美国大学生数学建模特等奖论文全集F43181
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!。
数学建模美赛一等奖优秀论文
52888
For office use only F1 ________________ F2 ________________ F3 ________________ F4 ________________
Team #52888
Page 1 of 23
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
2011年美国数学建模大赛一等奖获奖论文
Team #10076
Page 2 of 21
Contents
1 Introduction 2 Restatement of Problem 3 Basic Assumptions 4 Analysis of the Problem 4.1 Definitions of Zero-Potential Surface and “vertical 4.2 Turn Of Snowboarding . . . . . . . . . . . . . . . 4.3 Snowboarding on the Deck . . . . . . . . . . . . . 4.4 Snowboarding on the Quarter Arc . . . . . . . . . 5 “Vertical air” Model 6 Results and Sensitivity Analysis 6.1 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Model Applications And Analysis 7.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Maximum Twist Model 8.1 Analysis of Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Results and Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . 9 Tradeoffs to Develop a ”Practical” Course 9.1 Halfpipe Framework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Snow in the Halfpipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 4 6 6 7 10 10 14 16 16 16 17 17 17 18 18 18 19 20 20 20
美国大学生数学建模比赛的论文格式
ContentsⅠIntroduction (1)1.1Problem Background (1)1.2Previous Research (2)1.3Our Work (2)ⅡGeneral Assumptions (3)ⅢNotations and Symbol Description (3)3.1 Notations (4)3.2 Symbol Description (4)ⅣSpread of Ebola (5)4.1 Traditional Epidemic Model (5)4.1.1.The SEIR Model (5)4.1.2 (6)4.1.3 (6)4.2 Improved Model (7)4.2.1.The SEIHCR Model (8)4.2.2 (9)ⅤPharmaceutical Intervention (9)5.1 Total Quantity of the Medicine (10)5.1.1.Results from WHO Statistics (10)5.1.2.Results from SEIHCR Model (11)5.2 Delivery System (12)5.2.1.Locations of Delivery (13)5.2.2 (14)5.3 Speed of Manufacturing (15)ⅥOther Important Interventions (16)6.1 Safer Treatment of Corpses (17)6.2 Conclusion (18)ⅦControl and Eradication of Ebola (19)7.1 How Ebola Can Be Controlled (20)7.2 When Ebola Will Be Eradicated (21)ⅧSensitivity Analysis (22)8.1 Impact of Transmission Rate (23)8.2 Impact of the Incubation Priod (24)ⅨStrengths and Weaknesses (25)9.1 Strengths (26)9.2 Weaknesses (27)9.3 Future Work (28)Letter to the World Medical Association (30)References (31)ⅠIntroduction1.1.Promblem Background1.2.Previous Research1.3.Our WorkⅡGeneral Assumptions●●ⅢNotations and Symbol Description3.1. Notataions3.2. Symbol DescriptionSymbol DescriptionⅣSpread of Ebola4.1. Traditional Epidemic Model4.1.1. The SEIR Model4.1.2. Outbreak Data4.1.3. Reslts of the SEIR Model4.2. Improved Model4.2.1. The SEIHCR Model4.2.2. Choosing paametersⅤPharmaceutical Intervention 5.1. Total Quantity of the Medicine 5.1.1. Results from WHO Statistics5.2. Delivery System5.2.1. Locations of Delivery5.2.2. Amount of Delivery5.3. Speed of Manufacturong5.4. Medicine EfficacyⅥOther Important Interventions 6.1. Safer Treatment of Corpses6.2. ConclusionⅦControl and Eradication of Ebola 7.1. How Ebola Can Be Controlled7.2. When Ebola Will Be EradicatedⅧSensitivity Analysis8.1. Impact of Transmission Rate8.2. Impact of Incubation PeriodⅨStrengths and Weaknesses 9.1. Strengths●●●9.2. Weaknesses●●●9.3.Future WorkLetter to the World Medical AssociationTo whom it may concern,Best regards,Team #32150References ［1］［2］［3］［4］。
建模美赛获奖范文
建模美赛获奖范文标题：《探索与创新：建模美赛获奖作品范文解析》建模美赛（MCM/ICM）是全球大学生数学建模竞赛的盛事，每年都吸引了众多优秀的学生参与。
在这个舞台上，获奖作品往往展现了卓越的数学建模能力、创新思维和问题解决技巧。
本文将解析一份获奖范文，带您领略建模美赛获奖作品的风采。
一、背景与问题阐述（此处详细描述范文所针对的问题背景、研究目的和意义，以及问题的具体阐述。
）二、模型建立与假设1.模型分类与选择：根据问题特点，范文选择了适当的模型进行研究和分析。
2.假设条件：明确列出建模过程中所做的主要假设，并解释其合理性。
三、模型求解与结果分析1.数据收集与处理：介绍范文中所用数据来源、处理方法及有效性验证。
2.模型求解：详细阐述模型的求解过程，包括算法选择、计算步骤等。
3.结果分析：对求解结果进行详细分析，包括图表展示、敏感性分析等。
四、模型优化与拓展1.模型优化：针对原模型存在的问题，范文提出了相应的优化方案。
2.拓展研究：对模型进行拓展，探讨其在其他领域的应用和推广价值。
五、结论与建议1.结论总结：概括范文的研究成果，强调其创新点和贡献。
2.实践意义：分析建模结果在实际问题中的应用价值和意义。
3.建议：针对问题解决，提出具体的建议和措施。
六、获奖亮点与启示1.创新思维：范文在模型选择、求解方法等方面展现出创新性。
2.严谨论证：文章结构清晰，逻辑严密，数据充分，论证有力。
3.团队合作：建模美赛强调团队协作，范文体现了成员间的紧密配合和分工合作。
总结：通过分析这份建模美赛获奖范文，我们可以学到如何从问题背景出发，建立合理的模型，进行严谨的求解和分析，以及如何优化和拓展模型。
同时，也要注重创新思维和团队合作，才能在建模美赛中脱颖而出。
美国大学生数学建模大赛优秀论文一等奖摘要
SummaryChina is the biggest developing country. Whether water is sufficient or not will have a direct impact on the economic development of our country. China's water resources are unevenly distributed. Water resource will critically restrict the sustainable development of China if it can not be properly solved.First, we consider a greater number of Chinese cities so that China is divided into 6 areas. The first model is to predict through division and classification. We predict the total amount of available water resources and actual water usage for each area. And we conclude that risk of water shortage will exist in North China, Northwest China, East China, Northeast China, whereas Southwest China, South China region will be abundant in water resources in 2025.Secondly, we take four measures to solve water scarcity: cross-regional water transfer, desalination, storage, and recycling. The second model mainly uses the multi-objective planning strategy. For inter-regional water strategy, we have made reference to the the strategy of South-to-North Water Transfer[5]and other related strategies, and estimate that the lowest cost of laying the pipeline is about 33.14 billion yuan. The program can transport about 69.723 billion cubic meters water to the North China from the Southwest China region per year. South China to East China water transfer is about 31 billion cubic meters. In addition, we can also build desalination mechanism program in East China and Northeast China, and the program cost about 700 million and can provide 10 billion cubic meters a year.Finally, we enumerate the east China as an example to show model to improve. Other area also can use the same method for water resources management, and deployment. So all regions in the whole China can realize the water resources allocation.In a word, the strong theoretical basis and suitable assumption make our model estimable for further study of China's water resources. Combining this model with more information from the China Statistical Yearbook will maximize the accuracy of our model.。
