinds of food at same time.An advertisement is published in the end of the paper.WeaknessesPan’’s Thermal Conductivity Ignored.•PanThe heat comes from not only the electrical heating tubes,but also heat conduction of the pans themselves.But the pan’s thermal conduction is ignored in the model,which may cause little inaccuracy.•Thermal Conductivity of Electrical Heating Tubes IgnoredIgnored..it is assumed that there are two electrical heating tubes in the oven and placed in a specific location.The initial temperature of the tubes is a desired constant temperature. In other words,the time electrical heating tubes spend to heating themselves is ignored.The simplification can cause some inaccuracy.simplification..•Linear simplificationIn model2,the length of the oven is discretized,so that the number of pans will changes linearly.calculating through simple integer linear method.This will lead to the result of our model is not accurate enough.ApplicationWe have discussed the heat distribution in the oven in model1.The heat distributionis shown in figure1and figure2.Figure1Figure2As shown,the edges of the oven are distributed the most heat.Areas on both sides of the,is distributed the least heat.While the middle area absorbs little less than theedges.So,we can separate the oven area into six parts,as shown bellow.Part1and part2are distributed the least heat and located the furthest from the heat source(the electrical heating tubes locate on the bottom of the oven).So these two parts absorb the least heat.Part3and part4are distributed the least heat but locating the nearest to the heat source.Part5located far from the bottom but distributed the most heat.So simply,we regard the heat of part3,part4and part5as the same.Part6 is distributed the most heat,and locating nearest to the bottom.So,the heat part6 absorbs is the most in the oven.Based on our conclusion above,we invent the iPan,a new combined pan,which can bake three kinds of food at the same time.For example,one wants to have a little bread,pieces of sausage,a chicken wing and a pizza for lunch.He will have to wait 30minutes at least for his lunch,if he just has one oven.As the Chinese saying goes,‘Bear paws and fish never come together’.By using iPan can solve the issue for him,he could put the bread in pan1,pizza in pan2,sausage in pan5and chicken wing in pan6,and power on.Thus,he can have his delicious lunch in at least10minutes.So,bear paws and fish come together.We make an advertisement for Brownie Gourmet Magazine in the end of the paper.Advertising SheetsReferences[1]Heat Radiation,/view/f5ed1619cc7931b765ce1599.html, Page.4[2]G.S.Ranganath,Black-body Radiation,/article/10.1007%2Fs12045-008-0028-7?LI=true#,February, 2013[3]Kaiqing Lu,The Chemical Basis of Heat Transfer,Journal of Higher Correspondence Education(Natural Sciences Edition),Vol.3:p.33,1996[4]Mark M.Meerschaert,Mathematical Modeling(Third Edition),China:China Machine Press,May.2009[5]Jianzhong Zhang,Monte Carlo Method,Mathematics in Practice and Theory,Vol.1p.28,1974[6]Shallow water equations,/wiki/Shallow_water_equations。
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年数学建模一等奖论文
车道被占用对城市道路通行能力的影响模型摘要本文研究在信号灯下游车道被占用后对道路通行能力的影响。
对第1个问题，本文对附件一的视频进行实时数据采集，对通行能力的决定因素进行简化，以事故处横断面的单位时间车流量作为拥堵时刻的道路上的实际通行能。
运用matlab软件对数据进行统计，绘制出事故处横断面的实际通行能力的变化过程。
得出事故处横断面上的实际交通能力受交通信号灯的影响成周期性变化。
对第2个问题，同样本文对附件二的视频进行实时数据采集，绘制出事故处横断面上的实际通行能力的变化过程。
因为两个视频中车道被占用的情况不同，根据附件3的信息，分析出两组数据与实际通行能力变化过程的差异主要与不同车道上的车流量比率有关。
并且在模型改进中，提出了定量分析所占车道的位置与实际通行能力的关系。
在问题3的模型中，本文利用波的生成与传播理论，建立了车流波模型。
因为事故上游的红绿灯的影响，本文所建立的排队长度与实际通行和事故持续时间的函数关系为周期性变化的分段函数，在计算特定时间点的排队长有一定困难，通过运用计算机仿真的办法，编写matlab仿真程序，从而很容易得出特定时间点的排队长度。
在问题4 的模型中，本文通过分析问题4模型与问题3模型的区别，对模型3的车流量与每个周期形成排队的时间做适当的修改，很好的算出了解决了问题4，通过matlab算出经过148s后排队长度到达上游路口。
关键词：交通波模型排队论计算机仿真通行能力一、问题重述车道被占用是指因交通事故、路边停车、占道施工等因素，导致车道或道路横断面通行能力在单位时间内降低的现象。
由于城市道路具有交通流密度大、连续性强等特点，一条车道被占用，也可能降低路段所有车道的通行能力，即使时间短，也可能引起车辆排队，出现交通阻塞。
如处理不当，甚至出现区域性拥堵。
车道被占用的情况种类繁多、复杂，正确估算车道被占用对城市道路通行能力的影响程度，将为交通管理部门正确引导车辆行驶、审批占道施工、设计道路渠化方案、设置路边停车位和设置非港湾式公交车站等提供理论依据。
数学建模美赛一等奖优秀论文
52888
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Fall in love with your bathtub
Abstract It’s pleasant to go home to take a bath with the evenly maintained temperature of hot water throughout the bathtub. This beautiful idea, however, can not be always realized by the constantly falling water temperature. Therefore, people should continually add hot water to keep the temperature even and as close as possible to the initial temperature without wasting too much water. This paper proposes a partial differential equation of the heat conduction of the bath water temperature, and an object programming model. Based on the Analytic Hierarchy Process (AHP) and Technique for Order Preference by Similarity to Ideal Solution (TOPSIS), this paper illustrates the best strategy the person in the bathtub can adopt to satisfy his desires. First, a spatiotemporal partial differential equation model of the heat conduction of the temperature of the bath water is built. According to the priority, an object programming model is established, which takes the deviation of temperature throughout the bathtub, the deviation of temperature with the initial condition, water consumption, and the times of switching faucet as the four ob jectives. To ensure the top priority objective— homogenization of temperature, the discretization method of the Partial Differential Equation model (PDE) and the analytical analysis are conducted. The simulation and analytical results all imply that the top priority strategy is: The proper motions of the person making the temperature well-distributed throughout the bathtub. Therefore, the Partial Differential Equation model (PDE) can be simplified to the ordinary differential equation model. Second, the weights for the remaining three objectives are determined based on the tolerance of temperature and the