2012美赛金奖论文

SUMMARY

The problem requires us to schedule an optimal mix of trips with given trips X and campsites Y. Then we construct several models to achieve the goal.

In order to utilize the campsites in the best way, we establish a satisfaction model. An evaluated function related to sightseeing time is given to measure tourists ’ satisfaction. After balancing the satisfaction with the Tourism Company ’s income, we gain that: the company will earn the optimal income when tourists fully used the campsites. And 16 campsites should be set by analyzing the differential equations. With the same tour time, same speed and same cost assumptions, we construct model two, on the basis of the fixed campsites, to generate the detailed tour projects. The assumptions lead to the maximum of the campsites quantity, simultaneously optimize the number of travel trips. Results show that the tour period should be as short as possible. Because the shorter the tour period the more groups of tourist can be arranged.

However in reality, expenses vary from project to project. After collecting a mass of data, a mixed projects model is built. The objective function, maximum of Tour Company ’s income, approximately means fully use of campsites. Using LINGO software with various subjected conditions, and assign the initial value 225 for Y, we receive the arranged times of 8 different projects.

To further discuss the 8 tour projects, such as to determine the tour time, routes, rafting time in a day and so on, we set up an automatic simulated model with the help of simulated annealing algorithm, whose lightspot is the detailed self-acting output. All we need to do is assigning some values for the parameters. Then we do a sensitivity analysis for this model. Among parameters z , 0t and 3t , z has the biggest influence on the maximum groups of tourist the company can serve in a specific tour project.

Moreover, we explain the strengths and weakness of our models and further discuss this problem.

Contents

1. Restatement of this problem (3)

2. General Assumptions (3)

3. Symbols (4)

4. Solutions (5)

4.1 Model 1——Satisfaction Model (5)

4.1.1 Additional assumptions (5)

4.1.2 Model establishment and results (5)

4.2 Model 2——Single Project Model (8)

4.2.1 Additional assumptions (8)

4.2.2 Explaination of the dynamic system (8)

4.2.3 Analysis and model design (10)

4.3 Model 3——Mixed Projects Model (12)

4.3.1 Additional assumptions (12)

4.3.2 Model establishment and results (12)

4.4 Model 4——Automatic simulated based on Simulated Annealing Algorithm[5] (15)

①Establish coordinate system (15)

②Generate randomized tour project (type) (15)

③Arrange the tour routes automatically based on Simulated Annealing (16)

④Results and interpretation (18)

5. Further Discussion (18)

6. Sensitivity Analysis (19)

7. Strengths and Weaknesses (22)

●Strenghths (22)

●Weaknesses (22)

8.Memorandum (23)

9.References (24)

10. Appendix (24)

9.1 Appendix 1 (24)

9.2 Appendix 2 (25)

9.3 Appendix 3 (26)

9.4 Appendix 4 (27)

1. Restatement of this problem

Visitors 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.

2. General Assumptions

●The water flow rates in the Long Big River are uniform and when compared with the speeds of oar- powered rubber rafts or motorized boats, they’re too tiny to be considered. That is we neglect the water flow rates. So the actual onward speed of tourists equals to the speed of their vehicles (oar- powered rubber rafts travel on average 4 mph or motorized boats travel on average 8 mph).

●No two sets of campers can occupy the same site at the same time.

●Y campsites distributed fairly uniformly throughout the river’s corridor, in another words, the distances between any two adjacent sites are in common.

●Campsites only locate in one side of the river.

●There are 30 days in each month.

●Take into account safety, weather and other factors, all of the tourists must go to campsites for rests at night.

●Regardless of the case that travel plan is changed because of bad weather.

3. Symbols

4. Solutions

4.1 Model 1——Satisfaction Model

4.1.1 Additional assumptions

●X Trips and Y campsites given in the problem are adjustable, which means that in order to maximize the profit the park managers would like to adjust them.

●For the purpose of improving tourists ’ travel satisfaction, it ’s impossible for groups of boats to contact on the way (if any different groups of boats meet on the way then the satisfactions they feel turn to zero, which is too heavy a loss to burden for the park manager).

●Each tourist expenses the same amount of money on the travel, therefore the profit for the park only depends on the quantity of tourists.

●There must be a campsite at the entrance in case the tourists want to have a rest during their waiting time.

4.1.2 Model establishment and results

In this model we, at first, think about the situation that only one type of vehicle is employed (either oar- powered rubber rafts or motorized boats). Here just the situation of using oar- powered rubber rafts is explained and its running speed 4v mph =. As already stated in assumptions, no two sets of campers can occupy the same site at the same time, every group can ’t come across each other on the way, and all of the groups should go to campsites for rests in the evening, we then give the analysis of any two adjacent campsites n Y 、1n Y +.

The distance between these two adjacent campsites:

where 1n n d ->+ Considering that all of the groups should go to campsites for rests in the evening. And in the evening each group occupies a campsite. So the capacity between these two campsites at night is, at most, two groups. That is:

12n n s ->+≤ (4.1.2)

where 1n n s ->+ is the groups arranged between campsite n and campsite n+1. For any two adjacent campsites, formula (4.1.1) and formula (4.1.2) both fit. Because, in the case of maximizing tourists, each group occupies a campsite in the evening, we can tentatively think of that the groups of tourists equal to the number of campsites. Under this circumstance, the park manager, in order to chase maximum profit, would like to set as many campsites as possible. However, increasing the campsites number will lead to the decrease of distance between campsite n and campsite n+1. For a fixed traveling speed v , the sightseeing time tourists spend between these two campsites:

As for the sightseeing time that tourists spent between the adjacent campsites. It needs to meet this condition:

1024n n t ->+<< (4.1.4)

Then combine this condition with formula (4.1.2) and (4.1.3), we have:

① In order to maximize tourists ’ satisfactions, that is 18n n t ->+=, we can calculate the campsites number by function bellows:

1[225/]18n n Y vt ->+=+= (4.1.6)

as can be seen, the park manager have to set eight campsites along this river. ② To maximize tourism company ’s profit, the more campsites the more money be earned. But it can ’t be ignored that total sightseeing time must range from 6 to 18 days (618T ≤≤). As for tourists, they start their travel from a campsite and end at the adjacent campsite a day, so their travel time equals to the campsites along the river:

Y T = (4.1.7)

where T is the total sightseeing time. We now have max 18Y T ==, and the average sightseeing time tourists spent a day is 3.125 hours.