2016年美国大学生数学建模竞赛题论文
2016
MCM/ICM Summary Sheet (Your team's summary should be included as the first page of your electronic submission.) Type a summary of your results on this page. Do not include the name of your school, advisor, or team members on this page.
For office use only T1 ________________ T2 ________________ T3 ________________ T4 ________________
Team Control Number
52557
Problem Chosen
E
For office use only_ F3 ________________ F4 ________________
Team # 52557
i
Contents
1 Introduction ........................................................................................................................................... 1 1.1 Problem Statement..........................................................................................................................1 1.2 Problem Analysis............................................................................................................................1 1.2.1 Task 1 Analysis ........................................................................................................................ 1 1.2.2 Task 2 Analysis ........................................................................................................................ 1 1.2.3 Task 3 Analysis ........................................................................................................................ 2 1.2.4 Task 4 Analysis ........................................................................................................................ 2 1.2.5 Task 4 and 5 Analysis .............................................................................................................. 2
美国大学生数学建模竞赛2013 获奖论文
Team #111111
Page 3 of 22
General Assumptions
The heat can only transfer to the pan from its outer edge through the air. Since the food placed on it prevents the heat from conducting to it, this is a reasonable. The temperature in oven is even since the air is flowing. That there is only one kind of pans in the ovens. Initially there are two racks in the oven, evenly spaced. We suppose that the temperature and heat are equivalent and constant, so we just considerate one rack and the other one is the same with it The ratio of the oven plane’s width and length is W/L. Every pan shares the same area of A. The data we cited in the models are true. The area of the oven is S 750cm 2 , and the ratio between width and length is W / L 22 : 34 . Moreover, the area of pan is A 100cm2 .[2]
2012年美国大学生数学建模竞赛国际一等奖(Meritorious Winner)获奖论文
AbstractFirstly, we analyze the reasons why leaves have various shapes from the perspective of Genetics and Heredity.Secondly, we take shape and phyllotaxy as standards to classify leaves and then innovatively build the Profile-quantitative Model based on five parameters of leaves and Phyllotaxy-quantitative Model based on three types of Phyllotaxy which make the classification standard precise.Thirdly, to find out whether the shape ‘minimize’ the overlapping area, we build the model based on photosynthesis and come to the conclusion that the leaf shape have relation with the overlapping area. Then we use the Profile-quantitative Model to describe the leaf shape and Phyllotaxy-quantitative Model to describe the ‘distribution of leaves’, and use B-P Neural Network to solve the relation. Finally, we find that, when Phyllotaxy is determined, the leaf shape has certain choices.Fourthly, based on Fractal Geometry, we assume that the profile of a leaf is similar to the profile of the tree. Then we build the tree-Profile-quantitative Model, and use SPSS to analyze the parameters between Profile-quantitative Model and tree-Profile-quantitative Model, and finally come to the conclusion that the profile of leaves has strong correlation to that of trees at certain general characteristics.Fifthly, to calculate the total mass of leaves, the key problem is to find a reasonable geometry model through the complex structure of trees. According to the reference, the Fractal theory could be used to find out the relationship between the branches. So we build the Fractal Model and get the relational expression between the mass leaves of a branch and that of the total leaves. To get the relational expression between leaf mass and the size characteristics, the Fractal Model is again used to analyze the relation between branches and trunk. Finally get the relational expression between leaf mass and the size characteristics.Key words：Leaf shape, Profile-quantitative Model, Phyllotaxy-quantitative Model, B-P Neural Network , Fractal,ContentThe Leaves of a Tree ........................................................ 错误！未定义书签。
美国大学生数学建模竞赛二等奖论文
