2015年美国大学生数学建模竞赛论文之Exactly Search Crashed rashed Airplane

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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。

For office use onlyT1________________ T2________________ T3________________ T4________________Team Control Number 37090Problem ChosenAFor office use onlyF1________________F2________________F3________________F4________________2015 Mathematical Contest in Modeling (MCM) Summary SheetThe advent of licensed Ebola vaccines and drugs delights the whole world while also posing a dilemma of how to allocate the needed quantity among all Ebola outbreaks and deliver them with effectiveness and efficiency.We establish comprehensive Ebola response models in three most suffering countries (Guinea, Liberia and Sierra Leone) including a prediction model generating short-term estimates of the Ebola transmission situations, an allocation-and-delivery model planning the needed quantity of medicines and the optimal delivery route, and a cellular automaton model measuring the effect of effective isolation and treatment. Besides, we also give policy making suggestions to prevent international spread to some unaffected countries.Based on the special characteristic of Ebola, we create a modified SEIR epidemic model with an added intervention factor to stand for the effect of some forms of interventions other than vaccines and drugs. We predict the potential number of future Ebola cases with or without the use of effective medicine and the result also shows that if the transmission trends continue without effective interventions, countries will undergo worse and worse situations In the next model, we first classify all outbreaks into five levels due to the different Ebola case numbers. Then we apply minimum spanning tree method, Monte Carlo method and 0-1 programming to our model to locate an optimal number of medical center and sub-centers in each country aiming to eradicate Ebola. We set one medial center in each country and one more sub-center in Guinea, three more sub-centers in Liberia and four more sub-centers in Sierra Leone. The model also calculates the minimal needed number of vaccines and drugs in every manufacturing cycle.Then, we discuss the effect of isolation and treatment by cellular automaton model and find out that if only effective isolation is conducted, the retarding effect is limited.We present a comprehensive strategy to eradicate Ebola by conducting dynamic models and as time passes, we can update the statistic data to reality which adds accuracy to our models and optimal results.An Optimal Strategy to Eradicate EbolaIntroductionEbola virus disease (EVD) is a severe, often fatal illness in humans. It has become one of the most prevalent and devastating threat for its intense transmission. Since first cases of the current West African epidemic of Ebola virus disease were reported on March 22, 2014, over 20000 new cases have been found and about 9000 patients have died from it. The western Africa areas-Guinea, Liberia and Sierra Leone in particular-are outbreaks that have suffered most [1].With the help of licensed vaccines and drugs, we aim to stop Ebola transmission in affected countries within a short period and prevent international spread. Our objectives are:●to achieve full and fast coverage with vaccines for susceptible individuals and drugs for infectious individuals among three most suffering countries (Guinea, Liberia and Sierra Leone);●to ensure emergency and immediate application of comprehensive Ebola response interventions in countries with an initial case or with localized transmission;●to strengthen preparedness of all countries to rapidly detect and response to an Ebola exposure，especially those sharing land borders with an intense transmission area and those with international transportation hubs[1].For the first objective, we create a comprehensive Ebola response models in those three countries including a prediction model of Ebola transmission, an allocation-and-delivery model for vaccines and drugs used and a cellular automaton model measuring the effect of some crucial interventions. The last two objectives are closely related to policy making and in the following part of our paper we just present detailed information of our models.Basic Assumptions1. A patient can only progress forward through the four states and can never regress(e.g. go from the incubating to the susceptible) or skip a state (e.g. go from the incubating to the recovered state, skipping the infectious state).2.Once recovered from Ebola, an individual will not be infected again in a short time.3.Populations of each country remain the same over the prediction period.4.In absence of licensed vaccines or drugs, some other interventions are used, such as effective isolation for Ebola patients and safe burial protocol.5.When vaccines and drugs are introduced to the prediction model, the incubation period and the effect of interventions other than medicine will not change.6.Building a medical center is at a high cost (e.g. storage facilities of medicines, etc.) and every medical center are capable of delivering all needed medicines.7.We ignore the potential damage to medicines when delivering.8.We calculate the distance between two sites by measuring the spherical distance and ignore the actual traffic situation.9.Once received treatment with licensed drugs, patients will no longer be infectiousindividuals, which also means that we do not take the needed recovery period into account.10.The needed vaccines or drug for an individual is one unit.11.All the data searched from the Internet are of trustworthiness