③ Take into accout the situations from both sides, a general economical efficiency

function is constructed belows:

river, the company can obtain the maximize income, the degree of company ’s income equals to 1, that is ()=1f Y . Similarly, the degree of company ’s income equals to 0 when there are only 6 campsites along the river. And the value of Y ranges of [6,18] , so the result of ()f Y ranges from 0 to 1. Then we construct function use the normalized thought.

122251350

((),()),618

16961350

n n Y G f t f Y Y Y Y ->+-=

<≤-+

(4.1.11)

After dealing with it, we find out that 1((),())n n G f t f Y ->+ reaches the maximum (max =0.577G ) when 15.19Y =. So the general economical efficiency degree is 0.577 (ranges from 0 to 1), and 1 3.70n n t ->+=,114.82n n d ->+=.

Consequently, tourism company should set 16 campsites along the river, there must be 3.7 sightseeing hours for tourists between any two adjacent campsites, whose distance is 14.82 miles. And the general economical efficiency degree of this sightseeing route is 0.577.

Y

G

Fig 1. the general economical efficiency function vs. campsites quantity We plot a figure in MATLAB software to display the relation between the general economical efficiency function and the campsites quantity. As can be seen, the function increases progressively then reaches its peak, however decreases gradually later. It tells us in order to reaches the maximum, 15.19

Y should be choosen. The

corresponding maximum

max =0.577

G. That is the general economical efficiency degree of this sightseeing route is 0.577.

4.2 Model 2——Single Project Model

4.2.1 Additional assumptions

●The travelling time and the touring route is invariable. This means that the onward speed, entrance time and the touring route of the tourists now are the same as the tourists entranced previous.

●There is a campsite at the entrance in case the tourists want to have a rest during their waiting time.

●Tourists only use motorized boats for sightseeing, whose speed is 8 mph.

4.2.2 Explaination of the dynamic system

①The first phase

Assume that on the first day, there are n groups of tourist enter the tour on alternate fixed time which can be set by the manager. Hence, in the evening each group reaches their campsite and these groups of tourist occupy the front n campsites which are close to the First Launch. On the second day, tourists keep the same travelling speed as the first day at

max

v. Then there are another groups of

tourists, at the same speed max v , added, which are marked from group 1n + to group 2n . And similarly in the evening of the second day, these groups of tourists would occupy the front n campsites which are close to the First Launch. Besides, the group 1 to group n station in the 1n + to 21n + campsites counting from the First Launch. These course doesn ’t stop until group 1 to group n reach the Final Exit, at the same speed max v . ②The second phase

Supposed that tourists are in a large quantity, and each day there are n groups of tourists enter this tour and n groups of tourists leave at the same time. Note that in the evening all of the groups should fully use the campsites and no empty campsite exists.

③The third phase

On the last day of the travelling season, besides the last n groups of tourists, no more tourists are allowed to add in. In the evening, by the same token, these last n groups of tourists occupy the front n campsites which are close to the First Launch. Then the next day, all of the tourists move on by the same speed, which lead to the empty of the front n campsites closed to the First Launch in the evening. Later, there are n groups of tourist leave the tour each day continuously. And the campsites become fully empty when all of the tourists finish the tour.

For the purpose of providing a vivid explaination of our dynamic system, we plot a figure, whose ordinate is the groups of tourists in the tour ()x t and the abscissa is time t .

Fig 2. the relation between groups of tourists and time

As can be seen in figure 1, during period 10t -, the groups of tourist increase and there are n groups of tourist enter the tour every day, so the increase rate is n . During period 12t t -, all of the campsites are used and the quantity of tourist reaches the peak. As for period 2t T -, weather becomes colder and colder hence the rafting tour is stopped so the quantity of tourist in the travelling district decreases. When the overall tourist leave the Final Exit, then the travel ends.

4.2.3 Analysis and model design ①Restrained conditions:

The total travelling time for a group of tourist is 1t , as be stated in the problem it ranges from 6 to 18 days. Because tourist exits the tour the last day, so they don ’t rest in the campsites the last day ’s evening, then 16(1)18t ≤-≤ which leads to:

11719()t t N ≤≤∈ (4.2.1)

In view of that every day n groups of tourist added and all of the campsites will be fully used on 1t th day, so the total quantities of the campsite are:

1(1)()Y n t Y N =-∈ (4.2.2)

Because campsites are distributed fairly uniformly throughout the river’s corridor, and there is a campsite ar the entrance so the distance between two adjacent campsites is:

1225/n n d Y ->+= (4.2.3)

Each group of tourist moves on at the same speed max 8v mph =, so the travelling time a group spends a day can be calculated as belows:

211max /n n n n t nd v ->+->+= (4.2.4)

In addition,

21024n n t ->+<≤ (4.2.5)

②Object function

Holding that each group contains the same tourists a , so we approximate the maximum quantities of the campsites with the maximum of the groups of tourist. Hence the object function can be generated:

1()X n T t =- (4.2.6)

As stated in the problem, there are 6 months for travelling, each month is thought to have 30 days, then the total travelling days a year 180T =.

Here, we take formula (4.2.1)~(4.2.5) as the restrained conditions, and formula (4.2.6) as the objective function. However when we calculate the results using LINGO software, it shows that “it is unbounded ” and no results be yielded. Therefore, we modify our model, and abandon condition 1t N ∈and Y N ∈, the corresponding results can be gained (procedure and the results are presented in appendix 1). Unfortunately the results show that total groups of tourist 100.995107710X =?, campsites 90.345123110Y =?, and the groups of tourist in each day 80.575206810n =?, and the travelling time per day for each group of tourist 17t =, and the distance between any two adjacent campsites 1210n n n n d t ->+->+==. In this

case, the tour district is full of campsites, and tourists finish their vacation 7 days later which obviously leads to the maximum of tourists. However it is not realistic for that there isn ’t space for boats to move on in the river. So more restrained conditions should be set.