美国⼤学⽣数学建模竞赛⼆等奖论⽂The P roblem of R epeater C oordination SummaryThis paper mainly focuses on exploring an optimization scheme to serve all the users in a certain area with the least repeaters.The model is optimized better through changing the power of a repeater and distributing PL tones,frequency pairs /doc/d7df31738e9951e79b8927b4.html ing symmetry principle of Graph Theory and maximum coverage principle,we get the most reasonable scheme.This scheme can help us solve the problem that where we should put the repeaters in general cases.It can be suitable for the problem of irrigation,the location of lights in a square and so on.We construct two mathematical models(a basic model and an improve model)to get the scheme based on the relationship between variables.In the basic model,we set a function model to solve the problem under a condition that assumed.There are two variables:‘p’(standing for the power of the signals that a repeater transmits)and‘µ’(standing for the density of users of the area)in the function model.Assume‘p’fixed in the basic one.And in this situation,we change the function model to a geometric one to solve this problem.Based on the basic model,considering the two variables in the improve model is more reasonable to most situations.Then the conclusion can be drawn through calculation and MATLAB programming.We analysis and discuss what we can do if we build repeaters in mountainous areas further.Finally,we discuss strengths and weaknesses of our models and make necessary recommendations.Key words:repeater maximum coverage density PL tones MATLABContents1.Introduction (3)2.The Description of the Problem (3)2.1What problems we are confronting (3)2.2What we do to solve these problems (3)3.Models (4)3.1Basic model (4)3.1.1Terms,Definitions,and Symbols (4)3.1.2Assumptions (4)3.1.3The Foundation of Model (4)3.1.4Solution and Result (5)3.1.5Analysis of the Result (8)3.1.6Strength and Weakness (8)3.1.7Some Improvement (9)3.2Improve Model (9)3.2.1Extra Symbols (10)Assumptions (10)3.2.2AdditionalAdditionalAssumptions3.2.3The Foundation of Model (10)3.2.4Solution and Result (10)3.2.5Analysis of the Result (13)3.2.6Strength and Weakness (14)4.Conclusions (14)4.1Conclusions of the problem (14)4.2Methods used in our models (14)4.3Application of our models (14)5.Future Work (14)6.References (17)7.Appendix (17)Ⅰ.IntroductionIn order to indicate the origin of the repeater coordination problem,the following background is worth mentioning.With the development of technology and society,communications technology has become much more important,more and more people are involved in this.In order to ensure the quality of the signals of communication,we need to build repeaters which pick up weak signals,amplify them,and retransmit them on a different frequency.But the price of a repeater is very high.And the unnecessary repeaters will cause not only the waste of money and resources,but also the difficulty of maintenance.So there comes a problem that how to reduce the number of unnecessary repeaters in a region.We try to explore an optimized model in this paper.Ⅱ.The Description of the Problem2.1What problems we are confrontingThe signals transmit in the way of line-of-sight as a result of reducing the loss of the energy. As a result of the obstacles they meet and the natural attenuation itself,the signals will become unavailable.So a repeater which just picks up weak signals,amplifies them,and retransmits them on a different frequency is needed.However,repeaters can interfere with one another unless they are far enough apart or transmit on sufficiently separated frequencies.In addition to geographical separation,the“continuous tone-coded squelch system”(CTCSS),sometimes nicknamed“private line”(PL),technology can be used to mitigate interference.This system associates to each repeater a separate PL tone that is transmitted by all users who wish to communicate through that repeater. The PL tone is like a kind of password.Then determine a user according to the so called password and the specific frequency,in other words a user corresponds a PL tone(password)and a specific frequency.Defects in line-of-sight propagation caused by mountainous areas can also influence the radius.2.2What we do to solve these problemsConsidering the problem we are confronting,the spectrum available is145to148MHz,the transmitter frequency in a repeater is either600kHz above or600kHz below the receiver frequency.That is only5users can communicate with others without interferences when there’s noPL.The situation will be much better once we have PL.However the number of users that a repeater can serve is limited.In addition,in a flat area ,the obstacles such as mountains ,buildings don’t need to be taken into account.Taking the natural attenuation itself is reasonable.Now the most important is the radius that the signals transmit.Reducing the radius is a good way once there are more users.With MATLAB and the method of the coverage in Graph Theory,we solve this problem as follows in this paper.Ⅲ.Models3.1Basic model3.1.1Terms,Definitions,and Symbols3.1.2Assumptions●A user corresponds a PLz tone (password)and a specific frequency.●The users in the area are fixed and they are uniform distribution.●The area that a repeater covers is a regular hexagon.The repeater is in the center of the regular hexagon.●In a flat area ,the obstacles such as mountains ,buildings don’t need to be taken into account.We just take the natural attenuation itself into account.