and reliability.Model 1: Prediction ModelWe create a modified SEIR model [2] to estimate the potential number of future Ebola cases in countries with intense and widespread transmission- Guinea, Liberia and Sierra Leone. Not only useful in predicting future situation in absence of any licensed vaccine or drug, the modeling tool also can be used to estimate how control and prevention medicine can slow and eventually stop the epidemic.Terminology and definitionsdays is used from previous study. The resulting distribution has a mean incubation period of 6.3 days [3] and therefore, in our prediction model, patients are assumed to be infectious after a 6.3-day’s incubation period. Besides, in absence of licensed vaccines or drugs, Ebola is a disease with few cases of recovery. Thus, under this situation, we assume the recovery rate is 0.001, which is very close to zero.MethodA frightening characteristic of Ebola virus disease is that it has an incubation period ranging from 2 to 21 days before an individual exposed to the virus who finally become infectious. Thus, we create a SEIR epidemic model tracking individuals through the following four states: susceptible (at risk of contracting the disease), exposed (infected but not yet infectious), infectious (capable of transmitting the disease) and removed (recovered from the disease or dead).Moreover, based on Assumption 4, some forms of interventions other than vaccines and drugs may also reduce the spread of Ebola and death numbers, and therefore we introduce an intervention factor γ as a parameter to measure the effect. In those three intense-transmissioncountries(Guinea, Liberia and Sierra Leone),at least 20% of new Ebola infections occur during traditional burials of deceased Ebola patients when family and community members directly touching or washing the body. By conducting safe burial practice, the number of new Ebola cases may drop remarkably. Moreover, effective isolation with in-time treatment is also of significant importance in reducing transmission and deaths.In our modified SEIR model, we describe the flow of individuals between epidemiological classes as follows.Figure 1 A schematic representation of the flow of individuals between epidemiologicalclassesSusceptible individuals in class S in contact with the virus enter the exposed class E at the per-capita rate (λ-γ), where λ is transmission rate per infectious individual per day and γ is the intervention factor serves to retard the transmission. After undergoing an average incubation period of 1/α days, exposed individuals progress to the infectious class I. Infectious individuals (I) move to the R-class either recover or die at rate (μ+β+γ), where b stands for the recovery rate and d represents the fatality rate. Besides,The transmission process above is modeled by the following differential equation set: ()()()()()()()()()()()(++)()dS S t I t dt dE S t I t t E t dt dI E t I t dt dR I t dtλγλγααμβγμβγ⎧=--⎪⎪⎪=--⎪⎨⎪=-++⎪⎪⎪=⎩ (1.1)We modify SEIR model by adding intervention factor γ.Algorithm1. With known values of parameter α and μ, we solve the differential equation (1.1) by assigning certain value ranges and step values to parameter λ, β, and γ.2. We get the predicted numbers of exposed, infectious and dead individuals and these numbers can be fitted to real data by using the least square method to get the residual errors of each times’ loop iteration.3. By comparison every residual error, we find the least one and we use the corresponding values of parameters in our prediction for further prediction.ResultVia MA TLAB programming, we obtain the optimal values for parameters λ,μ,α,βand γ(Table1)and then get the estimated cumulative number of cases in Guinea, Liberia and Sierra Leone separately(Figure 2, 3 and 4). The result shows that if Ebola transmission trends continue without effective drugs and vaccines, countries will undergo worse and worse situations.Sierra Leone 0.101 0.001 0.1587 0.03 0.02Figure 2 Cumulative numbers of cases in LiberiaFigure 3 Cumulative numbers of cases in GuineaFigure 4 Cumulative numbers of cases in Sierra LeoneStability testDefinition of stabilityAn aggregation of all possible parameters’ values resulting in a downwards trend of the total number of exposed individuals and infectious individuals are defined as the stability range in our model [4].Stability range First, we draw two equations from the differential equation set (1.1):()()()()()()()()dE S t I t t E t dt dI E t I t dtλγααμβγ=--=-++ As ()E t and ()R t is relatively small, we assume that ()1()S t I t =-. Then, we sum the two equations up and get:()()[1()]()()()E I d I t I t I t tλγμβγ+=---+- In order to prevent the spread of Ebola, the total percentage of E(t) and I(t) has to present a decline trend from the first day of taking action with the licensed medicine, which also means()[]0[()]d E I d dt d I t +< . When I(t)=I(0),the inequality is equivalent to(2)()()0I t λμβγλγ-----<As ()0I t ≈, the relationship of parameters λ, μ, β and γ are(2)0λμβγ---<To conclude, the stability range for model one is (2)0λμβγ---<. When parameters’ values satisfy this inequality, the model is of stability.Model 2: Allocation-and-delivery ModelWe create an allocation-and-delivery model for vaccines and drugs used in three most suffering countries (Guinea, Liberia and Sierra Leone) and the optimal strategy is assumed to have significant effect of eradicating Ebola in 180 days.In our allocation-and-delivery model, we set medical centers and sub-centers, which serve to treat Ebola patients, inject vaccines to susceptible individuals and also store needed amount of drugs