1
Topic 2 – Network Screening
CEE 763
CEE 763
Fall 2011
2
OBJECTIVES

Identify locations for further study which
have both


A high risk of crash losses
An economically justifiable opportunity for
reducing the risk

Identify countermeasure options and
priorities which maximize the economic
benefits

It is as much about exclusion of sites from
consideration as it is about inclusion
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NETWORK SCREENING

Key tool in a highway safety improvement
program

Definition–

A process which aims to identify locations
within the road system where correctable
crashes are found in order to develop
appropriate and cost-effective treatments to
reduce the frequency or severity of crashes
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EFFECTIVENESS

It is important to identify sites with the
most “promise” for improvement as
engineering studies are expensive.
Agencies have limited budgets, and if a
site with potential is not identified, an
opportunity to substantially improve
safety is missed.
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SOME TYPICAL NAMES

High crash location

High accident
potential

Black spot

High risk location

Top 5%

Crash
concentration
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Terms: Site and Facility

Site – a basic safety study location, e.g., a
segment (homogeneous), an intersection, and a
freeway ramp

Facility – a contiguous set of sites


Freeway (segments, ramps)

Urban and suburban arterials (segments, intersections): divided, undivided,
signalized, TWSC etc.

Rural highway (segments): two-lane, multi-lane
HSM only covers predictive methods for certain
facility types
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Fall 2011
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NETWORK SCREENING PROCESS



Establish focus

Sites with potential to reduce crash frequency

Specific crash types or severity
Identify sites and reference population

Type of site: segments, intersections, ramps

Sites of similar characteristics
Select performance measures


Select screening method


Frequency, rate, severity, etc.
Ranking, sliding window, peak searching etc.
Screen and evaluate results
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Fall 2011
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ESTABLISH FOCUS
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PERFORMANCE MEASURES

Crash frequency*

Crash rate*

Quality control*



Excess predicted crash frequency using method of moments
Critical rate
Crash severity*


Equivalent property damage only (EPDO) crash frequency
Relative severity index

Level of service of safety

Excess predicted average crash frequency using SPFs*

Probability of specific crash types exceeding threshold proportion

Excess proportion of specific crash types

Expected crash frequency with EB adjustment*

Excess expected crash frequency with EB adjustment
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CRASH FREQUENCY

Method


Benefits



Rank locations with highest count of crashes for
investigation
Simple
Focuses on areas with most crashes
limitations



Does not account for exposure
Favors high-volume, urban locations
Engineering fix may not be present
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CRASH RATE

Method


Benefits




Rank locations by rate of crashes
Accounts for exposure
Relatively simply
Efforts focused on potential problem not just high volume
locations
Limitations


Favors low volume, low collision sites
Cannot compare cross different volumes
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INTERSECTION RATES

Crashes per million entering vehicles
(MEV)
N
Ri 
MEV
TEV  n  365
MEV 
1,000,000
Ri = intersection crash rate
N = number of crashes in the study period
n = number of years in the study period
TEV = the sum of volumes entering from all approaches,
in Average Daily Traffic
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EXAMPLE
Observed 46 crashes in two years. The ADT
for the minor approach was 3000 and the
major approach was 6000. Note - volumes
includes both directions. What is the crash
rate?
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SEGMENT RATES

Crashes per million vehicle miles of
travel (MVMT)
N
Rs 
MVMT

V  L  n  365
MVMT 
1,000,000
Example

Observed 40 crashes on a 17.5 mile segment
in one year. The ADT was 5,000.
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CRASH AND VOLUME
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FREQUENCY-RATE CRITERIA

Method



Rank by combination of frequency and rate based methods
Various ways to combine rankings for composite rankings
Benefits

Simple
 Address drawbacks of both the frequency and rate methods

Drawbacks

Final ranking dependent of combination
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EXAMPLE

Five intersections have the following crash frequency and
crash rate.
Crash
Data
Intersections
1
2
3
4
5
Frequency
7
12
4
14
10
Rate
0.5
1.5
2.1
1.0
1.8
If a critical frequency is set at 10, and a critical crash rate is
set at 1.5, which intersection(s) should be ranked as high
crash locations?
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QUALITY CONTROL
Rate or Frequency