●The power of a repeater is fixed.3.1.3The Foundation of ModelAs the number of PLz tones (password)and frequencies is fixed,and a user corresponds a PLz tone (password)and a specific frequency,we can draw the conclusion that a repeater can serve the limited number of users.Thus it is clear that the number of repeaters we need relates to the density symboldescriptionLfsdfminrpµloss of transmission the distance of transmission operating frequency the number of repeaters that we need the power of the signals that a repeater transmits the density of users of the areaof users of the area.The radius of the area that a repeater covers is also related to the ratio of d and the radius of the circular area.And d is related to the power of a repeater.So we get the model of function()min ,r f p µ=If we ignore the density of users,we can get a Geometric model as follows:In a plane which is extended by regular hexagons whose side length are determined,we move a circle until it covers the least regular hexagons.3.1.4Solution and ResultCalculating the relationship between the radius of the circle and the side length of the regular hexagon.[]()()32.4420lg ()20lg Lfs dB d km f MHz =++In the above formula the unit of ’’is .Lfs dB The unit of ’’is .d Km The unit of ‘‘is .f MHz We can conclude that the loss of transmission of radio is decided by operating frequency and the distance of transmission.When or is as times as its former data,will increase f d 2[]Lfs .6dB Then we will solve the problem by using the formula mentioned above.We have already known the operating frequency is to .According to the 145MHz 148MHz actual situation and some authority material ,we assume a system whose transmit power is and receiver sensitivity is .Thus we can conclude that ()1010dBm mW +106.85dBm ?=.Substituting and to the above formula,we can get the Lfs 106.85dBm ?145MHz 148MHz average distance of transmission .()6.4d km =4mile We can learn the radius of the circle is 40mile .So we can conclude the relationship between the circle and the side length of regular hexagon isR=10d.1)The solution of the modelIn order to cover a certain plane with the least regular hexagons,we connect each regular hexagon as the honeycomb.We use A(standing for a figure)covers B(standing for another figure), only when As don’t overlap each other,the number of As we use is the smallest.Figure1According to the Principle of maximum flow of Graph Theory,the better of the symmetry ofthe honeycomb,the bigger area that it covers(Fig1).When the geometric centers of the circle andthe honeycomb which can extend are at one point,extend the honeycomb.Then we can get Fig2,Fig4:Figure2Fig3demos the evenly distribution of users.Figure4Now prove the circle covers the least regular hexagons.Look at Fig5.If we move the circle slightly as the picture,you can see three more regular hexagons are needed.Figure 52)ResultsThe average distance of transmission of the signals that a repeater transmit is 4miles.1000users can be satisfied with 37repeaters founded.3.1.5Analysis of the Result1)The largest number of users that a repeater can serveA user corresponds a PL and a specific frequency.There are 5wave bands and 54different PL tones available.If we call a code include a PL and a specific frequency,there are 54*5=270codes.However each code in two adjacent regular hexagons shouldn’t be the same in case of interfering with each other.In order to have more code available ,we can distribute every3adjacent regular hexagons 90codes each.And that’s the most optimized,because once any of the three regular hexagons have more codes,it will interfere another one in other regular hexagon.2)Identify the rationality of the basic modelNow we considering the influence of the density of users,according to 1),90*37=3330>1000,so here the number of users have no influence on our model.Our model is rationality.3.1.6Strength and Weakness●Strength:In this paper,we use the model of honeycomb-hexagon structure can maximize the use of resources,avoiding some unnecessary interference effectively.It is much more intuitive once we change the function model to the geometric model.●Weakness:Since each hexagon get too close to another one.Once there are somebuildingsor terrain fluctuations between two repeaters,it can lead to the phenomenon that certain areas will have no signals.In addition,users are distributed evenly is not reasonable.The users are moving,for example some people may get a party.3.1.7Some ImprovementAs we all know,the absolute evenly distribution is not exist.So it is necessary to say something about the normal distribution model.The maximum accommodate number of a repeater is 5*54=270.As for the first model,it is impossible that 270users are communicating in a same repeater.Look at Fig 6.If there are N people in the area 1,the maximum number of the area 2to area 7is 3*(270-N).As 37*90=3330is much larger than 1000,our solution is still reasonable to this model.Figure 63.2Improve Model3.2.1Extra SymbolsSigns and definitions indicated above are still valid.Here are some extra signs and definitions.symboldescription Ra the radius of the circular flat area the side length of a regular hexagon3.2.2Additional AdditionalAssumptionsAssumptions ●The radius that of a repeater covers is adjustable here.●In some limited situations,curved shape is equal to straight line.