and vaccines. Besides, countries manufacturing medicines (e.g., America, Canada, etc.) are not where in need of medicines, so we set one medical center to receive drugs and vaccines from the manufacturing country and then delivers the needed amount to every sub-center once a month. For sake of the inconvenience might face when delivering medicines across borders, we model three countries desperately. In another word, we set one medical center in Guinea, one in Liberia and one in Sierra Leone respectively and drugs and vaccines are delivered from every center to the sub-centers within borders.The Figure 5 below demonstrates the model with a hypothetical scenario. The dotted arrow lines show that individuals from every Ebola outbreak (E) will go to the nearest medical center (MC) or sub-center (MSC) for treatment or injection, while the solid arrow lines represent the delivery process of medicines from manufacturing county to each medical center and then to sub-centers.Figure 5 The allocation-and-delivery mode lInstead of building new treating places, we locate our medical centers and sub-centers in some existing Ebola Treating Units (ETUs) [1]. The model shows how we choose from current ETUs, including deciding the optimal number and location.Table 3 existing ETUs their locationTerminology and definitionsGoalWe determine the number and location of medical center and sub-centers on the basis of ● Minimizing the total time-cost that an infectious individual from one outbreak spends on the way to the corresponding medical center or sub-center, while locating those center and sub-centers as few as possible, also means0min N nN ij i o j C d ===∑∑● Minimizing the total distance among one medical center to other sub-centers, also meansmin ()Nij i o D i j =≠∑● Averaging the workloads of medical center and sub-centers, also meansmin N N NSV CV AV =AlgorithmFigure 6 the flow chart for model 2Initialize parameters in previous prediction model●We do not change the value of α and γ used in Model 1.●We have deduced the relationship of parameters λ, μ, β and γ in the stability test of model 1.Estimate daily added number of infectious individualsWe use the prediction model to simulate the situation of daily added number of infectious individuals DI i in 6 months(180 days) for 10 times and choose the worst case(maximal numbers) as the final estimation of daily added number.Build geographical distribution of new added infectious individualsWe categorize all outbreaks into five levels as level I, II, III, IV and V according to the number of confirmed cases and then calculate each level’s probability of a new occurring case. According to the number of new added infectious individuals and the probability of occurring in every outbreak, we build geographical distribution among all outbreaks of new added infectious individuals.Table 5 Outbreaks and classificationSet n from 1 to kWe set n from 1 to k to conduct the process for k times and compare each optimal result as N changes.Locate sub-centers randomlyWe locate sub-centers randomly and for each sub-center, the corresponding outbreaks represent all those outbreaks with a nearer distance to this sub-center compared to others.Calculate total time-costWe define the time-cost as the period that an infectious individual from one outbreak spends on the way to the corresponding medical center or sub-center, and we add up the corresponding distance as the measurement of the time-cost. When calculating the total time-cost, the number of all potential patients is taken into account.Make comparisonWe compare the total time-cost calculated in 400 times’ loop and choose the minimal one as the optimal result.Output optimal n, C n, A V n, CV nLocate medical centerWe calculate the total distance of every medical sub-center to others and locate the one with minimal total distance as the medical center which serve to receive all needed medicine from manufacturing country and deliver the required amount to every sub-center [5].ResultWe locate medical centers and sub-centers separately in three countries as shown in Table 7 and Figure7. We get the different values of indicators (shown in Table 6) and taking total distance and margin distance into account, we choose the optimal number and location of medical sub-centersTable 6 Values of indicatorsTable 7 Location of medical center and sub-centers and their corresponding outbreaksFigure 7 Locations of medical center and sub-centers and the routesWe determine the needed amount of vaccines and drugs.We assume that the successful immune rate is 90%, the recovery rate when drugs are used is 60% and the manufacturing cycle of the licensed drug is 30 days. These rates and cycle-days can be adjusted according to reality. VaccinesIndividuals having received vaccine injection can be protected from being infectious. The larger proportion of population being injected, the lower the transmission rate is. This relationship can be measured as 1'(1)dk λλ=- and we solve this equation and get thenumber of needed vaccines (1k ) is'1dλλ-DrugsPatients will have a higher recovery rate and lower fatality rate. The shorter the course of treatment is, the greater the impact on recovery rate and fatality rate. We rewrite therelationship in mathematic equations as 2'rk D μμ=+or 2'rkDββ=-. Thus, the number of needed drugs (2k ) is (')D r μμ- or (')Drββ- .The resultWe calculate an allocation plan for vaccines and drugs in 6 months and the detailed number are present in table 8 and 9. We can see that the demand for vaccine is much larger than that of drugs because there is a wider range of individuals who need vaccine injections as an effective protection.Table 8 Allocation plan for vaccines in 6 monthsTable 9 Allocation plan for drugs in 6 monthsStability testWe make 10 times’ simulation for the three countries by the following procedures.First, we estimate the needed number of medicines for one month and supply at the first day of that month.Then, we generate added numbers of infectious individuals randomly and calculate the consumed and remaining amount of medicines.Finally, we get the line of daily reaming amount of medicines as shown in Figure 8.