③Additonal restrained conditions

According to the analysis aforementioned, we know that it is unrealistic to set the campsites extraodinary density. So we restrict the campsites quantity for different degrees, append 1t N ∈and Y N ∈ again, then the corresponding results can be received (procedure and detailed results is showed in appendix 2). Here are the results varies with limited campsites quantity Y :

Table 1. results of the changing limited Y

Additionally, we can plot a figure to observe the relation between the limited campsites quantity Y (abscissa) and tourists quantity X (ordinate). Here it is:

Fig 3. results of the changing limited Y

1 stands for 225Y ≤,

2 stands for 200Y ≤,

3 stands for 175Y ≤,

4 stands for 150Y ≤,

5 stands for 125Y ≤,

6 stands for 100Y ≤,

7 stands for 75Y ≤. From table 1, we can easily find that the real campsites quantity Y , the whole groups of tourists X and the groups of tourists each day n reduce with the decrease limited campsites quantity, which is conform to reality. From figure 3, a linear relation between the limited campsites quantity and the total groups of tourist is gained. The travelling time for a group of tourist 1t , remaining unchanged, equals to 7 days. It

means that each group of tourist travels in the shortest period ——7 days, which allows the park manager to serve the most groups of tourist. Simultaneously, the distance between two adjacent campsites, reduces with the decrease of campsites quantity. And the travelling time by boats from the second day on merely changes, which is in the vicinity of 4.6 hours. When we select different value of Y , 1[37.10,37.34]n n nd ->+∈, where 1n n nd ->+ hardly changes, so we define 137n n nd miles ->+≈. That ’s the reason why 21n n t ->+ merely changes when Y differs. After that we analyze the case of first day in detail. With the same travelling speed max 8v mph =, the travelling time which the first group used 1max /[4.63,4.66]n n nd v ->+∈, the time which the second group need 1max (1)/[4.33,4.55]n n n d v ->+-∈, and the third group ’s time 1max (2)/[4.05,4.41]n n n d v ->+-∈……The time each group used is close to each other. Therefore, if each group of tourist departs on alternate fixed time (for instance, manager sets it as 20 minutes), then it can not only gurantee tourists reaching the campsites but also avoid contact between each group.

Then in the following days, tourists increase and move on at the same speed, which is also avoiding different groups ’ encounter. Note that it, in a sense, is the ideal condition which needs all of the tourists to set out at the stated time and move on at the stated speed and so on.

4.3 Model 3——Mixed Projects Model

4.3.1 Additional assumptions

●The whole travelling time and expenses vary from tourist to tourist, and tourists can and only can choose the travelling projects belows: 6-day tour, 7-day tour, 8-day tour, 9-day tour, 13-day tour, 14-day tour, 15-day tour and 16-day tour [1].

●Once a tourist schedules his or her tour length, and then the tourism company will arrange the specific date for him or her. However the period from being scheduled the project to be notified the starting of tour can not be too long.

●The Tourism Company arranges the projects in order to maximize its income. ●Each group has the same quantity of tourist, which is given as a . ●Campsites quantity Y is known as 225. 4.3.2 Model establishment and results ●Ascertain the object function

According to the first assumption, we impose (1,2,8)i x i =??? to, individually, represents 6 days, 7 days, 8 days, 9 days, 13 days, 14 days, 15 days and 16 days. And

imposing (1,2,8)i X i =??? to express the arranged times of each corresponding project in a travelling season (6 months).

There are Y campsites along the river, and in the evening all of the groups have to go to the campsites for rest, So in a single day, the group number of tourist equals to the campsites quantity Y , which leads to the maximum of tourists. In order to achieve the maximum income, the tourism company has to make sure that during the travelling season (6 months), there must be Y groups of tourist going to the campsites for rests in the evening everyday. As already given, there are 30 days a month, so the total travelling days in the travelling season is 180 days. That is, there are 180 days for the company to set tour project. Hence the total quantities of tourists one a day are aY , where a is the tourists ’ number in each group .

We know there are 8 types of tour projects which last for different periods. As for all of the tourists, the total travelling time can be determined:

T aYT =∑ (4.3.1)

It can also be determined by:

8

1

i

i i T

aX

x ==

∑∑ (4.3.2)

That is, plusing 11aX x , 22aX x ,……, 88aX x we gain tourists ’ total travelling time under each project. Adding these 8 results, we have the total travelling time of the whole tourists.

Combining formula (4.3.1) and formula (4.3.2) yields:

8

1

i i

i X x

YT ==∑ (4.3.3)

From the perspective of tourism company, its income, the object function, in the travelling season is:

8

8

1

1

i i i i i i S s aX a s X ====∑∑

(4.3.4)

where i s is the expense of project i , which can be looked up from the data we collected [1], (https://www.wendangku.net/doc/5f7509661.html,/trip_popup_dates_2012.htm) ●Related restrained conditions

First, the times of tour projects arranged by company should be nature number.

i X N ∈ (4.3.5)

Second, considering that in reality, there must be mixed tour project, as opposed to single project. So the total times of an single project arranged by company must have its rational boundaries, we define it as:

i P X Q ≤≤ (4.3.6)

where P is the rational minimum and Q is the rational maximum . Both of these two values can be set by empirical knowledge.

Now problem here is to calculate P and Q . If we take no account of other tour projects and only think about the single tour project, then the whole time in the travelling season can fully be arranged for this single tour project. So the maximum times can be arranged in the travelling season max /i i X YT x =. However there is a huge gap between this situation and the reality on the condition that tourists have multiple choices. As given in the assupmtion, there are 225 campsites(225Y =). The distance between two adjacent campsites is 1 miles. On the other hand, we find that if the tourism company divides their total travelling time for every tour project equally, the relation can be expressed as:

11227788aX x aX x aX x aX x ==???== (4.3.7)

Then we have:

/8i i X YT x = (4.3.8)

Consequently,

max /12.5%i i X X = (4.3.9)

We suppose that the value of max /i i X X fluctuates in the vicinity of 12.5%, for example increases and decreases 2.5% (of course 5%, 7.5%, 10%, 12.5% and other fluctuating degrees are allowed), then:

max 10%/15%i i X X ≤≤ (4.3.10)

That is

max max 10%15%i i i X X X ≤≤ (4.3.11)

Comparing formula (4.3.11) with formula (4.3.6) we find:

max 10%i P X =，max 15%i Q X =

Then we take maximum of formula (4.3.4) as the objective function and regard formula (4.3.5) and formula (4.3.11) as the restrained conditions. Put them into LINGO software and run for the results (procedure and detailed results are showed in appendix 3):

Table 2. results of tour projects

Under this situation, the total income of the tour company 11371770S a =. And the total travelling times of trips =4437X .

