2012年美国国际大学生数学建模竞赛(MCM ICM)题目 翻译
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The Leaves of a Tree"How much do the leaves on a tree weigh?" How might one estimate the actual weight of the leaves (or for that matter any other parts of the tree)? How might one classify leaves? Build a mathematical model to describe and classify leaves. Consider and answer the following:• Why do leaves have the various shapes that they have?• Do the shapes “minimize” overlapping individual shadows that are cast, so as to maximize exposure? Does the distribution of leaves within the “volume” of the tree and its branches effect the shape?• Speaking of profiles, is leaf shape (general characteristics) related to tree profile/branching structure?• How would you estimate the leaf mass of a tree? Is there a correlation between the leaf mass and the size characteristics of the tree (height, mass, volume defined by the profile)?In addition to your one page summary sheet prepare a one page letter to an editor of a scientific journal outlining your key findings.Now that all our deciduous trees are in leaf, and roadside ditches and forest floors in the Georgia Piedmont are green with life, it is a marvelous time to take a walk and see how many different leaf shapes you can find. Many shapes are particular to a certain kind of tree, shrub, or vine; some, like that of the maple, have even been celebrated on a country's flag. There are the three jagged leaflets of poison ivy, and five of Virginia creeper; the many lobes (rounded or jagged) of the oaks; the tulip-shaped leaves of the tulip tree tree; the five-pointed star of the sweet gum; and the heart-shaped leaves of the wild yam. Some plants cannot settle upon one leaf shape, but instead have several. Leaves of the sassafras tree can be simple ovals, shaped like mittens, or have three broad, blunt lobes. As it is known to us, different countries and districts have different criterions for mercury toxicity. In our case, we adopt LD50 as the toxic criterions(LD50 is the dosage at which 50% of the humans exposed to a particular chemical will die. The LD50 for methylmercury is 50 mg/kg.). We speculate mercury toxicity has effect on the ability of eliminating mercury, therefore, we set up variable-elimination model at the basis of the first model. According to the first model, the amount of methylmercury in human body is 50 ug/kg, far less than 50 mg/kg, so we reach the conclusion that the fish consumption restrictions put forward by the reservoir advisories can protect the average adult. If the amount of methylmercury ingested increases, the amount of bioaccumulation will go up correspondingly. If 50 mg/kg is the maximum amount of methylmercury in human body, we can obtain the maximum number of fish that people consume safely permonth is 997.Keywords: methylmercury discrete dynamical system model variable-elimination modeldiscrete uniform random distribution model random-ingestion modelIntroductionIt is one of those wonders of the natural world that there is such diversity inthe shapes of leaves. As one website on leaf shapes remarks, "Plants have leaves in many different shapes - the thicker the book you refer to, the more leaf shapes they seem to find." The various classifications and permutations of shape form an arcane language, limited to a handful of botanists and elementary Montessori school students: terms such as runcinate, trifoliate, cordate, digitate, and deltoid. Although the words may be unfamilar, they describe the shapes of leaves encountered all around us: dandelion, clover, morning glory, maple, cottonwood. Beyond terms for general shape are further classifications for leaf form: toothed or untoothed, simple or compound, entire or lobed. Finally, even though one set of terms might be used to classify a leaf from a particular species of plant, some plants, particularly oaks, have leaves that vary considerably while keeping to generally the same overall shape.Problem Onediscrete dynamical system modelWhy? Why is there such diversity in leaves? What makes one leaf angular andanother rounded, one leaf wide and another narrow? These are the kinds of questions wondering children might ask, after being satisfied with a general explanation for the color of the sky and the forces causing the wind to blow. And, like most questions children generate, the answer is not an easy one. Indeed, there is no one clear explanation out there. Only a few days ago, a physics blog reported a new theoretical model that purports to explain all leaf shape variation as an incidental effect of the different patterns of veins in the leaf.Assumptions● The amount of methylmercury in fish is absorbed completely and instantly bypeople.● The elimination of mercury is proportional to the amount remaining.● People absorb fixed amount of methylmercury at fixed term per month. ● We assume the half-life of methylmercury in human body is 69.3 days. SolutionsLet 1α denote the proportion of eliminating methylmercury per month, 1β denote the accumulation proportion. As we know, methylmercury decays about 50 percent every 65 to 75 days, if no further methylmercury is ingested during that time. Consequently,111,βα=-69.3/3010.5.β=Through calculating, we get10.7408.β=L et’s define the following variables :0ω denotes the amount of methylmercury at initial time,n denotes the number of month,n ω denotes the amount of methylmercury in human body at the moment people have just ingested the methylmercury in the month n ,1x denotes the amount of methylmercury that people ingest per month and113000.7910x ug ug =⨯=.Moreover, we assume0=0.ωThough,111,n n x ωωβ-=⋅+we get1011x ωωβ=⋅+2201111x x ωωββ=⋅+⋅+⋅⋅⋅10111111n n n x x x ωωβββ-=⋅+⋅+⋅⋅⋅+⋅+121111(1)n n n x ωβββ--=++⋅⋅⋅++⋅11111.1n n x βωβ--=- With the remaining amount of methylmercury increasing, the elimination of methylmercury is also going up. We know the amount of ingested methylmercury per mouth is a constant. Therefore, with time going by, there will be a balance between absorption and elimination. We can obtain the steady-state value of remaining methylmercury as n approaches infinity.1*1111111lim 3505.11n n n x x ug βωββ-→∞-===-- The value of n ωis shown by figure 1.Figure 1. merthylmercury completely coming from fish and ingested at fixed term per monthIf the difference of the remaining methylmercury between the month n and 1n - is less than five percent of the amount of methylmercury that people ingest per month, that is,115%.n n x ωω--<⋅Then we can getAt the same time, we can work out the time that people have taken to achieve 3380 ug is 11 months.From our model, we reach the conclusion that the maximum amount of methylmercury the average adult human will bioaccumulate in their lifetime is 3505 ug.If people ingest methylmercury every half of a month, however, the sum of methylmercury ingested per month is constant, consequently,11910405,0.86.2x ug β=== As a result, we obtain the maximun amount of methylmercury in human body is 3270ug. When the difference is within 5 %, we get the time people have taken to achieve it is 11 months.Similarly, if people ingest methylmercury per day, we get the maximum amount is 3050ug, and the time is 10 months.Revising ModelSome aspects of leaf shape have been explained based upon comparing tropical forest plants with temperate forest ones, such as those here in Georgia. A visit to a tropical rainforest, or a greenhouse full of tropical plants, will reveal that most tropical leaves are thicker than temperate ones, as well as more rounded and smooth-edged Assumptions● The amount of methylmercury in the seafood is absorbed completely andinstantly by people.● The elimination of methylmercury is proportional to the amount remaining. ● People ingest fixed amount of methylmercury from other seafood every day. ● We assume the half-life of methylmercury in human body is 69.3 days. SolutionsLet 0ωdenote the amount of methylmercury at initial time, t denote the number of days, t ω denote the remaining amount on the day t , and 2x denote the amountof methylmercury that people ingest per day. Moreover, we assume0=0.ωIn addition, we work out2x =50.4/30=1.68 ug.The proportion of remaining methylmercury each day is 2β, then69.320.5.β=Through calculating, we get20.99.β=Because of12221,1t t x βωβ--=- we obtain steady-state value of methylmercury1*2222211lim 168.11t t t x x ug βωββ-→∞-===-- If the difference of remaining methylmercury between the day t and 1t - is less than five percent of the amount of methylmercury that people ingest every day, that is,125%.t t x ωω--<⋅We have301= 160 ug.ωSo we can reach the conclusion that the maximum amount of methylmercury the average adult human will bioaccumulate from seafood is 160 ug and the time that people take to achieve the maximum is 301 days.Let 1x denote the amount of methylmercury people ingest through bass at fixedterm per month, so the amount of methylmercury an average adult accumulate on the day t is1221221if t is a positive integer and not divisible by 30if t is a positive integer and divisible by 30.t t t t x x x ωωβωωβ--=⋅+⎧⎨=⋅++⎩ The value of t ωis shown by figure 2.Figure 2. merthylmercury coming from fish and other seafood and ingested at fixed term per dayThe change of t ω reflects the change of the amount of methylmercury in humanbody. Through revising model, we can figure out the maximum amount of methylmercury the average adult human will bioaccumulate in their lifetime is 3679 ug.Problem TwoRandom-ingestion modelAlthough people consume one fish per month, the consuming time has great randomness. We speculate the randomness has effect on the bioaccumulation of methylmercury, therefore, we construct a new model.Assumptions● The amount of methylmercury in fish is absorbed completely and instantly bypeople.● The elimination of methylmercury is proportional to the amount remaining. ● People consume one fish per month, but the consuming time has randomness.We assume the half-life of methylmercury in human body is 69.3 days.Let 0L denote the amount of methylmercury at initial time, n L denote theamount of methylmercury at the moment people have just ingested methylmercury in the month n , and x denote the amount of methylmercury that people absorb each time.We assume0=0.LWe have910.x ug =We define 1β the proportion of remaining methylmercury every day.Through69.310.5,β=we can get10.99.β=Let i obey discrete uniform random distribution with maximum 30 and minimum 1 and n t denote the number of days between the day 1n i - of the month 1n - and theday n i of the month n , then we have-130-,n n n t i i =+(1)1.n t n n L L x β-=⋅+The value of n L is shown by figure 3.Figure 3. merthylmercury completely coming from fish and ingested at random per monthFigure 3 shows the amount of methylmercury in human body has a great change due to the randomness of consuming time. Through the computer simulation, if we have numberless samples, n L will achieve the maximum value.That is,4261.n L ug =Revising modelUltimately, we simply do not know why there are so many leaf shapes are out there, and how (if at all) a star-shaped leaf might better serve a sweet gum tree's needs than the compound structure of ovate leaflets of a hickory. It is humbling and perhaps even a bit comforting, though, that something so commonplace as the formsIn that situation, we have1212.30(-1)30(-1)n n n nn n L L x if n n i L L x x if n n i ββ=⋅+≠⨯+⎧⎨=⋅++=⨯+⎩ Through the computer simulation, we can get a set of data about n L shown by figure4.Figure 4. remaining merthylmercury coming from fish consumed at random per month and otherfood consumed at fixed term per dayThough the revised model, we reach the conclusion that if we have numberless samples, n L will achieve the maximum value.That is,4420.n L ug =Variable-eliminateion modelAs a matter of fact, the state of human health can affect metabolice rate so that the ability of eliminating methylmercury is not constant. We have koown the amount of methylmercury in human body will affect human health. So we can draw the conclusion that the amount of methylmercury in human body will affect the abilitity of eliminating methylmercury.Assumptions● The amount of methylmercury in fish is absorbed completely and instantly bypeople.● the elimination of methylmercury is not only proportional to the amountremaining, but also affected by the change of human health which are caused by the amount of methylmercury.● People absorb fixed amount of methylmercury at fixed term per month throughconsuming bass.● We assume the half-life of methylmercury in human body is 69.3 days.● In condition that no further methylmercury is ingested during a period of time, welet χ denote the eliminating proportion per month. We have known methylmercury decays about 50 percent every other day 5 to a turn 5 days, so we determine the half-life of methylmercury in human body is 69.3 days. Then we have69.3/301(1)0.5χ⋅-=.By calculating, we getχ=0.2592.We adopt LD50 as the toxic criterions, then we get the maximum amount of methylmercury in human body is 63.510⨯ ug.L et’s define the following variables :0ω denotes the amount of methylmercury at initial time,n denotes the number of month,n ω denotes the amount of methylmercury in human body at the moment peoplehave just ingested the methylmercury in the month n ,n χ denotes the ability of eliminating methylmercury in the month n .γ denotes the effect on human health caused by methylmercury toxicity.1161 3.510r n n ωχχ-⎛⎫⎡⎤=⋅- ⎪⎢⎥ ⎪⨯⎣⎦⎝⎭ 1(1)n n n ωωχϕ-=⋅-+Hence, we have101(1)ωωχϕ=⋅-+20212(1)(1)(1)ωωχχϕχϕ=⋅-⋅-+⋅-+[]01233(1)...(1)(1)(1)...(1)(1)...(1)...