Method


Rank location if the crash rate or frequency at a site is
statistically significantly higher than a predetermined rate or
frequency for locations of similar characteristics
Benefits

Based on Poisson distribution
 Seems to identify locations with possible treatments

Drawbacks

More data is required
 Categorization is key
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QUALITY CONTROL

Method

1) Select average rate or frequency for similar facility
 2) Calculate the critical rate or frequency
 3) Compare actual rate or frequency
 4) Flag or rank if exceeds
Ra
1
RC  Ra  P

M 2M
RC = critical rate or critical frequency
Ra = the average rate or frequency for similar facility
P = probability constant based on desired level of
significance (1.645 for 95%)
M = millions of VMT or entering vehicles
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EXAMPLE

There were 40 observed crashes on a 17.5
mile segment in one year. The ADT was
5,000. Given the average rate for similar
segments is 1.02 MVMT, does the subject
segment exceed the critical rate at 95%
confidence?
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SEVERITY

Method


Rank locations by weighting the severity of crashes
Benefits

Adds severity to the frequency method
 Usually relates to benefit/cost selection

Drawbacks

Dependent on weighting, may concentrate on fatal collisions
 Weights are essentially arbitrary since it assigned from global
crash costs
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EQUIVALENT PROPERTY DAMAGE ONLY
(EPDO) CRASH FREQUENCY
EPDO   f i Ni
i
EPDO = Equivalent property damage only crashes
fi
= weight for crash type I
Ni
= number of crashes of type i
Severity
Cost
Weight
Fatal (K)
$4,008,900
542
Injury (A,B,C)
$82,600
11
PDO (O)
$7,400
1
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EXAMPLE

A location has experienced 2 fatal, 12 injury A,
30 injury B, 40 injury C, and 140 PDO crashes in
5 years. What is the EPDO crashes?
• Fatal = $3,400,000
• A = $260,000
• B = $56,000
• C = $27,000
• PDO = $4,000
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RELATIVE SEVERITY INDEX (RSI)
n
RSIi 
RSIi
 RSI
j 1
j
Ni
= relative severity index cost for intersection i
RSIj
= relative severity index cost for crash type j
Crash Type
Number of
Cost per Crash
Crashes
Rear End
19
$13,200
Sideswipe
7
$34,000
Angle
5
$61,100
Fixed Object
3
$94,700
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RSI EXAMPLE

An intersection has the following crashes.
Determine the RSI for this intersection
Crash Type
Number of
Crashes
Cost per Crash
Rear End
19
$13,200
Sideswipe
7
$34,000
Angle
5
$61,100
Fixed Object
3
$94,700
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SAFETY INDICES

Method


Benefits


Rank locations by creating an index which includes a number of
factors such as rates, frequencies, severities, and possibly site
data. A weighted average or scores are then combined to
calculate a composite index. The “Relative Severity Index”
discussed earlier is one of these types.
Simple and attempts to combine criteria
Drawbacks

Rank is sensitive to weights of scores which are usually assigned
“arbitrarily”
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*ODOT SAFETY PRIORITY INDEX SYSTEM
(SPIS)



Composite score assigned for frequency,
severity, and rate
3 years data, 0.10 mile sections
Maximum index is 100
• 25 points max for frequency
• 25 points max rate
• 50 points max severity

Total score = Sum of Indicator values (IV) of
Frequency, Rate, and Severity
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*SAFETY PRIORITY INDEX SYSTEM
IVFreq
  LOGTotalCrashes  1

25
 min 25, 

LOG150 1

 

Note: Max SPIS score is 100
 









TotalCrash
es
1
,
000
,
000
  LOG 

  1 


 

 3 yr 365days ADT    

25
IVRate  min 25,


LOG7  1
 


 


 

 100FATAL INJ A   10INJ B  INJC   PDO


50
IVSeverity  min 50, 

300

 

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EXAMPLE

0 Fatal, 1 A, 0 B, 3 C, 4 PDO. ADT 14,200.
IVFreq
  LOGTotalCrashes  1

25
 min 25, 

LOG150 1

 

 









TotalCrash
es
1
,
000
,
000
  LOG 

  1 


 