-100100200300400500600700Dates u r p l u sFigure 8 Surplus of medicine in Guinea, Liberia and Sierra LeoneThe figures demonstrate that the supply of medicine is sufficient except a small probability (less than 10%) of deficit at the end of the first month. Thus, the model is of high stability.Sensitivity analysisWe have estimated the cumulative number of infectious individuals based on the optimal number and location of medical center and sub-centers in model 2. Then we change the values of parameters to conduct sensitivity analysis. The results are shown in the following table. Table 10 result of sensitivity analysisThe result shows the optimal result will not change unless there is some big fluctuation of parameters’ values. Besides, the fluctuation of transmission rate will result in more significant changes to the number of infectious individuals and therefore, we should put emphasis on the generalization of vaccine injections.Dates u r p l usDates u r p l u sFigure 9 Number of daily added infectious Figure 10 Present number of infectious, exposed individuals in Sirrea,Liberia,Guinea and dead individuals in Sirrea,Liberia,GuineaModel 3: the cellular automaton modelIn model 1, we estimate the transmission trends of Ebola and then in model 2, we measure the trends when licensed vaccines and drugs are used and make an allocation-and-delivery plan of medicines. We now introduce a cellular automaton model to present a clearer dynamic simulation of the spread of Ebola in one area.Cellular automaton[6] is a model in which time, space and other variables are all discrete. lt can be expressed asCA = (Ld, S, N, f)Where Ld represents a d-dimensional cellular spaces and we set d=2, L ×L=1000×1000, S represents all finite discrete set of cell stateN represents t he set of a cell’s eight neighbors’ statef represents the transfer function of one cell and it is expressed as S t+1f(S t,N t)Figure 11 A cell and its eight neighborsThere are five states{S, E, I, Q, D, R} in our model which represent susceptible, exposed, infectious, quarantined, dead and recovered individuals. We assign them as{0, 1, 2, 3, 4, 5}. Initialize all cells state value Si j = 0, which means that all cells are susceptible individuals. We select a proportion of 0.0005’s cells in the cellular spaces randomly and set their state value Si j =2, which represent the initial infectious individuals.From t=0, we scan all cells in the cellular spaces and compare the effect of treatment and isolation. We set three situations as no treatment and no isolation, only isolation but no treatment and both isolation and treatment, and then simulate all these situations.Take the third situation (both isolation and treatment) as an example to show the renewing rules.When Si j=0, we calculate the probability p i j that a single cell C ij become infectious when contacting with its neighbors. Then we judge weather susceptible individuals will become exposed individuals with the probability p i j. If it is not the probability, they remain susceptible individuals.When S ij=1, cell C ij is exposed individuals with a probability of e to become infectious individuals (S ij=2).When S ij=2, cell C ij is infectious individuals with a probability of r1 to be isolated (S ij=3) and a probability of d to dead(S ij=4 and are moved out of the transfer).When S ij=3, cell C ij is quarantined individuals with a probability of r3 to be cured (S ij=5 andare moved out of the transfer because of high immune ability).We update the states of all cells in the cellular spaces at the same time and use the result as the initial state in the next time’s simulation.ResultWe use Matlab to realize a simulation process of 200 days and the following figures show the results.Figure 12No isolation and no treatment2040608010012014016018020020406080100120140160180200204060801001201401601802002040608010012014016018020020406080100120140160180200204060801001201401601802002040608010012014016018020020406080100120140160180200DAY 50DAY 100DAY 150DAY 200Figure 13 Only isolation and no treatmentFigure 14 Both treatment and isolation20406080100120140160180200204060801001201401601802002040608010012014016018020020406080100120140160180200204060801001201401601802002040608010012014016018020020406080100120140160180200204060801001201401601802002040608010012014016018020020406080100120140160180200204060801001201401601802002040608010012014016018020020406080100120140160180200204060801001201401601802002040608010012014016018020020406080100120140160180200DAY 50DAY 100DAY 150DAY 200DAY 50DAY 100DAY 150DAY 200The results shows that the transmission accelerates with no isolation and treatment, while slows down significantly when effective isolation is added. However, simple isolation as intervention cannot stop the spread of Ebola. Only with effective isolation and treatment, the transmission can be limited and the fatality rate is reduced.We use the cellular automaton model to simulate the spread of Ebola in three situations and illustrate that effective isolation and treatment is of significant importance,Sensitivity analysisWe assign different values to parameters λ, 12r r ⨯ and μand simulate the situation of the 100th day. The results are as follows.Figure 15 Result of sensitivity analysisThe figure demonstrates that the model is not sensitive to isolation level while sensitive to r transmission and recovery rate. The results indicate that the eradication of Ebola is rely heavily on the control of transmission and recovery rate. Besides, isolation is more effective with a relatively small scale of infectious individuals.Evaluation of the modelStrengths●The prediction model is a modified one adjusted to the unique characteristic of Ebola and this model is much more suitable for the prediction of Ebola transmission than the traditional SEIR epidemic model.