Transcript
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 1/25
1
1. IntroductionTraffic along with highways can be described by their probabilities or statistical aspects
in addition to theoretical models such as the car following and hydrodynamic models
which deal with deterministic characteristics of traffic flow. Many variables such as
arrival pattern, headway, gaps can be dealt with by the ordinary statistical aspects. As the
vehicle pattern and inter arrival times are very difficult to predict, some distribution
techniques can be applied to illustrate them properly. The study of spacing and headway
characteristics are very important topics of study as they determine operating
characteristics of headway and level of service. Vehicle arrival pattern is not at all
predictable to study the arrival pattern we must use any mathematical distribution
techniques. There are two methods of vehicle arrival study.
1. Continuous distribution technique or headway modeling..2. Discrete or Poisson distribution
Poisson distribution technique deals with arrival rate of the vehicle which is a discrete
variable. Continuous distribution technique deals with the headway or gap which will be
discussed in detail in this study. There various traffic scenarios based on which there are
different distribution techniques used to illustrate them namely Poisson, normal, Pearson
type III distribution. They are discussed in detail in next chapter. Discrete distribution is
also another important phenomenon which is kept as an advanced topic after the
discussion. After analyzing the vehicle arrival patterns and different vehicle headway
modeling still it is not possible to predict the random nature of the vehicle arrival pattern.
The vehicle behavior is independent and also a sequence of vehicle does not follow any
systematic pattern. Thus vehicles almost behave like random numbers. Keeping this fact
in mind in this report following Monte Carlo method vehicles are generated assuming
they are following some distribution techniques with some random numbers.
2. Headway modeling
2.1. Basics of headway modeling
A microscopic view of traffic flow is shown in fig 2.1 as several individual vehicles
traverse a length of roadway for a certain period of time. The arrival time of each vehicle
at the observation point is noted as t1, t2 Etc. The elapsed time between the arrival of pairs
of vehicle defined as the time headways can be shown as
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 2/25
2
(h) 1-2=t2-t1, (h)2-3=t3-t2,….etc.
Observe the time headway actually consists of two time intervals: the occupancy
time for the physical vehicle to pass the observation point and the time gap between
the rear of the lead vehicle and the front of the following vehicle.
Fig 1: Definition of headway
Source: May.A,D. “Traffic flow fundamentals” Prentice hall, 1990 , pp-12
Time headway distribution generally varies directly with the traffic condition or flowrate. We can divide the headway distribution pattern depending on traffic in 3 parts:
1. Under low flow condition.
2. Under intermediate flow condition.
3. Under heavy flow condition.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 3/25
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 4/25
4
Now m is defined as the avg. no. of vehicles arriving in time interval t. The hourly flow
rate is V and t in seconds. Then,
m= (V/3600) t (5)
eqn (3.3) becomes
P (h ≥t) = e-(V/3600)t(6)
The mean time headway can be (μ) can be determined easily so,
P (h ≥t) = e-t/ μ (7)
To calculate probability of a time headway between t and t+Δt
P (t ≤ h ≤ t+Δt)= P (h ≥ t)-P (h≥ t+Δt) (8)
Following the above mentioned procedure it is possible to fit this distribution into
different flow levels to show the characteristics of the distribution. The theoretical results
are superimposed on the measured time headway distribution. Careful study will give
some characteristics of the random distribution to observed one.
Fig 2- Random time headway distributionSource: May.A,D. “Traffic flow fundamentals” Prentice hall, 1990 , pp-19
Some of the important observations are
1. The random distribution has a characteristic of smallest headways occurring most
likely, probabilities continuously decrease with the increase with time headway.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 5/25
5
2. The comparison is best under lowest flow level.
2.3 Constant headway state
When the flow is heavy then, the driver is attempting to maintain a constant headway
But some error is making the headway vary about the mean time headway, and then we
can apply normal distribution,
The lowest theoretical headway, α=µ-2s
where, µ=mean time headway,
s=standard deviation of the time headway distribution
Now the pdf of the normal distribution is
f(t)=(1/√2πσ) exp(-(t-µ)2/2σ2) (9)
From this eqn we can write
P(h≤t)=-∞∫t (1/√2πσ) exp(-(t-µ)2/2σ2) (10)
In order to get the probability when headway is less than t+δt
P( h≤ t+δt)=-∞∫t+δt (1/√2πσ) exp(-(t-µ)2/2σ2) (11)
In order to get the probability in between t and t+δt
P(t≤ h≤ t+δt)= P( h≤ t+δt)- P(h≤t) (12)
The integration of normal distribution is not available in closed form solution. That’s why
one has to numerically integrate or normalize the given distribution having mean µ and
standard deviation σ to a standard normal distribution (µ=0, sd σ =1). Whose values are
available in standard table.