(1)1n n n n n ωωχχϕχχχχχχ=⋅--+⋅-⋅--+--++-+ We define the value of γ is 0.5, then we get the maximum amount of maximum in human body is3567 ug, that is,*=3567 ug n ωNot taking the effect on the ability of eliminating maximum caused by methylmercury toxicity into account in model one,we obtain the maximum amount is 3510 ug. The difffference proves methylmercury toxicity has effect on eliminating methylmercury. We find out through calculating when r increases, the amount of methylmercury go up correspondingly. The reason for it is that methylmercury toxicity rises as a result of r increasing. Correspondingly, the effect on human health will increase, which is in accordance with fact.Problem ThreeAccording to the first model revised, we can get the maximum amount of bioaccumulation methylmercury is 3679 ug. We assume the average weight of an adult is 70 kg and the amount of methylmercury in human body is 53 ug/kg, far less than 50 mg/kg. Therefore, according to our model, the fish consumption restrictions put forward by the reservoir advisories can protect the average adult from reaching the LD50(LD50 is the dosage at which 50% of the humans exposed to a particular chemical will die. The LD50 for methylmercury is 50 mg/kg).We assume the lethal dosage of methylmercury is not gradually increasing. If the amount of methylmercury people ingests goes up rapidly, the bioaccumulation amount will reach to a higher value. Moreover, the value probably endangers human safety. Let LD50 be the maximum amount of methylmercury in human body, that is,*n =50 mg/kg 70 kg=3500 mg.ω⨯Let 1x denote the amount of methylmercury people ingest per month. According to the first model,1*1111111lim .11n n n x x βωββ-→∞-==-- We can figure out1 x =907.2 mg.We know the mean value of methylmercury in bass samples is 1.3 mg/kg, hence, we can obtain the maximum amount of fish that people consume safely per month is1max 698.1.3x M kg =≈ The maximum number of fish is 698/0.7=997.ConclusionSome aspects of leaf shape have been explained based upon comparing tropical forest plants with temperate forest ones, such as those here in Georgia. A visit to a tropical rainforest, or a greenhouse full of tropical plants, will reveal that most tropical leaves are thicker than temperate ones, as well as more rounded and smooth-edged (untoothed). Tropical plants retain their leaves for years, while deciduous plants in Georgia forests all keep their leaves only for one season. As a result, tropical leaves are sturdier than temperate forest ones, and therefore thicker. Thinner leaves require less energy to produce, and are more effective at gas exchange needed for photosynthesis. However, there is a cost of having thinner leaves: they are not as sturdy, particularly in areas distant from the major leaf veins (which provide structural support for the leaf). So those distant areas are simply done away with, resulting in lobed leaves like those of the white oak. The lobes (or teeth) of many leaves of Georgia plants also help to reduce wind resistance (and the damage that could result from it). Also, botanists have noticed that lobed or toothed leaves can permit sunlight to reach leaves beneath them, so perhaps the indentations help the plants filter sunlight down to leaves on their lower branches or stems.References[1] Dr.D.N.Rahni, PHD. Airborne Mercury Contamination and theNeversinkReservoir./dnabirahni/rahnidocs/Envsc/Airborne%20Mercury%20C ontamination%20and%20the%20Neversink%20Reservoir.doc[2] Hu Dong Bai Ke. Bass. /wiki%E9%B2%88%E9%B1%BC .[3] Centre for Food Safety Food and Environmental Hygiene Department TheGovernment of the Hong Kong Special Administrative Region. Mercury in Fish and Food Safety..hk/english/Programmme/programme_rafs/Programme_rafs_fc _01_19_mercury_in_fish.html.。
We develop a model to schedule trips down the Big Long River. The goalComputing Along the Big Long RiverChip JacksonLucas BourneTravis PetersWesternWashington UniversityBellingham,WAAdvisor: Edoh Y. AmiranAbstractis to optimally plan boat trips of varying duration and propulsion so as tomaximize the number of trips over the six-month season.We model the process by which groups travel from campsite to campsite.Subject to the given constraints, our algorithm outputs the optimal dailyschedule for each group on the river. By studying the algorithm’s long-termbehavior, we can compute a maximum number of trips, which we define asthe river’s carrying capacity.We apply our algorithm to a case study of the Grand Canyon, which hasmany attributes in common with the Big Long River.Finally, we examine the carrying capacity’s sensitivity to changes in thedistribution of propulsion methods, distribution of trip duration, and thenumber of campsites on the river.IntroductionWe address scheduling recreational trips down the Big Long River so asto maximize the number of trips. From First Launch to Final Exit (225 mi),participants take either an oar-powered rubber raft or a motorized boat.Trips last between 6 and 18 nights, with participants camping at designatedcampsites along the river. To ensure an authentic wilderness experience,at most one group at a time may occupy a campsite. This constraint limitsthe number of possible trips during the park’s six-month season.We model the situation and then compare our results to rivers withsimilar attributes, thus verifying that our approach yields desirable results.Our model is easily adaptable to find optimal trip schedules for riversof varying length, numbers of campsites, trip durations, and boat speeds.No two groups can occupy the same campsite at the same time.Campsites are distributed uniformly along the river.Trips are scheduled during a six-month period of the year.Group trips range from 6 to 18 nights.Motorized boats travel 8 mph on average.Oar-powered rubber rafts travel 4 mph on average.There are only two types of boats: oar-powered rubber rafts and motorizedTrips begin at First Launch and end at Final Exit, 225 miles downstream.*simulates river-trip scheduling as a function of a distribution of trip*can be applied to real-world rivers with similar attributes (i.e., the Grand*is flexible enough to simulate a wide range of feasible inputs; andWhat is the carrying capacity of the riverÿhe maximum number ofHow many new groups can start a river trip on any given day?How should trips of varying length and propulsion be scheduled toDefining the Problemmaximize the number of trips possible over a six-month season?groups that can be sent down the river during its six-month season?Model OverviewWe design a model thatCanyon);lengths (either 6, 12, or 18 days), a varying distribution of propulsionspeeds, and a varying number of campsites.The model predicts the number of trips over a six-month season. It alsoanswers questions about the carrying capacity of the river, advantageousdistributions of propulsion speeds and trip lengths, how many groups canstart a river trip each day, and how to schedule trips.ConstraintsThe problem specifies the following constraints:boats.AssumptionsWe can prescribe the ratio of oar-powered river rafts to motorized boats that go onto the river each day.There can be problems if too many oar-powered boats are launched with short trip lengths.The duration of a trip is either 12 days or 18 days for oar-powered rafts, and either 6 days or 12 days for motorized boats.This simplification still allows our model to produce meaningful results while letting us compare the effect of varying trip lengths.There can only be one group per campsite per night.This agrees with the desires of the river manager.Each day, a group can only move downstream or remain in its current campsiteÿt cannot move back upstream.This restricts the flow of groups to a single direction, greatly simplifying how we can move groups from campsite to campsite.Groups can travel only between 8 a.m. and 6 p.m., a maximum of 9hours of travel per day (one hour is subtracted for breaks/lunch/etc.).This implies that per day, oar-powered rafts can travel at most 36 miles, and motorized boats at most 72 miles. This assumption allows us to determine which groups can reasonably reach a given campsite.Groups never travel farther than the distance that they can feasibly travelin a single day: 36 miles per day for oar-powered rafts and 72 miles per day for motorized boats.We ignore variables that could influence maximum daily travel distance, such as weather and river conditions.There is no way of accurately including these in the model.Campsites are distributed uniformly so that the distance between campsites is the length of the river divided by the number of campsites.We can thus represent the river as an array of equally-spaced campsites.A group must reach the end of the river on the final day of its trip:A group will not leave the river early even if able to.A group will not have a finish date past the desired trip length.This assumption fits what we believe is an important standard for theriver manager and for the quality of the trips.MethodsWe define some terms and phrases:Open campsite: Acampsite is open if there is no groupcurrently occupying it: Campsite cn is open if no group gi is assigned to cn.Moving to an open campsite: For a group gi, its campsite cn, moving to some other open campsite cm ÿ= cn is equivalent to assigning gi to the new campsite. Since a group can move only downstream, or remain at their current campsite, we must have m ÿ n.Waitlist: The waitlist for a given day is composed of the groups that are not yet on the river but will start their trip on the day when their ranking onthe waitlist and their ability to reach a campsite c includes them in theset Gc of groups that can reach campsite c, and the groups are deemed “the highest priority.” Waitlisted groups are initialized with a current campsite value of c0 (the zeroth campsite), and are assumed to have priority P = 1 until they are moved from the waitlist onto the river.Off the River: We consider the first space off of the river to be the “final campsite” cfinal, and it is always an open campsite (so that any number of groups can be assigned to it. This is consistent with the understanding that any number of groups can move off of the river in a single day.The Farthest Empty CampsiteOurscheduling algorithm uses an array as the data structure to represent the river, with each element of the array being a campsite. The algorithm begins each day by finding the open campsite c that is farthest down the river, then generates a set Gc of all groups that could potentially reach c that night. Thus,Gc = {gi | li +mi . c},where li is the groupÿs current location and mi is the maximum distance that the group can travel in one day.. The requirement that mi + li . c specifies that group gi must be able to reach campsite c in one day.. Gc can consist of groups on the river and groups on the waitlist.. If Gc = ., then we move to the next farthest empty campsite.located upstream, closer to the start of the river. The algorithm always runs from the end of the river up towards the start of the river.. IfGc ÿ= ., then the algorithm attempts tomovethe groupwith the highest priority to campsite c.The scheduling algorithm continues in this fashion until the farthestempty campsite is the zeroth campsite c0. At this point, every group that was able to move on the river that day has been moved to a campsite, and we start the algorithm again to simulate the next day.PriorityOnce a set Gc has been formed for a specific campsite c, the algorithm must decide which group to move to that campsite. The priority Pi is a measure of how far ahead or behind schedule group gi is:. Pi > 1: group gi is behind schedule;. Pi < 1: group gi is ahead of schedule;. Pi = 1: group gi is precisely on schedule.We attempt to move the group with the highest priority into c.Some examples of situations that arise, and how priority is used to resolve them, are outlined in Figures 1 and 2.Priorities and Other ConsiderationsOur algorithm always tries to move the group that is the most behind schedule, to try to ensure that each group is camped on the river for aFigure 1. The scheduling algorithm has found that the farthest open campsite is Campsite 6 and Groups A, B, and C can feasibly reach it. Group B has the highest priority, so we move Group B to Campsite 6.Figure 2. As the scheduling algorithm progresses past Campsite 6, it finds that the next farthest open campsite is Campsite 5. The algorithm has calculated that Groups A and C can feasibly reach it; since PA > PC, Group A is moved to Campsite 5.number of nights equal to its predetermined trip length. However, in someinstances it may not be ideal to move the group with highest priority tothe farthest feasible open campsite. Such is the case if the group with thehighest priority is ahead of schedule (P <1).We provide the following rules for handling group priorities:?If gi is behind schedule, i.e. Pi > 1, then move gi to c, its farthest reachableopen campsite.?If gi is ahead of schedule, i.e. Pi < 1, then calculate diai, the number ofnights that the group has already been on the river times the averagedistance per day that the group should travel to be on schedule. If theresult is greater than or equal (in miles) to the location of campsite c, thenmove gi to c. Doing so amounts to moving gi only in such a way that itis no longer ahead of schedule.?Regardless of Pi, if the chosen c = cfinal, then do not move gi unless ti =di. This feature ensures that giÿ trip will not end before its designatedend date.Theonecasewhere a groupÿ priority is disregardedisshownin Figure 3.Scheduling SimulationWe now demonstrate how our model could be used to schedule rivertrips.In the following example, we assume 50 campsites along the 225-mileriver, and we introduce 4 groups to the river each day. We project the tripFigure 3. The farthest open campsite is the campsite off the river. The algorithm finds that GroupD could move there, but GroupD has tD > dD.that is, GroupD is supposed to be on the river for12 nights but so far has spent only 11.so Group D remains on the river, at some campsite between 171 and 224 inclusive.schedules of the four specific groups that we introduce to the river on day25. We choose a midseason day to demonstrate our modelÿs stability overtime. The characteristics of the four groups are:. g1: motorized, t1 = 6;. g2: oar-powered, t2 = 18;. g3: motorized, t3 = 12;. g4: oar-powered, t4 = 12.Figure 5 shows each groupÿs campsite number and priority value foreach night spent on the river. For instance, the column labeled g2 givescampsite numbers for each of the nights of g2ÿs trip. We find that each giis off the river after spending exactly ti nights camping, and that P ÿ 1as di ÿ ti, showing that as time passes our algorithm attempts to get (andkeep) groups on schedule. Figures 6 and 7 display our results graphically.These findings are consistent with the intention of our method; we