2010美国大学生数学建模comap#7541

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Problem Chosen

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________________ In the paper, we try to reach the goal of finding out the "center of mass" point. We

build up the probability model, which can be used for the crime problem. Using two different methods of the probability model, the paper gives the satisfying results. Moreover, we adopt a case to test our model, which shows the model is reliable.

Our model consists of four main aspects: marking the locations of the crime on the map, analyzing the geographical profile and finding out the points which have the similar environmental conditions, equating the "center of mass" points through the two probability methods, and combining the results of the two methods and predicting the area of crime.

We used two methods to generate the geographical profile and calculate the probability of the six predicted points.

The first method makes use of the Distance Decay Function to calculate the probability. First, we calculate the distance decay value between the predicted points and the crime points. Second, to sum up all the distance decay value, we obtain the probability of the predicted points. Finally, we conclude that the point which have the maximum of all the values is the "center of mass" point.

The second method is based on the Bayes Theorem equation to calculate the probability. In the beginning, we calculate the prior distribution by means of the ratio between the number of the predicted points and the number of the crime locations. Then we calculate the distribution probability which comes from the multiplication of the independent probability. In addition, we divide the crime region into four small areas to equate the overall probability. By calculating each probability of the four areas, we obtain the overall probability. Finally, we get the probability of the "center of mass" point with the four above probability.

In order to check the reliable of the accomplished model, the paper adopt a case to test it. The case is very similar to the problem the paper faced with. Therefore, it has strong comparability. With the same model, we use two different methods to calculate the "center of mass" point. Compared with the two outcomes, the paper gets the satisfying results. It indicates that the reliability of our model is relatively high.

The Probability Model for Predicting the Area of A Serial Crime Key words: the Bayes Theorem equation; "center of mass"; crime; the Distance Decay Function; reliability; model; the geographical profile.

Contents

1 Introduction 3

1.1 Comprehension to the Problem (3)

1.2 The Goal of the Model (4)

1.3 The Analysis of the Problem (4)

2 Notation and Definitions 4

3 Simplifying Assumptions 4

4 The Model

4.1 Mark the Locations of Crime (5)

4.2 Predict the Locations (7)

4.3 Two Methods to find the "center of mass" (8)

4.3.1 The First Method (8)

4.3.2 The Result of the First Method (8)

4.3.3 The Second Method (10)

4.3.4 The Result of the Second Method (12)

4.3.5 The Combination of the Two Results (13)

5 Case Study: The Test of the Model 14

6 Model Evaluation 16

7 Executive Summary 17

8 Literature References 19

9 Additional Lists 20

1 Introduction

After getting the problem, we search the following sites for relevant information. https://www.wendangku.net/doc/2717758391.html,/wiki/Peter_Sutcliffe

https://www.wendangku.net/doc/2717758391.html,/hiper22/sutcliffe_cf.htm

https://www.wendangku.net/doc/2717758391.html,/group/topic/6142810/

Peter William Sutcliffe (born 2 June 1946 in Bingley, West Riding of Yorkshire) is an English serial killer who was dubbed The Yorkshire Ripper. Sutcliffe committed his first documented assault on the night of 5 July 1975 in Keighley.On 2 January 1981, Sutcliffe was stopped by the police with 24 year old prostitute Olivia Reivers in the driveway of Light Trades House, Melbourne Avenue, Broomhill, Sheffield, South Yorkshire.A police check revealed the car was fitted with false number plates and Sutcliffe was arrested for this offence and transferred to Dewsbury Police Station, Sutcliffe was convicted in 1981 of murdering 13 women and attacking several others.Details are as follows:

Murder Type/Practices - Serial Killer / Sadism.Method/Weapons Used - Stabbing, Strangulation, Bashing / Ball-Pein Hammer, Knives, Claw Hammer, Hacksaw, Screwdrivers, Rope.

Organization - Disorganized

Mobility - Stable..

