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Transportation Engineering - I
Traffic Flow Characteristics.
Dr. Attaullah Shah
Traffic Flow
• Traffic flow: No of vehicle passing per unit time at a road section
q = n/t
n: No of vehicles t: duration of time, usually an hour
• Time head way: The time between passing of successive vehicles i.e.
bumpers
– t = Σ hi t; duration of time interval hi; time head way of ith vehicle.
– q = n/ Σ hi = Average mean time headway
• The average speed is defined in two ways:
– Arithmetic mean of total speeds at a particular point also called time mean
speed: Mean Ut = Σui/n
– Space Mean Speed: Assumed that the average speed for all vehicles is
measured at the same length of roadway.
– Space mean speed in unit distance per unit time = length of strip/mean time
Example
• The speed of five vehicles is measured with radar at the midpoint of 0.8Km
strip (0.5 mi) The speeds are measured as 44,42,51,49 and 46 mi/h.
Assuming the vehicles were travelling constant speed, determine the time
mean speed.
• Tim Mean speed Ut = Σui/n = (44+42+51+49+46)/5 = 46.4 mi/h
L
nL
n


• The space mean =
t
t
1 / v = 5/ [1/44+1/42+1/51+1/49+1/46]

n
= 46.17 mi/h
Space mean speed is always smaller than time mean speed.
v 
s
i
i
i
i
i
i
Traffic density
• K = n/l
The No of vehicles per unit length.
• The density is also related to individual spacing of the vehicles;
 The total length of roadway l = Σ si
 k= 1/( Mean space)
 Traffic flow = q = uk
 q: flow of traffic No of vehicles/hour
 Speed ( Space mean speed)
 k = density units of vehicle/mile.
 Example 5.2
Traffic Flow
• Complex: between vehicles and drivers,
among vehicles
• Stochastic: variability in outcome, cannot
predict with certainty
• Theories and models
– Macroscopic: aggregate, steady state
– Microscopic: disaggregate, dynamics
– Human factor: driver behavior
Speed (v)
• Rate of motion
• Individual speed
L
L
v
, vavg 
T
T
• Average speed
– Time mean speed
Arithmetic mean
– Space mean speed
Harmonic mean
vt 
v
i
i
n
L
nL
vs 

ti  ti
i n i
vt  vs
Individual Speed
(1)
Time of
Location (ft) Speed
Passing (sec) 600
700 (ft/sec)
700 600
v1 
 50 ft/sec
2.0  0.0
 50* 3600/ 5280 34.09 (mi/hr or mph)
Vehicle
Spot
Speed
1
0.0
2.0
50.0
2
4.4
6.7
45.0
3
6.0
8.0
50.0
4
11.4
14.3
35.0
5
15.0
17.5
40.0
6
17.5
20.0
40.0
7
21.1
23.3
45.0
8
23.3
25.0
60.0
Time Mean Speed
Mile
post
Observation Period
vt 
50  45  50  35  40  40  45  60
 45.6 (ft/sec)
8
Observation Distance
Space Mean Speed
Observation Period
100* 8
vs 
 44.2(ft/sec)  30.1(mi/hr)
2  2.3  2  2.9  2.5  2.5  2.2  1.7
Volume (q)
• Number of vehicles passing a point during
a given time interval
• Typically quantified by Rate of Flow
(vehicles per hour)
Time-Space Plot
1000
900
800
Distance (ft)
700
q
600
500
400
300
200
100
0
0
5
10
15
Time (sec)
20
25
8
 1152 (veh/hr or vph)
25 / 3600
Volume (q)
Density (k)
• Number of vehicles occupying a given
length of roadway
• Typically measured as vehicles per mile
(vpm),
or vehicles per
mile per lane
(vpmpl)
Density (k)
Density (k)
q(veh/hr)  v(mi/hr) * k(veh/mi)
1152 (veh/hr)  30 .1(mi/hr) * k
k  1152 / 30 .1  38 .22 (veh/mi)
Spacing (s)
• Front bumper to front bumper distance
between successive vehicles
S2-3
S1-2
Headway (h)
• Time between successive vehicles passing
a fixed point
T=0 sec T=3sec
h1-2=3sec
Spacing and Headway
spacing
headway
Spacing and Headway
What are the individual headways and the average headway measured at
location A during the 25 sec period?
A
Spacing and Headway
What are the individual headways and the average headway measured at
location A during the 25 sec period?
Time of
Location (ft)
Passing (sec) 600
700
Vehicle
A
1
0.0
2.0
2
4.4
6.7
3
6.0
8.0
4
11.4
14.3
5
15.0
17.5
6
17.5
20.0
7
21.1
23.3
8
23.3
25.0
h1-2
h2-3
Lane Occupancy
• Ratio of roadway occupied by vehicles
L2
L3
LO 
L
i
i
D
L1
D
Clearance (c) and Gap (g)
• Front bumper to back bumper distance
and time
Clearance (ft) or Gap (sec)
g avg  havg 
Lavg
vavg
cavg  gavg * vavg
Spacing (ft) or headway (sec)
Basic Traffic Stream models
1. Speed density Model: Relationship between speed and density
– Initially when there is only one vehicle, the density is very low and driver will
move freely at the speed closely to the design speed of the highway. This is
called free flow speed.
– When more and more vehicles would come, the density of the road will increase
and the speed of vehicles would reduce. The speed of the vehicle may ultimately
reduce to u=0
– This high traffic density is referred as “Jam Density”
– This model can be expressed as: U = Uf ( 1- k/kj)
•
•
•
•
U= Space mean speed in mi/h (km/h)
Uf = Free flow speed in mi/h
k = density in veh/mi ( veh/km)
This gives relationship between traffic flow speed and density
– The speed density relationship tends to be non linear at low density and
very high density.
• Non linear relationship at low densities that has speed slowly declining from free flow
to value
• Linear relationship over the medium density region ( speed declining linearly with the
density).
• Non linear relationship near the traffic jam density relationship
Speed vs. Density