see inthis small-scale simulation that our algorithm produces desirable results.Case StudyThe Grand CanyonThe Grand Canyon is an ideal case study for our model, since it sharesmany characteristics with the Big Long River. The Canyonÿs primary riverrafting stretch is 226 miles, it has 235 campsites, and it is open approximatelysix months of the year. It allows tourists to travel by motorized boat or byoar-powered river raft for a maximum of 12 or 18 days, respectively [Jalbertet al. 2006].Using the parameters of the Grand Canyon, we test our model by runninga number of simulations. We alter the number of groups placed on thewater each day, attempting to find the carrying capacity for the river.theFigure 7. Priority values of groups over the course of each trip. Values converge to P = 1 due to the algorithm’s attempt to keep groups on schedule.maximumnumber of possible trips over a six-month season. The main constraintis that each trip must last the group’s planned trip duration. Duringits summer season, the Grand Canyon typically places six new groups onthe water each day [Jalbert et al. 2006], so we use this value for our first simulation.In each simulation, we use an equal number of motorized boatsand oar-powered rafts, along with an equal distribution of trip lengths.Our model predicts the number of groups that make it off the river(completed trips), how many trips arrive past their desired end date (latetrips), and the number of groups that did not make it off the waitlist (totalleft on waitlist). These values change as we vary the number of new groupsplaced on the water each day (groups/day).Table 1 indicates that a maximum of 18 groups can be sent down theriver each day. Over the course of the six-month season, this amounts to nearly 3,000 trips. Increasing groups/day above 18 is likely to cause latetrips (some groups are still on the river when our simulation ends) and long waitlists. In Simulation 1, we send 1,080 groups down river (6 groups/day?80 days) but only 996 groups make it off; the other groups began near the end of the six-month period and did not reach the end of their trip beforethe end of the season. These groups have negligible impact on our results and we ignore them.Sensitivity Analysis of Carrying CapacityManagers of the Big Long River are faced with a similar task to that of the managers of the Grand Canyon. Therefore, by finding an optimal solutionfor the Grand Canyon, we may also have found an optimal solution forthe Big Long River. However, this optimal solution is based on two key assumptions:?Each day, we put approximately the same number of groups onto theriver; and?the river has about one campsite per mile.We can make these assumptions for the Grand Canyon because they are true for the Grand Canyon, but we do not know if they are true for the Big Long River.To deal with these unknowns,wecreate Table 3. Its values are generatedby fixing the number Y of campsites on the river and the ratio R of oarpowered rafts to motorized boats launched each day, and then increasingthe number of trips added to the river each day until the river reaches peak carrying capacity.The peak carrying capacities in Table 3 can be visualized as points ina three-dimensional space, and we can find a best-fit surface that passes (nearly) through the data points. This best-fit surface allows us to estimatethe peak carrying capacity M of the river for interpolated values. Essentially, it givesM as a function of Y and R and shows how sensitiveM is tochanges in Y and/or R. Figure 7 is a contour diagram of this surface.The ridge along the vertical line R = 1 : 1 predicts that for any givenvalue of Y between 100 and 300, the river will have an optimal value ofM when R = 1 : 1. Unfortunately, the formula for this best-fit surface is rather complex, and it doesn’t do an accurate job of extrapolating beyond the data of Table 3; so it is not a particularly useful tool for the peak carrying capacity for other values ofR. The best method to predict the peak carrying capacity is just to use our scheduling algorithm.Sensitivity Analysis of Carrying Capacity re R and DWe have treatedM as a function ofR and Y , but it is still unknown to us how M is affected by the mix of trip durations of groups on the river (D).For example, if we scheduled trips of either 6 or 12 days, how would this affect M? The river managers want to know what mix of trips of varying duration and speed will utilize the river in the best way possible.We use our scheduling algorithm to attempt to answer this question.We fix the number of campsites at 200 and determine the peak carrying capacity for values of R andD. The results of this simulation are displayed in Table 4.Table 4 is intended to address the question of what mix of trip durations and speeds will yield a maximum carrying capacity. For example: If the river managers are currently scheduling trips of length?6, 12, or 18: Capacity could be increased either by increasing R to be closer to 1:1 or by decreasing D to be closer to ? or 12.?12 or 18: Decrease D to be closer to ? or 12.?6 or 12: Increase R to be closer to 4:1.ConclusionThe river managers have asked how many more trips can be added tothe Big Long Riverÿ season. Without knowing the specifics ofhowthe river is currently being managed, we cannot give an exact answer. However, by applying our modelto a study of the GrandCanyon,wefound results which could be extrapolated to the context of the Big Long River. Specifically, the managers of the Big Long River could add approximately (3,000 - X) groups to the rafting season, where X is the current number of trips and 3,000 is the capacity predicted by our scheduling algorithm. Additionally, we modeled how certain variables are related to each other; M, D, R, and Y . River managers could refer to our figures and tables to see how they could change their current values of D, R, and Y to achieve a greater carrying capacity for the Big Long River.We also addressed scheduling campsite placement for groups moving down the Big Long River through an algorithm which uses priority values to move groups downstream in an orderly manner.Limitations and Error AnalysisCarrying Capacity OverestimationOur model has several limitations. It assumes that the capacity of theriver is constrained only by the number of campsites, the trip durations,and the transportation methods. We maximize the river’s carrying capacity, even if this means that nearly every campsite is occupied each night.This may not be ideal, potentially leading to congestion or environmental degradation of the river. Because of this, our model may overestimate the maximum number of trips possible over long periods of time. Environmental ConcernsOur case study of the Grand Canyon is evidence that our model omits variables. We are confident that the Grand Canyon could provide enough campsites for 3,000 trips over a six-month period, as predicted by our algorithm. However, since the actual figure is around 1,000 trips [Jalbert et al.2006], the error is likely due to factors outside of campsite capacity, perhaps environmental concerns.Neglect of River SpeedAnother variable that our model ignores is the speed of the river. Riverspeed increases with the depth and slope of the river channel, makingour assumption of constant maximum daily travel distance impossible [Wikipedia 2012]. When a river experiences high flow, river speeds can double, and entire campsites can end up under water [National Park Service 2008]. Again, the results of our model don’t reflect these issues. ReferencesC.U. Boulder Dept. of Applied Mathematics. n.d. Fitting a surface to scatteredx-y-z data points. /computing/Mathematica/Fit/ .Jalbert, Linda, Lenore Grover-Bullington, and Lori Crystal, et al. 2006. Colorado River management plan. 2006./grca/parkmgmt/upload/CRMPIF_s.pdf .National Park Service. 2008. Grand Canyon National Park. High flowriver permit information. /grca/naturescience/high_flow2008-permit.htm .Sullivan, Steve. 2011. Grand Canyon River Statistics Calendar Year 2010./grca/planyourvisit/upload/Calendar_Year_2010_River_Statistics.pdf .Wikipedia. 2012. River. /wiki/River .Memo to Managers of the Big Long RiverIn response to your questions regarding trip scheduling and river capacity,we are writing to inform you of our findings.Our primary accomplishment is the development of a scheduling algorithm.If implemented at Big Long River, it could advise park rangerson how to optimally schedule trips of varying length and propulsion. Theoptimal schedule will maximize the number of trips possible over the sixmonth season.Our algorithm is flexible, taking a variety of different inputs. Theseinclude the number and availability of campsites, and parameters associatedwith each tour group. Given the necessary inputs, we can output adaily schedule. In essence, our algorithm does this by using the state of theriver from the previous day. Schedules consist of campsite assignments foreach group on the river, as well those waiting to begin their trip. Given knowledge of future waitlists, our algorithm can output schedules monthsin advance, allowing managementto schedule the precise campsite locationof any group on any future date.Sparing you the mathematical details, allow us to say simply that ouralgorithm uses a priority system. It prioritizes groups who are behindschedule by allowing them to move to further campsites, and holds backgroups who are ahead of schedule. In this way, it ensures that all trips willbe completed in precisely the length of time the passenger had planned for.But scheduling is only part of what our algorithm can do. It can alsocompute a maximum number of possible trips over the six-month season.We call this the carrying capacity of the river. If we find we are below ourcarrying capacity, our algorithm can tell us how many more groups wecould be adding to the water each day. Conversely, if we are experiencingriver congestion, we can determine how many fewer groups we should beadding each day to get things running smoothly again.An interesting finding of our algorithm is how the ratio of oar-poweredriver rafts to motorized boats affects the number of trips we can send downstream. When dealing with an even distribution of trip durations (from 6 to18 days), we recommend a 1:1 ratio to maximize the river’s carrying capacity.If the distribution is skewed towards shorter trip durations, then ourmodel predicts that increasing towards a 4:1 ratio will cause the carryingcapacity to increase. If the distribution is skewed the opposite way, towards longer trip durations, then the carrying capacity of the river will always beless than in the previous two cases—so this is not recommended.Our algorithm has been thoroughly tested, and we believe that it isa powerful tool for determining the river’s carrying capacity, optimizing daily schedules, and ensuring that people will be able to complete their trip as planned while enjoying a true wilderness experience.Sincerely yours,Team 13955。
MCM: The Mathematical Contest in ModelingICM: The Interdisciplinary Contest in ModelingMCM:数学建模竞赛ICM:交叉学科建模竞赛Contest Registration and Instructions竞赛注册和指导(All instructions and rules apply to ICM as well as to MCM, except where otherwise noted.)(所有MCM的说明和规则除特别说明以外都适用于ICM)To participate in MCM a team must be sponsored by a faculty advisor f rom their institution. The registration process must be completed by the advisor.每个MCM的参赛队需有一名所在单位的指导教师负责。
整个注册报名过程需由该指导教师完成。
IMPORTANT CHANGE TO CONTEST RULES FOR MCM/ICM 2009:2009年MCM/ICM规则的重要改变:Teams (Student or Advisor) are now required to submit an electronic copy of their solution paper by email to solutions@. Your email MUST be received at COMAP by the submission deadline of 8:00 PM EST, February 9, 2009.要求参赛队(由学生或者指导教师)通过Email提交一份解决方案的电子版拷贝,发到solutions@。
Email邮件必须在美国东部时间2009年2月9日上午8点前发到COMAP。
2012 MCM ProblemsPROBLEM A:The Leaves of a Tree"How much do the leaves on a tree weigh?" How might one estimate the actual weight of the leaves (or for that matter any other parts of the tree)? How might one classify leaves? Build a mathematical mode l to describe and classify leaves. Consider and answer the following:• Why do leaves have the various shapes that they have?• Do the shapes “minimize” overlapping individual shadows that are cast, so as to maximize exposure? Does the distribution of leaves within the “volume” of the tree and its branches effect the shape?• Speaking of profiles, is leaf shape (general characteristics) related to tree profile/branching structure?• How would you estimate the leaf mass of a tree? Is there a correlation between the leaf mass and the size characteristics of the tree (height, mass, volume defined by the profile)?In addition to your one page summary sheet prepare a one page letter to an editor of a scientific journal outlining your key findings.“多少钱树的叶子有多重?”怎么可能估计的叶子(或树为此事的任何其他部分)的实际重量?会如何分类的叶子吗?建立了一个数学模型来描述和分类的叶子。