4.4 Model 4——Automatic simulated based on Simulated Annealing Algorithm [5]

①Establish coordinate system

First we construct a coordinate system, whose t axis stands for time. Each day occupies a coordinate point. As regulated before that there are 180 days in 6 months so 0180,t t N ≤≤∈. The p axis stands for the marked number of campsite. Suppose, similarly, there are 225 campsites along the river and each campsite occupies a coordinate point so 0225,p p N ≤≤∈. For instance, 11(,)t p means that tourists have to sleep in campsite that is marked 1p on on 1t th day ’s evening. Each point is unique in this coordinate system hence if the point is occupied, then we say that the campsite is used in an evening of a specific tour project. No more tourists are allowed to occupy this campsite. In view of this thought, the use ratio of campsite can be showed by whether the coordinate points are be occupied or not. The more occupancy the higher use ratio of campsite. ②Generate randomized tour project (type)

Consider that tourists can and only can choose the travelling projects belows: 6-day tour, 7-day tour, 8-day tour, 9-day tour, 13-day tour, 14-day tour, 15-day tour and 16-day tour [1]. We create a random matrix whose elements distribute equally ranges from 0.0 to 1.0 in MATLAB software. After the random matrix generates a number it has to multiply 10 and take the integer part, so that the generated number will range from 0 to 10. Let 1 stands for 6-day tour, 2 stands for 7-day tour, 3 stands for 8-day tour, 4 stands for 9-day tour, 5 stands for 13-day tour, 6 stands for 14-day tour, 7 stands for 15-day tour and 8 stands for 16-day tour. If computer produce 0, 9 or 10, then reproduce another one.

i x is the days spend in project i . With MATLAB software, we can generate i x random numbers range from 0 to 225 (mutiply 225 then take the integer part after producing a random value) in the same way. And the random number presents that tourists rest in campsites which have different mark. According to assumptions aforementioned, tourists all use motorized boats, which travel on 8 mph. There are 225 campsites along the river. And there is, at most, only 8 hours for tourists to move on a day. So the distance, at most, tourists can travel a day is: 8864h mph miles ?=. That is tourists can cross no more than 64 campsites a day and rest in the 65th campsite. In other words, ranking the numbers generated from small to large we find that the value difference between any two adjacent number can not surpass 64. Simultaneously, the number generated first has to be no more than 64 and can not be 0. The number generated last has to be more than 161

(22564161-=) and repeated numbers are not allowed.

Therefor, according to the i x generated number we know the tourists ’ route. For example, on the first day they rest in the campsite which is marked by the number generated first. On the second day they rest in campsite which is marked by the number generated second, and the course continues. The travelling time by boats, in addition, can be determined.

③Arrange the tour routes automatically based on Simulated Annealing Figure 3 plots the detailed diagram of our algorithm:

Fig 4. diagram of the algorithm

●Generate the initial axises.

Create a zero matrix 180225tp ? which has 180 rows and 225 lists. Approximate it with the coordinate system and note that all of the campsites are empty now.

●Generate the initial result and initial assignment.

Regard the tour project 1x i '= which is generated first and the corresponding marked number of campsite as initial values. Assume that the marked number are 11121,,,i n n n ???. Here we take the initial values as the best solution and current solution, and then assign a current objective function e_current and the optimal objective function e_best factitiously. Give the parameter of cooling temperature funcion a initial value 0.90z =. And assign the initial temperature 709.910t =?, the initial value of controlled temperature 31t =. All of these parameters are adjustable when we run the program.

Give zero matrix the initial answer by subsituting 0 in 11121(1,),(2,),,(,)i n n i n ??? with 1 to show that the points have been arranged and the corresponding campsite is occupied. Simutaneously, collect all of the initial results in matrix sol_new, which included the tour project 1x i '= and the corresponding marked numbers 11121,,,i n n n ??? ●Produce new result.

Allow computer to generate another tour project 2x k '= and the corresponding marked numbers 21222,,,k n n n ???. First of all, judge whether the values of 21222(1,),(2,),,(,)k n n k n ??? in the judging matrix equal to 0. If the whole values equal to 0, then arrange these tourists into the tour on the first day. Otherwise judge whether the value of 21222(2,),(3,),,(1,)k n n k n ???+ equal to 0. If the whole values equal to 0, then arrange these tourists into the tour on the second day. Otherwise, judge whether the value of 21222(3,),(4,),,(2,)k n n k n ???+ equal to 0……This course doesn ’t stop until the suitable days is found, whose marked numbers of campsites are all equel to 0.

Similarly, when a new result is generated, subsititute 0 with 1 in matrix 180225tp ?. And add them into matrix sol_new, which also included tour project 2x k '= and the corresponding marked number of campsites 21222,,,k n n n ???. ●Judging terms.

Add all of the values in 180225tp ?, and take the maximum of the sum as our objective function. Because point in matrix 180225tp ? is unique enough to express whether a campsite is used or not in a specific period. And the more 1 in matrix 180225tp ?, the higher use ratio of the campsites. So do tourists ’ quantity.

Compared 2s with 1s , if 21s s >, which means results generated at the second time is superior to results generated at the first time, that is e_new>e_current, then subsititute the new result for the current result: e_current=e_new. And subsititute the new solution for the current solution: sol_current=sol_new. At the same time, judge whether the new result is superior to the optimal result: e_new>e_best. If so then save the new result: e_best=e_new ，sol_best=sol_new. On the other hand, If 21s s <, then accept the new result in probability of (__/)e new e current t e --, otherwise the current result remains unchanged. And the new result here still equals to the current result.

●Circulate calculation and output the results.

After finishing a judgement of new result, flame’s temperature will decrease gradually at the parameter 0.90

z . This process doesn’t stop until temperature reaches the controled temperature

t, then output the final optimal results and

3

solutions.

As for the total groups of tourist in a specific tour project that tourism company served and the detailed arrangement can be received in the output at the same time.

④Results and interpretation

Because the results covers lots of pages, so we only display parts of the results.