k
u  u f 1 
 k
j

Speed (mph)
uf
Free Flow Speed
kj
Density (veh/mile) Jam Density




Basic Traffic Stream models
• Flow Density Model:
–
–
–
–
As we know that q = uk and U = Uf ( 1- k/kj)
Parabolic density model q = Uf ( k – k2/Kj )
The max flow rate qcap: The highest flow rate
The traffic density that corresponds to this capacity flow rate is kcap
and the corresponding values qcap ,Kcap ,Ucap
• dq/dk = Uf (1-2k/kf )= 0 Since the free flow speed cannot be zero.
• Kcap = Kj / 2
• Substituting the value: Ucap = ( k – Kj /2kj ) = Uf /2
q
c ap
 u
c ap
k
c ap
 u k
f
j
/4
Flow vs. Density
2 

k
q  uf k  


k
j 

Congested Flow
FLow (veh/hr)
Optimal flow,
capacity, qm
Uncongested Flow
km
Optimal density
Density (veh/mile)
kj
Jam Density
Speed Flow Model

k 
u u 
1  k 




u 
k  k 1 

u



u 
q  uk  k u 

u


f
j
j
f
2
j
f
• This is parabolic relationship
• Two speeds are possible as it is quadratic equation, up
to the highway capacity q
• Similarly two densities are possible for given flow
cap
Speed vs. Flow
Speed (mph)
uf
Free Flow Speed
2

u
q  k j u 

u
f





Uncongested Flow
um
Congested Flow
Optimal flow,
Flow (veh/hr) capacity, qm
Example
At a section of highway in Islamabad the free flow speed is 90km/h and capacity of
5000 veh/h. During traffic census at a particular section 2500 vehicles were
counted. Determine the space mean speed of the vehicle.
Solution: The traffic flow is related to the space mean speed and jam density as
The jam
2