SummaryMany scholars conclude that leaf shape is highly related with the veins. Based on this theory, we assume the leaf growth in each direction satisfies a function. For the leaves in the same tree, the parameters are different; for those of separate trees, the function mode is different. Thus the shape of leaf differs from that of another. In the end of section 3, we simulate one growing period and depict the leaf shape.Through thousands of years of evolution, the leaves find various wa ys to make a full use of natural resources, including minimizing overlapping individual shadows. In order to find the main factors promoting the evolution of leaves, we analyze the distribution of adjacent leaves and the equilibrium point of photosynthesis and respiration. Besides, we also make a coronary hierarchical model and transmission model of the solar radiation to analyze the influence of the branches.As to the tree structure and the leaf shape, first we consider one species. Different tree shapes have different space which is built up by the branch quality and angle, effect light distribution, ventilation and humidity and concentration of CO2 in the tree crown. These are the factors which affect the leaf shape according to the model in section 1. Here we analyze three typical tree shapes: Small canopy shape, Open center shape and Freedom spindle shape, which can be described by BP network and fractional dimension model. We find that the factors mainly affect the function of Sthat affects the additional leaf area. Factors are assembled in different ways to create different leaf shapes. So that the relationship between leaf shape and tree profile/branching structure is proved.Finally we develop a model to calculate the leaf mass from the basic formula of . By adjusting the crown of a tree to a half ellipsoid, we first define thefunction of related factors,such as the leaf density and the effective ratio of leaf area. Then we develop the model using calculus. With this model, weapproximately evaluate the leaf mass of a middle-sized tree is 141kg.Dear editor,How much the leaves on a tree weigh is the focus of discussion all the time. Our team study on the theme following the current trend and we find something interesting in the process.The tree itself is component by many major elements. In our findings, we analyze the leaf mass with complicated ones, like leaf shape, tree structure and branch characteristics, which interlace with each other.With the theory that leaf shape is highly related with the veins, we assume theleaf growth in each direction satisfies a function . For the leaves in thesame tree, the parameters are different; for those of separate trees, the function mode is di fferent. That’s why no leaf shares the same shape. Also, we simulate one growing period and depict the leaf shape.In order to find the main factors promoting the evolution of leaves, we analyze the distribution of adjacent leaves and the equilibrium point of photosynthesis and respiration. Besides, we also make a coronary hierarchical model and transmission model of the solar radiation to analyze the influence of the branches.As to the tree structure and the leaf shape, different tree shapes have differe nt space which is built up by the branch quality and angle, effect light distribution, ventilation and humidity and concentration of CO2 in the tree crown that affect leaf shapes. Here we analyze three typical tree shapes which can be described by BP network and fractional dimension model. We find that the factors mainly affect the function of Sthat affects the additional leaf area. Factors are assembled in different ways to create different leaf shapes. So that the relationship between leaf shape and tree profile or branching structure is proved.Finally we develop the significant model to calculate the leaf mass from thebasic formula of . By adjusting the crown of a tree to a half ellipsoid,we first define the function of related factors and then we develop the model using calculus. With this model, we approximately evaluate the leaf mass of a middle-sized tree is 141kg.We are greatly appreciated that if you can take our findings into consideration. Thank you very much for your precious time for reading our letter.Yours sincerely,Team #14749Contents1. Introduction (4)2. Parameters (4)3. Leaves have their own shapes (5)3.1 Photosynthesis is important to plants (5)3.2 How leaves grow? (6)3.3 Build our model (7)3.4 A simulation of the model (10)4. Do the shapes maximize exposure? (14)4.1 The optimum solution of reducing overlapping shadows (14)4.1.1 The distribution of adjacent leaves (14)4.1.2 Equilibrium point of photosynthesis and respiration (15)4.2 The influence of the “volume” of a tree and its branches (17)4.2.1 The coronary hierarchical model (17)4.2.2 Spatial distribution model of canopy leaf area (18)4.2.3 Transmission model of the solar radiation (19)5. Is leaf shape related to tree structure? (20)5.1 The experiment for one species (21)5.2 Different tree shapes affect the leaf shapes (23)5.2.1 The light distribution in different shapes (23)5.2.2 Wind speed and humidity in the canopy (23)5.2.3 The concentration of carbon dioxide (24)5.3 Conclusion and promotion (25)6. Calculus model for leaf mass (26)6.1 How to estimate the leaf mass? (26)6.2 A simulation of the model (28)7. Strengths and Weakness (29)7.1 Strengths (29)7.2 Weaknesses (30)8. Reference (30)1. IntroductionHow much do the leaves on a tree weigh? Why do leaves have the various shapes that they have? How might one estimate the actual weight of the leaves? How might one classify leaves?We human-beings have never stopped our steps on exploring the natural world. But, as a matter of fact, the answer to those questions is still unresolved. Many scientists continue to study on this area. Recently , Dr. Benjamin Blonder (2010) achieved a new breakthrough on the venation networks and the origin of the leaf econo mics spectrum. They defined a standardized set of traits – density , distance and loopiness and developed a novel quantitative model that uses these venation traits to model leaf-level physiology .Now, it is commonly thought that there are four key leaf functional traits related to leaf economics: net carbon assimilation rate, life span, leaf mass per area ratio and nitrogen content.2. Parametersthe area a leaf grows decided by photosynthesisthe additional leaf area in one growing periodthe leaf growing obliquity Pthe total photosynthetic rate 0d R the dark respiration rate of leavesn P the net photosynthetic rateh the height of the canopyd the distance between two branchesdi the illumination intensity of scattered light from a given directionthe solar zenith angleh the truck highh the crown high13. Leaves have their own shapes3.1 Photosynthesis is important to plantsIt is widely accepted that two leaves are different, no matter where they are chosen from; even they are from the very tree. To understand how leaves grow is helpful to answer why leaves have the various shapes that they have.The canopy photosynthesis and respiration are the central parts of most biophysical crop and pasture simulation models. In most models, the acclamatory responses of protein and the environmental conditions, such as light, temperature and CO2 concentration, are concerned[1].In 1980, Farquhar et al developed a model named FvCB model to describe photosynthesis[2]:The FvCB model predicts the net assimilation rate by choosing the minimum between the Rubisco-limited net photosynthetic rate and the electron transport-limited net photosynthetic rate.Assume A n, A c, A j are the symbols for net assimilation rate, the Rubisco-limited net photosynthetic rate and the electron transport-limited net photosynthetic rate respectively, and the function can be described as:(1)(2)where and are the intercellular partial pressures of CO2 and O2,respectively, and are the Michaelis–Menten coefficients ofRubisco for CO2and O2, respectively, is the CO2compensation point inthe absence of (day respiration in andis the photosystem II electron transport rate that is used for CO2fixation and photorespiration[3][4].We apply the results of this model to build the relationship between the photosynthesis and the area a leaf grows during a period of time. It can be released as:(3)and are the area of the target leaf and the period of time it grows.is a function which can transfer the amount ofCO2into the area the leaf grows and the are parameters which affect S p. S p can be used as a constraint condition in our model.3.2 How leaves grow?As the collocation of computer hardware and software develops, people can refer to bridging biology, morphogenesis, applied mathematics and computer graphics to simulate living organisms[5], thus how to model leaves is of great challenge. In 2001, Dengler and Kang[6]brought up the thought that leaf shape is highly related to venation patterns. Recently, Runions[7] brought up a method to portray the leaf shape by analyzing venation patterns. Together with the Lindenmayer system (L-system), an advanced venation model can adjust the growth better that it solved the problem occurred in the previous model that the secondary veins are retarded.We knowleaves have various shapes.For example, leaves can be classified in to simple leaves which have an undivided blade and compound leaves whose blade is divided into two or more distinctleaflets such as the Fabaceae. As to the shape of a leaf, it may have marginal dentations of the leaf blades or not, and like a palm with various fingers or an elliptical cake. Judd et al defined a set of terms which describe the shape of leaves as follows [8]:We chose entire leaves to produce this model as a simplification. What’s more, they confirmed again that the growth of venations relates with that of the leaf.To disclose this relationship, Relative Elementary Rate of Growth (RERG) can be introduced to depict leaves growth [9]. RERG is defined as the growth rate per distance, in the definitive direction l at a point p of the growing object, yielding(4)Considered RERG , the growth patterns of leaves are also different. Roth-Nebelsick et al brought up four styles in their paper [10]:3.3 Build our modelWe chose marginal growth to build our model. Amid all above-mentioned studies, weFigure 3.1 Terms pertinent to the description of leaf shapes.Figure 3.2 A sample leaf (a) and the results of its: (b) marginal growth, (c) uniform isotropic (isogonic) growth, (d) uniform anisotropic growth, and (e) non-uniform anisotropic growth.Figure 3.3 The half of a leaf is settled in x-y plane like this with primary vein overlapping x-axis. The leaf grows in the direction of .assume that the leaf produce materials it needs to grow by photosynthesis to expand its leaf area from its border and this process is only affected by what we have discussed in the previous section about photosynthesis. The border can be infinitesimally divided into points. Set as the angle between the x-axis and thestraight line connecting the grid origin and one point on the curve,andas the growth distance in the direction of . To simplify the model, we assume that the leaf grows symmetrical. We put half of the leaf into the x-y plane and make the primary vein overlap x-axis.This is how we assume the leaf grows.In one circle of leaf growth, anything that photosynthesis provided transfers into theadditional leaf area, which can be described as:(5)while in the figure.In this case, we can simulate leaf growth thus define the leaf shape by using iterative operations the times N a leaf grow in its entire circle ①.① For instance, if the vegetative circle of a leaf is 20 weeks on average, the times of iterative operations N can beset as 20 when we calculate on a weekly basis.Figure 3.4 The curves of the adjacent growing period and their relationship.First, we pre-establish the border shape of a leaf in the x-y plane, yielding . where , the relevant satisfies:(6)(7) In the first growing period, assume , the growth distance in the direction of ,satisfies:(8)(9) It releases the relationship of the coordinates in the adjacent growing period. In this case, we can use eq.(9)to predict the new border of the leaf after one period of growth②:(10)And the average simple recursions are③:②That means, in the end of period 1.③As we both change the x coordinate and the y coordinate, in the new period, these two figures relate through those in the last period in the functions.(11)After simulate the leaf borders of the interactive periods, use definite integral ④ to settle parametersin the eq.(8) then can be calculated in each direction of , thus the exact shape of a leaf in the next period is visible.When the number of times N the leaf grows in its life circle applies above-mentioned recursions to iterate N times and the final leaf shape can be settled.By this model, we can draw conclusions about why leaves have different shapes. For the leaves on the same tree, they share the same method of expansion which can be described as the same type of function as Eq.(8). The reason why they are different, not only in a sense of big or small, is that in each growing period they acquire different amount of materials used to expand its own area. In a word, the parameters in the fixedly formed eq.(8) are different for any individual leaf on the same tree. For the leaves of different tree species, the corresponding forms of eq.(8) are dissimilar. Some are linear, some are logarithmic, some are exponential or mixtures of that, which settle the totally different expansion way of leaf, are related with the veins. On that condition, the characters can be divided by a more general concept such as entire or toothed.3.4 A simulation of the modelWe set the related parameters by ourselves to simulate the shape of a leaf and to express the model better.First we initialize the leaf shape by simulating the function of a leaf border at the④The relationship must meet eq.(5).Figure 3.5 and the recurrence relations.beginning of growing period 1 in the x-y plane. By observation, we assume that themovement of the initial leaf border satisfied:(12)Suppose the curve goes across the origin of coordinates, then the constraint conditionscan be:(13)Thus the solution to eq.(13) is:(14)By using Mathematica we calculatewhereandthe area of the half leaf is: (15)Wesettleaccording to the research by S. V . Archontoulisin et al [11] in eq.(3). On a weekly basis, theparameter. In eq.(15), we have .Figure 3.6When assume ,thus constraint condition eq.(5) becomes: (16)When eq.(8) is linear and after referring to Runions’s paper, we assume that Y -valuedecreases when X-value increases, which means the leaf grows faster at the end of theprimary vein. If the grow rate at the end of the primary vein is 0,as ,eq.(8) can be described as: (17)where b is decided by eq.(16).To settle the value of b , we calculate multiple sets of data by Excel then use a planecurve to trace them and get the approximation of b . In this method, we use grid toapproximate .Apparently,is monotone.When b is 2: The square of each square is 0.0139cm 2, and the total number of the squares in theadditional area is about 240. SoFigure 3.7When b is 2.5:Thesquare of each squareis 0.0240cm 2, and the total number of the squares in theadditional area is about 201. SoWhen b is 3:The square of each square is 0.0320cm 2, and the total number of the squares in the additional area is about 193. SoAfter comparing, we can draw a conclusion that fit eq.