Victim Vicinity - Yorkshire, Northern England.(Leeds,Bradford,Huddersfield, Manchester)

Murder Time Span - July 1975 - January1981.

Victim Type - Prostitutes.

Victims - Wilma McCann (Died 30 Oct 1975), Emily Jackson (Died 20 Jan 1976), Irene Richardson (Died 5 Feb 1977), Patricia Atkinson (Died 23 Apr 1977), Jayne MacDonald (Died 26 Jun 1977), Jean Jordan (Died 1 Oct 1977), Yvonne Pearson (Died 21 Jan 1978), Helen Rytka (Died 31 Jan 1978), Vera Millward (Died 16 May 1978), Josephine Whitaker (Died 4 Apr 1979), Barbara Leach (Died 2 Sep 1979), Marguerite Walls (Died 18 Aug 1980), Jacqueline Hill (Died 17 Nov 1980) .

There are a lot of murders like this, which lead to plenty of victims everyday. We have developed a number of ways to predict the murder based on these problems in order to make the innocent people away from the suffering.

1.1 Comprehension to the problem

The given problem requires us to solve serial crimes. On one hand, we have to predict the possible locations in the next stages based on the time and locations of the past crime scenes. On the other hand, estimating the reliability of our prediction is what we

have to do. In addition, the given problem also asks us to accomplish additional two-page executive summary which will be read by the chief of police. After the above analysis of the problem's requirements, we understand that how to determine the locations and environment of the past crime scenes is the key factor to solve the problem. The model we have established should find a way to solve this problem. Only in this way can we make the best prediction.

1.2 The goal of the model

The goal of our model is to predict the possible locations of the crime. In order to achieve this goal, we take the following steps.

Step 1: Analyzing the locations and environments of the past crime scenes.

Step 2: Making use of two different schemes to generate a geographical profile. Step 3: Combining the results of the schemes and generate a useful prediction. Step 4: Estimate the reliability of our prediction and accomplish the summary.

1.3 The analysis of the problem

First, we mark the locations of murders and attacks on the map and find the murders and their attack time. Then, using longitude and latitude to represent these points, and regard them as the coordinates of these points. What's more, finding out the “center of mass” is what we have to do. We also have to use two different methods to form a geographical profile. Finally, the probability of next murdering in each region can be predicted. In addition to that, we have to test and verify our model. Thus, the problem analysis is completed.

2 Notation and Definitions

Notation meaning

Xi The past crime points

Yi The predicted points

d(x,y) The distance between x and y

f(d) The distance decay function

S(y) The hit score function

Lon Longitude

Lat Latitude

Won The latitude's actual distance

Wat The longitude's actual distance

Δon The longitude difference

Δat The latitude difference

d Th

e expected value

Zi The divided area

m The "center of mass" point

3 Simplifying Assumptions

1.During a short period, the criminal's mental feature doesn't alter.

2.The criminal doesn't suffer from the unexpected things, such as the accident, the illness and so on.

3.Nature disasters don't appear in this region.

4.The criminal has a stable residence.

4 The model

4.1 Mark the Locations of crime

In order to easily make a forecast, we find out the detailed locations and time of the murders and the attacks. Then we mark them out on the map.

Sutcliffe is convicted of murdering the following 13 victims:

Table.1 locations and time of the crime

Date Name of

victim

Age

at

deat

h

Body found

Loca

tion

on

map

30 October 1975 Wilma

McCann

28 Prince Phillip Playing Fields, Leeds 1

20 January 1976 Emily

Jackson

42 Manor Street, Sheepscar, Leeds at 2

5 February 1977 Irene

Richardson

28 Roundhay Park, Leeds at 3

23 April 1977 Patricia

Atkinson

32 Flat 3, 9 Oak Avenue, Bradford 4

26 June 1977 Jayne

MacDonald

16 Adventure playground, Reginald Street, Leeds 5

1 October 1977 Jean Jordan 20

Allotments next to Southern Cemetery,

Manchester

6

21 January 1978 Yvonne

Pearson

21

Under a discarded sofa on waste ground off

Arthington Street, Bradford

7

31 January 1978 Helen Rytka 18

Timber yard in Great Northern Street,

Huddersfield

8

16 May 1978 Vera

Millward

40 Grounds of Manchester Royal Infirmary 9

4 April 1979 Josephine

Whitaker

19 Savile Park, Halifax 10

2 September 1979 Barbara

Leach

20 Back of 13 Ashgrove, Bradford 11

20 August 1980 Marguerite

Walls

47

Garden of a house called "Claremont", New

Street, Farsley, Leeds

12

17 November 1980 Jacqueline

Hill

20

Waste ground off Alma Road, Headingley，

Leeds

13

13 locations are marked on the following map:

Fig.1 The locations marked on the map

Taking into account the complexity of the locations on the map, we can use the longitude and latitude to represent these points. Taking the locations of attacks into account, we receive a total of 21 points. Then the coordinates of these 21 points can be obtained.

Table.2 The coordinates of 21 locations

point latitude longitude

1 53.7996374 -1.54911

2 53.807214836 -1.522035598

3 53.837335272 -1.49538517

4 53.810002329 -1.7658376694

5 53.81818635 -1.5337085724

6 53.429531854 -2.2586345673

7 53.800403705 -1.7713308334

8 53.653413662 -1.7797207832

9 53.460702343 -2.2254180908

10 53.710348595 -1.8726539612

11 53.780800484 -1.8473768234

12 53.808051104 -1.6720247269

13 53.822955402 -1.5755295753

14 53.8656925 -1.9097452

15 53.914662 -1.9379458

16 53.8375092 -1.4962726

17 53.793853 -1.7524422

18 53.7996374 -1.54911

19 53.7996374 -1.54911

20 53.6451179 -1.784876

21 53.72438 -1.8615766

4.2 Predict the locations

After we have marked out these points on the map, we can find out the environmental conditions near the points.

Table.3 The environmental conditions

point geographical profile

1 park

2 business center

3 near the park,residential area

4 residential area

5 small playground

6 near the cemetery

7 near the Recycle bin

8 lumber yard

9 hospital

10 park

11 residential area,private courtyard

12 residential areas, gardens

13 residential areas, wasteland

Through the analysis of the 13 points, we find that the majority of offenders commit crimes are in the places where there are few people and a relatively open space, such as parks, gardens, wasteland, pile wood field, the cemeteries and so on. Therefore, very few people and open place is the common point of the locations of the crime. Then we can take the park for an example. We can find the six parks that have the similar environmental conditions.

Then we can find out the Longitudes and latitudes of the six places.

Table.4 Six locations predicted

place longitude latitude

Lister Park -1.7781543731689453 53.81083854068482

Pollard Park -1.7348957061767578 53.803438755384576

Victoria Park -2.3321914672851562 53.505558395542735

Middleton park -1.549673080444336 53.75424262648326

Norman Park -1.785020828247070 53.665391639441296

Lister park -1.7774677276611328 53.81585546024039

4.3 Two methods to find the "center of mass"

4.3.1 The first method

Here is a total of 21 points of the crime locations. We use x 1, x 2, x 3, x 4, x 5......x 21 to represent them.

Then the six predicted points can be represented by y 1, y 2, y 3, y 4, y 5, y 6. The distance between the point x and point y will be d(x,y).

We found that the distances are linked to the probability of occurrence of these crimes. We can regard one location as the center. If one point is farther away from the center, the probability of occurrence is smaller. So we can use the distance decay function to represent these relationships. distance decay function [1].