 3 yr 365days ADT    

25
IVRate  min 25,


LOG7  1
 


 


 

 100FATAL INJ A   10INJ B  INJC   PDO

50
IVSeverity  min 50, 

300

 

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EXAMPLE

0 Fatal, 1 A, 0 B, 3 C, 4 PDO. ADT 14,200.
 LOG8  1 
25  10.95
IVFreq  

 LOG150 1


81,000,000   1 


LOG

  3 yr 365days14,200   
 

25  4.99
IVRate  


LOG7  1




1000  1  103  4
IVSeverity  
50  22.33


300


Answer: SPIS Score = 38.27
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POTENTIAL ACCIDENT REDUCTION

Method


Benefits



Rank or flag locations where the difference between observed
and expected crash experience will maximize benefits if their
crash history can be reduced to the expected value.
Most uses frequency rather than rates
Can account for “regression to the mean”
Drawbacks

Data hungry, expected values must be predicted
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EXCESS PREDICTED CRASH FREQUENCY USING
METHOD OF MOMENTS

Calculate average crash frequency per reference population
 Calculate crash frequency variance
n
VAR 

i 1
obs ,i
 N pa ) 2
n 1
Calculate adjusted observed crash frequency per site
N adj  N obs 

 (N
N pa
VAR
( N pa  N obs )
N adj  adjusted
N pa  populationaverage
N obs  observed
Calculate potential for improvement (PI) per site
PIi  Nadj  N pa

Rank site according to PI (highest to lowest)
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EXAMPLE

An unsignalized intersection has observed 11
crashes in a year. Suppose among all the
unsignalized intersections, the average crashes
per year is 8, and the standard deviation of
crash for all the intersections is 3. Calculate the
PI for this intersection.
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EXCESS PREDICTED CRASH FREQUENCY USING
SAFETY PERFORMANCE FUNCTIONS

Calculate expected crash frequency using SPF
 Calculate excess predicted average crash
frequency
Nexcess,i  Nobs ,i  Nexpected

Rank site according to the excess frequency
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EXAMPLE

An unsignalized intersection has observed 11
crashes in a year. According to the SPF
developed for all the unsignalized intersections,
the predicted crash frequency per year is 8.
What is the excess predicted crash frequency?
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EMPIRICAL BAYES METHODS
E{ k / K }  E( k )  ( 1   )K
1

VAR{ k }
1 Y
E{ k }
1

1  YE(k ) / 
Crash Frequence
K - Observed # of crashes
E{k/K} is best
estimate for the
expected # of
crashes
SPF
E(k) -Modeled # of crashes
E(k) is the predicted value at similar sites,
in crash/year
Y is the analysis period in number of years
Volume
φ is over-dispersion factor
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SAMPLE DATA
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SAMPLE DATA
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Fall 2011
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CRASH FREQUENCY WITH EB ADJUSTMENT

Step 1 – Calculate the predicted average crash
frequency using an SPF

Step 2 – Calculate annual correction factor
Cn 
N predicted,n
N predicted,1
Year
Predicted Average
Correction factor
1
2
3
2.5
2.5
2.7
1.0
1.0
?
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CRASH FREQUENCY WITH EB ADJUSTMENT

Step 3 – Calculate EB weighting factor, 
Note: rely on dispersion factor or variance.

1

1  YE( k ) / 
1
N
1  1 /   N predicted ,n
n 1
Year
Predicted Average
1
2
3
2.5
2.5
2.7

Dispersion factor  1 /   0.49
1
N
1  1 /   N predicted ,n
?
n 1
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CRASH FREQUENCY WITH EB ADJUSTMENT

Step 4 – Calculate first year EB adjusted average
crash frequency.
N
N expected,1  N predicted,1  ( 1   )
N
n 1
observed ,n
N
C
n 1
n
Year
Predicted Average
Observed Crashes
1
2
3
2.5
2.5
2.7
11
9
14
Nexpected,1  ?
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CRASH FREQUENCY WITH EB ADJUSTMENT

Step 5 – Calculate final year EB adjusted average
crash frequency.
Nexpected,n  Nexpected,1 * Cn

Step 6 – Calculate the variance (optional)
Cn
Var( n year )  N expected,n * ( 1   )
 Cn
th

Step 7 - Rank sites based on the EB adjusted
expected average crash frequency for the final year.
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OTHER CRITERIA

Level of service safety (LOSS)
 Konokov et al. (Colorado DOT)

Method of moments
 PIARC manual

Proportions testing
 Exceeding a particular crash type

Rank locations bases on the current annual cost of
crashes based on average cost of crash by
accident type
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WHICH CRITERIA TO USE?