(16)best; accordingly, in period 1: (18)In each growing period, may be different for the amount of material produced isFigure 3.9Figure 3.8related with various factors, such as the change of relative location and CO 2 or O 2concentration, and other reasons. By using the same method, the leaf shape in period2 or other period can be generated on the basis of the previous growing period untilthe end of its life circle.What’s more, when eq.(8) is remodeled, the corresponding leaf shape can be changed.With , the following figure shows the transformationof a leaf shape in the firstgrowing period with the samesettled above and we can see that the shape will be dissected in the end.4. Do the shapes maximize exposure?4.1 The optimum solution of reducing overlapping shadows4.1.1 The distribution of adjacent leavesV ein is the foundation of the leaves. With the growing of veins, the leaves graduallyexpand around. The distribution of main and lateral veins plays an important decisiverole in the shapes of leaves. The scientists created a mathematical model which usesthree decisive factors - the relationship between the rate of photosynthesis, leaf life,carbon consumption or nitrogen consumption, to simulate the leaves’ shape. Becauseof carbon consumption is a constant for one tree, and we take the neighboring leavesin the same growth cycle to observe. So, we can only focus on one factor - the rate ofphotosynthesis.Figure 3.10The shape differs from figure 3.7-3.9, as the kind offunction of is different. In this case, it is a cubic model while a linear model in figure 3.7-3.9.Through the observation of dicotyledon, leaves on a branch will grow in a staggered way that can reduce the overlapping individual shadows of adjacent leaves and make them get more sunlight (Figure 4.1).Figure 4.1 The rotation distribution of leavesBase on the similar environment, we assume that adjacent leaves nearly have the same shape. From the perspective of looking down, the leaves grow from a point on the branch. So, we can simplify the vertical view of leaves as a circle of which the radius is the length of a vein which is represented with r. The width of the leaf is represented with w. The angle between two leaves is represented with β(Figure4.2).Figure 4.2 The vertical view of leavesThe leaves should use the space as much as possible, and for the leaf with one main vein, oval is the best choice. In general, βis between 15°and 90°. In this way, effectively reduce the direct overlapping area. According to the analysis of the first question, r and w are determined by the rate of photosynthesis and respiration. Besides, the width of leaf is becoming narrower when the main vein turns to be thinner.4.1.2 Equilibrium point of photosynthesis and respirationThe organism produced by photosynthesis firstly satisfies needs of leaf itself. Then the remaining organism delivered to the root to meet the growth needs of the tree. As we know, respiration needs to consume organism. If the light is not sufficient, organism produced by photosynthesis may no longer be able to afford the materials required for the growth of leaves. There should be an equilibrium point so as toprevent the leaf is behindhand in its circumstances.The formula [12] that describes the photosynthetic rate in response to light intensity with gradual exponential growth index can be expressed as:(19) Where P is total photosynthetic rate, m ax P is maximum photosynthetic rate ofleaves, a is initial solar energy utilization and I is photosynthetic photon quanta flux density .The respiration rate is affected by temperature ,using the formal of index to describe as follows:025102T d d R R -=⋅ (20)Where T is temperature and 0d R is the dark respiration rate of leaves, when the temperature is 25 degrees. As a model parameter, 0d R can be determined bynonlinear fitting. Therefore, the net photosynthetic rate can be expressed in the indexform as follows:(21)Where n P is the net photosynthetic rate, which does not include the concentration ofcarbon dioxide and other factors. The unit of n P is 21m ol m s μ--⋅⋅.We assume that the space for leaf growth is limited, the initial area of the leaf is 0S . At this point, the entire leaf happens to be capable of receiving sunlight. If the leaf continue to grow, some part of the leaf will be in the shadows and the area in the shadows is represented with x . Ignore the fluctuation cycle of photosynthesis and respiration, on average, the duration of photosynthesis is six hours per day . In the meanwhile, the respiration is ongoing all the time. When the area increased to S , we established an equation as follows:(22)where μ is the remaining organism created by the leaf per day . And from where 0μ=, which means the organism produced by photosynthesis has all been broken down completely in the respiration, we get the equilibrium point.4n n d S P x P R ⋅=+ (23) The proportion of the shaded area in the total area:4nn d P xS P R =+ (24)Cite an example of oak trees, we found the following data (Figure 4.3), which shows the relationship between net photosynthesis and dark respiration during 160 days. Thehabitat of the trees is affected by the semi-humid monsoon climate.Figure 4.3 The diurnal rate of net photosynthesis and dark respiration [13]In general, 217n P m ol m s μ--=⋅⋅,214d R m ol m s μ--=⋅⋅. Taking the given numbersinto the equation, we can get the result.0025xS =From the result, we can see that in the natural growth of leaves, with the weakening of photosynthesis, the leaves will naturally stop growing once they come across the blade between blocked. Therefore, the leaves can always keep overlapping individual shadows about 25%, so as to maximize exposure.4.2 The influence of the “volume” of a tree and its branches4.2.1 The coronary hierarchical modelMr. Bōken and Dr. J. Fischer (1987) found that in order to adequate lighting, leaves have different densities and the branches is distributed according to certain rules. They observed tropical plants in Miami, found that the ratio of main branch and two side branches is 1:0.94:0.87, and the angles between them are 24.4°and 36.9°. According to the computer simulation, the two angles can maximize the exposure of leaves.For simply, we use hemisphere to simulate the shape of the canopy. According to the light transmittance rate, we can divide the canopy into outer and inner two layers(Figure 4.3). The volume of the hemisphere depends on the size and distribution of branches.Figure 4.3 The coronary hierarchical modelThe angle between the branches will affect the depth of penetration of sun radiation and the distribution of leaves. Besides, thickness of the branches will affect the transfer of nutrients to the leaves.4.2.2 Spatial distribution model of canopy leaf areaAssume that the distribution of leaves is uniform in the section xz but not uniform in the y direction, as the figure 4.3. Then, we can get the formula of leaf area index (LAI) as follows [14]:/20/21(,)h d d dz x z dx LAI d -∂=⎰⎰(25)Where h is the height of the canopy , d is the distance between two branches, and (,)x z ∂ is the leaf area density function of the micro-body at the point (,)x z . x a is the function of leaf area which means the distribution of cross-section of the X direction, and Its value is a dimensionless, defined as follows: /2/21()d x d a x dx LAI d -=⎰ (26)()()x s a x d LAI C x =⋅⋅ (27)For the canopy which is not uniform in the horizontal direction, the distribution of its leaves is not entirely clear. We use s C to represent the distribution function of thedensity of leaves. s C can be expressed as a quadratic function or a Gaussiandistribution function.4.2.3 Transmission model of the solar radiationAssume that the attenuation of the solar radiation accords with the law of Beer-Lambert. The attenuation value at a point of canopy where the light arrives with a certain angle of incidence and azimuth is proportional to the length of the path, and the length can be calculated by Goudriaan function. Approximate function of the G function is shown as follows [15]:00(12)cos G G k G θ=+- (28)where k is a parameter decided by different plants.Take a micro unit in the canopy . Direct sunlight intercepted in this micro unit can be expressed in the following form:()(,)b b L dI I z G z dLAI =-Θ (29)Where b I means the direct sunlight, Θ is the solar zenith angle and L dL A I is the LAI on the path of light.Figure 4.4 Micro unit of the canopyThrough the integral and the chain rule, we can get the transmittance at the point (',')x z as follows:(,)[()]tan sin ()(',')exp()()h b b z G z a x z a z I x z I h ΘΘ∆Φ+=-⎰ (30)Average direct transmission rate:/2/21(')(',')'d b b d t z t x z dx d --=⎰(31)Different with the direct light, the scattered light in all directions is intercepted by theleaf surface from the upper hemisphere. The irradiance d dI of scattering at the point(',')x z can be expressed as follows:'cos d d b dI i t d ωθ=⋅⋅ (32) where d i is the illumination intensity of scattered light from a given direction.The transmission rate of the sun scattered light is as follows:2/2'00(',')1sin cos ()d b d I x z t d d I h ππθθθϕπ=⎰⎰ (33) Average scattering transmission rate:/2/21(')(',')'d d d d t z t x z dx d --=⎰(34)Global solar radiation reaching the canopy with a given depth z :(')()[(1)(')(')]d b d d I z I h k t z k t z ---=-+ (35)According to eq.(30) and eq.(33), the maximum depth the solar radiation can reach has a major link with h and θ. The shape of leaves have a relationship with photosynthesis. So, the h and θ of branches do have an influence on the leaves. Results showed that the light level and light utilization of high stem and open ce nter shape as well as small and sparse canopy shape were better than others .Double canopy shape ,spindle shape and center shape took second place ,while big canopy shape had the lowest light distribution [16].5. Is leaf shape related to tree structure?Due to internal and external factors, there are many kinds of tree shapes in the nature. For example, the apple's tree shape is semi-ellipsoidal, the willow's is hemispherical, the peach's likes a cup, the pine's likes cone and so on. The shape of their leaves varies. The leaf shape of apple is oval, pine's is needle, and the Indus's is palm. Is leaf shape related to tree shape? Even for the same species, there are many kinds of tree shapes. For instance, Small canopy shape, Open center shape, Freedom spindle shape, high stem and open center shape, Double canopy shape and so on. The sizes of leaves are different. Does the tree profile/branching structure effects the leaf shape?5.1 The experiment for one speciesTo solve this problem, we first consider the relationship of one tree species such as apple, which is semi-ellipsoidal. According to the second question, the hemispherical model is further extended to the semi-ellipsoidal model. Every tree individual shows irregularly because of natural and man-made factors, which reflects on the diversity of the crown in the canopy height direction of the changing relationship between the crown and canopy height. Figure 5.1 shows the relationship.Figure5.1 Schematic diagram of tree crown contourWhere 0h means the truck high, 1h means the crown high, 8710,,,,d d d d ⋅⋅⋅ meanscrown diameter. Tree shape with the scale change can be characterized in fractal dimension. Across the same scales, a fixed fractal dimension indicates the boundary shape's self-similarity; on different scales, the change of fractal dimension means that different processes or limiting factor has superiority. (Wiens, 1989)[17] According to the application of BP network and fractional dimension, we describe the tree shapes.[18]According to Li Guodong and Zhang Junke's work which detected and evaluated in different tree shapes of ‘Fuji’ apple. We choose three kinds of tree shapes to study; they are Small canopy shape, Open center shape and Freedom spindle shape as figure5.2 shows.。
美赛题目翻译————————————————————————————————作者:————————————————————————————————日期:2012美赛A题:一棵树的叶子(数学中国翻译)“一棵树的叶子有多重?”怎么能估计树的叶子(或者树的任何其它部分)的实际重量?怎样对叶子进行分类?建立一个数学模型来对叶子进行描述和分类。
模型要考虑和回答下面的问题:•为什么叶子具有各种形状?•叶子之间是要将相互重叠的部分最小化,以便可以最大限度的接触到阳光吗?树叶的分布以及树干和枝杈的体积影响叶子的形状吗?•就轮廓来讲,叶形(一般特征)是和树的轮廓以及分枝结构有关吗?•你将如何估计一棵树的叶子质量?叶子的质量和树的尺寸特征(包括和外形轮廓有关的高度、质量、体积)有联系吗?除了你的一页摘要以外,给科学杂志的编辑写一封信,阐述你的主要发现B:沿着Big Long River野营【数学中国翻译】0 i" k1 T3 h' B# u" ]游客在“大长河”(225英里)可以享受到秀丽的风光和令人兴奋的白色湍流。
这条河对于背包客来说是进不去的,因此畅游这条长河的唯一办法就是在这条河上露营上几天。
这次旅行从开始的下水点到最终结束点,共225英里,且是顺流而下的。
乘客可以选择平均4英里/小时的以浆作为动力的橡胶筏或者平均8英里/小时的机动帆船旅行。
整个旅行从开始到结束会经历6至18个夜晚。
负责管理这条河的政府机构希望到这里的每一次旅行都能够享受到野外经历,以最少的接触到在河上其它的船只。
目前,每年在六个月期间(一年的其余部分的天气对于河流旅行来说太冷),共有X次旅行,有Y 处露营地,露营地均匀的分布整个河道。
由于漂流的受欢迎程度的上升,公园管理者已经被要求允许更多的旅行次数。
所以他们想确定怎样可能安排一个最优的混合的旅行方案,不同的时间(单位为夜)和推动方式(马达或浆),最大限度的利用露营地。
1985 年美国大学生数学建模竞赛MCM 试题1985年MCM:动物种群选择合适的鱼类和哺乳动物数据准确模型。
模型动物的自然表达人口水平与环境相互作用的不同群体的环境的重要参数,然后调整账户获取表单模型符合实际的动物提取的方法。
包括任何食物或限制以外的空间限制,得到数据的支持。
考虑所涉及的各种数量的价值,收获数量和人口规模本身,为了设计一个数字量代表的整体价值收获。
找到一个收集政策的人口规模和时间优化的价值收获在很长一段时间。
检查政策优化价值在现实的环境条件。
1985年MCM B:战略储备管理钴、不产生在美国,许多行业至关重要。
(国防占17%的钴生产。
1979年)钴大部分来自非洲中部,一个政治上不稳定的地区。
1946年的战略和关键材料储备法案需要钴储备,将美国政府通过一项为期三年的战争。
建立了库存在1950年代,出售大部分在1970年代初,然后决定在1970年代末建立起来,与8540万磅。
大约一半的库存目标的储备已经在1982年收购了。
建立一个数学模型来管理储备的战略金属钴。
你需要考虑这样的问题:库存应该有多大?以什么速度应该被收购?一个合理的代价是什么金属?你也要考虑这样的问题:什么时候库存应该画下来吗?以什么速度应该是画下来吗?在金属价格是合理出售什么?它应该如何分配?有用的信息在钴政府计划在2500万年需要2500万磅的钴。
美国大约有1亿磅的钴矿床。
生产变得经济可行当价格达到22美元/磅(如发生在1981年)。
要花四年滚动操作,和thsn六百万英镑每年可以生产。
1980年,120万磅的钴回收,总消费的7%。
1986 年美国大学生数学建模竞赛MCM 试题1986年MCM A:水文数据下表给出了Z的水深度尺表面点的直角坐标X,Y在码(14数据点表省略)。
深度测量在退潮。
你的船有一个五英尺的草案。
你应该避免什么地区内的矩形(75200)X(-50、150)?1986年MCM B:Emergency-Facilities位置迄今为止,力拓的乡牧场没有自己的应急设施。
问题B:大量物种灭绝已经发生
亚马逊雨林是这个世界上最大的雨林,它拥有世界上最多的野生动物。
它座落在南美洲北面被巴西,玻利维亚等9个国家所分享。
因为乱砍乱伐的在雨林造成了致命的影响,所以这条信息需要普及。
虽然短期内砍伐能产生经济效益,但是长期下去会损害亚马逊雨林。
砍伐已经到达了历史的最低点,而在巴西仅仅只有50%的雨林处于被保护的状态。
但是栖息地遗失所产生的影响将在一段时间后会显示出来。
“砍伐在很短的时间内并不能杀死鸟类,长时间就会死亡,它们被挤在狭小的剩下的栖息地里,这样死亡率从就会逐渐提高”,生物学家这样说。
要求:
一.建立砍伐效应的模型,物种大灭绝是否会在未来近期发生。
二.用你的模型区预估多少物种会在未来20年会灭绝,考虑四种情景:
1.商业自由放任砍伐
2.出台一些规定
3.按照政府的规定目标,到2020年为止坚决减少80%的砍伐
4. 2020年终止采伐
三.写一份20页的报告(不包括摘要)表现出你对关于砍伐效应猜想的分析,保证包括政府在确保生态安全方面所扮演的角色。