,)(2d

a

d f = (1) Next w

e need to find the reliability o

f the six predicted points. Through the distance decay function, we find the predicted point to these 21 points' f value. Then we add all the values together and we get the construct a hit score function S(y) by computing. This function can be the probability of the predicted point. hit score function:

∑=++==21

1211)).,((...)),(()),(()(i i y x d f y x d f y x d f y S (2)

)(y S reflects possibility of the murder in y point. If )(y S is greater, the possibility

of the murder is greater. If smaller, the possibility is smaller.

There will exist a maximum )(y S of these six predicted points. Then, the point that has the maximum )(y S is the "center of mass". Then we get the "center of mass".

4.3.2 The result of the first method

To solve it through the distance decay function, we have to compute the distances between points. Our coordinates is represented by latitude and longitude. So we have to convert the latitude and longitude into the actual distance. The following is the way to the conversion.

We use Lon to represent the longitude and Lat to represent the latitude.

The latitude's actual distance can be represented by Won and the longitude's actual distance can be represented by Wat.

Δon and Δat represent the longitude difference and the latitude difference. Now we can equate the Wat and Won by means of Lon, Lat, Δon and Δat.

)cos(111Lat on Won ??=Δ. (3)

111?=at Wat Δ. (4)

Through the above equations, we can equate the Wat and Won by means of Lon, Lat, Δon and Δat. Now take the data of the Table.2 the Table.4 into the equations. We

can get the f and )(y S .

Taking the complexity of the data into account, now we take the y 1 point for an example. We put the other points' data into the additional list.(list.1)

Table.5 The data of the y 1 point

point Δat Δon Won Wat d(x i ,y 1) f(d(x i ,y 1)) X 1 -0.011201 0.229044 -23.48998 -1.243327 23.522856 0.0018072 X 2 -0.003623 0.2561187 -26.18346 -0.402231 26.186554 0.0014582 X 3 0.0264967 0.2827692 -28.52664 2.9411372 28.677866 0.0012159 X 4 -0.000836 0.012316 -1.257668 -0.092819 1.2610890 0.6287941 X 5 0.0073478 0.2444458 -24.87264 0.8156068 24.886017 0.0016146 X 6 -0.381306 -0.480480 53.319853 -42.32504 68.076545 0.0002157 X 7 -0.010434 0.0068235 -0.699575 -1.158266 1.3531400 0.5461532 X 8 -0.157424 -0.001566 0.1686226 -17.47416 17.474975 0.0032746 X 9 -0.350136 -0.447263 49.574901 -38.86511 62.993398 0.0002520 X 10 -0.100489 -0.094499 10.010757 -11.15438 14.987846 0.0044518 X 11 -0.030038 -0.069222 7.1533152 -3.334224 7.8922095 0.0160547 X 12 -0.002787 0.1061296 -10.84597 -0.309405 10.850388 0.0084939 X 13 0.0121168 0.2026247 -20.57421 1.3449716 20.618127 0.0023523 X 14 0.0548539 -0.131590 13.097272 6.0887894 14.443353 0.0047936 X 15 0.1038234 -0.159791 15.500586 11.524403 19.315280 0.0026803 X 16 0.0266706 0.2818817 -28.43485 2.9604431 28.588547 0.0012235 X 17 -0.016985 0.0257121 -2.643220 -1.885395 3.2467411 0.0948647 X 18 -0.011201 0.229044 -23.48997 -1.243326 23.522856 0.0018072 X 19 -0.011201 0.2290443 -23.48997 -1.243326 23.522856 0.0018072 X 20 -0.165720 -0.006721 0.7250615 -18.39499 18.409275 0.0029507 X 21 -0.086458 -0.083422 8.7976123 -9.596898 13.01915 0.0058997

We add the f values together and then we get the 1)(y S =0.111932. Using this method, we can equate the other )(y S value.

Table.6 )(y S value point )(y S

Y 1 0.111932 Y 2 0.725086 Y 3 0.001451 Y 4 0.017765 Y 5 0.060981 Y 6

0.082784

From the table.6, we can see that the 2)(y S value of the y 2 point is the maximum. So we can conclude that the possibility of the murder in y 2 point is the greatest. The y 2

point is the "center of mass".