●The allocation-and-delivery model is based on the real location of outbreaks and ETUs, and the resulting locations of medical centers and sub-centers are of high practical value.●The value of parameters in the allocation-and-delivery model is highly adjustable. Policy makers can change the value according to the reality or determined goals and this will not affect the modeling process.●The cellular automaton model presents a brief picture of the transmission trends. The result shows the limited retarding effect of simple isolation and indicates the crucial role of effective vaccines and drugs.Weaknesses●We use previous data and probability distribution to determine the value of some parameters in our model. Maybe they deviate from the current situation.●The models fail to take some emergent cases and their effect into account. For example, we ignore the real traffic situations and potential congestions when delivering medicines.Conclusions●We estimate the transmission trend of Ebola in (Guinea, Liberia and Sierra Leone) and present a comprehensive strategy to eradicate Ebola by planning the allocation and delivery system.●The model also presents the different effect of three kinds of interventions-injecting vaccines, treating with drugs, isolation. The best retarding method is to inject vaccines and treating with drugs can reduce deaths in a short period, while isolation is the least choice in absence of other forms of interventions.●To prevent international transmission to unaffected counties, immediate supply of vaccines and drugs should be delivered to any new initial outbreaks from the nearest available place and all unaffected counties have to establish a full Ebola surveillance preparedness and response plan.References[1] http://www.who.int/en/, Feb 2015[2] Ma J L,Ma Z E.Epidemic threshold condition for seasonally forced SEIR models. Mathematical Bio-sciences and Engineering . 2006[3] Chowell G, Hengartner NW, Castillo-Chavez C, Fenimore PW, Hyman JM. The basic reproductive number of Ebola and the effects of public health measures: the cases of Congo and Uganda. J Theor Biol 2004;229:119-26 [4]Katsuaki Koike,Setsuro Matsuda. New Indices for Characterizing Spatial Models of Ore Deposits by the Use of a Sensitivity Vector and an Influence Factor[J]. Mathematical Geology . 2006 (5)[5] Peter Kovesi.MA TLAB and Octave Functions for Computer Vision and Image Processing. Digital Image Computing:Techniques and Applications . 2012[6] rraga,,J.A.delRio,,L.Alvarez-lcaza.Cellularautomationforonelanetrafficmodeling.Transportatio researchpartC . 2005ReportTo whom it may concern:Ebola virus disease (EVD) are posing a threat to all human beings but the advent of licensed vaccines and drugs enable us to fight with Ebola. We have studied out a comprehensive strategy to stop Ebola transmission in affected countries within a short period and prevent international spread.For those unaffected countries and light Ebola outbreaks, immediate response actions to a new initial case are of significant importance. According to our model, effective isolation and treatment can prevent the widespread transmission of Ebola. Thus, immediate supply of vaccines and drugs should be delivered to any new initial outbreaks from the nearest available place and all unaffected counties have to establish a full Ebola surveillance preparedness and response plan including isolation and treatment of infectious individuals and injection of vaccines to susceptible individuals.For countries with intense and widespread transmission- Guinea, Liberia and Sierra Leone- besides the immediate isolation and treatment, a plan of allocating and delivering medicines is also crucial. We model the potential number of future Ebola cases in these three countries and estimate the goal number of transmission rate, recovery rate and fatality rate with which we can control the spread of Ebola. Meanwhile, we classify all the outbreaks in those three countries according to the number of cumulative confirmed cases. Outbreaks with different level will have a different probability of a new occurring case and we use our model to predict the possible new outbreak.Classification of outbreaksAccording to our prediction, 567 units of drugs and 2069139 units of vaccines are needed in the first manufacturing cycle, and therefore, we model an optimal delivering system with the highest efficiency. For sake of the inconvenience might face when delivering medicines across borders, we model three countries desperately. We set one medical center (MC) and a certain number of medical sub-centers (MSC), in each country which serve to treat Ebola patients, inject vaccines to susceptible individuals and also store needed amount of drugs and vaccines. Besides, the medical center serves to receive drugs and vaccines from the manufacturing country and then delivers the needed amount to every sub-center once a month.。