●Assumptions concerning the anterior process are the same as the Basic Model3.2.3The Foundation of ModelThe same as the Basic Model except that:We only consider one variable(p)in the function model of the basic model ;In this model,we consider two varibles(p and µ)of the function model.3.2.4Solution and Result1)SolutionIf there are 10,000users,the number of regular hexagons that we need is at least ,thus according to the the Principle of maximum flow of Graph Theory,the 10000111.1190=result that we draw needed to be extended further.When the side length of the figure is equal to 7Figure 7regular hexagons,there are 127regular hexagons (Fig 7).Assuming the side length of a regular hexagon is ,then the area of a regular hexagon is a .The area of regular hexagons is equal to a circlewhose radiusis 22a =1000090R.Then according to the formula below:.221000090a R π=We can get.9.5858R a =Mapping with MATLAB as below (Fig 8):Figure 82）Improve the model appropriatelyEnlarge two part of the figure above,we can get two figures below (Fig 9and Fig 10):Figure 9AREAFigure 10Look at the figure above,approximatingAREA a rectangle,then obtaining its area to getthe number of users..The length of the rectangle is approximately equal to the side length of the regular hexagon ,athe width of the rectangle is ,thus the area of AREA is ,then R ?*R awe can get the number of users in AREA is()，2**10000 2.06R a R π=????????9.5858R a =As 2.06<<10,000,2.06can be ignored ,so there is no need to set up a repeater in.There are 6suchareas(92,98,104,110,116,122)that can be ignored.At last,the number of repeaters we should set up is,1276121?=2)Get the side length of the regular hexagon of the improved modelThus we can getmile=km 40 4.1729.5858a == 1.6* 6.675a =3)Calculate the power of a repeaterAccording to the formula[]()()32.4420lg ()20lg Lfs dB d km f MHz =++We get32.4420lg 6.67520lg14592.156Los =++=32.4420lg 6.67520lg14892.334Los =++=So we get106.85-92.156=14.694106.85-92.334=14.516As the result in the basic model,we can get the conclusion the power of a repeater is from 14.694mW to 14.516mW.3.2.5Analysis of the ResultAs 10,000users are much more than 1000users,the distribution of the users is more close toevenly distribution.Thus the model is more reasonable than the basic one.More repeaters are built,the utilization of the outside regular hexagon are higher than the former one.3.2.6Strength and Weakness●Strength:The model is more reasonable than the basic one.●Weakness:Repeaters don’t cover all the area,some places may not receive signals.And thefoundation of this model is based on the evenly distribution of the users in the area,if the situation couldn’t be satisfied,the interference of signals will come out.Ⅳ.Conclusions4.1Conclusions of the problem●Generally speaking,the radius of the area that a repeater covers is4miles in our basic model.●Using the model of honeycomb-hexagon structure can maximize the use of resources,avoiding some unnecessary interference effectively.●The minimum number of repeaters necessary to accommodate1,000simultaneous users is37.The minimum number of repeaters necessary to accommodate10,000simultaneoususers is121.●A repeater's coverage radius relates to external environment such as the density of users andobstacles,and it is also determined by the power of the repeater.4.2Methods used in our models●Analysis the problem with MATLAB●the method of the coverage in Graph Theory4.3Application of our models●Choose the ideal address where we set repeater of the mobile phones.●How to irrigate reasonably in agriculture.●How to distribute the lights and the speakers in squares more reasonably.Ⅴ.Future WorkHow we will do if the area is mountainous?5.1The best position of a repeater is the top of the mountain.As the signals are line-of-sight transmission and reception.We must find a place where the signals can transmit from the repeater to users directly.So the top of the mountain is a good place.5.2In mountainous areas,we must increase the number of repeaters.There are three reasons for this problem.One reason is that there will be more obstacles in the mountainous areas. The signals will be attenuated much more quickly than they transmit in flat area.Another reason is that the signals are line-of-sight transmission and reception,we need more repeaters to satisfy this condition.Then look at Fig11and Fig12,and you will know the third reason.It can be clearly seen that hypotenuse is larger than right-angleFig11edge(R>r).Thus the radius will become smaller.In this case more repeaters are needed.Fig125.3In mountainous areas,people may mainly settle in the flat area,so the distribution of users isn’t uniform.5.4There are different altitudes in the mountainous areas.So in order to increase the rate of resources utilization,we can set up the repeaters in different altitudes.5.5However,if there are more repeaters,and some of them are on mountains,more money will be/doc/d7df31738e9951e79b8927b4.html munication companies will need a lot of money to build them,repair them when they don’t work well and so on.As a result,the communication costs will be high.What’s worse,there are places where there are many mountains but few persons. Communication companies reluctant to build repeaters there.But unexpected things often happen in these places.When people are in trouble,they couldn’t communicate well with the outside.So in my opinion,the government should take some measures to solve this problem.5.6Another new method is described as follows(Fig13):since the repeater on high mountains can beFig13Seen easily by people,so the tower which used to transmit and receive signals can be shorter.That is to say,the tower on flat areas can be a little taller..