Table 3. results of model 4

From table 3, the matrix which is used to store the results consists of two parts. For example, let’s talk about line 1 and line 2. The first value in line 1 means that the first group of tourists enters on the first day. The second value in line 1 means that these tourists have a 7-day tour. And values in line 2, individually, present the marked numbers of campsites to which tourists are going to have rests. In the first day’s evening, this group of tourist will sleep at campsite 26. In the second day’s evening ,they are going to sleep at campsite 49……So we can know the detailed arranged routes. We find that projects are vivid and clear in such a table.

5. Further Discussion

We are able to succesfully solve the problem with the limitation that no two sets of campers can occupy the same site at the same time. And each model in our paper

meets this condition, that is in the evening each group of tourist goes to the campsite which has the unique marked number. However, as for tourists meet on the way when rafting, we find it too difficult to find a satisfied solution. In this paper, we tentatively solve this problem by following means: ①tourists can ’t meet each other any time ②regardless of tourists ’ contacts on the way when rafting. There is a definite gap between what we done and thought, for which we felt sense of disability.

We have try our best to limit the maximum contact times at a fixed number, however no satisfied model can be construct with the limitation. And at the same time, we find that if we restrain the maximum contact times, then the automatic simulated model in our paper will take no effect. So if there must be something we eager to modify and improve, it must be working on this problem. Maybe WUSM [2] is useful for us to gain a more ideal simulated process. However, considering our limited time, energy, resource and other factors, what we can do are limited too. But we will continue our research in the future in the hope of obtaining a satisfied and more perfect model.

6. Sensitivity Analysis

In order to run model four which is base on simulated annealing algorithm, we assign initial values for parameters z (parameter of the cooling temperature function), 0t (the initial temperature) and 3t (the controled temperature). As we know, results vary with the values of parameters, so we make sensitivity analysis for this model and analyze the three parameters individually.

The initial values for these parameters are: 0.90z =, 709.910t =?, 31t =. First of all, we keep 0t and 3t unchanged. Then alter the value of z . and plug the changing z into the program, we obtain table 4:

Note that inf stands for no best results can be found in the program. X is the maximum groups of tourist the company can serve in a specific tour project. X ? is the value difference between two adjacent X . According to table 4, the value of parameter z has a big impact on the final results, which means that our results are not robust enough. In addition to the conclusion, we also find that the optimal results

can be gained when 0.99z =.

So we select 0.99z =, and keep 31t = as before, then discuss about 0t . Here we built table 5 for changing 0t .

Table 5. results of changing parameter

0t

From table 5, X grows with the increase of 0t . We, in some degree, approximate the relation with linearity. As can be seen, 0t has the weaker influence on X than z does.

Known that 1509.910t =? leads to the maximum X , so we set 0.99z =，1509.910t =?. Similarly, alter the value of 3t and plug it into the program. And the corresponding results are presented in table 6.

Table 6. results of changing parameter

3t

数学建模国家一等奖优秀论文

２01４高教社杯全国大学生数学建模竞赛 承诺书 我们仔细阅读了《全国大学生数学建模竞赛章程》和《全国大学生数学建模竞赛参赛规则》(以下简称为“竞赛章程和参赛规则”，可从全国大学生数学建模竞赛网站下载)。 我们完全明白,在竞赛开始后参赛队员不能以任何方式(包括电话、电子邮件、网上咨询等）与队外的任何人（包括指导教师)研究、讨论与赛题有关的问题。 我们知道，抄袭别人的成果是违反竞赛章程和参赛规则的，如果引用别人的成果或其他公开的资料(包括网上查到的资料),必须按照规定的参考文献的表述方式在正文引用处和参考文献中明确列出。 我们郑重承诺，严格遵守竞赛章程和参赛规则，以保证竞赛的公正、公平性。如有违反竞赛章程和参赛规则的行为,我们将受到严肃处理。 我们授权全国大学生数学建模竞赛组委会，可将我们的论文以任何形式进行公开展示(包括进行网上公示，在书籍、期刊和其他媒体进行正式或非正式发表等)。 我们参赛选择的题号是(从Ａ/B/C/Ｄ中选择一项填写)：B 我们的报名参赛队号为（8位数字组成的编号）: 所属学校（请填写完整的全名)： 参赛队员(打印并签名) :1. 2． 3．

指导教师或指导教师组负责人(打印并签名): ?（论文纸质版与电子版中的以上信息必须一致,只是电子版中无需签名。以上内容请仔细核对，提交后将不再允许做任何修改。如填写错误，论文可能被取消评奖资格。) 日期： 20１4 年 9 月15日 赛区评阅编号(由赛区组委会评阅前进行编号）:

２０14高教社杯全国大学生数学建模竞赛 编号专用页 赛区评阅编号（由赛区组委会评阅前进行编号）：赛区评阅记录(可供赛区评阅时使用）：

09年美赛A题优秀论文翻译

A题设计一个交通环岛 在许多城市和社区都建立有交通环岛，既有多条行车道的大型环岛（例如巴黎的凯旋门和曼谷的胜利纪念碑路口），又有一至两条行车道的小型环岛。有些环岛在进入口设有“停车”标志或者让行标志，其目的是给已驶入环岛的车辆提供行车优先权；而在一些环岛的进入口的逆向一侧设立的让行标志是为了向即将驶入环岛的车辆提供行车优先权；还有一些环岛会在入口处设立交通灯（红灯会禁止车辆右转）；也可能会有其他的设计方案。 这一设计的目的在于利用一个模型来决定如何最优地控制环岛内部，周围以及外部的交通流。该设计的目的在于可利用模型做出最佳的方案选择以及分析影响选择的众多因素。解决方案中需要包括一个不超过两页纸，双倍行距打印的技术摘要，它可以指导交通工程师利用你们模型对任何特殊的环岛进行适当的流量控制。该模型可以总结出在何种情况之下运用哪一种交通控制法为最优。当考虑使用红绿灯的时候，给出一个绿灯的时长的控制方法（根据每日具体时间以及其他因素进行协调）。找一些特殊案例，展示你的模型的实用性。 标题：一个环来控制一切:优化交通圈。 安德里亚?利维亚伦 安德烈娅?利维 拉塞尔?梅里克 哈维姆德学院 顾问:苏珊 摘要 我们的目的是利用车辆动力学考虑在圆形交叉路口的道路情况。我们主要根据进入圆形道路的速度决定最好的方式来控制车流量。我们假设在一个车道通过圆形道路循环，这样交通输入量能够被调节。（也就是，不会有优先的交通输入量） 对于我们的模型，可改变的参数是排队等候进入的速率，进入圆形道路的速率（服务速率），这个圆形道路最大的容量和离开这个道路的速率。我们使用带有队列和交通圈的隔室模型作为隔间。来自外界的车辆首先进行排队等候，然后进入圆环交叉路口，最后离开到外界。我们把服务速率和离开速率作为在圆环交叉路口的车辆数量参考。 另外，我们利用计算机来拟态一个可见表示，发生在不同情形下的圆环交叉路口。允许我们检验不同的情况，例如不平等的交通流量由于不同的队列，一些十字路口比其他车辆有一个更高的概率。这个拟态模仿实施栩栩如生，例如如何当前面是空道路时进行加速，而当前面有其他车辆时进行减速。大多数情况下，我们发现：一个高服务效率能够保持交通顺畅的最佳方式，这意味着对于进入交通的效率是最有效的。然而，当交通变得拥堵时，较低的服务率更好的适应了交通，这指示应该使用一个红绿灯。所以，在不同时间段，依靠预测中的交通流量，一个信号灯应该被安装进行循环实现。