基于非线性曲线拟合的经纬度测量方法摘要本文首先基于天体物理学知识，构造出地球上某处直杆的影长与时间的函数关系式；然后运用非线性曲线拟合的方法，求解缺省参数，再根据直杆影长的变化规律，推算出测量点的地理位置及所处的日期。

在问题一中，本文以北京时间为参考时间，对地球上某一点处直杆影长的影响因素进行分析，发现其与直杆所处纬度、太阳直射点处纬度、所处时刻及经度等因素有关，结合地理知识构造出影长与影响因素的函数关系式。

在各项参数均已给定的情况下，即可作出题目所要求的影长-时间变化曲线。

对于问题二，本文由附件1给定的时刻及其影长，运用非线性曲线拟合的方法，利用问题一中建立的关系式，将时间与影长作为已知参数，利用lsqcurvefit函数拟合求解经纬度参数。

联系实际，筛选出可能的4个位置，并认为海南省白沙黎族自治县是最有可能的地点。

问题三与问题二基本相似，本文仍然在附件所得的数据基础上进行lsqcurvefit非线性曲线拟合，得到经度、纬度以及赤纬的可行解，根据所求赤纬，通过查表可以得到可能的日期。

由附件2得到3个可能的地点与6个可能的日期，并认为其中新疆维吾尔自治区喀什地区巴楚县是最有可能的地点，5月24日或7月20日是最有可能的日期；由附件3同样得到3个可能的地点与6个可能的日期，认为湖北省十堰市郧西县与陕西省商洛市山阳县均是可能的地点，可能的日期为2月6日或11月6日前后。

对于问题四，首先用MATLAB进行图像处理并得到等时间间隔的图片，然后经过筛选得到21张图片。

经滤镜处理后，由所得帧的图像得到影长与杆长的比例关系，进而得到不同时刻下的影长。

在日期已知的情况下，问题四应用非线性拟合函数fit得到可行解，筛选后得到最可能地点为内蒙古自治区乌兰察布市丰镇市；若未给日期条件，在本题上一问的基础上，将太阳赤纬设为未知，利用fit函数求出可行解，经筛选得到最可能的地点为内蒙古自治区乌兰察布市，日期为6月6日或7月8日，与准确日期相差无几。

比赛时间：美国东部时间：2015年2月5日（星期四）下午8点-2月9日下午8点（共4天）北京时间：2015年2月6日（星期五）上午9点-2月10日上午9点农历：十二月十八～十二月廿二重要说明：●COMAP是所有的规则和政策的最后仲裁者，对不遵循竞赛规则和程序的任何队伍，拥有唯一的自由裁量权，取消参赛资格或拒绝登记。

●评委、竞赛组织者、以及UMAP杂志的编辑拥有最终裁定权。

●如果参赛队伍违反竞赛规则，其指导老师一年内将不能指导其他团队，其所在参赛单位将被处以一年的察看处理。

●如果同一机构第二次被抓到违反规则的队伍，该学校将至少不被允许参加下一年度的赛事。

●以下所有时间都是美国东部时间EST（北京时间比美国东部时间早13个小时）●递交参赛论文后，意味参赛者同意以下条款：⏹论文提交后，出版权归COMAP, Inc所有；⏹COMAP可以使用，编辑，引用和出版论文，用于宣传或任何其他目的，包括在线展示，出版电子版，在UMAP杂志刊登或其他方式，并且没有任何形式的补偿；⏹COMAP可以在没有进一步的通知，许可，或补偿的情形下，使用这次比赛相关材料，团队成员、指导老师的名字，以及和他们的背景资料。

●递交参赛论文后，意味参赛者作出以下承诺：⏹论文中出现的所有的图像，数据，照片，图表，图画，如果未注明，都是由参赛者创建；如果引用其它资源，都在参考文献中列出，并在引用的具体位置标注来源。

⏹不论是直接，还是转述方式的文字引用，都在参考文献中列出，并在引用的具体位置标注来源；直接的文字引用使用引号标注。

比赛之前注册报名1.报名截至时间：2015年2月5日下午2：00 EST。

截止日期后，注册系统将自动关闭，不再接受任何新的注册，没有例外。

2.每支参赛队伍都必须有一位来自参赛机构（institute）的教师担任导师（faculty advisor），不允许学生担任导师。

由指导老师负责为其指导队伍注册报名，每位指导老师可注册的队伍数目没有限制。

1. ICM aims to identify the risk of churn in its early stages, as it is cheaper to gain the loyalty of anemployee early in their carreer rather than have to improve the culture once it has soured. It is more productive to have a motivated workforce from the start rather than having to provide incentives to prevent people from leaving.ICM旨在识别生产处于早期阶段的风险,因为它是便宜的忠诚员工在他们的事业上,而不是改善早期文化一旦恶化。