2012 Contest ProblemsMCM PROBLEMSPROBLEM A: The Leaves of a Tree"How much do the leaves on a tree weigh?" How might one estimate the actual weight of the leaves (or for that matter any other parts of the tree)? How might one classify leaves? Build a mathematical model to describe and classify leaves. Consider and answer the following:• Why do leaves have the various shapes that they have?• Do the shapes “minimize” overlapping individual shadows that are cast, so as to maximize exposure? Does the distribution of leaves within the “volume” of the tree and its branches effect the shape?• Speaking of profiles, is leaf shape (general characteristics) related to tree profile/branching structure?• How would you estimate the leaf mass of a tree? Is there a correlation between the leaf mass and the size characteristics of the tree (height, mass, volume defined by the profile)?In addition to your one page summary sheet prepare a one page letter to an editor of a scientific journal outlining your key findings.“一棵树的叶子有多重?”怎么能估计树的叶子(或者树的任何其它部分)的实际重量?怎样对叶子进行分类?建立一个数学模型来对叶子进行描述和分类。
Section A“文化撞击”这一术语最早出现于1985年,人们用这一术语来描述人在进入全新的环境之后所产生的焦虑。
这种焦虑表现为:处于新环境之中不知道该做什么事,也不知道如何做事。
可以说,“文化撞击”是人们在进入一个不同于以往的新环境时所经历的精神与身体上的不适应的感觉。
通常情况下,新环境中的人们并不接受新成员原来的生活习惯,人们认为那些生活习惯是奇怪的。
所以对于新成员来说,一切都是新的,如:新的语言,不知道怎么使用的取款机,还有不会用的电话,等等。
在很多时候人们都会表现出“文化撞击”的症状。
虽然人们可能由于“文化撞击”而感受到真实的痛楚,但这也是成长和学习的机会,它可以帮助人更好地了解自己,并激发人们的创造力。
“文化撞击”可以分为许多阶段,有些阶段是彼此相连的,而有的阶段则是间隔开来,而且只在特定的时间表现出来。
第一阶段,又叫做“蜜月阶段”。
在这个阶段中,人们初次接触这个新的环境,这种体验让他们觉得很快乐。
随之而来的是第二个阶段,人们有可能会在日常生活中遇到一些困难,例如:交流上的障碍,其他人有可能无法听懂他们的话。
在这个阶段,人们有可能会失去耐心,变得忧伤,愤怒起来。
这个阶段中,人们从旧的生活方式向新的生活方式转变,而这是一个既耗时又费力的过程。
到了第三阶段,人们的一个特点是:他们开始初步了解新环境中的文化了,因此,他们会产生一种愉悦的感觉,还有可能伴随一点幽默感。
从心理上说,人们对新旧环境的感觉趋向于一致,这是因为:人们进一步熟悉了新环境并且开始对新环境产生了归属感。
此外,人们还会将新的生活方式与原来的生活方式加以对比和评价。
在第四阶段,人们认识到新的环境不仅有好的一面也有不好的一面,因此他们会将曾经体验过的两种或者三种文化融合起来。
这种融合的一大特点就是:这个过程中人们有一种强烈的归属感。
人们对开始对自我有了明确的定位,对生活也有了明确的目标。
第五阶段,又叫“二次撞击阶段”,这一阶段是人们回到原来的国家时经历的一个阶段。
2011国际大学生数学建模竞赛参赛规则中英文对照MCM: The Mathematical Contest in Modeling(数学建模竞赛)ICM: The Interdisciplinary Contest in Modeling(交叉学科建模竞赛)Contest Rules, Registration and Instructions(比赛规则,比赛注册和指导)(All rules and instructions apply to both ICM and MCM contests, except where otherwise noted.)(所有MCM的说明和规则除特别说明以外都适用于ICM)To participate in a contest, each team must be sponsored by a faculty advisor from its institution.(每个MCM的参赛队需有一名所在单位的指导教师负责。
)Team Advisors: Please read these instructions carefully. It is your responsibility to make sure that teams are correctly registered and that all of the following steps required for participation in the contest are completed:(指导老师:请认真阅读这些说明,确保完成了所有相关的步骤。
每位指导教师的责任包括确保每个参赛队正确注册并正确完成参加MCM/ ICM所要求的相关步骤。
)Please print a copy of these contest instructions for reference before, during, and after the contest. Click here for the printer friendly version. (请在比赛前做一份《竞赛注册和指导》的拷贝,以便在竞赛时和结束后作为参考。
2012年美国大学生数学建模竞赛MCM-A题-树与树叶-一等奖论文Leaf-Mass & Leaves ClassificationFebruary 14, 2012AbstractIn this paper, we present a statistical model to estimate the leaf mass and classify various leaf shapes based on the data obtained through tree harvest and leaf removal of 21 urban trees. We first explain the diversification of leaf shapes and describe the relationship between leaf shapes and tree profile in order to define the correlation between tree profile and leaf shapes. In order to estimate leaf mass, we apply the regression analysis with four observed independent variables, which allows us to estimate the leaf-mass of a whole-tree. In order to classify leaves, we use hierarchical clustering method according to the dry leaf-mass by Ward's Method. In this method we use the standard Euclidean distance as our principle of classification. Finally we obtain a regression model with vary high coefficient of determination and a good way to classify leaves.For office use onlyT1 ________________T2 ________________T3 ________________T4 ________________ Team Control NumberProblem Chosen AFor office use only F1 ________________ F2 ________________ F3 ________________ F4 ________________2012 Mathematical Contest in Modeling (MCM) Summary Sheet(Attach a copy of this page to each copy of your solution paper.)Type a summary of your results on this page. Do not include the name of your school, advisor, or team members on this page.Dear Editors,By operating and analyzing the experimental data, we have set up an optimizedregression model to estimate leaf mass, and have classified leaves into several specificgroups depending on a reasonable criterion. More importantly, we have noticed somereasonable results.To begin with, a classic regression model is cited to uncover the ambiguousrelationship between tree forms and leaf shapes. As a matter of fact, the leaf shapesare negatively correlated to the tree forms, and the result enables to expose the truthwhy narrower leaves on lower-crown trees are comparatively prolific.Moreover, compared with other regression models, an optimal regression modelestimating leaf mass was proposed. Under the specific significance level, tree mass isclosely relevant to some tree characteristics including crown height (m), crown radius(m) and leaf mass per area of crown projection (2/m g ). Therefore, if theseparameters are measured, the actual tree mass can beestimated.In addition, hierarchical clustering method is applied in leaf classification. In this way,leaves can be reasonably classified relying on the different dry mass.Contents1 Introduction (3)1.1 Outline of Our Approach (3)2 Leaves and the Leaf Shapes (4)2.1 The Diversification of Leaf Shapes (4)2.2 The Relationship between Leaf Shapes and Tree Profile (5)2.3 Leaf Mosaic (7)3 Estimating Leaf Mass (8)3.1 Assumptions of Regression Analysis (8)3.2 Basic Regression Model (8)3.2.1 Independent and Dependent Variables (9)3.2.2 Regression Equation (9)3.2.3 Correlation Matrix and Multicollinearity (9)3.2.4 A Global Test on the Set of Independent Variables (10)3.2.5 Checking the Significance of Each Independent Variable andANOVA Table (11)3.2.6 Diagnostic Plots for the Regression Model (12)3.3 Improved Regression Model (16)3.3.1 Solution and Results Analysis (17)4 Hierarchical Clustering in Leaves (19)4.1 Data Description (19)4.2 Methodology (20)4.3 Results of Clustering (21)4.4 Conclusion (21)5 References (22)。
PROBLEM B: Camping along the Big Long RiverVisitors to the Big Long River (225 miles) can enjoy scenic views and exciting white water rapids. The river is inaccessible to hikers, so the only way to enjoy it is to take a river trip that requires several days of camping. River trips all start at First Launch and exit the river at Final Exit, 225 miles downstream. Passengers take either oar- powered rubber rafts, which travel on average 4 mph or motorized boats, which travel on average 8 mph. The trips range from 6 to 18 nights of camping on the river, start to finish.. The government agency responsible for managing this river wants every trip to enjoy a wilderness experience, with minimal contact with other groups of boats on the river. Currently, X trips travel down the Big Long River each year during a six month period (the rest of the year it is too cold for river trips). There are Y camp sites on the Big Long River, distributed fairly uniformly throughout the river corridor. Given the rise in popularity of river rafting, the park managers have been asked to allow more trips to travel down the river. They want to determine how they might schedule an optimal mix of trips, of varying duration (measured in nights on the river) and propulsion (motor or oar) that will utilize the campsites in the best way possible. In other words, how many more boat trips could be added to the Big Long River’s rafting season? The river managers have hired you to advise them on ways in which to develop the best schedule and on ways in which to determine the carrying capacity of the river, remembering that no two sets of campers can occupy the same site at the same time. In addition to your one page summary sheet, prepare a one page memo to the managers of the river describing your key findings.Nowadays the heavy metal pollution is so common that people pay more and more attention to it. The aim of this paper is to calculate the maximum of methylmercury in human body during their lifetime and the maximum number of fish the average adult can safely eat per month. From City Officials research[1], we get information that the mean value of methylmercury in bass samples of the Neversink Reservoir is 1300 ug/kg and the average weight of bass people consume per month is 0.7 kg. According to the different consuming time in every month, we construct a discrete dynamical system model for the amount of methylmercury that will be bioaccumulated in the average adult body. In ideal conditions, we assume people consume bass at fixed term per month. Based on it, we construct fixed-ingestion model and we reach the conclusion that the maximum amount of methylmercury the average adult human will bioaccumulate in their lifetime is 3505 ug. As methylmercury ingested is not only coming from bass but also from other food, hence, we make further revise to our model so that the model is closer to the actual situation.As a result, we figure out the maximum amount of methylmercury the average adult human will bioaccumulate in their lifetime is 3679 ug. As a matter of fact, although we assume people consume one fish per month, the consuming time has great randomness. Taking the randomness into consideration, we construct a random-ingestion model at the basis of the first model. Through computer simulations, we obtain the maximum of methylmercury in human body is 4261 ug. We also calculate the maximum amount is 4420 ug after random-ingestion model is revised. As it is known to us, different countries and districts have different criterions for mercury toxicity. In our case, we adopt LD50 as the toxic criterions(LD50 is the dosage at which 50% of the humans exposed to a particular chemical will die. The LD50 for methylmercury is 50 mg/kg.). We speculate mercury toxicity has effect on the ability of eliminating mercury, therefore, we set up variable-elimination model at the basis of the first model. According to the first model, the amount of methylmercury in human body is 50 ug/kg, far less than 50 mg/kg, so we reach the conclusion that the fish consumption restrictions put forward by the reservoir advisories can protect the average adult. If the amount of methylmercury ingested increases, the amount of bioaccumulation will go up correspondingly. If 50 mg/kg is the maximum amount of methylmercury in human body, we can obtain the maximum number of fish that people consume safely per month is 997.Keywords: methylmercury discrete dynamical system model variable-elimination modeldiscrete uniform random distribution model random-ingestion modelIntroductionWith the development of industry, the degree of environmental pollution is also increasing. Human activities are responsible for most of the mercury emitted into the environment. Mercury, a byproduct of coal, comes from acid rain from the smokestack emissions of old, coal-fired power plants in the Midwest and South. Its particles rise on the smokestack plumes and hitch a ride on prevailing winds, whichoften blow northeast. After colliding with the Catskill mountain range, the particles drop to the earth. Once in the ecosystem, micro-organisms in the soil and reservoir sediment break down the mercury and produce a very toxic chemical form known as methylmercury . It has great effect on human health.Public officials are worried about the elevated levels of toxic mercury pollution in reservoirs providing drinking water to the New Y ork City . They have asked for our assistance in analyzing the severity of the problem. As a result of the bioaccumulation of methylmercury , if the reservoir is polluted, we can make sure that the amount of methylmercury in fish is also increasing. If each person adheres to the fish consumption restrictions as published in the Neversink Reservoir advisory and consumes no more than one fish per month, through analyzing, we construct a discrete dynamical system model of time for the amount of methylmercury that will bioaccumulate in the average adult person. Then we can obtain the maximum amount of methylmercury the average adult human will bioaccumulate in their lifetime. At the same time, we can also get the time that people have taken to achieve the maximum amount of methylmercury . As we know, different countries and districts have different criterions for the mercury toxicity . In our case, we adopt the criterion of Keller Army Community Hospital. If the maximum amount of methylmercury in human body is far less than the safe criterion, we can reach the conclusion that the reservoir is not polluted by mercury or the polluted degree is very low, otherwise we can say the reservoir is great polluted by mercury . Finally, the degree of pollution is determined by the amount of methylmercury in human body .Problem Onediscrete dynamical system modelThe mean value of methylmercury in bass samples of the Neversink Reservoir is 1300 ug/kg and the average weight of bass is 0.7 kg. According to the subject, people consume no more than one fish per month. For the safety of people, we must consider the bioaccumulation of methylmercury under the worst condition that people absorb the maximum amount of methylmercury . Therefore, we assume that people consume one fish per month.Assumptions● The amount of methylmercury in fish is absorbed completely and instantly bypeople.● The elimination of mercury is proportional to the amount remaining.● People absorb fixed amount of methylmercury at fixed term per month. ● We assume the half-life of methylmercury in human body is 69.3 days. SolutionsLet 1α denote the proportion of eliminating methylmercury per month, 1β denote the accumulation proportion. As we know, methylmercury decays about 50 percent every 65 to 75 days, if no further methylmercury is ingested during that time. Consequently ,111,βα=-69.3/3010.5.β=Through calculating, we get10.7408.β=L et’s define the following variables :ω denotes the amount of methylmercury at initial time, n denotes the number of month,n ω denotes the amount of methylmercury in human body at the moment people have just ingested the methylmercury in the month n ,1x denotes the amount of methylmercury that people ingest per month and113000.7910x ug ug=⨯=. Moreover, we assume0=0.