Ⅵ.References[1]YU Fei,YANG Lv-xi,"Effective cooperative scheme based on relay selection",SoutheastUniversity,Nanjing,210096,China[2]YANG Ming,ZHAO Xiao-bo,DI Wei-guo,NAN Bing-xin,"Call Admission Control Policy based on Microcellular",College of Electical and Electronic Engineering,Shijiazhuang Railway Institute,Shijiazhuang Heibei050043,China[3]TIAN Zhisheng,"Analysis of Mechanism of CTCSS Modulation",Shenzhen HYT Co,Shenzhen,518057,China[4]SHANGGUAN Shi-qing,XIN Hao-ran,"Mathematical Modeling in Bass Station Site Selectionwith Lingo Software",China University of Mining And Technology SRES,Xuzhou;Shandong Finance Institute,Jinan Shandon,250014[5]Leif J.Harcke,Kenneth S.Dueker,and David B.Leeson,"Frequency Coordination in the AmateurRadio Emergency ServiceⅦ.AppendixWe use MATLAB to get these pictures,the code is as follows:1-clc;clear all;2-r=1;3-rc=0.7;4-figure;5-axis square6-hold on;7-A=pi/3*[0:6];8-aa=linspace(0,pi*2,80);9-plot(r*exp(i*A),'k','linewidth',2);10-g1=fill(real(r*exp(i*A)),imag(r*exp(i*A)),'k');11-set(g1,'FaceColor',[1,0.5,0])12-g2=fill(real(rc*exp(i*aa)),imag(rc*exp(i*aa)),'k');13-set(g2,'FaceColor',[1,0.5,0],'edgecolor',[1,0.5,0],'EraseMode','x0r')14-text(0,0,'1','fontsize',10);15-Z=0;16-At=pi/6;17-RA=-pi/2;18-N=1;At=-pi/2-pi/3*[0:6];19-for k=1:2;20-Z=Z+sqrt(3)*r*exp(i*pi/6);21-for pp=1:6;22-for p=1:k;23-N=N+1;24-zp=Z+r*exp(i*A);25-zr=Z+rc*exp(i*aa);26-g1=fill(real(zp),imag(zp),'k');27-set(g1,'FaceColor',[1,0.5,0],'edgecolor',[1,0,0]);28-g2=fill(real(zr),imag(zr),'k');29-set(g2,'FaceColor',[1,0.5,0],'edgecolor',[1,0.5,0],'EraseMode',xor';30-text(real(Z),imag(Z),num2str(N),'fontsize',10);31-Z=Z+sqrt(3)*r*exp(i*At(pp));32-end33-end34-end35-ezplot('x^2+y^2=25',[-5,5]);%This is the circular flat area of radius40miles radius 36-xlim([-6,6]*r) 37-ylim([-6.1,6.1]*r)38-axis off;Then change number19”for k=1:2;”to“for k=1:3;”,then we get another picture:Change the original programme number19“for k=1:2;”to“for k=1:4;”,then we get another picture:。
历年美赛数学建模优秀论文大全
2008国际大学生数学建模比赛参赛作品---------WHO所属成员国卫生系统绩效评估作品名称：Less Resources, more outcomes参赛单位：重庆大学参赛时间：2008年2月15日至19日指导老师：何仁斌参赛队员：舒强机械工程学院05级罗双才自动化学院05级黎璨计算机学院05级ContentLess Resources, More Outcomes (4)1. Summary (4)2. Introduction (5)3. Key Terminology (5)4. Choosing output metrics for measuring health care system (5)4.1 Goals of Health Care System (6)4.2 Characteristics of a good health care system (6)4.3 Output metrics for measuring health care system (6)5. Determining the weight of the metrics and data processing (8)5.1 Weights from statistical data (8)5.2 Data processing (9)6. Input and Output of Health Care System (9)6.1 Aspects of Input (10)6.2 Aspects of Output (11)7. Evaluation System I : Absolute Effectiveness of HCS (11)7.1Background (11)7.2Assumptions (11)7.3Two approaches for evaluation (11)1. Approach A : Weighted Average Evaluation Based Model (11)2. Approach B: Fuzzy Comprehensive Evaluation Based Model [19][20] (12)7.4 Applying the Evaluation of Absolute Effectiveness Method (14)8. Evaluation system II: Relative Effectiveness of HCS (16)8.1 Only output doesn’t work (16)8.2 Assumptions (16)8.3 Constructing the Model (16)8.4 Applying the Evaluation of Relative Effectiveness Method (17)9. EAE VS ERE: which is better? (17)9.1 USA VS Norway (18)9.2 USA VS Pakistan (18)10. Less Resources, more outcomes (19)10.1Multiple Logistic Regression Model (19)10.1.1 Output as function of Input (19)10.1.2Assumptions (19)10.1.3Constructing the model (19)10.1.4. Estimation of parameters (20)10.1.5How the six metrics influence the outcomes? (20)10.2 Taking USA into consideration (22)10.2.1Assumptions (22)10.2.2 Allocation Coefficient (22)10.3 Scenario 1: Less expenditure to achieve the same goal (24)10.3.1 Objective function： (24)10.3.2 Constraints (25)10.3.3 Optimization model 1 (25)10.3.4 Solutions of the model (25)10.4. Scenario2: More outcomes with the same expenditure (26)10.4.1Objective function (26)10.4.2Constraints (26)10.4.3 Optimization model 2 (26)10.4.4Solutions to the model (27)15. Strengths and Weaknesses (27)Strengths (27)Weaknesses (27)16. References (28)Less Resources, More Outcomes1. SummaryIn this paper, we regard the health care system (HCS) as a system with input and output, representing total expenditure on health and its goal attainment respectively. Our goal is to minimize the total expenditure on health to archive the same or maximize the attainment under given expenditure.First, five output metrics and six input metrics are specified. Output metrics are overall level of health, distribution of health in the population,etc. Input metrics are physician density per 1000 population, private prepaid plans as % private expenditure on health, etc.Second, to evaluate the effectiveness of HCS, two evaluation systems are employed in this paper:●Evaluation of Absolute Effectiveness(EAE)This evaluation system only deals with the output of HCS，and we define Absolute Total Score (ATS) to quantify the effectiveness. During the evaluation process, weighted average sum of the five output metrics is defined as ATS, and the fuzzy theory is also employed to help assess HCS.●Evaluation of Relative Effectiveness(ERE)This evaluation system deals with the output as well as its input, and also we define Relative Total Score (RTS) to quantify the effectiveness. The measurement to ATS is units of output produced by unit of input.Applying the two kinds of evaluation system to evaluate HCS of 34 countries (USA included), we can find some countries which rank in a higher position in EAE get a relatively lower rank in ERE, such as Norway and USA, indicating that their HCS should have been able to archive more under their current resources .Therefore, taking USA into consideration, we try to explore how the input influences the output and archive the goal: less input, more output. Then three models are constructed to our goal:●Multiple Logistic RegressionWe model the output as function of input by the logistic equation. In more detains, we model ATS (output) as the function of total expenditure on health system. By curve fitting, we estimate the parameters in logistic equation, and statistical test presents us a satisfactory result.