数学建模美赛参考文献

数学建模美赛参考文献 Since 1982, the official publication of the teaching of mathematical modeling contest, translations and guidance materials, and related with the mathematical modeling of mathematics experiment teaching material ( only according to statistics all told ): E. A. Bender, an introduction to mathematical model, Zhu Yaochen, Xu Weixuan translation, popular science press, 1982 Kondo Jiro, Miya Eiaki, et al, mathematical model, mechanical industry press, 1985 C. L. Daimler, E. S. Ai Wei, mathematical modeling principle, Ocean Press, 1985 Jiang Qiyuan, mathematical model, higher education press, 1987 Ren Shanqiang, mathematical model, Chongqing University press, 1987 M. Braun, C. S. Coleman, D. A. Drew, the differential equation model, Zhu Yumin, Zhou yu-hun translation, National University of Defense Technology press, ( the book for the W. F.Lucas editor of the Modules in Applied Mathematics a book first volume ), 1988 Chen Anqi, mathematical model of scientific and technical engineering, China Railway Publishing House, 1988 Jiang Yuzhao, Xin Peiqing, mathematical model and computer simulation, University of Electronic Science and Technology Press, 1989 Yang Qifan, Bian Fu Ping, mathematical model, Zhejiang University press, 1990 Dong Jiali, Cao Xudong, Shim Hito, mathematical model, Beijing University of Technology press, 1990 Tang Huanwen, Feng Enmin, sun Yuxian, Sun Lihua, an introduction to the mathematics model, Dalian University of Technology press, 1990 Jiang Qiyuan, the mathematical model (the Second Edition ), higher education press, 1991 H. P. Williams, the mathematical model and computer application, National Defence Industry Press, 1991

数学建模国赛一等奖论文

电力市场输电阻塞管理模型 摘要 本文通过设计合理的阻塞费用计算规则，建立了电力市场的输电阻塞管理模型。 通过对各机组出力方案实验数据的分析，用最小二乘法进行拟合，得到了各线路上有功潮流关于各发电机组出力的近似表达式。按照电力市场规则，确定各机组的出力分配预案。如果执行该预案会发生输电阻塞，则调整方案，并对引起的部分序内容量和序外容量的收益损失，设计了阻塞费用计算规则。 通过引入危险因子来反映输电线路的安全性，根据安全且经济的原则，把输电阻塞管理问题归结为：以求解阻塞费用和危险因子最小值为目标的双目标规划问题。采用“两步走”的策略，把双目标规划转化为两次单目标规划：首先以危险因子为目标函数，得到其最小值；然后以其最小值为约束，找出使阻塞管理费用最小的机组出力分配方案。 当预报负荷为982.4MW时，分配预案的清算价为303元/MWh，购电成本为74416.8元，此时发生输电阻塞，经过调整后可以消除，阻塞费用为3264元。 当预报负荷为1052.8MW时，分配预案的清算价为356元/MWh，购电成本为93699.2元，此时发生输电阻塞，经过调整后可以使用线路的安全裕度输电，阻塞费用为1437.5元。 最后，本文分析了各线路的潮流限值调整对最大负荷的影响，据此给电网公司提出了建议；并提出了模型的改进方案。

一、问题的重述 我国电力系统的市场化改革正在积极、稳步地进行，随着用电紧张的缓解，电力市场化将进入新一轮的发展，这给有关产业和研究部门带来了可预期的机遇和挑战。 电网公司在组织电力的交易、调度和配送时，必须遵循电网“安全第一”的原则，同时按照购电费用最小的经济目标，制订如下电力市场交易规则： 1、以15分钟为一个时段组织交易，每台机组在当前时段开始时刻前给出下一个时段的报价。各机组将可用出力由低到高分成至多10段报价，每个段的长度称为段容量，每个段容量报一个段价，段价按段序数单调不减。 2、在当前时段内，市场交易-调度中心根据下一个时段的负荷预报、每台机组的报价、当前出力和出力改变速率，按段价从低到高选取各机组的段容量或其部分，直到它们之和等于预报的负荷，这时每个机组被选入的段容量或其部分之和形成该时段该机组的出力分配预案。最后一个被选入的段价称为该时段的清算价，该时段全部机组的所有出力均按清算价结算。 电网上的每条线路上有功潮流的绝对值有一安全限值，限值还具有一定的相对安全裕度。如果各机组出力分配方案使某条线路上的有功潮流的绝对值超出限值，称为输电阻塞。当发生输电阻塞时，需要按照以下原则进行调整： 1、调整各机组出力分配方案使得输电阻塞消除； 2、如果1做不到，可以使用线路的安全裕度输电，以避免拉闸限电，但要使每条 线路上潮流的绝对值超过限值的百分比尽量小； 3、如果无论怎样分配机组出力都无法使每条线路上的潮流绝对值超过限值的百分 比小于相对安全裕度，则必须在用电侧拉闸限电。 调整分配预案后，一些通过竞价取得发电权的发电容量不能出力；而一些在竞价中未取得发电权的发电容量要在低于对应报价的清算价上出力。因此，发电商和网方将产生经济利益冲突。网方应该为因输电阻塞而不能执行初始交易结果付出代价，网方在结算时应该适当地给发电商以经济补偿，由此引起的费用称之为阻塞费用。网方在电网安全运行的保证下应当同时考虑尽量减少阻塞费用。 现在需要完成的工作如下： 1、某电网有8台发电机组，6条主要线路，附件1中表1和表2的方案0给出了各机组的当前出力和各线路上对应的有功潮流值，方案1~32给出了围绕方案0的一些实验数据，试用这些数据确定各线路上有功潮流关于各发电机组出力的近似表达式。 2、设计一种简明、合理的阻塞费用计算规则，除考虑电力市场规则外，还需注意：在输电阻塞发生时公平地对待序内容量不能出力的部分和报价高于清算价的序外容量出力的部分。 3、假设下一个时段预报的负荷需求是982.4MW，附件1中的表3、表4和表5分别给出了各机组的段容量、段价和爬坡速率的数据，试按照电力市场规则给出下一个时段各机组的出力分配预案。 4、按照表6给出的潮流限值，检查得到的出力分配预案是否会引起输电阻塞，并在发生输电阻塞时，根据安全且经济的原则，调整各机组出力分配方案，并给出与该方案相应的阻塞费用。 5、假设下一个时段预报的负荷需求是1052.8MW，重复3~4的工作。 二、问题的分析

