Lecture 02 Traffic Flow Characteristics (Traffic Engineering هندسة المرور & Dr. Usama Shahdah)

1. TRAFFIC ENGINEERING COURSE
(PWE 8322)
Instructor: Usama Elrawy Shahdah, PhDLecture # 02

2. Contact Information
2
 Email: usama.shahdah@mans.edu.eg,
usama.elrawy@gmail.com
 Office Location: 2nd floor next to the Production
Engineering Block
 Office hours: 10:00 AM – 3:00 PM
On Tuesdays by appointment

3. Course Website
3
 Google Group
https://groups.google.com/d/forum/traffic2015_2016
“Subscribe to this group”
 You will need a Google (i.e., Gmail) Account. You
can create one via:
https://accounts.google.com/SignUp?service=mail&co
ntinue=https%3A%2F%2Fmail.google.com%2Fmail%
2F&ltmpl=default

4.  Two vehicles A and B
 y-axis is distance or
space, and the x-axis
is time.
 Afront represents the
position of the front
bumper of vehicle A
as a function of time
 Lines Arear represent
the position of the
rear bumper of
vehicle A
Traffic Flow Characteristics:
Time-Space Diagram4

5. Headway and Gap
5

6. Headway and Gap
6
 Headway (t): the elapsed time between the front of a
vehicle (vehicle A ) passing a location and the front of the
following vehicle (vehicle B ) passing the same location
 t = tB,f – tA,f
 Time headway is usually measured in seconds.
 Separation Time or Gap (tg): the elapsed time between the
rear of a lead vehicle passing a location and the front of
the following vehicle passing the same location
 tg = tB,f – tA,r
 Separation time is usually measured in seconds.
 Separation time and time headway are different by an amount
equal to the time taken for the lead vehicle’s length to pass a
fixed location
 t = tg + LA/sA where sA is the velocity of vehicle A

7. Spacing and clearance
7
 Distance headway (or Spacing): the distance
between the front of the lead vehicle and the front
of the following vehicle measured at a specific time.
 Distance headway is usually measured in meters.
 Separation distance (or Clearance): the distance
between the rear of the lead vehicle and the front
of the following vehicle measured at a specific time.

8. Vehicle Arrival Patterns
8
 t: time headway
 Time headway distribution is linked to the distribution of the vehicle arrivals
 Poisson distribution is used to model the vehicles’ arrival

9. Poisson Process
9
 The number of events occurring in one segment of time
or space is independent of the number of events in any
previous segment (i.e. a Poisson process has no
memory).
 The mean process rate (e.g. vehicles per hour) is
denoted λ and must remain constant for the entire time
or space span considered
 The shorter the segment of time (or smaller the segment
of space), the less likely it is for more than one event to
occur during that segment.

10. Poisson Distribution (Discrete Distribution)
10
 Probability Mass Distribution Function (PMF)
 A significant property of the Poisson distribution is that:
Mean = Variance

11. Example: Using the Poisson distribution
11
Assume that the average traffic flow rate on a single
lane one-way street is 300 vph. Compute the
probability of observing 0, 1, and ≥ 2 vehicles in any
30 second period.

12. Solution
12
30 Sec

13. Solution: Probability of X vehicles arriving
during a 30 second period
13

14. Solution
14
 What if we did not convert the units of t and ?
 This solution is incorrect and quite different from the
correct solution.

15. Distribution of Time Headways
15
 In the gap acceptance process we are interested in
the availability of time headways, not the number
of vehicles arriving during a fixed time period.
 Fortunately, when the Poisson distribution describes
the arrival of vehicles, then the distribution of time
headways between vehicles can be described by
the Exponential distribution.
 Probability of headways being larger than some
value (Exponential dist.)

16. Exponential Distribution
16
 Continuous distribution
 Probability density function (PDF):
 The distribution of time headways:
( ) x
f x e 
 

 the average arrival rate:
 then the exponential distribution can also be written as

17. Difficulties with using exponential Dist.
17
 Very small time headways are possible
 For a traffic stream with an average arrival rate of 300
vph:
 there is a 4% probability that time headways will be less
than 0.5 seconds and
 an 8% probability that headways will be less than 1.0
seconds.
 In reality, very small headways are not possible as the
headway must be at least as long as the time required
for the length of the lead vehicle to pass the observation
location.