ωThough,111,n n x ωωβ-=⋅+we get1011x ωωβ=⋅+2201111x x ωωββ=⋅+⋅+ ⋅⋅⋅10111111n n n x x x ωωβββ-=⋅+⋅+⋅⋅⋅+⋅+121111(1)n n n x ωβββ--=++⋅⋅⋅++⋅11111.1n n x βωβ--=-With the remaining amount of methylmercury increasing, the elimination of methylmercury is also going up. We know the amount of ingested methylmercury per mouth is a constant. Therefore, with time going by, there will be a balance between absorption and elimination. We can obtain the steady-state value of remaining methylmercury as n approaches infinity.1*1111111lim 3505.11n n n x x ug βωββ-→∞-===--The value of n ω is shown by figure 1.Figure 1. merthylmercury completely coming from fish and ingested at fixed term per monthIf the difference of the remaining methylmercury between the month n and 1n - is less than five percent of the amount of methylmercury that people ingest per month, that is,115%.n n x ωω--<⋅Then we can get11=3380ug.ωAt the same time, we can work out the time that people have taken to achieve 3380 ug is 11 months.From our model, we reach the conclusion that the maximum amount of methylmercury the average adult human will bioaccumulate in their lifetime is 3505 ug.If people ingest methylmercury every half of a month, however, the sum of methylmercury ingested per month is constant, consequently,11910405,0.86.2x ug β===As a result, we obtain the maximun amount of methylmercury in human body is 3270ug. When the difference is within 5 %, we get the time people have taken to achieve it is 11 months.Similarly, if people ingest methylmercury per day , we get the maximum amount is 3050ug, and the time is 10 months.Revising ModelAs a matter of fact, the amount of methylmercury in human body is not completely coming from fish. According to the research of Hong Kong SAR Food andEnvironmental Hygiene Department [1], under normal condition, about 76 percent of methylmercury comes from fish and 24 percent comes from other seafood. In order to make our model more and more in line with the actual situation, it is necessary for us to revise it. The U.S. environmental Protection Agency (USEPA) set the safe monthly dose for methylmercury at 3 microgram per kilogram (ug/kg) of body weight. If we adopt USEPA criterion, we can calculate the amount of methylmercury that the average adult ingest from seafood is 50.4 ug per month.Assumptions● The amount of methylmercury in the seafood is absorbed completely andinstantly by people.● The elimination of methylmercury is proportional to the amount remaining. ● People ingest fixed amount of methylmercury from other seafood every day . ● We assume the half-life of methylmercury in human body is 69.3 days. SolutionsLet 0ωdenote the amount of methylmercury at initial time, t denote the number of days, t ω denote the remaining amount on the day t , and 2x denote theamount of methylmercury that people ingest per day . Moreover, we assume0=0.ωIn addition, we work out2x =50.4/30=1.68 ug.The proportion of remaining methylmercury each day is 2β, then69.320.5.β=Through calculating, we get20.99.β=Because of12221,1t t x βωβ--=- we obtain steady-state value of methylmercury1*2222211lim 168.11t t t x x ug βωββ-→∞-===--If the difference of remaining methylmercury between the day t and 1t - is less than five percent of the amount of methylmercury that people ingest every day, that is,125%.t t x ωω--<⋅We have301= 160 ug.ωSo we can reach the conclusion that the maximum amount of methylmercury the average adult human will bioaccumulate from seafood is 160 ug and the time that people take to achieve the maximum is 301 days.Let 1x denote the amount of methylmercury people ingest through bass at fixedterm per month, so the amount of methylmercury an average adult accumulate on the day t is1221221if t is a positive integer and not divisible by 30if t is a positive integer and divisible by 30.t t t t x x x ωωβωωβ--=⋅+⎧⎨=⋅++⎩ The value of t ωis shown by figure 2.Figure 2. merthylmercury coming from fish and other seafood and ingested at fixed term per dayThe change of t ω reflects the change of the amount of methylmercury inhuman body . Through revising model, we can figure out the maximum amount of methylmercury the average adult human will bioaccumulate in their lifetime is 3679 ug.Problem TwoRandom-ingestion modelAlthough people consume one fish per month, the consuming time has great randomness. We speculate the randomness has effect on the bioaccumulation of methylmercury , therefore, we construct a new model.AssumptionsThe amount of methylmercury in fish is absorbed completely and instantly bypeople.● The elimination of methylmercury is proportional to the amount remaining. ● People consume one fish per month, but the consuming time has randomness. ● We assume the half-life of methylmercury in human body is 69.3 days.Let 0L denote the amount of methylmercury at initial time, n L denote theamount of methylmercury at the moment people have just ingested methylmercury in the month n , and x denote the amount of methylmercury that people absorb each time.We assume0=0.LWe have910.x ug =We define 1β the proportion of remaining methylmercury every day .Through69.310.5,β=we can get10.99.β=Let i obey discrete uniform random distribution with maximum 30 and minimum 1 and n t denote the number of days between the day1n i - of the month 1n - and the day n i of the month n , then we have-130-,n n n t i i =+(1)1.n tn n L L x β-=⋅+ The value of n L is shown by figure 3.Figure 3. merthylmercury completely coming from fish and ingested at random per monthFigure 3 shows the amount of methylmercury in human body has a great change due to the randomness of consuming time. Through the computer simulation, if we have numberless samples, n L will achieve the maximum value.That is,4261.n L ug =Revising modelIn order to make our model more accurate, we need to make further revise. We take methylmercury coming from other seafood into consideration. We know the amount of methylmercury that people ingest from other seafood every day is 1.68 ug.In that situation, we have1212.30(-1)30(-1)n n n n n n L L x if n n i L L x x if n n i ββ=⋅+≠⨯+⎧⎨=⋅++=⨯+⎩ Through the computer simulation, we can get a set of data about n L shown by figure4.Figure 4. remaining merthylmercury coming from fish consumed at random per month and otherfood consumed at fixed term per dayThough the revised model, we reach the conclusion that if we have numberless samples, n L will achieve the maximum value.That is,4420.n L ug =Variable-eliminateion modelAs a matter of fact, the state of human health can affect metabolice rate so that the ability of eliminating methylmercury is not constant. We have koown the amount of methylmercury in human body will affect human health. So we can draw the conclusion that the amount of methylmercury in human body will affect the abilitity of eliminating methylmercury .Assumptions● The amount of methylmercury in fish is absorbed completely and instantly bypeople.● the elimination of methylmercury is not only proportional to the amountremaining, but also affected by the change of human health which are caused by the amount of methylmercury .● People absorb fixed amount of methylmercury at fixed term per month throughconsuming bass.● We assume the half-life of methylmercury in human body is 69.3 days.● In condition that no further methylmercury is ingested during a period of time, welet χ denote the eliminating proportion per month. We have known methylmercury decays about 50 percent every other day 5 to a turn 5 days, so we determine the half-life of methylmercury in human body is 69.3 days. Then we have69.3/301(1)0.5χ⋅-=. By calculating, we getχ=0.2592.We adopt LD50 as the toxic criterions, then we get the maximum amount ofmethylmercury in human body is 63.510⨯ ug. L et’s define the following variables :ω denotes the amount of methylmercury at initial time,ndenotes the number of month,nω denotes the amount of methylmercury in human body at the moment people have just ingested the methylmercury in the monthn,n χ denotes the ability of eliminating methylmercury in the month n . γ denotes the effect on human health caused by methylmercury toxicity .1161 3.510r n n ωχχ-⎛⎫⎡⎤=⋅- ⎪⎢⎥ ⎪⨯⎣⎦⎝⎭1(1)n n n ωωχϕ-=⋅-+Hence, we have101(1)ωωχϕ=⋅-+20212(1)(1)(1)ωωχχϕχϕ=⋅-⋅-+⋅-+[]01233(1)...(1)(1)(1)...(1)(1)...(1)...(1)1n n n n n ωωχχϕχχχχχχ=⋅--+⋅-⋅--+--++-+We define the value of γ is 0.5, then we get the maximum amount of maximum in human body is 3567 ug, that is,*=3567 ug n ωNot taking the effect on the ability of eliminating maximum caused by methylmercury toxicity into account in model one,we obtain the maximum amount is 3510 ug. The difffference proves methylmercury toxicity has effect on eliminating methylmercury . We find out through calculating when r increases, the amount of methylmercury go up correspondingly. The reason for it is that methylmercurytoxicity rises as a result of r increasing. Correspondingly, the effect on human health will increase, which is in accordance with fact.Problem ThreeAccording to the first model revised, we can get the maximum amount of bioaccumulation methylmercury is 3679 ug. We assume the average weight of an adult is 70 kg and the amount of methylmercury in human body is 53 ug/kg, far less than 50 mg/kg. Therefore, according to our model, the fish consumption restrictions put forward by the reservoir advisories can protect the average adult from reaching the LD50(LD50 is the dosage at which 50% of the humans exposed to a particular chemical will die. The LD50 for methylmercury is 50 mg/kg).We assume the lethal dosage of methylmercury is not gradually increasing. If the amount of methylmercury people ingests goes up rapidly , the bioaccumulation amount will reach to a higher value. Moreover, the value probably endangers human safety . Let LD50 be the maximum amount of methylmercury in human body , that is,*n =50 m g/kg 70 kg=3500 m g.ω⨯Let 1x denote the amount of methylmercury people ingest per month. According to the first model,1*1111111lim.11n n n x x βωββ-→∞-==--We can figure out1 x =907.2 m g.We know the mean value of methylmercury in bass samples is 1.3 mg/kg, hence, we can obtain the maximum amount of fish that people consume safely per month is1m ax 698.1.3x M kg =≈The maximum number of fish is 698/0.7=997.ConclusionIn problem one, the paper calculates the final steady-state value at the same time interval per month, per half a month and per day . Through comparing the results, we get the final bioaccumulation amount of methylmercury is less, when discrete time unit is smaller. It shows when the interval of consuming fish is smaller and the sum of methylmercury ingested is constant for a period of time, the possibility of poisoning is lower.In problem two, we analyze the change of the amount of methylmercury under the condition that consuming time is random. We find out the amount o f methylmercury in human body is changing constantly in fixed range, when people have just consumed fish. Moreover, the maximum is 4261 ug, which is far bigger than3505 ug. So we can reach the conclusion that people are more endangered when the consuming time is irregular.In order to closer to the actual situation, we construct a model in which the half-life of methylmercury in human body is not constant. Through analyzing the data of computer simulation, the maximum amount of methylmercury will increase, that is, the risk of poisoning will be higher.References[1] Dr.D.N.Rahni, PHD. Airborne Mercury Contamination and the NeversinkReservoir./dnabirahni/rahnidocs/Envsc/Airborne%20Mercury%20C ontamination%20and%20the%20Neversink%20Reservoir.doc[2] Hu Dong Bai Ke. Bass. /wiki%E9%B2%88%E9%B1%BC.[3] Centre for Food Safety Food and Environmental Hygiene Department TheGovernment of the Hong Kong Special Administrative Region. Mercury in Fish and Food Safety..hk/english/Programmme/programme_rafs/Programme_rafs_fc _01_19_mercury_in_fish.html.。
星际犯罪塑型(ICM)正在调查串谋犯犯罪行为。
调查是非常有信心,他们知道的几名成员的阴谋,但他们进行逮捕之前,希望能找出其他成员和领导人。
阴谋和所有可能涉嫌同谋为同一家公司在一个大办公室复杂的工作。
“公司一直快速增长,并为自己的名称,开发和销售计算机银行和信用卡公司的软件。
ICM最近发现了一个消息从一个小集组82个工人,他们认为在公司将帮助他们找到最有可能的候选人身份不明的同谋者和未知的领导人。
由于信息流量是所有的办公在该公司的工人,它很可能是一些(或许很多)在确定的传播者消息流量不涉及阴谋。
事实上,他们是一定的,他们知道有些人谁是不是在阴谋。
建模工作的目标将是确定人们在办公室复杂谁是最有可能的同谋。
一个优先列表将是理想的,因此ICM可以调查,监视之下的地方,和/或询问最有可能的候选人。
一个判别线分离从非同谋的同谋也将是有益的,以明显的分类,在每个人组。
这也将是有用的模型,如果提名的阴谋领导人检察官办公室。
在当前情况下的数据是给你的犯罪建模团队,你的上司给你以下情形(称为调查的EZ),她曾在几年前在另一座城市。
甚至虽然她是她对简易案件的工作感到非常自豪,她说,这是一个非常小的,简单的例子,但它可以帮助你了解自己的任务。
她的数据如下:她考虑为同谋的十人分别为:安妮#,鲍勃,卡罗尔,大卫*,艾伦,弗雷德,乔治·哈利,伊内兹和JAYE#的。
(*表示已知的同谋者,#表示事先已知nonconspirators)28消息,她为她的案件有编号为每个主题年表的消息,她分配的基础上分析她的消息:安妮对鲍勃:你为什么今天迟到了吗?(1)鲍勃对卡罗尔:这该死的安妮总是看着我。
我是不是晚了。
(1)卡罗尔对戴夫:安娜和鲍勃再次战斗是鲍勃的迟到。
(1)戴夫对艾伦:我要看看你今天早晨。
当你能来吗?带来的预算文件。
(2)戴夫对弗雷德:我能来,随时随地今天看到你。
让我知道什么时候是一个好时机。
我应该带来的预算文件吗?(2)戴夫对乔治:我会看到你以后---说不完的话。
2012 Contest ProblemsPROBLEM A: The Leaves of a Tree"How much do the leaves on a tree weigh?" How might one estimate the actual weight of the leaves (or for that matter any other parts of the tree)? How might one classify leaves? Build a mathematical model to describe and classify leaves. Consider and answer the following:• Why do leaves have the various shapes that they have?• Do the shapes “minimize” overlapping individual shadows that are cast, so as to maximize exposure? Does the distribution of leaves within the “volume” of the tree and its branches effect the shape?• Speaking of profiles, is leaf shape (general characteristics) related to tree profile/branching structure?• How would you estimate the leaf mass of a tree? Is there a correlation between the leaf mass and the size characteristics of the tree (height, mass, volume defined by the profile)?In addition to your one page summary sheet prepare a one page letter to an editor of a scientific journal outlining your key findings.2012美赛A题:一棵树的叶子(数学中国翻译)“一棵树的叶子有多重?”怎么能估计树的叶子(或者树的任何其它部分)的实际重量?怎样对叶子进行分类?建立一个数学模型来对叶子进行描述和分类。