●Linear Optimization Model on minimizing the total expenditure on healthWe try to minimize the total expenditure and at the same time archive the same, that is to get a ATS of 0.8116. We employ software to solve the model, and by the analysis of the results. We cut it to 2023.2 billion dollars, compared to the original data 2109.8 billion dollars.●Linear Optimization Model on maximizing the attainment. We try to maximize the attainment (absolute total score) under the same total expenditure in2007.And we optimize the ATS to 0.8823, compared to the original data 0.8116.Finally, we discuss strengths and weaknesses of our models and make necessary recommendations to the policy-makers。
美国大学生数学建模大赛模拟1论文
生猪年末存栏量及猪肉价格 周期性波动研究
摘要
本文是关于时间序列数据的预测探究的，我们利用灰色系统理论和自己建立 的模拟函数对生猪的年末存栏量和 36 个大中城市的猪肉价格进行了模拟和预 测。 在对第一问生猪年末存栏量的预测中，由于 1997 年以前的统计数据不真实， 从而导致可用信息贫乏，据此特点我们采用了灰色系统理论中的 GM(1,1)模型。 通过对已知数据和生猪养殖业的分析我们发现生猪养殖随供求关系的变化呈周 期性的波动， 波动周期为 3 至 4 年，又鉴于我国经济社会和人民生活水平的不断 发展，生猪养殖业总体上呈现上升趋势。GM(1,1)并无对周期性变化数据的预测 能力， 但对单调趋势的小信息量数据有较好的预测能力，据此我们剔除数据中的 波动成分，即选取原始序列时每隔 3 年取一点，得出 2010 的生猪存栏量为 44770.6 （万头） 。 最后对预测结果的检验发现其与原数据的偏差不会超过 3.03%， 从而保证了预测结果的可靠性。 利用生猪年末存栏量随时间增长但呈现周期性波 动的特性， 我们还建立了与之适应的模拟函数模型，模拟函数包括由一次函数表 征的增长部分、由正弦函数表征的波动部分和由正态函数表征的冲击部分（如 03 年和 06 年生猪养殖业遭到了重大变故，反应在存栏量上有很大波动） 。以最 小二乘法确定此模拟函数的各项参数，进而预测出 2010 年末的生猪存栏量为 46813 万头。由于模拟函数是在机理分析的基础上建立的，所以很好的吻合了已 知数据， 从而对未来数据的预测也就有了保证。两种预测模型预测结果还是有较 大差异的，但我们认为机理分析在预测中是更有效的方式，所以更倾向于把 46813 万头作为最终预测结果。 第二问要求对 36 个大中城市的猪肉价格做出预测。通过对猪肉市场和已有 数据的分析研究， 我们发现猪肉价格在总体上仍然呈现 3 到 4 年为一个周期的波 动，而在每一年中，猪肉价格受节假日、经济波动和病疫等诸多因素的影响，呈 现小幅度短周期的波动情况。 鉴于此变化规律，我们构造的模拟函数包括表示初 始价格的常数、 表示总体波动的周期为 4 年的正弦函数和表示众多繁杂因素影响 的傅里叶级数（出于计算考虑取级数前 50 项） 。通过最小二乘法确定各参数，得 出的模拟函数对已知数据完全贴合。在模型检验中，我们只用部分已知数据来确 定模拟函数， 发现模拟结果与未使用的数据也有很好的吻合程度。 为了进一步检 验模型我们查阅了新的猪肉价格信息发现其与预测价格有相近的走势。 这些都验 证了模拟方法的正确性和模拟函数的有效性。
2014美国大学生数学建模特等奖优秀论文
Page 1 of 25
Best all time college coach Summary
In order to select the “best all time college coach” in the last century fairly, We take selecting the best male basketball coach as an example, and establish the TOPSIS sort - Comprehensive Evaluation improved model based on entropy and Analytical Hierarchy Process. The model mainly analyzed such indicators as winning rate, coaching time, the time of winning the championship, the number of races and the ability to perceive .Firstly ， Analytical Hierarchy Process and Entropy are integratively utilized to determine the index weights of the selecting indicators Secondly，Standardized matrix and parameter matrix are combined to construct the weighted standardized decision matrix. Finally, we can get the college men's basketball composite score, namely the order of male basketball coaches, which is shown in Table 7. Adolph Rupp and Mark Few are the last century and this century's "best all time college coach" respectively. It is realistic. The rank of college coaches can be clearly determined through this method. Next, ANOVA shows that the scores of last century’s coaches and this century’s coaches have significant difference, which demonstrates that time line horizon exerts influence upon the evaluation and gender factor has no significant influence on coaches’ score. The assessment model, therefore, can be applied to both male and female coaches. Nevertheless, based on this, we have drawn coaches’ coaching ability distributing diagram under ideal situation and non-ideal situation according to the data we have found, through which we get that if time line horizon is chosen reasonably, it will not affect the selecting results. In this problem, the time line horizon of the year 2000 will not influence the selecting results. Furthermore, we put the data of the three types of sports, which have been found by us, into the above Model, and get the top 5 coaches of the three sports, which are illustrated in Table10, Table 11, Table12 and Table13 respectively. These results are compared with the results on the Internet[7], so as to examine the reasonableness of our results. We choose the sports randomly which undoubtedly shows that our model can be applied in general across both genders and all possible sports. At the same time, it also shows the practicality and effectiveness of our model. Finally, we have prepared a 1-2 page article for Sports Illustrated that explains our results and includes a non-technical explanation of our mathematical model that sports fans will understand. Key words: TOPSIS Improved Model; Entropy; Analytical Hierarchy Process; Comprehensive Evaluation Model; ANOVA
2009美国大学生数学建模竞赛特等奖论文全集
163
On
Jargon
Ptolemyto
Fourier:
Epicycles
Fawaz
Hjouj
...............................................................169
Strivingfor
Balance:
Why
ReintroducingMore
Speciesto
Fish
Farm
EcosystemYields
BiggerProfits
Sean
Clement,
Timothy
Newlin,
andJoseph
Lucas
and
empowerone’s
life.
I
also
Байду номын сангаасagree
with
Dudley
that
even
though
some
students
donotfullyunderstandtheconceptsandtheoriesofcalculus,itshouldstillbetaught—and
Associate
Editors
Don
AdolphsonChris
ArneyAaron
Archer
Ron
Barnes
Arthur
BenjaminRobert
Bosch
James
M.
CargalMurray
K.