k
q  uf k  

k j 

k  4q / u  4  5000 / 90  222veh / km
To determine the jam density
j
f
From above equation of traffic flow we can rearrange the equation as
k
u k u q 0
u
j
2
j
f
222
u  222u  2500  0
90
2.47u  222u  2500  0
u  80.53km / h
2
2
28.36km / h
Both the speeds are feasible as shown on previous slide
Traffic Flow Models-Poisson Model
• The Poisson distribution:
– Is a discrete (as opposed to continuous)
distribution
– Is commonly referred to as a ‘counting
distribution’
– Represents the count distribution of random
events
Poisson Distribution
( t ) e
P ( n) 
n!
n
 t
P(n) = probability of having n vehicles arrive in time t
λ = average vehicle arrival rate in vehicles per unit time
t= duration of time interval over which vehicles are
counted
e= base of the natural logarithm (e=2.718)
Example
• An observer counts 360veh/h at a specific highway location. Assuming
the arrival of vehicles at this point follows Poisson’s distribution.
Estimate the probabilities of 0,1,2,3,4 and 5 vehicles arriving in 20 sec
time interval.
• Solution:
– The average arrival rate =λ= 360 veh/h = 360/60x60=0.1 veh/sec
(  t ) n e  t
P ( n) 
n!
– t=20 sec
( 20  0.1) e
P ( n) 
n!
– For probability of no vehicle
n
0.120
( 20  0.1) e
P (0) 
0!
– P(1) = 0.271 P(2) =0.271 P(3) = 0.180
0
0.120
 0.135
P(4) = 0.090
– For 5 or more vehicles:
– P(n> 5) =1-P(n<5)= 1-(0.135+0.271+0.271+0.18+0.090 = 0.053
– Draw the histogram of the distribution
Poisson Example
• Example:
– Consider a 1-hour traffic volume of 120
vehicles, during which the analyst is
interested in obtaining the distribution of 1minute volume counts
Poisson Example

What is the probability of more than
6 cars arriving (in 1-min interval)?
Pn  6  1  Pn  6
6
 1   Pn  i 
i 0
Pn  6  1  (0.135 0.271 0.271 0.180 0.090 0.036 0.012)
 1  0.995
 0.005or (0.5%)
Poisson Example

What is the probability of between 1
and 3 cars arriving (in 1-min
interval)?
P1  n  3  Pn  1  Pn  2  Pn  3
P1  n  3  27.1%  27.1%  18.0%
 72.2%
Poisson distribution
• The assumption of Poisson distributed
vehicle arrivals also implies a
distribution of the time intervals
between the arrivals of successive
vehicles (i.e., time headway)
Negative Exponential
• To demonstrate this, let the average arrival
rate, , be in units of vehicles per second, so
that
q

veh h veh


sec h sec
3600
Substituting into Poisson equation
yields
( t ) e
P ( n) 
n!
n
 t
n
 qt
3600
 qt 

 e
3600 

P ( n) 
n! (Eq. 5.25)
Negative Exponential
• Note that the probability of having no
vehicles arrive in a time interval of
length t [i.e., P (0)] is equivalent to
the probability of a vehicle headway,
h, being greater than or equal to the
time interval t.
Negative Exponential
• So from Eq. 5.25,
P(0)  P(h  t )

1e

 qt
3600
1
e
(Eq. 5.26)
 qt
3600
This distribution of vehicle headways is known as the
negative exponential distribution.
Note:
x0  1
0!  1
Negative Exponential Example
• Assume vehicle arrivals are Poisson distributed
with an hourly traffic flow of 360 veh/h.
Determine the probability that the headway
between successive vehicles will be less than 8
seconds.
Determine the probability that the headway
between successive vehicles will be between 8
and 11 seconds.
Negative Exponential Example
• By definition, Ph  t   1  Ph  t 
Ph  8  1  Ph  8
Ph  8  1  e
 1 e
 qt
3600
360 ( 8 )
3600
 1  0.4493
 0.551
Negative Exponential Example
P8  h  11  Ph  11  Ph  8
 1  Ph  11  Ph  8
360 (11)
3600
 1 e
 0.551
 1  0.3329 0.551
 0.1161
Negative Exponential
1.0
Prob (h >= t)
0.8
e^(-qt/3600)
0.6
0.4
0.2
0.0
0
5
10
15
20
Time (sec)
For q = 360 veh/hr
25
30
35
Negative Exponential
c.d.f.
1.0
Probability (h < t)
0.8
1 - e^(-qt/3600)
0.6
0.551
0.4
0.2
0.0
0
5
8 10
15
20
Time (sec)
25
30
35
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