Little consensus on methods

The key issue is how the criteria adopted direct the
analyst to consider sites which contributes to the
overall road safety goal, namely the maximization of
benefits of road safety treatments
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Fall 2011
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METHOD USAGE

All of the methods are in use either alone
or in combination

In US states
• Crash frequency by 15%
• Crash rate or RQC by 15% of agencies
• Crash severities by 50% of agencies
• Indices by 18%
• Other by 16%
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MORE PRECISE DEFINATION OF SITE

Three alternatives (Hauer et al., TRR 1784
– Screening the road network for sites
with promise)
Based on “Section”
 Based on a uniform length of a roadway, e.g.,
0.1 mi
 Based on a minimum segment that identifies
the highest accident frequency while
satisfying the statistical limits (i.e., CV).

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Fall 2011
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SEARCHING ALGORITHMS
Expected
Segment average
Segment average does not
correspond to the highest
Expected
Segment average
CEE 763
Segments of different length
with the highest crash
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48
SLIDING WINDOW
0.3-mile window with 0.1 increment
Roadway Segment
0.0 mi
0.1 mi
0.2 mi
0.3 mi
0.4 mi
0.5 mi
0.6 mi
Win # 1
Win # 2
Win # 3
Win # 4
*The window that has the highest risk is used to rank the segment.
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EXAMPLE

A roadway network has ten segments composed of
three types of facilities. Using the sliding window
method and the crash rate to rank Segments 1 and 2.
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More Data

Segment 1 starts at mile post 1.2 and ends at 2.0.
Segment 2 starts at mile post 2.0 and ends at 2.4.
Segment 1
1.2
1.3
1.4
CEE 763
1.5
Segment 2
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
Fall 2011
2.4
51
SLIDING WINDOW
0.3-mi window with 0.1-mi increment
Site No. 1
Second Sliding Window
W = 0.3 mi
MP 1.0
0.1 mi
MP 2.6
Sliding window is moved incrementally
by 0.1 mi along the roadway segment.
0.2 mi
0.3 mi
0.4 mi
0.5 mi
First Sliding Window
W = 0.3 mi
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SLIDING WINDOW
0.3-mi window with 0.1-mi increment
Site No. 11
MP 21.4
Site No. 12
MP 22.2
MP 23.0
Sliding Window
W = 0.3 mi
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SLIDING WINDOW
0.3-mi window with 0.1-mi increment
Site No. 23
MP 35.4
Site No. 24
MP 36.2
0.1 mi
MP 36.37
Site No. 25
MP 36.7
0.17 mi 0.03 mi
Sliding Window Concepts: Bridging Three Contiguous Roadway Segments
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SLIDING WINDOW
0.3-mi window with 0.1-mi increment
A
Site No. 31
Site No. 32
MP 53.5
MP 54.3
0.2 mi
B
Site No. 32
MP 54.3
0.1 mi
C
Site No. 31
MP 53.5
MP 54.48
0.1 mi
Site No. 31
MP 53.5
Site No. 33
Site No. 33
MP 54.48
0.18 mi
Site No. 32
MP 54.3
Site No. 33
MP 54.48
0.3 mi
Sliding Window Concepts: Window Positions at the End of Contiguous
Roadway Segments When Window is Moved Incrementally by 0.1 Miles
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SLIDING WINDOW
0.3-mi window with 0.1-mi increment
Site No. 22
MP
35.5
Site No. 24
Site No. 23
MP
35.6
MP
36.2
Window No. 1
Site No. 25
MP
36.7
Window No. 9
Window No. 2
Window No. 10
Window No. 3
Window No. 11
Window No. 4
Window No. 12
Window No. 5
Window No. 13
Window No. 6
Window No. 7
Window No. 8
Sliding Window Concepts: Example of Position and Location of Sliding Windows and Subsegments
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SLIDING WINDOW
0.3-mi window with 0.1-mi increment
Site No. 22
MP
35.5
Site No. 24
Site No. 23
MP
35.6
Site No. 25
MP
36.2
Sum( X Y ( EPDO ) )  36.74
Window No. 1
Window No. 2
Window No. 12
Sum( X Y ( EPDO ) )  34.51
Window No. 6
Sum( X Y ( EPDO ) )  33.11
Window No. 11
Sum( X Y ( EPDO ) )  45.96
Window No. 5
Sum( X Y ( EPDO ) )  39.28
Window No. 10
Sum( X Y ( EPDO ) )  50.43
Window No. 4
Sum( X Y ( EPDO ) )  44.85
Window No. 9
Sum( X Y ( EPDO ) )  37.50
Window No. 3
Sum( X Y ( EPDO ) )  24.11
Window No. 13
Sum( X Y ( EPDO ) )  34.51
Sum( X Y ( EPDO ) )  39.28
Window No. 7
Note: Sum(XY(EPDO))
expressed as acc/mi
MP
36.7
Sum( X Y ( EPDO ) )  36.25
Window No. 8
Sum( X Y ( EPDO ) )  46.85
limiting value: 40 acc/mi/yr
Sliding Window Concepts: Ranking Example
CEE 763
Fall 2011
57
EXAMPLE