As we know that,
If X∼N (μ,σ2),
Then Z = X−μ / σ
is a normal random variable with mean 0 and variance 1.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 6/25
6
. Such a random variable Z is said to have a standard, or unit, normal distribution. Let
Ф(・) denote its distribution function as
Ф (t)=-∞∫t (1/√2πσ) exp(-(t-µ)2/2σ2)
This result that Z = (X−μ)/ σ has a standard normal distribution when X is normal with
parameters μ andσ² is quite important, it enables us to write all probability statements
about X in terms of probabilities for Z. So we can write,
P(t≤ h≤ t+δt)=P[((t-µ)/σ)≤(h-µ/σ)≤((t+δt)-µ/σ)]
= P(h≤ (t+δt)-µ/σ)- P(h≤ (t-µ)/σ) (13)
Fig 3- Standardized normal distribution
Source: Ross S.M. “Introduction to probability and statistics” Elsevier press, 2003, pp-171
Considering, this we can find out the values using normal table.
Following the above mentioned procedure it is possible to fit this distribution into
different flow levels to show the characteristics of the distribution. The theoretical results
are superimposed on the measured time headway distribution. Careful study will give
some characteristics of the normal distribution to observed one.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 7/25
7
Fig 4- Normal time headway distribution
Source: May.A,D. “Traffic flow fundamentals” Prentice hall, 1990 , pp-23
Some of the important observations are
1. The normal distribution has a characteristic of symmetrical about the mean time
headway and a bell shaped distribution.
2. The comparison is best under highest flow level.
2.3. Intermediate headway state
The intermediate headway state lies between the two boundary conditions random,
constant headway states. This is the situation encountered almost everyday. In this
section it is tried to describe Pearson type III distribution by which we can easily
illustrate this headway state.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 8/25
8
Pearson type III distribution is a generalized mathematical model approach. The pdf of
this distribution is
f(t)=λ/ Г(k)[(t-α)λ]k-1e
-λ(t-α) (14)
f(t)= probability density function
λ= parameter, a function of α, µ,k.
Г(k)= gamma function is equivalent to (k -1)!=(k-1 Г(k -1)
Let k= 3.785
Г(3.785)= 2.785Г(2.785)= 2.785*1.785Г(1.785)= 2.785*1.785*.927(from gamma
distribution table attached)= 0.461
,α= shift parameter
K= shape parameter
Probability of headway greater than t is given as (fig 3.5(b))
P (h≥t) =∞∫t f (t) dt (15)
Thus probability of a headway lying between t and t+δt is (fig 3.5(c))
P(t≤h≤t+δt)=∞∫t f(t) dt-∞∫t+δt f(t)dt (16)
The formulation can be simplified into (fig 3.5(d))
P(t≤h≤t+δt)={[f(t)+f(t+δt)]/2}dt
Fig 5- Probability function of a Pearson type III distribution
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 9/25
9
Following the above mentioned procedure it is possible to fit this distribution into
different flow levels to show the characteristics of the distribution. The theoretical results
are superimposed on the measured time headway distribution. Careful study will give
some characteristics of the random distribution to observed one.
Fig 6- Pearson type III distribution
Source: May.A,D. “Traffic flow fundamentals” Prentice hall, 1990 , pp-26
Some of the important observations are
1. The probability of the theoretical and measured distribution are most inconsistent
when time headways between 1 and 4 sec.
2. The comparison between this and measured distribution at four flow levels
indicates that qualitatively the two are about the same.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 10/25
10
2.4. Numerical illustratation
Fit a (i) negative exponential distribution, (ii) normal distribution, ( iii) Pearson type III
distribution for the following data with scan interval 2 sec, mean time headway 5 sec,
standard deviation 3.9 sec and compute the estimated no. of headways.
No. h h+2.0 observed no. of
headways
1 0 2 20
2 2 4 16
3 4 6 14
4 6 8 5
5 8 10 2
Total no. of vehicles =57
Solution:
(i) Negative exponential distribution
First observed probability of headway= 20/57=0.35
By this procedure we can find out all other observed probabilities.
As we know P (t≤ h) = e-t/μ
So, P (t+2.0≥h)= e-t+2/μ
When t=0 sec, t=2.0 sec, P (t≤ h) = e-t/μ=e
0=1.