4.3.3 The second method

Suppose only that the offender chooses to offend at the location x with probability density P(x).

The probability density P(x) depends on the following three variables. 1. The "center of mass" point m of the murder.

2. The "center of mass" point must have a well-defined meaning- e.g. the murder ’s place of residence.

3. The "center of mass" point needs to be stable during the crime series. The average distance d is the expected value of the locations of murder.

∑=?=21

1

),(211)(i i y x d y d (5)

We are left with the assumption that a murder with "center of mass" point m

and mean murder distance d commits an offense at the location x with probability density ),(d m x P .

Our method then can be summarized as follows.

Given a sample x 1, x 2, . . . , x 21 (the crime sites) from a probability distribution

),(d m x P , and estimate the parameter m (the "center of mass"). This is a well-studied mathematical problem.

Our approach is the theory of the maximum likelihood. Construct the likelihood function:

),()...,(),(),(21121

1d y x P d y x P d y x P d y F i i ==∏= (6)

This is a Joint Conditional Probability function. Another likelihood function is:

),(ln ...),(ln ),(ln ),(21121

1

d y x P d y x P d y x P d y i i ++==∑=λ (7)

Then the best choice of m is the choice of y that makes the likelihood as large as possible. Then this is the "center of mass".

We use Δi to represent the subtraction between the )(y d and ),(y x d i . Δi reflects the difference between the locations of crime and the expected value.

)(),(y d y x d i i -=Δ (8)

1. Δi is greater than zero.

)

,()

(),(y x d y d d y x P i i =

(9)

2. Δi is smaller than zero.

)

()

,(),(y d y x d d y x P i i =

(10) Put the Eq.9 and Eq.10 into the Eq.6 and Eq.7, we can get the maximal F value and the λ value. Then this point is the "center of mass".

After we have found the "center of mass" point, we should calculate the probability of the points. Here we can use the Bayes Theorem equation [2].

)

,...,()

(),,...,(),...,(211211211x x P m H d m x x P x x d m P ?=

(11)

)(m H is the prior distribution of "center of mass" points.

It represents our knowledge of the probability density for the "center of mass"point m and average murder distance d before we incorporate information about the serial crime.

We will assume t hat the offender’s choice s of crime sites are mutually independent, so that:

),()...,(),,...,(211211d m x P d m x P d m x x P = (12)

To equate the probability ),...,(211x x P , we can divide the crime areas into several small areas. The longitude and latitude is divided into 16 equal portions. By means of the environmental conditions near the conditions of crime and the time of crime, we narrowed the region to four main areas A, B, C and D.

Fig.2 Locations of crime in the areas

Then we equate each point's probability of crime in these four areas. We use )(i x P to represent each area's probability of crime. Then we multiply the four area's probability. So the ),...,(211x x P can be equated.

)()()()()(),...,(432121

1211z P z P z P z P z P x x P i i ???==∏= (13)

By means of these above equations, we can equate the ),...,(211x x d m P . Then we equate the probability of the "center of mass" point.

4.3.4 The result of the second method

In the beginning, we have to equate the expected value )(i y d . ),(y x d i can be acquired from the Table.5 and List.1. So we can equate the )(i y d by means of Eq.5. Data are as following:

Table.7 The expected value

Y 1 Y 2 Y 3 Y 4 Y 5 Y 6 )(i y d 23.4726 22.8237 67.3768

25.781

25.3911

21.7063

Then by means of Eq.8, Eq.9 and Eq.10, we can equate the ),(d y x P i . The data are put into the additional list.(list 2)

By means of Eq.6 and Eq.7, we can equate the F and λ as followings.

Table.8 F and λ

Y 1 Y 2 Y 3 Y 4 Y 5 Y 6

F 1.81E-6 1.96E-8 1.19E-3 1.72E-7 7.7E-6 1.02E-6 λ -13.22 -17.75 -6.73 -15.58 -11.77 -13.8

From the Table.8, we can see that y 3 point's F value is the greatest. So the y 3 point is the "center of mass".