美赛论文要点

摘要： 第一段：写论文解决什么问题 1．问题的重述 a. 介绍重点词开头： 例1：“Hand move” irrigation, a cheap but labor-intensive system used on small farms, consists of a movable pipe with sprinkler on top that can be attached to a stationary main. 例2:……is a real-life common phenomenon with many complexities. 例3：An (effective plan) is crucial to……… b. 直接指出问题： 例 1：We find the optimal number of tollbooths in a highway toll-plaza for a given number of highway lanes: the number of tollbooths that minimizes average delay experienced by cars. 例2：A brand-new university needs to balance the cost of information technology security measures with the potential cost of attacks on its systems. 例3：We determine the number of sprinklers to use by analyzing the energy and motion of water in the pipe and examining the engineering parameters of sprinklers available in the market. 例4: After mathematically analyzing the …… problem, our modeling group would like to pres ent our conclusions, strategies, (and recommendations )to the ……. 例5：Our goal is... that (minimizes the time )………. 2．解决这个问题的伟大意义 反面说明。如果没有…… Without implementing defensive measure, the university is exposed to an expected loss of $8.9 million per year. 3．总的解决概述 a．通过什么方法解决什么问题 例：We address the problem of optimizing amusement park enjoyment through distributing Quick Passes (QP), reservation slips that ideally allow an individual to spend less time waiting in line. b．实际问题转化为数学模型

2012年国赛A葡萄酒获奖论文带附录(完整版)

2012高教社杯全国大学生数学建模竞赛 编号专用页 赛区评阅编号（由赛区组委会评阅前进行编号）： 赛区评阅记录（可供赛区评阅时使用）： 评 阅 人 评 分 备 注 全国统一编号（由赛区组委会送交全国前编号）： 全国评阅编号（由全国组委会评阅前进行编号）：

A 题葡萄酒的评价 摘要：确定葡萄酒质量时一般是通过聘请一批有资质的评酒员进行品评。一方面由于每个品酒员间存在评价尺度、评价位置和评价方向等方面的差异，导致不同品酒员对同一酒样的评价存在差异，从而不能真实地反映不同酒样间的差异。另一方面葡萄酒的质量和酿酒葡萄的好坏又有直接的关系，于是根据题中所给的条件和问题提出相关的约束条件和目标函数，建立合理的数学模型。 对于问题一，在分析附件1中所给的数据后，首先根据每组的10名评酒员对其中的一种酒进行品尝后确定葡萄的质量，然后在进行分析评酒员评27种红葡萄酒的差异，最后运用方差分析对两组评酒员的评价结果进行测定，得出两组评酒员存在是否有显著性差异的结果，看其哪组评酒员的技术水平更高些。 问题二是为了对酿酒葡萄进行分级，要从酿酒葡萄的理化指标和葡萄酒的质量进行分级，在附件2、3中，发现酿酒葡萄的成分数据中有很多因素，首先对酿酒葡萄的理化指标经过查找资料、专家咨询进行了较为有效的分类，我们从中选取一些有效因素，例如：氨基酸总量、糖、单宁、色差值、酸、芳香物质等。然后再采取系统聚类分析法对酿酒葡萄进行分级。等级大致分为优、良、中、差四个级别。 在解决问题三时，不仅要考虑酿酒葡萄还要考虑葡萄酒的理化指标，因而采用多元回归模型，模型如下： 其中，b0为常数项，为回归系数，错误！未找到引用源。是随机误差。

美赛论文模板(强烈推荐)

Titile Summary During cell division, mitotic spindles are assembled by microtubule-based motor proteins1, 2. The bipolar organization of spindles is essential for proper segregation of chromosomes, and requires plus-end-directed homotetrameric motor proteins of the widely conserved kinesin-5 (BimC) family3. Hypotheses for bipolar spindle formation include the 'push?pull mitotic muscle' model, in which kinesin-5 and opposing motor proteins act between overlapping microtubules2, 4, 5. However, the precise roles of kinesin-5 during this process are unknown. Here we show that the vertebrate kinesin-5 Eg5 drives the sliding of microtubules depending on their relative orientation. We found in controlled in vitro assays that Eg5 has the remarkable capability of simultaneously moving at 20 nm s-1 towards the plus-ends of each of the two microtubules it crosslinks. For anti-parallel microtubules, this results in relative sliding at 40 nm s-1, comparable to spindle pole separation rates in vivo6. Furthermore, we found that Eg5 can tether microtubule plus-ends, suggesting an additional microtubule-binding mode for Eg5. Our results demonstrate how members of the kinesin-5 family are likely to function in mitosis, pushing apart interpolar microtubules as well as recruiting microtubules into bundles that are subsequently polarized by relative sliding. We anticipate our assay to be a starting point for more sophisticated in vitro models of mitotic spindles. For example, the individual and combined action of multiple mitotic motors could be tested, including minus-end-directed motors opposing Eg5 motility. Furthermore, Eg5 inhibition is a major target of anti-cancer drug development, and a well-defined and quantitative assay for motor function will be relevant for such developments