hobby of the person by applying Analytic Hierarchy Process (AHP) and Technique for Order Preference by Similarity to Ideal Solution (TOPSIS). Therefore, the evaluation model of the synthesis score of the strategy is proposed to determine the best one the person in the bathtub can adopt. For example, keeping the temperature as close as the initial condition results in the fewer number of switching faucet while attention to water consumption gives rise to the more number. Third, the paper conducts the analysis of the diverse parameters in the model to determine the best strategy, respectively, by controlling the other parameters constantly, and adjusting the parameters of the volume, shape of the bathtub and the shape, volume, temperature and the motions and other parameters of the person in turns. All results indicate that the differential model and the evaluation model developed in this paper depends upon the parameters therein. When considering the usage of a bubble bath additive, it is equal to be the obstruction between water and air. Our results show that this strategy can reduce the dropping rate of the temperature effectively, and require fewer number of switching. The surface area and heat transfer coefficient can be increased because of the motions of the person in the bathtub. Therefore, the deterministic model can be improved as a stochastic one. With the above evaluation model, this paper present the stochastic optimization model to determine the best strategy. Taking the disparity from the initial temperature as the suboptimum objectives, the result of the model reveals that it is very difficult to keep the temperature constant even wasting plentiful hot
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Keywords: Brownie Pan Heat Conduction Equation ANSYS
Temperature Variance Combinatorial Optimization Model
Team # 22063
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Contents
I. Introduction....................................................................................................................................3 II. Models.......................................................................................................................................... 3
Team # 22063
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Team Control Number
22063
Problem Chosen
A
1. Model 1..................................................................................................................................3 1.1 Symbols and Definitions............................................................................................. 3 1.2 Assumptions................................................................................................................ 4 1.3 The foundation of model 1.......................................................................................... 4 1.4 Solution to the problem............................................................................................... 5 1.5 Verification with simulation diagram..........................................................................7 1.6 Strength and Weakness................................................................................................9 1.7 Conclusion:..................................................................................................................9
The Ultimate Brownie Pan
Abstract
Brownie Cake is famous for its fantastic taste, but how can we cook Brownie Cake better? Apparently, it needs to be baked in a suitable pan. Now, let’s take a close look at Ultimate Brownie Pan designed in this article.
2. Model 2..................................................................................................................................9 2.1 Symbols and Definitions............................................................................................. 9 2.2 Assumptions.............................................................................................................. 10 2.3 The foundation of model 2........................................................................................ 10 2.4 Solution to the problem............................................................................................. 12 2.5 Result and Analysis to the problem........................................................................... 12 2.6 Strength and Weakness..............................................................................................18 2.7 Conclusion.................................................................................................................18
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2013 Mathematical Contest in Modeling (MCM) Summary Sheet
In order to study the best shape of pan under given conditions, model 2 is a mixed one of two-dimensional packing model and combination optimization model. Temperature variance of pan and utilization radio of area are selected to quantize two indexes given.Six basic shapes of pans are discussed.We get the best appropriate pan in the different shapes. In particular, a curve have been fitted to reflect which shape of pan is suitable for the oven under certain condition.
Teuction
Product gets overcooked at the corners when baked in a rectangular pan, but not overcooked at the edges when baked in a round pan. However, using round pan is not efficient with respect to using space in a rectangular oven. So it is necessary to design pans of new shape satisfied for maximizing both even distribution of heat for the pan and number of pans that can fit in the oven.
IV A letter........................................................................................................................................20 V References................................................................................................................................... 21