Now we have to equate the probability of the "center of mass" point. Since the y 3 point is the "center of mass", we can use m point to replace y 3 point. By means of Eq.12 and the additional list.2, we can equate the probability

001193.0),,...,(211=d m x x P .

Then 2857.021/6),(==d m H .

The following is what we can conclude from the Fig.2:

Table.9 P(z i )

area The number of crimes P(z i ) A 9 0.42857142857 B 5 0.2380952381 C 5 0.2380952381 D 2 0.0952********

Then 000624372.0)()()()()(),...,(432121

1211=???==∏=z P z P z P z P z P x x P i i

So that,

54594834

.0000624372

.028*******

.0001193062.0)

,...,()

(),,...,(),...,(211211211=?=

?=

x x P m H d m x x P x x d m P Then we get the probability of the "center of mass" point is 0.546.

4.3.5 The combination of the two results

By means of the first method, we know that y 2 point is the "center of mass". But through the other method, we equate that the y3 point is the "center of mass". So we have to combine the two results of the methods.

Taking advantage of the longitude and latitude, we can calculate the distance between

the y2 point and the y3 point. It is about 995m.

Now we use the distance between the two points as the diameter to generate a circle. The area inside this circle can be regarded as the locations of the next possible crime. At this point our model building is completed.

5 Case Study: The Test of Model

Taking the following case for example.

Nickname:The Night Stalker

Reign of terror:17/3/85 - 31/8/85

Motive:He enjoyed it.

Crimes:Ramirez was charged with 63 crimes: including 13 murders, and 30 other offences including attempted murder, rape and sexual assaults.

Table.10 The offences

Date Victim Crime Method Place June 1984 Jennie Vincow Rape and

murder

- -

17/3/85 Maria

Hernandez Attempted

murder

Shot Rosemead, Los

Angeles

17/3/85 Dayle Okasaki Murder Shot Rosemead, Los

Angeles 17/3/85 Tsai Lian Yu Murder Shot Monterey

Park, LA

27/3/85 Vincent

Zazzara Murder Shot Near the San

Gabriel

freeway

27/3/85 Maxine

Zazzara Murder Shot, stabbed

and her eyes

were gouged

out

Near the San

Gabriel

freeway

14/4/85 William Doi Murder Shot in the

head Monterey Park, LA

14/4/85 Lillie Doi Attempted

murder Savagely

beaten

Monterey

Park, LA

30/5/85 Carol Kyle Rape - Burbank, LA 27/6/85 Patty Elaine

Higgins

Murder Slashed throat Arcadia, LA

2/7/85 Mary Louise

Cannon

Murder Slashed throat Arcadia, LA

5/7/85 Whitney

Bennet Attempted

murder

Beaten with a

crowbar

Arcadia, LA

7/7/85 Joyce Lucille

Nelson Murder Beaten to

death and

strangled

Monterey

Park, LA

7/7/85 Sophie

Dickman Rape and

Robbery

- Monterey Park

20/7/85 Max and Lela

Kneiding Murder Shot. Lela was

also stabbed

repeatedly.

Glendale, LA

20/7/85 Chainarong

Khovananth

Murder Shot Sun Valley, LA

20/7/85 Somkid

Khovananth

GBH and rape - Sun Valley, LA

5/8/85 Christopher

and Virginia

Peterson Attempted

murder

- -

9/8/85 Elyas Abowath Murder Shot San Gabriel

Valley

9/8/85 Sakina

Abowath GBH and rape Beaten San Gabriel

Valley

17/8/85 Barbara and

Peter Pan Murder Shot in the

head

Lake Merced

24/8/85 William Carns Attempted

murder Shot 3 times in

the head

Mission Viego

24/8/85 Inez Erickson Rape - Mission Viego The areas of crime on the map are as follows.