A segment with 2 lanes, rural

ADT= 6000

Limiting frequency: 10

SPF:
Intercept:-3.63
ADT coefficient: 0.53
Over dispersion Parameter: 0.5
CEE 763
Fall 2011
58
EXAMPLE
0.043
0.147
0.231
0.231
0.240
0.251
0.251
0.287
0.287
0.287
0.310
0.311
0.325
0.329
0.433
0.434
0.440
0.440
0.440
0.441
0.452
0.454
0.483
0.493
CEE
763
Accident locations
(mile)
0.533
0.598
0.636
0.636
0.658
0.743
0.806
Site A: 0-0.4 mile
0.806
0.808
0.822
0.823
0.848
0.862
Site C: 0.9-1 mile
Non contiguous
0.862
Site B: 0.4-0.9 mile
Contiguous
0.901
0.948
0.983
Fall 2011
59
PEAK SEARCHING
0.1-mile window
Roadway Segment
0.0 mi
0.1 mi
0.2 mi
0.3 mi
0.4 mi
0.5 mi
Win # 2
Win # 3
Note:
Window length = 0.1 mi
0.07 mi
0.03 mi
Win # 1
0.6 mi 0.67 mi
Win # 4
Win # 5
Win # 6
Win # 7
CEE 763
Fall 2011
60
PEAK SEARCHING
0.2-mile window
Roadway Segment
0.0 mi
0.1 mi
0.2 mi
0.3 mi
0.4 mi
0.5 mi
0.6 mi 0.67 mi
0.07 mi
0.03 mi
Win # 1
Win # 2
Win # 3
Note:
Window length = 0.2 mi
Win # 4
Win # 5
Win # 6
CEE 763
Fall 2011
61
PEAK SEARCHING
0.4-mile window
Roadway Segment
0.0 mi
0.1 mi
0.2 mi
0.3 mi
0.4 mi
0.5 mi
0.6 mi 0.67 mi
Win # 1
Win # 2
Win # 3
Win # 4
Note:
Window length = 0.4 mi
CEE 763
Fall 2011
62
EXAMPLE

A roadway segment is 0.47 miles long. Using a window
length of 0.1 miles, the following crash data were
obtained for each sub-segment. Calculate the CV for
each sub-segment, and determine whether the search
should continue with longer window sizes (assume the
limiting CV is 0.25).
CEE 763
Fall 2011
63
EXAMPLE-continued
Sub-segment
Position
Excess Expected Crash
Frequency
B1
0.00-0.20
6.50
B2
0.10-0.30
4.45
B3
0.20-0.40
3.80
B4
0.27-0.47
7.15
CEE 763
C.V.
Fall 2011