是更有生产力从一开始就积极的劳动力而不必提供激励措施阻止人们离开。

2.A worker is more likely to churn if he or she was connected to other former employees who have churned. Thus churn seems to diffuse from employee to employee, so identifying those that are likely to churn is valuable information to prevent further churning.一个工人更有可能生产如果他或她与其他前员工有搅拌。

因此从员工流失似乎弥漫员工,因此识别那些可能产生有价值的信息防止进一步的生产。

3.One HR issue is matching employees to the right position such that their knowledge and abilities can be maximized. Currently each employee gets an annual evaluation based on performance as judged by the supervisor. These ratings are currently not used by the HR office.一个人力资源问题是匹配到正确的位置,这样员工知识和能力可以最大化。

信息学院师生在美国大学生数学建模竞赛中再创佳绩
2015年美国大学生MCM/ICM（数学建模/交叉学科建模）竞赛于美国东部时间2015年2月5日至2月9日举行，本次比赛吸引了全球十几个国家的41915支队伍参加，获奖结果日前揭晓。

我校信息学院经过精心组织、培训和指导，本科组和研究生组都取得优异成绩，研究生组5支参赛队伍中，1支参赛队获一等奖、1支参赛队获二等奖、3支参赛队获成功参赛奖，本科生组11支参赛队队伍中，4支参赛队获二等奖、7支参赛队获成功参赛奖。

云南大学是云南参赛高校中唯一获得一等奖的学校，并在各层次奖项，较其他高校都有很大优势。

美国大学生数学建模竞赛是培养大学生创新能力和实践能力的一项重要赛事活动，一直受到各高校的高度重视。

近年来，信息学院以《数学建模与数学实验》校级精品课程的建设为依托，至上而下形成了积极参加课外科技实践和创新活动的学术氛围与合力，培养了一支经验丰富的指导教师团队，通过课堂教学和各类选拔赛不断提高学生的数学建模能力，并以此推动教学改革。

教练组全体成员放弃节假日的休息，组织辅导学生参加竞赛，为我校大学生数学建模能力的培养和提高付出了辛勤的劳动，我们向他们表示热烈的祝贺和衷心的感谢！
感谢学校领导和教务处、学生处、研工部、校团委等部门对信息学院长期以来的大力支持和帮助，特别是学生公寓管理中心的领导和值班老师，在寒假期间为学生成功参赛提供的优质服务，再次表示感谢！
信息学院
2015.4.13。

太阳影子定位摘要太阳与地球的运转规律造就了太阳在地球上的阴影规律，本文将根据其规律，通过太阳的变化确定阴影的位置。

本文问题探究由浅到深，最终可通过视频中的阴影判断出视频的拍摄位置和拍摄时间。

针对问题1，本文基于对太阳与地球的运转规律和太阳光在地球上的阴影变化规律分析，考虑到太阳高度角和经纬度及北京时间与当地时间等转换，建立了直杆影子长度和直杆杆长、直杆所在地经纬度、日序数、北京时间之间关系的空间解析几何模型，并最终通过已知数据计算并绘制出直杆在2015年10月22日北京时间9：00-15：00之间天安门广场3米高的直杆影子长度变化曲线。

针对问题2，本文根据问题1得出的影子长度变化规律，将问题转换为寻找最优未知参数集{},,P P H δλ使得所给实测影子长度和理论影长的最小二乘偏差最小。

由于计算的复杂度，我们考虑“大小步长套用搜索”算法并通过合理地分析计算优化了搜索范围，最终通过相应Matlab 程序计算出一组最可能参数集，即最可能地点为东经84.9950, ，南纬4.3170 。

针对问题3，相对问题2增加了未知参数赤纬角，因此利用与问题二类似的思想建立了相应的最小二乘模型，针对附件2和附件3给出的两种不同情况给出了相应的搜索算法，并最终各拟合出两组最可能地点，四个最可能日期，如附件2给出的数据一组最可能的地点为东经79.85, 北纬39.6, 相应日期为5月2日或7月21日。

针对问题4，先对视频进行了去帧和图片的灰度处理，从而提取出了影子的变化数据，推算出了真实的影子变化数据。

进而按照问题一所建立的关系式通过最小二乘法拟合参数。

最后推算出的视频拍摄地点东经为110.48 ，北纬40.245 ，并在拍摄日期未知的情况下对模型进行了验证。

本文严格推导了太阳光阴影变化规律，探究问题层层深入，最终解决了根据视频上的阴影变化确定视频拍摄地点及日期，同时也验证了我们建立的物体影子和物体所在经纬度之间关系的正确性。