18. Difficulties with using exponential Dist.
18

19. Difficulties with using exponential Dist.
19
 A tractor-trailer truck configuration having:
 a length of 25m and
 travelling at 60 km/h
 requires 1.5 seconds for the vehicle length to pass a fixed
location.
 Naturally, the time for the vehicle length to pass a
location decreases with shorter vehicles and faster
travel speed.
 For example, a car having a length of 6m and travelling at
60 km/h requires only 0.36 seconds to pass a location.
 Drivers require some reaction time and therefore travel
at a separation distance that provides sufficient time to
respond to the actions of the lead vehicle.

20. Notes
20
 This constrain (i.e., very short headway) is only true
when examining time headways for a single lane.
 If data are collected for traffic traveling in more
than one lane and time headways are measured
between consecutive vehicles passing a point
regardless of the lane they are in,
 then very small headways are possible as the
headway is computed between two vehicles that
are not physically acting as a lead and following
vehicle.

21. Shifted Exponential Distribution of Time
Headways
21
 We can modify the exponential distribution to prevent very small
(unrealistic) headways by introducing another parameter, α.
The Mean and the Variance:

22. Example: Using the shifted exponential distribution
22
Consider a traffic stream with a mean arrival rate of
300 vph. If the time headway distribution can be
modeled using a shifted exponential distribution with
α= 2 seconds, then determine
(a) the probability of a time headway being greater
than or equal to 2 seconds; and
(b) the probability of a time headway being greater
than or equal to 3 seconds.

23. Solution
23

24. Shifted Exponential Distribution:
α = 0s, α = 2s, α = 4s
24

25. Hypothesis on an underlying Distribution
25
 How do we confirm that vehicle arrivals follow
Poisson distribution?
 How do we confirm that headway time follow
exponential distribution?
 Solution: use Chi‐Square test
 Null hypothesis: H0
 Use Chi-Square test to accept/reject H0
 Example:
 H0:headways can be assumed to follow Exponential
distribution

26. Type I and Type II Errors
26
 In hypothesis testing, there are four outcomes possible, two of which
lead to incorrect decisions.
Decision
True Situation Do not reject Ho Reject Ho
Ho is true Correct decision Wrong decision
(No error) (Type I error)
Ho is false Wrong decision Correct decision
(Type II error) (No error)
α = P (Type I error) = P (reject H0 / H0 is true)
β = P (Type II error) = P (Do not reject H0 / H0 is false)

27. How to Perform Chi-Square Test
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 Compute the theoretical frequencies (fi, t) for each
category
 Compute the Chi-Square statistics for all categories
( )
 Refer to table of Chi-‐Square distribution
 is Chi-Squared distributed
 We expect low values for if our hypothesis is
correct
2

2

2


28. Chi-square test
28
 The Chi Square statistic is computed as follows,

29. Chi-Square test
29
 When using the Chi Square test, the intervals
for the distributions must be chosen so that fi,t
≥ 5.
 There is no requirement that each interval
represent the same range of values for the time
headways.

30. 30
 The table value for the Chi Square statistic is
dependent on two parameters;
 the selected level of significance (usually 5%) and
 the degrees of freedom

31. Example: Chi Square test
31
 Time headway data from one direction of a two
lane road have been collected over a period of 1
hour and are summarized in table 1 in terms of the
frequency distribution.
 A uniform bin size of 2 seconds has been arbitrarily
chosen.
 The average flow rate during the period of
observation was 300 vph.

32. 32

33. Chi-square example
33
 We assume that the observed headways can be
modelled using an exponential distribution with λ =
300 vph.
 Consequently, we can use the exponential
distribution to estimate the theoretical frequency
distribution (column 4 of Table 1) for each headway
interval.
 Note that the theoretical frequency for the last row
(interval 20) reflects the frequency of headway
greater than 38 seconds.

