
Need to understand interaction of
macroscopic measures…
 Speed vs Density
 Flow vs Density
CE 322
Transportation Engineering
Dr. Ahmed Abdel-Rahim, Ph. D., P.E.
There are several
different forms
 Speed vs Flow

Equation 5.14 helps generalize
 If you have model for u vs k can estimate q
Given u vs k…

k 
u  u f 1  
 k 
j 

q vs k and q vs u
are then…

k2 
q  uf k  

k j 


Given

k 
u  u f 1  
 k 
j 

•Solve for q
•What is qcap ?
 Calibration parameters
▪ uf = free flow speed (75 mph)
▪ kj = jam density (60 vpmpl
 Input parameter
▪ k = density (45 vpmpl)
 q=uxk

u2 
q  k j u  

u f 

1
Macroscopic relationships and analyses are
very valuable, but
 Large portion of traffic analysis occurs at the
microscopic level
 Elapsed time between the arrival of
successive vehicles (i.e., time headway)


Simplest approach to modeling vehicle arrivals

Gives deterministic, uniform arrival pattern

Assuming uniform pattern is usually unrealistic

Need a model to represent a random arrival
process
 uniform spacing
 constant time headway between all vehicles
 vehicle arrivals typically follow a random process
 Vehicle queues
 Wait time
What is meant by ‘random’?


For a sequence of events to be considered truly random,
two conditions must be met:
1.
2.
Any point in time is as likely as any other for an event to occur
(e.g., vehicle arrival)
Arrival times are independent of each other


Poisson distribution fits this description
The Poisson distribution:
 Discrete distribution (not continuous)
 Referred to as a ‘counting distribution’
 Represents the count distribution of random
events
time
2
Equation for Poisson dist. is:
P ( n) 




(t ) n e  t
n!

(Eq. 5.23)
P (n) = probability of exactly n vehicles arriving in a time interval t
 = average arrival rate (veh/unit time)
n = # of vehicles arriving in a specific time interval
t = selected time interval (duration of each counting period (unit time))
Interpretation: If you wanted to know the probability of a certain number of
vehicles arriving during a certain length of time then “t” would be the length
of time and “n” would be the number of vehicles. The flow rate would be given
as “λ”. Make sure that you specify λ and t using the same time units.

Setup
Consider a highway with an hourly
flow rate of 120 vehicles per hour,
during which the analyst is
interested in obtaining the
distribution of 1-minute volume
counts.

What is the distribution?
What is the likelihood of seeing 3 vehicles
arrive?
 What is the cumulative likelihood of
seeing fewer than 4 vehicles arrive?
 What is the frequency that you would
expect to see each count for a 60 minute
period?


  = (120 veh/hr) / (3600 sec/hr) = 0.0333 veh/s
 t = 0.0333 veh/sec  60 sec = 2 veh
 OR
  = (120 veh/hr) / (60 min/hr) = 2 veh/min
 t = 2 veh/min  1 min = 2 veh
P(3) 
What is the probability of between 0
and 3 cars arriving (in 1-min interval)?
Pn  4   Pn  0   P (n  1)  Pn  2   Pn  3  0.857
23 e 2
 0.271
3!
3
Examples of using Poisson Distribution
On an intersection approach with a left turn volume of 120
vph. (Poisson), what is the probability of skipping the
green phase for the left turn traffic? The intersection is
controlled by am actuated signal with an average cycle
length of 90 seconds.
m = Average number of left turn vehicles per Cycle
number of cycles per hour = 3600/90 = 40
m = 120 vph / 40 = 3 vehicles/cycle
Probability of x = 0
x m
P(x)  m e
x!
P(0) = 0.049787 = 4.9 % (1.96 cycles/hours)
Examples of using Poisson Distribution
An intersection is controlled by a fixed time signal having a
cycle length of 55 seconds. From the northbound, there is a
permitted left turn movement of 175 vph. If two vehicles can
turn each cycle without causing delay, on what percent of the
cycles will delay occur?
m = Average number of left turn vehicles per Cycle
number of cycles per hour = 3600/55 = 65.46
m = 175 vph / 65.46 = 2.67 LT vehicles/cycle
Probability of x > 2
x m
P(x)  m e
x!

What is the probability of more than 3
cars arriving (in 1-min interval)?
Pn  3  1  Pn  3
3
 1   Pn  i 
i 0
Pn  3  1  0.857
 0.143 or (14.3%)
P (x>2) = 1 – [P(0) + P(1) + P (2)]
= 1 – [0.069 + 0.185 + 0.247] = 49.9% [ 32.67 Cycles/hour]
4
 Poisson distributed vehicle arrivals

▪ implies a distribution of the time between vehicle
arrivals (i.e., time headway)

 Described by negative exponential
distribution.
 qt

P(h  t )  e 3600


P(0)  P(h  t )
Note:
1e

 qt
3600
1
e
veh h veh

sec h sec
Substituting into Poisson equation yields
n

Derive Negative exponential
distribution from Poisson,
q
3600
 qt
 qt  3600

 e
3600 
P ( n)  
n!
Probability of no vehicles
arriving in time t is…
 P(h≥t).
To demonstrate this, let the average arrival rate, ,
be in units of vehicles per second, so that
 qt
3600
x0  1
0!  1
(Eq. 5.25)
Assume vehicle arrivals are Poisson distributed with an hourly traffic flow
of 120 veh/h.
Determine the probability that the headway between successive vehicles
will be less than 8 seconds P(h<8).
Determine the probability that the headway between successive vehicles
will be between 8 and 11 seconds P(8<h<11).
5

By definition, Ph  t   1  Ph  t 
P8  h  11  Ph  11  Ph  8
P h  8  1  Ph  8
Ph  8  1  e
 qt
120(11)
3600
 1 e
 0.234
 0.307  0.234
 0.073
3600
120(8)
3600
 1 e
 1  0.766
 0.234
0.766
0.693
For q = 120 veh/hr
0.307
0.234
6