Teams (Student or Advisor) are now required to submit an electronic copy (summary sheet and solution) of their solution paper by email too solutions@ as a Word or PDF attachment. Your email MUST be received at COMAP by the submission deadline of 8:00 PM EST, February 9, 2015. Note you will not receive an auto response.Subject lineCOMAP your control numberExample: COMAP 11111Click here to download a PDF of the complete contest instructions.Click here to download a copy of the Summary Sheet in Microsoft Word format.*Be sure to change the control number and problem selected before printing out the page. You must still login to the website and choose your problem.Teams are free to choose between MCM Problem A, MCM Problem B, ICM Problem C or ICM Problem D.COMAP Mirror Site: For more in:/undergraduate/contests/mcm/MCM: The Mathematical Contest in ModelingICM: The Interdisciplinary Contest in Modeling2015 Contest ProblemsMCM PROBLEMSPROBLEM A: Eradicating EbolaThe world medical association has announced that their new medication could stop Ebola and cure patients whose disease is not advanced. Build a realistic, sensible, and useful model that considers not only the spread of the disease, the quantity of the medicine needed, possible feasible delivery systems, locations of delivery, speed of manufacturing of the vaccine or drug, but also any other critical factors your team considers necessary as part of the model to optimize the eradication of Ebola, or at least its current strain. In addition to your modeling approach for the contest, prepare a 1-2 page non-technical letter for the world medical association to use in their announcement. PROBLEM B: Searching for a lost planeRecall the lost Malaysian flight MH370. Build a generic mathematical model that could assist "searchers" in planning a useful search for a lost plane feared to have crashed in open water such as the Atlantic, Pacific, Indian, Southern, or Arctic Ocean while flying from Point A to Point B. Assume that there are no signals from the downed plane. Your model should recognize that there are many different types of planes for which we might be searching and that there are many different types of search planes, often using different electronics or sensors. Additionally, prepare a 1-2 page non-technical paper for the airlines to use in their press conferences concerning their plan for future searches.ICM PROBLEMSPROBLEM C: Managing Human Capital in OrganizationsClick the title below to download a PDF of the 2015 ICM Problem C.Your ICM submission should consist of a 1 page Summary Sheet and your solution cannot exceed 20 pages for a maximum of 21 pages.Managing Human Capital in OrganizationsPROBLEM D: Is it sustainable?Click the title below to download a PDF of the 2015 ICM Problem D.Your ICM submission should consist of a 1 page Summary Sheet and your solution cannot exceed 20 pages for a maximum of 21 pages.Is it sustainable?© 2015 COMAP, The Consortium for Mathematics and Its ApplicationsMay be reproduced for academic/research purposesFor More information on COMAP and this project visit 问题一：根除病毒世界医学协会已经宣布他们的新药物能阻止埃博拉病毒和治愈患者的疾病不先进。

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:。

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。

承诺书我们仔细阅读了五一数学建模联赛的竞赛规则。

我们完全明白，在竞赛开始后参赛队员不能以任何方式（包括电话、电子邮件、网上咨询等）与本队以外的任何人（包括指导教师）研究、讨论与赛题有关的问题。

我们知道，抄袭别人的成果是违反竞赛规则的，如果引用别人的成果或其它公开的资料（包括网上查到的资料），必须按照规定的参考文献的表述方式在正文引用处和参考文献中明确列出。

我们郑重承诺，严格遵守竞赛规则，以保证竞赛的公正、公平性。

如有违反竞赛规则的行为，我们愿意承担由此引起的一切后果。

我们授权五一数学建模联赛赛组委会，可将我们的论文以任何形式进行公开展示（包括进行网上公示，在书籍、期刊和其他媒体进行正式或非正式发表等）。

我们参赛选择的题号为（从A/B/C中选择一项填写）： _________________ B ____________ 我们的参赛报名号为：________________________________________________________________ 参赛组别（研究生或本科或专科）：_____________________________________________________ 所属学校（请填写完整的全名）_______________________________________________________ 参赛队员（打印并签名）：1. _____________________________________________2. ________________________________________3. __________________________________________日期：2015 年5 月 3 日获奖证书邮寄地址：__________________________________邮政编码：—收件人姓名: ____________________________________ 联系电话：__________________编号专用页竞赛评阅编号（由竞赛评委会评阅前进行编号）：评阅记录竞赛评阅编号（由竞赛评委会评阅前进行编号）：参赛队伍的参赛号码：（请各参赛队提前填写好）：摘要本文研究了空气污染问题的主要污染因素及其模型刻画。