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承诺书 我们仔细阅读了中国大学生数学建模竞赛的竞赛规则. 我们完全明白，在竞赛开始后参赛队员不能以任何方式（包括、电子、网上咨询等）与队外的任何人（包括指导教师）研究、讨论与赛题有关的问题。 我们知道，抄袭别人的成果是违反竞赛规则的, 如果引用别人的成果或其他公开的资料（包括网上查到的资料），必须按照规定的参考文献的表述方式在正文引用处和参考文献中明确列出。 我们重承诺，严格遵守竞赛规则，以保证竞赛的公正、公平性。如有违反竞赛规则的行为，我们将受到严肃处理。 我们参赛选择的题号是（从A/B/C/D中选择一项填写）： D 我们的参赛报名号为（如果赛区设置报名号的话）： 所属学校（请填写完整的全名）： 参赛队员(打印并签名) ：1. （此部分容不便公开，见谅） 2. 3. 指导教师或指导教师组负责人(打印并签名)： 日期： 2012 年 9 月 10 日赛区评阅编号（由赛区组委会评阅前进行编号）：

编号专用页 赛区评阅编号（由赛区组委会评阅前进行编号）： 全国统一编号（由赛区组委会送交全国前编号）：全国评阅编号（由全国组委会评阅前进行编号）：

机器人避障问题 摘要 针对机器人避障问题，本文分别建立了机器人从区域中一点到达另一点的避障的最短路径、最短时间路径的非线性0-1整数规划模型。同时，本文为求带有NP属性的非线性0-1整数规划模型，构建了有效启发式算法，利用MATLAB软件编程，求得了O→A、O→B、O→C、O→A→B→A→C的最短路径，同时得到了O→A的最短时间路径，求得的各类最短路径均是全局最优。 针对区域中一点到达另一点的避障的最短路径问题，首先，本文证明了圆弧位置设定在需要绕过障碍物的顶角上，且圆弧半径为10个单位时，能够使得机器人从区域中一点到达另一点的行进路径最短；其次，本文将最短路径选择问题转化成了最短路径的优选问题，根据避障条件，建立了具有较高普适性的避障最短路径的优化模型。为便于求解，本文巧妙地将此优化模型转化成了以可行路径不与障碍物边界相交、不与圆弧相交为约束条件，以机器人从区域中一点达到另一点避障路径最短为目标的0-1规划模型；再次，本文构建了两种有效的启发式算法，利用MATLAB软件编程求得了O→A、O→B、O →C、O→A→B→A→C的最短路径，最短路径长分别为471.0372、853.7001、1088.1952、2725.1596，其中O-->A的最短路径为(0,0)→(70.5063,213.1405) →(75.975,219.1542)→（300,300)，对应圆弧的圆心坐标为(80,210)，O→B的最短路径，对应圆弧的圆心坐标：(60,300)、(150,435)、（220、470）、(220,530)、(150,600)， O→C经过的圆心：(410,100)、(230,60)、(720,520)，（720,600），(500,200)， O→A→B→C→O经过的圆心：(410,100)，(230,60)， (80,210)，(220,530)，(150,600)，(270,680)，(370,680)，(430,680)，(670,730)，(540,730)，(720,520)，(720,600)，(500,200)。 针对最短时间路径问题，我们建立了从o点出发到任意目标点的0-1非线性整数规划模型，同时针对题意要求，具体构建了从o点出发到A的最短时间路径的0-1非线性整数规划模型，利用LINGO软件求解，获得了机器人从o点出发，到达A的最短时间路径，求得最短时间路径下转弯半径为12.9885 ，同时最短时间路径时间长为94.2283个单位。相应圆弧的圆心坐标为（82.1414，207.9153），两切点坐标分别为（69.8045，211.9779）、（77.7492,220.1387）。 本文确定路线思路循序渐进，先建立了有计算避障约束公式的普适性模型，再建立了以不取相交点来简化0-1变量取值关系的简化模型；给出了二种启发式算法，最短路径即最短时间路径具有一定可信度。同时第一个启发算法可以求得全局最优解，第二个启发算法是针对问题的NP属性减少求解时间而构建的，两个算法都具有较重要的意义。 【关键词】机器人避障最短路径启发算法 0-1规划模型

美赛-数学建模-写作模版(各部分)

摘要 第一段：写论文解决什么问题 1．问题的重述 a. 介绍重点词开头： 例1：“Hand move” irrigation, a cheap but labor-intensive system used on small farms, consists of a movable pipe with sprinkler on top that can be attached to a stationary main. 例2:……is a real-life common phenomenon with many complexities. 例3：An (effective plan) is crucial to……… b. 直接指出问题： 例1：We find the optimal number of tollbooths in a highway toll-plaza for a given number of highway lanes: the number of tollbooths that minimizes average delay experienced by cars. 例2：A brand-new university needs to balance the cost of information technology security measures with the potential cost of attacks on its systems. 例3：We determine the number of sprinklers to use by analyzing the energy and motion of water in the pipe and examining the engineering parameters of sprinklers available in the market. 例4: After mathematically analyzing the ……problem, our modeling group would like to present our conclusions, strategies, (and recommendations )to the ……. 例5：Our goal is... that (minimizes the time )………. 2．解决这个问题的伟大意义 反面说明。如果没有…… Without implementing defensive measure, the university is exposed to an expected loss of $8.9 million per year. 3．总的解决概述 a．通过什么方法解决什么问题 例：We address the problem of optimizing amusement park enjoyment through distributing Quick Passes (QP), reservation slips that ideally allow an individual to spend less time waiting in line. b．实际问题转化为数学模型 例1 We formulate the problem as a network flow in which vertices are the locations of escorts and wheelchair passengers. 例2 : A na?ve strategy would be to employ the minimum number of escorts to guarantee that all passengers reach their gates on time. c.将问题分阶段考虑 例3：We divide the jump into three phases: flying through the air, punching through the stack, and landing on the ground. 第二、三段：具体分析 1．在什么模型中/ 建立了什么模型 a. 主流模型 例1：We formulate a differential model to account for the rates of change of these uses, and how this change would affect the overall consumption of water within the studied region.