Fig.3 The map

The steps of establishing the model are as follows.

Step.1 We mark the points on the map.

Step.2 We analyze the environmental conditions of these points and find out the common features of these points.

Step.3 Then we find out the other points which have the similar environmental conditions.

Step.4 By means of the model, we calculate the "center of mass".

Step.5 We calculate the probability of the "center of mass" point.

Through the comparison of the data, the "center of mass" point is suited with the practical conditions. So the reliability of our model is relevantly high.

6 Model Evaluation

The model we established has advantages and disadvantages. The advantages are as follows:

1.Our model has a very clear structure and a good level. By means of two different schemes to generate a geographic profile, we predict the locations of the crime. So a simple structure is a major feature of this model.

2.Starting from the issues, we find a great deal of information and analyse various situations that we should consider into our model. With model, we also make a case study, which can make our model more practical.

3.We use Excel software to deal with the data, which make our model more objective. Moreover, the calculation of the probability enhances the explanatory power of our model power of the persuasion. So our model is also relatively objective.

4.In order to establish the model, we use many tables and figures, which could make

our model more simple and clear.

Meanwhile, there are disadvantage of our model:

1.During our prediction of the locations of the crime based on the time and locations in the past, we only choose six points to make the prediction. Lack of the predicting points is one of shortcomings of our model.

2.In order to simplify the model, we make some idealized assumptions which might make the results less reliable.

7 Executive Summary

Our model is established based on the background of serial murders. The purpose of this model is to assist the police in forecasting the locations of the possible serial murders.

Firstly, according to the time and locations of the past crime scenes, we analyze the environmental conditions near the locations of murder and investigate the characteristics of the crime time. Secondly, with these characteristics, we find out the locations of the similar environment in the areas where there are frequent murders. These locations of the similar environment are the places where he might make a murder. Thirdly, we use two different methods to predict the locations of the crime. The first method takes advantage of the Distance Decay Function, and the other method is a probability model. Using the model, policeman can predict the next locations of committing the crime as long as the relevant crime data are given. However, it is necessary to pay attention to the following several points when using the model:

a. The model can be used to make forecasts under some conditions.

1. The crime is a serial attacks or murders.

2. The locations of the crime have the common features and have the similar environmental conditions of the crime locations. This is the basement of the model. If the feature has not existed, the model will not work effectively.

3. The locations of the crime must be located between several cities or small areas because the model is based on the small areas between several cities and large areas can't be suit with the model.

4. The model can be used in the conditions where single people's crime as well as the group's crime.

5. The model can be used in the conditions where terrain is not complicated. Remote places suit with the model best.

b. On some conditions, the model can't be used to make forecasts

1.The crime is not a serial attack or murder because single crime can't be satisfied with the conditions the model requirements.

2.The locations of the crime have not the common features or the similar environmental conditions. If the environmental conditions have great differences, the model can't be used to predict.

3.If the area of the crime is very large, this model can't work effectively. The small area of the crime is the premise of our model. So the large area is beyond the model.

4.The model can not be used when many different people commit a crime at the same time. Since different people have different mental activity and habit of crime, this condition can't make the model working effectively.

5.Our model can't be used in the conditions if the terrain is very complicated. The downtown area is not included in the model.

Overview of the potential issues

1.To seek the points.

When establishing the model, we have to seek the points where have the similar environmental conditions in the crime locations. There are several points that we must pay attention to. Firstly, it is very important to analyze the locations of the past crime because it could supply the common features of the environmental conditions. In addition, we have to analyze the environmental conditions near the locations of crime. Secondly, we list the table of the environmental conditions of the past time crime and select one representative environmental condition. Then we should find out the similar environmental conditions based on this representative point, which requires us to select these points not only on the map but also in the actual terrain. Only by doing this can we make an accurate prediction Thirdly, we mark these points on the map and compare the environmental conditions of these points, find out the characteristics of these points and use these them to make a prediction.