34. Estimating the theoretical frequency
34
t t/average t P(T>=t) P(t ≤ T ≤ t+Δt)
F(t ≤ T ≤ t+Δt) =
P(t ≤ T ≤ t+Δt) * N
0 0.00 1
2 0.17 0.846482 0.153518275 46.06
4 0.33 0.716531 0.129950414 38.99
6 0.50 0.606531 0.110000651 33.00
8 0.67 0.513417 0.093113541 27.93
10 0.83 0.434598 0.078818911 23.65
12 1.00 0.367879 0.066718767 20.02
14 1.17 0.311403 0.056476217 16.94
16 1.33 0.263597 0.047806086 14.34
18 1.50 0.22313 0.040466978 12.14
20 1.67 0.188876 0.034254557 10.28
22 1.83 0.15988 0.028995857 8.70
24 2.00 0.135335 0.024544463 7.36
26 2.17 0.114559 0.020776439 6.23
28 2.33 0.096972 0.017586876 5.28
30 2.50 0.082085 0.014886969 4.47
32 2.67 0.069483 0.012601547 3.78
34 2.83 0.058816 0.01066698 3.20
36 3.00 0.049787 0.009029403 2.71
38 3.17 0.042144 0.007643225 2.29
>38 0 0.042143844 12.64
N = the total
number of
observed
headways

35. Chi-square example
35
 theoretical frequency shows that in intervals 15 to
19, the estimated frequency is < 5
 therefore we must aggregate some of the intervals
 combine intervals 15 through 19 into a single
interval (representing headways between 28 and
38 seconds)
 The observed frequency for this interval is 18 and
the estimated frequency is 16.4.

36. Observed versus expected frequencies
36

37. Chi-square example
37

38. Chi-square table value
38
 Degrees of freedom
n = (I - 1) – p = (16 – 1) – 1 = 14
 Assume level of significance of 5%
 Chi-square critical value = 23.685

39. Chi-Square table
39

40. Recall acceptance/rejection regions for
Hypothesis test40
Reject Ho
1-
Do not
reject Ho Reject Ho
Two-tailed
1-
Do not
reject Ho
Reject Ho
Left-tailed
1- 
Do not
reject Ho
Reject Ho
Right-tailed
Null Hypothesis: H0 : μ = μo
Alternative Hypothesis
1. Two-tailed test:
H1 : μ  μo
2. Left-tailed test:
H1: μ < μo
3. Right-tailed test:
H1: μ > μo

41. Conclusion
41
 The calculated Chi Square value is less than the
table value
 implying that there is no evidence to reject the null
hypothesis and
 we can safely use the exponential distribution with λ
= 300 vph to model the observed data.

42. Notes on Chi-Square Test
42
 It is possible to show that different distributions could
present the same data
 The test does not prove a distribution
 It does not oppose the desire to assume a distribution
 The test is not directly on the hypothesized distribution
 The test is on the expected versus observed number of
samples
 The actual test is between the histograms
 The size of each category (interval) has a significant role

43. Generating Exponential Headways
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 exponential distribution
 We want to solve for the headway t. set
Then and
 U1, U2, ..., UN are independent uniform random variables between 0 and
1.0.
 Use RAND() function in Microsoft Excel
 To generate exponentially distributed time headways

44. Generating Shifted Exponential
Headways44
 Use the same procedure

45. Macroscopic Measures of a Traffic Stream
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 Speed
 Unit: km/h
 Variable: S
 Flow
 Unit: Veh/h
 Variable: V
 Density
 Unit: Veh/km
 Variable: D

46. Speed
46
 Rate of motion expressed as distance
unit per unit of time
Space mean speed (harmonic average)
Time mean speed (arithmetic average)
Time mean speed >= Space mean
speed
The relationship …

47. Rate of Flow / Volume
47
 Volume: Total number of vehicles passing a given
point during a given time interval
 Interval may be an hour, day, week, or even a year
 Rate of Flow: Equivalent hourly rate at which
vehicles pass a given point during a given time
interval less than one hour

48. Peak-Hour Factor
48

49. Density
49
 Number of vehicles occupying a given
length of roadway, averaged over
time

50. Fundamental Traffic Flow Relationship
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Flow= Density × Speed(SMS)

51. 51

52. Macroscopic Speed-Flow-Density Relationship
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 There is a maximum speed at which vehicles will travel
(free flow speed, Sf)
 The minimum speed vehicles can travel at is zero
 There is a maximum number of vehicles that can
occupy a given segment of roadway (jam density, Dj).
 There must be a condition at which the maximum flow
occurs (Capacity, Vc)
 If vehicles are very close together, people tend to drive
more slowly (as density increases, speeds decrease)

53. Home Reading
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 Chapter #5
Traffic engineering book 3rd edition by Roess et. al

54. Questions
54
Thanks for your time