IMPORTANT CHANGE TO CONTEST RULES FOR MCM/ICM 2012:Teams (Student or Advisor) are now required to submit an electronic copy (summary sheet and solution) of their solution paper by email to solutions@. Your email MUST be received at COMAP by the submission deadline of 8:00 PM EST, February 13, 2012. Teams are free to choose between MCM Problem A, MCM Problem B or ICM Problem C.COMAP Mirror Site: For more in:/undergraduate/contests/mcm/MCM: The Mathematical Contest in ModelingICM: The Interdisciplinary Contest in Modeling2012 Contest ProblemsMCM PROBLEMSPROBLEM A: The Leaves of a Tree"How much do the leaves on a tree weigh?" How might one estimate the actual weight of the leaves (or for that matter any other parts of the tree)? How might one classify leaves? Build a mathematical model to describe and classify leaves. Consider and answer the following:• Why do leaves have the various shapes that they have?• Do the shapes “minimize” overlapping individual shadows that are cast, so as to maximize exposure? Does the distribution of leaves within the “volume” of the tree and its branches effect the shape?• Speaking of profiles, is leaf shape (general characteristics) related to tree profile/branching structure?• How would you estimate the leaf mass of a tree? Is there a correlation between the leaf mass and the size characteristics of the tree (height, mass, volume defined by the profile)?In addition to your one page summary sheet prepare a one page letter to an editor of a scientific journal outlining your key findings.2012美赛A题:一棵树的叶子(数学中国翻译)“一棵树的叶子有多重?”怎么能估计树的叶子(或者树的任何其它部分)的实际重量?怎样对叶子进行分类?建立一个数学模型来对叶子进行描述和分类。
模型要考虑和回答下面的问题:• 为什么叶子具有各种形状?• 叶子之间是要将相互重叠的部分最小化,以便可以最大限度的接触到阳光吗?树叶的分布以及树干和枝杈的体积影响叶子的形状吗?就轮廓来讲,叶形(一般特征)是和树的轮廓以及分枝结构有关吗?• 你将如何估计一棵树的叶子质量?叶子的质量和树的尺寸特征(包括和外形轮廓有关的高度、质量、体积)有联系吗?除了你的一页摘要以外,给科学杂志的编辑写一封信,阐述你的主要发现。
PROBLEM B:Camping along the Big Long River Visitors to the Big LongRiver (225 miles) can enjoy scenic views and exciting white water rapids. The river is inaccessible to hikers, so the only way to enjoy it is to take a river trip that requires several days of camping. River trips all start at First Launch and exit the river at Final Exit, 225 miles downstream. Passengers take either oar- powered rubber rafts, which travel on average 4 mph or motorized boats, which travel on average 8 mph. The trips range from 6 to 18 nights of camping on the river, start to finish.. The government agency responsible for managing this river wants every trip to enjoy a wilderness experience, with minimal contact with other groups of boats on the river. Currently, X trips travel down the Big Long River each year during a six month period (the rest of the year it is too cold for river trips). There are Y camp sites on the Big Long River, distributed fairly uniformly throughout the river corridor. Given the rise in popularity of river rafting, the park managers have been asked to allow more trips to travel down the river. They want to determine how they might schedule an optimal mix of trips, of varying duration (measured in nights on the river) and propulsion (motor or oar) that will utilize the campsites in the best way possible. In other words, how many more boat trips could be added to the Big Long River’s rafting season? The river managers have hired you to advise them on ways in which to develop the best schedule and on ways in which to determine the carrying capacity of the river, remembering that no two sets of campers can occupy the same site at the same time. In addition to your one page summary sheet, prepare a one page memo to the managers of the river describing your key findings.问题B:沿长江大露营到Big Long River(225英里)游玩的游客可以享受那里的风景和振奋人心的急流。
远足者没法到达这条河,唯一去的办法是漂流过去。
这需要几天的露营。
河流旅行始于First Launch,在 Final Exit结束,共225英里的顺流。
旅客可以选择依靠船桨来前进的橡皮筏,它的速度是4英里每小时,或者选择8英里每小时的摩托船。
旅行从开始到结束包括大约6到18个晚上的河中的露营。
负责管理这条河的政府部门希望让每次旅行都能尽情享受野外经历,同时能尽量少的与河中其他的船只相遇。
当前,每年经过Big Long河的游客有X组,这些漂流都在一个以6个月长短的时期内进行,一年中的其他月份非常冷,不会有漂流。
在Big Long上有Y处露营地点,平均分布于河廊。
随着漂流人数的增加,管理者被要求应该允许让更多的船只漂流。
他们要决定如何来安排最优的方案:包括旅行时间(以在河上的夜晚数计算)、选择哪种船(摩托还是桨船),从而能够最好地利用河中的露营地。
换句话说,Big Long River在漂流季节还能增加多少漂流旅行数?管理者希望你能给他们最好的建议,告诉他们如何决定河流的容纳量,记住任两组旅行队都不能同时占据河中的露营地。
此外,在你的摘要表一页,准备一页给管理者的备忘录,用来描述你的关键发现。
ICM PROBLEMPROBLEM C: Modeling for Crime BustingClick the title below to download a ZIP file containing the 2012 ICM Problem.Your ICM submission should consist of a 1 page Summary Sheet and your solution cannot exceed 20 pages for a maximum of 21 pages.2012 ICM ProblemModeling for Crime BustingYour organization, the Intergalactic Crime Modelers (ICM), is investigating a conspiracy to commit acriminal act. The investigators are highly confident they know several members of the conspiracy, buthope to identify the other members and the leaders before they make arrests. The conspirators andthe possible suspected conspirators all work for the same company in a large office complex. Thecompany has been growing fast and making a name for itself in developing and marketing computersoftware for banks and credit card companies. ICM has recently found a small set of messages from agroup of 82 workers in the company that they believe will help them find the most likely candidates forthe unidentified co‐conspirators and unknown leaders. Since the message traffic is for all the officeworkers in the company, it is very likely that some (maybe many) of the identified communicators in themessage traffic are not involved in the conspiracy. In fact, they are certain that they know some peoplewho are not in the conspiracy. The goal of the modeling effort will be to identify people in the officecomplex who are the most likely conspirators. A priority list would be ideal so ICM could investigate,place under surveillance, and/or interrogate the most likely candidates. A discriminate line separatingconspirators from non‐conspirators would also be helpful to distinctly categorize the people in eachgroup. It would also be helpful to the DA’s office if the model nominated the conspiracy leaders.Before the data of the current case are given to your crime modeling team, your supervisor gives youthe following scenario (called Investigation EZ) that she worked on a few years ago in another city. Eventhough she is very proud of her work on the EZ case, she says it is just a very small, simple example, butit may help you understand your task. Her data follow:The ten people she was considering as conspirators were: Anne#, Bob, Carol, Dave*, Ellen, Fred,George*, Harry, Inez, and Jaye#. (* indicates prior known conspirators, # indicate prior known nonconspirators)Chronology of the 28 messages that she had for her case with a code number for the topic of eachmessage that she assigned based on her analysis of the message:Anne to Bob: Why were you late today? (1)Bob to Carol: That darn Anne always watches me. I wasn't late. (1)Carol to Dave: Anne and Bob are fighting again over Bob's tardiness. (1)Dave to Ellen: I need to see you this morning. When can you come by? Bring the budget files.(2)Dave to Fred: I can come by and see you anytime today. Let me know when it is a good time. Should Ibring the budget files? (2)Dave to George: I will see you later ‐‐‐ lots to talk about. I hope the others are ready. It is important toget this right. (3)Harry to George: You seem stressed. What is going on? Our budget will be fine. (2) (4)Inez to George: I am real tired today. How are you doing? ( 5)Jaye to Inez: Not much going on today. Wanna go to lunch today? (5)Inez to Jaye: Good thing it is quiet. I am exhausted. Can't do lunch today ‐‐‐ sorry! (5) George to Dave: Time to talk ‐‐‐ now! (3)Jaye to Anne: Can you go to lunch today? (5)Dave to George: I can't. On my way to see Fred. (3)George to Dave: Get here after that. (3)Anne to Carol: Who is supposed to watch Bob? He is goofing off all the time. (1)Carol to Anne: Leave him alone. He is working well with George and Dave. (1)George to Dave: This is important. Darn Fred. How about Ellen? (3)Ellen to George: Have you talked with Dave? (3)George to Ellen: Not yet. Did you? (3)Bob to Anne: I wasn't late. And just so you know ‐‐‐ I am working through lunch. (1)Bob to Dave: Tell them I wasn't late. You know me. (1)Ellen to Carol: Get with Anne and figure out the budget meeting schedule for next week and help mecalm George. (2)Harry to Dave: Did you notice that George is stressed out again today? (4)Dave to George: Darn Harry thinks you are stressed. Don't get him worried or he will be nosing around.(4)George to Harry: Just working late and having problems at home. I will be fine. (4)Ellen to Harry: Would it be OK, if I miss the meeting today? Fred will be there and he knows thebudget better than I do. (2)Harry to Fred: I think next year's budget is stressing out a few people. Maybe we should take time toreassure people today. (2) (4)Fred to Harry: I think our budget is pretty healthy. I don't see anything to stress over. (2) END of MESSAGE TRAFFICYour supervisor points outs that she assigned and coded only 5 different topics of messages: 1) Bob'stardiness, 2) the budget, 3) important unknown issue but assumed to be part of conspiracy, 4) George'sstress, and 5) lunch and other social issues. As seen in the message coding, some messages had twotopics assigned because of the content of the messages.The way your supervisor analyzed her situation was with a network that showed the communicationlinks and the types of messages. The following figure is a model of the message network that resultedwith the code for the types of messages annotated on the network graph.Figure 1: Network of messages from EZ CaseYour supervisor points out that in addition to known conspirators George and Dave, Ellen and Carolwere indicted for the conspiracy based on your supervisor's analysis and later Bobself‐admitted hisinvolvement in a plea bargain for a reduced sentence, but the charges against Carol were later dropped.Your supervisor is still pretty sure Inez was involved, but the case against her was never established.Your supervisor's advice to your team is identify the guilty parties so that people like Inez don't get off,people like Carol are not falsely accused, and ICM gets the credit so people like Bob do not have theopportunity to get reduced sentences.The current case:Your supervisor has put together a network‐like database for the current case, which has the samestructure, but is a bit larger in scope. The investigators have some indications that a conspiracy is takingplace to embezzle funds from the company and use internet fraud to steal funds from credit cards ofpeople who do business with the company. The small example she showed you for case EZ had only 10people (nodes), 27 links (messages), 5 topics, 1 suspicious/conspiracy topic, 2 known conspirators, and 2known non‐conspirators. So far, the new cas e has 83 nodes, 400 links (some involving more than 1topic), over 21,000 words of message traffic, 15 topics (3 have been deemed to be suspicious), 7 knownconspirators, and 8 known non‐conspirators. These data are given in the attached spreadsheet files:Names.xls, Topics.xls, and Messages.xls. Names.xls contains the key of node number to the officeworkers' names. Topics.xls contains the code for the 15 topic numbers to a short description of thetopics. Because of security and privacy issues, your team will not have direct transcripts of all themessage traffic. Messages.xls provides the links of the nodes that transmitted messages and the topiccode numbers that the messages contained. Several messages contained up to three topics. To helpvisualize the message traffic, a network model of the people and message links is provided in Figure 2.In this case, the topics of the messages are not shown in the figure as they were in Figure 1. These topicnumbers are given in the file Messages.xls and described in Topics.xls.Figure 2: Visual of the network model of the 83 people (nodes) and 400 messages between thesepeople (links).Requirements:Requirement 1: So far, it is known that Jean, Alex, Elsie, Paul, Ulf, Yao, and Harvey are conspirators.Also, it is known that Darlene, Tran, Jia, Ellin, Gard, Chris, Paige, and Este are not conspirators. Thethree known suspicious message topics are 7, 11, and 13. There is more detail about the topics in fileTopics.xls. Build a model and algorithm to prioritize the 83 nodes by likelihood of being part of theconspiracy and explain your model and metrics. Jerome, Delores, and Gretchen are the seniormanagers of the company. It would be very helpful to know if any of them are involved in theconspiracy.Requirement 2: How would the priority list change if new information comes to light that Topic 1 isalso connected to the conspiracy and that Chris is one of the conspirators?Requirement 3: A powerful technique to obtain and understand text information similar to thismessage traffic is called semantic network analysis; as a methodology in artificial intelligence andcomputational linguistics, it provides a structure and process for reasoning about knowledge orlanguage. Another computational linguistics capability in natural language processing is text analysis.For our crime busting scenario, explain how semantic and text analyses of the content and context ofthe message traffic (if you could obtain the original messages) could empower your team to developeven better models and categorizations of the office personnel. Did you use any of these capabilities onthe topic descriptions in file Topics.xls to enhance your model?Requirement 4: Your complete report will eventually go to the DA so it must be detailed and clearlystate your assumptions and methodology, but cannot exceed 20 pages of write up. You may includeyour programs as appendices in separate files that do not count in your page restriction, but includingthese programs is not necessary. Your supervisor wants ICM to be the world's best in solving whitecollar,high‐tech conspiracy crimes and hopes your methodology will contribute to solving importantcases around the world, especially those with very large databases of message traffic (thousands ofpeople with tens of thousands of messages and possibly millions of words). She specifically asked youto include a discussion on how more thorough network, semantic, and text analyses of the messagecontents could help with your model and recommendations. As part of your report to her, explain thenetwork modeling techniques you have used and why and how they can be used to identify, prioritize,and categorize similar nodes in a network database of any type, not just crime conspiraciesand messagedata. For instance, could your method find the infected or diseased cells in a biological network whereyou had various kinds of image or chemical data for the nodes indicating infection probabilities andalready identified some infected nodes?2012美赛C题:犯罪克星数学中国翻译你的组织,银河犯罪建模中心(ICM),正在调查一个实施犯罪行为的阴谋。