美国大学生数学建模竞赛优秀论文翻译
优化和评价的收费亭的数量景区简介由於公路出来的第一千九百三十，至今发展十分迅速在全世界逐渐成为骨架的运输系统，以其高速度，承载能力大，运输成本低，具有吸引力的旅游方便，减少交通堵塞。
以下的快速传播的公路，相应的管理收费站设置支付和公路条件的改善公路和收费广场。
然而，随着越来越多的人口密度和产业基地，公路如花园州公园大道的经验严重交通挤塞收费广场在高峰时间。
事实上，这是共同经历长时间的延误甚至在非赶这两小时收费广场。
在进入收费广场的车流量，球迷的较大的收费亭的数量，而当离开收费广场，川流不息的车辆需挤缩到的车道数的数量相等的车道收费广场前。
因此，当交通繁忙时，拥堵现象发生在从收费广场。
当交通非常拥挤，阻塞也会在进入收费广场因为所需要的时间为每个车辆付通行费。
因此，这是可取的，以尽量减少车辆烦恼限制数额收费广场引起的交通混乱。
良好的设计，这些系统可以产生重大影响的有效利用的基础设施，并有助于提高居民的生活水平。
通常，一个更大的收费亭的数量提供的数量比进入收费广场的道路。
事实上，高速公路收费广场和停车场出入口广场构成了一个独特的类型的运输系统，需要具体分析时，试图了解他们的工作和他们之间的互动与其他巷道组成部分。
一方面，这些设施是一个最有效的手段收集用户收费或者停车服务或对道路，桥梁，隧道。
另一方面，收费广场产生不利影响的吞吐量或设施的服务能力。
收费广场的不利影响是特别明显时，通常是重交通。
其目标模式是保证收费广场可以处理交通流没有任何问题。
车辆安全通行费广场也是一个重要的问题，如无障碍的收费广场。
封锁交通流应尽量避免。
模型的目标是确定最优的收费亭的数量的基础上进行合理的优化准则。
主要原因是拥挤的随着经济的发展，交通系统逐渐形成和完善自己。
不同种类的车辆已迅速改善的数量，质量，速度，和类型。
为了支付维修费用的高速公路，收费站系统的建立。
然而，费时费给我们带来的拥塞，高度增加烦恼的司机。
一般来说，在收费亭的数量大于数量的车道。
美国数学建模竞赛成绩揭晓
2021美国数学建模竞赛成绩揭晓近日，2021美国大学生数学建模竞赛成绩揭晓。
西交利物浦大学92个队参赛共获一等奖4个、二等奖21个、三等奖67个，获奖总数再创新高。
值得一提的是，今年西交利物浦大学首次有7个队参加穿插学科建模竞赛〔ICM〕，该赛事要求参赛者具有运用数学模型解决跨学科问题的才能，同时对利用计算机处理大批量信息的才能也有着更高的要求。
最终西交利物浦大学有6支参赛队伍获得ICM类二等奖，该获奖比例在全国各高校中名列前茅。
美国大学生数学建模竞赛〔MCM〕是数学建模领域内的国际性权威赛事，由美国自然基金协会和美国数学应用协会共同主办，美国数学学会、运筹学学会、工业与应用数学学会等多家机构协办。
自1985年以来，美国大学生数学建模竞赛已经成功举办28届，大赛每年吸引了包括哈佛大学、麻省理工学院、北京大学、清华大学等著名高校的优秀学生参与奖项角逐，2021年吸引了来自全球各高校的3000余支队伍参加。
美国大学生数学建模竞赛分一般类型〔MCM〕和穿插学科〔ICM〕两类，今年的赛事于美国东部时间2月9日晚上8点在全球通过网络准时开场。
MCM今年共设两个题目：关于树叶分类与重量计算的A题?TheLeavesofaTree?，关于漂流时间安排与河流容量计算的B题?CampingalongtheBigLongRiver?。
ICM的赛题是C题：从网络通讯记录中找出潜在作案同谋?ModelingforCrimeBusting?。
参赛学生3人组成一队，任选其中一题，利用建立数学模型的方法解决实际问题。
大赛不仅要求参赛选手具备扎实的数学、计算机和论文写作功底，同时对其学术英文表达程度等也提出了相当高的要求。
西交利物浦大学本着“自愿参与、自行组队、不选拔、不排除〞的原那么，积极为学生搭建参与此高程度国际赛事的平台。
“美国数学建模竞赛为学生们提供了一个培养数学学习与应用才能的时机，相较于比赛成绩，我们更看重的是学生在比赛过程中得到的锻炼时机。
美赛特等奖获得者的文章
编者按：国际数学建模竞赛可以说是数学建模竞赛级别最高的赛事，每年有来自全球各著名高校2千多个参赛队参加但获得率只有不到1%的特等奖更是含金量极高的奖励。
特等奖获得者不但向世界展示了大学生自身高超的科研素质，同时也为其所在的学校在世界大学生中闻名。
从下面的文章中，我们可以看到本届竞赛特等奖获得者吉昱茜、卢磊和孟凡东的成长经历，希望他们的经历能为在校北京交通大学学子的成长提供帮助。
我们的数学建模经历2010国际数学建模竞赛特等奖获得者北京交通大学吉昱茜卢磊孟凡东国际大学生数学建模竞赛是世界范围内检验大学生科研素质的竞技平台，每年有世界范围内约2千多个参赛队参赛，特别是由于有世界很多著名学校挑选出来的建模尖子参加，使这个竞赛的挑战性极大。
由于该竞赛的特等奖比例不到全体竞赛队的1%，获得这个奖一直是我们梦寐以求的愿望。
在2010年4月7日，这是我们终生难忘的日子，因为在这一天，我们被告知获得本年度国际大学生数学建模竞赛特等奖，而且我们的论文将发表在国际著名杂志上！得到此消息，我们激动无比，幸福至极！时间有若苍狗流云，白驹过隙。
数学建模竞赛的时间虽然仅有三四天，但是要获得好成绩，在参加比赛的准备时间是一个很漫长而且重要的时段，我们在大学的学习可以说是伴随者数学建模一起成长的。
自2007年我们进入北京交通大学以来，就知到交大的数学建模竞赛活动开展的很好，而且每年在国内外的竞赛中交大的成绩都在前列。
从学校的数学建模网上，我们对数学建模有了初步的了解。
为了也能尽快进入参加数学建模竞赛的行列，我们从大一下学期开始就选修学了学校数学建模I和数学建模II课程。
这些课程的内容不需要太深的数学知识，我们大学一年级就能学懂。
在学习的过程中我们对数学建模产生了很大兴趣，并愿意阅读更多更深的数学建模书籍，同时我们有了参加数学建模竞赛的冲动。
学校每年有4个不同层次的数学建模竞赛，这些竞赛给我们提供了很好的锻炼自己的机会。
出于兴趣和爱好，我们已陆续的参加了2008年的校内建模、2009年全国大学生高教社杯数学建模竞赛、2009年全国电子电工杯数学建模竞赛和2010年国际数学建模竞赛。