2.To establish the probability model.

To establish the probability model is the major method in the model. The model is based on the following points. Firstly, we should take the time of the crime into account. The characteristics of the crime time are the key to our probability model. The interval time between two points should comply with these characteristics. Only by finding the characteristics can we solve this problem. Secondly, the murder's mental activity should be taken into account. What the murder is thinking has a direct impact on the probability of the model. According to the evidence that the murder has left, we can make a conclusion of his or her mental activity. Thirdly, taking into the totality is also very important. The probability of the each places are not isolated. They are linked together. So the totality is the basis of the establishment of the probability model. Only by paying attention to the points above can we establish a good model.

The technical details

1. The time and the locations of crime should be combined together when finding the crime points.

2. All the points should be represented by longitude and latitude.

3. Having a comprehension of the terrain is very important when finding points.

The above is the outline of the model we have established, which can assist the police

to predicate the locations of the serial murder.

8 Literature References

[1] Diao Mingbi: . 1998.

[2] Mao Shisong, Cheng Yiming and Pu Xiaolong: . 2004.7 P42.

9 Additional Lists

List.1

y2 point:

point ΔatΔon Won Wat d(x i,y1) f(d(x i,y1)) X1-0.00380 0.18579 -19.05352 -0.42195 26.18687 0.00146 X20.00378 0.21286 -21.76106 0.41914 28.52973 0.00123 X30.03390 0.23951 -24.16258 3.76251 3.96714 0.06354 X40.00656 -0.03094 3.15951 0.72856 24.88332 0.00162 X50.01475 0.20119 -20.47103 1.63698 53.34498 0.00035 X6-0.37391 -0.52374 58.12036 -41.50367 41.50956 0.00058 X7-0.00304 -0.03644 3.73547 -0.33689 0.37673 7.04578 X8-0.15003 -0.04483 4.82538 -16.65279 52.29709 0.00037 X9-0.34274 -0.49052 54.36971 -38.04374 39.33880 0.00065 X10-0.09309 -0.13776 14.59334 -10.33301 12.56746 0.00633 X11-0.02264 -0.11248 11.62358 -2.51285 11.13327 0.00807 X120.00461 0.06287 -6.42513 0.51197 20.58058 0.00236 X130.01952 0.15937 -16.18179 2.16635 13.27517 0.00567 X140.06225 -0.17485 17.40275 6.91017 16.97111 0.00347 X150.11122 -0.20305 19.69690 12.34578 30.99934 0.00104 X160.03407 0.23862 -24.07113 3.78182 4.61398 0.04697 X17-0.00959 -0.01755 1.80379 -1.06402 23.51406 0.00181 X18-0.00380 0.18579 -19.05352 -0.42195 23.49376 0.00181 X19-0.00380 0.18579 -19.05352 -0.42195 0.83890 1.42095 X20-0.15832 -0.04998 5.39137 -17.57361 19.65273 0.00259 X21-0.07906 -0.12668 13.35962 -8.77552 8.77552 0.01299 Y3 point:

point ΔatΔon Won Wat d(x i,y1) f(d(x i,y1)) X10.29408 0.78308 -80.31 32.64277 86.69056 0.00013 X20.30166 0.81016 -82.8236 33.48386 89.33602 0.00013 X30.33178 0.83681 -84.4196 36.82723 92.10279 0.00012 X40.30444 0.56635 -57.8308 33.79328 66.98053 0.00022 X50.31263 0.79848 -81.2465 34.70170 88.34713 0.00013 X6-0.07603 0.07356 -8.16276 -8.43895 11.74080 0.00725 X70.29485 0.56086 -57.5016 32.72783 66.16302 0.00023 X80.14786 0.55247 -59.4729 16.41193 61.69592 0.00026 X9-0.04486 0.10677 -11.8348 -4.97902 12.83952 0.00607

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