(Given, µ=mean headway=5 sec) P (t+2.0≥h)= e-t+2/μ
= e-2/5
=0.67
Therefore, P(t≤h≤t+2.0)= P (t≤ h)-P(t+2.0≥h)=1-0.67=0.33
Now, the no. of headways estimated by negative exponential distribution =0.33*57=19
By this method we can find the estimated no. of headways by negative exponential
distribution which is solved in a tabular manner.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 11/25
11
No. h h+2 no. of
observed
headways
obs
prob.
P (t≤ h) P(t≤h≤t+2.0) Actual
no of
h/w
1 0 2 20 0.35 1 0.33 19
2 2 4 16 0.28 0.67 0.22 12
3 4 6 14 0.25 0.45 0.15 9
4 6 8 5 0.09 0.30 0.10 6
5 8 10 2 0.03 0.20 0.20 11
Total=57 1.00 Total=57
(ii)Normal distribution
Given, α=0.5, mean headway= 5 sec
Standard deviation( for minimum headway)=(5-0.5)/2=2.25 sec
As we know,. P(t≤ h≤ t+2) = P(h≤ (t+2)-µ/ σ)- P(h≤ (t- µ)/ σ)= P( h≤2)-P(h≤0)
=P(h≤(2-5)/2.25)- P(h≤(0-5)/2.25)
= P(h≤-1.33)- P(h≤-2.22)
=0.091-0.013 (from table 1)
= 0.078
Actual no. of headways observed during the first interval is= 0.078*57=4
By the above procedure we can find all the estimated no. of headways. The complete
calculation is done in the following tabular form.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 12/25
12
No. h h+2 Observed
no. of
headways
Observed
prob.
P(t≤h) P(t≤h≤t+2) Actual
no. of
headways
1 0 2 20 0.35 0.01 0.08 4
2 2 4 16 0.28 0.09 0.24 14
3 4 6 14 0.25 0.33 0.34 19
4 6 8 5 0.09 0.67 0.24 14
5 8 10 2 0.03 0.91 0.10 6
Total=57 1.00 Total=57
(iii)Pearson type III distribution
As we know from the pdf of Pearson type III distribution that,
f(t)= λ/ Г(k)[(t-α)λ]k-1
e-λ(t-α)
Now, k=(µ-α)/σ =(5-0.5)/3.9=1.154
,λ=k/(µ-α)=0.256
Г(k)= 0.933 (from table 2)
Putting respective values in the given equation we get,
f(t=2)=0.256/0.933((2-0.5)0.256)1.154-1
e-0.256(2-0.5)
=0.161f(t=4)=(putting t=t+2 in the above eqn)=0.109
P(2≤h≤4)= P(t≤h≤t+δt)={[f(t)+f(t+δt)]/2}dt
=(0.161+0.109)/2*2
= 0.068
Actual no. of headways observed in this interval=0.068*57=4
Thus by applying the above method we can find out all the actual no. of headways found
by this distribution technique. The complete solution is given in the following tabular
manner.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 13/25
13
No. h h+2 Observed
no. of
headways
Observed
prob.
f(t) P(t≤h≤t+2) Actual
no. of
headways
1 0 2 20 0.35 0.16 10
2 2 4 16 0.28 0.16 0.26 14
3 4 6 14 0.25 0.09 0.15 9
4 6 8 5 0.09 0.07 0.12 7
5 8 10 2 0.03 0.04 0.31 17
Total=57 Total=57
The derived probabilities with different distributions are plotted in y axis and shown in a
graphical manner
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
1 2 3 4 5
pearson type IIIobserved
normal
negative exponential
.Fig 7. Comparison of three different types of distributions fitted into same headway data
In fig 7 as we can see the different distributions are shown properly with respect to
observed distribution
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 14/25
14
2.5.Numerical illustratation
Fit a (i) negative exponential distribution, (ii) normal distribution, ( iii) Pearson type III
distribution for the following data with scan interval 0.5 sec, mean time headway 5 sec,
standard deviation 3.9 sec and compute the estimated no. of headways.Total no. of
headways observed = 1320.
No.
h
h+0.5
observed no. of
headways
1 0.0 0.5 81
2 0.5 1.0 873 1.0 1.5 90
4 1.5 2.0 1025 2.0 2.5 87
6 2.5 3.0 1027 3.0 3.5 83
8 3.5 4.0 819 4.0 4.5 65
10 4.5 5.0 36
11 5.0 5.5 4012 5.5 6.0 41
13 6.0 6.5 3314 6.5 7.0 32
15 7.0 7.5 2616 7.5 8.0 20
17 8.0 8.5 2218 8.5 9.0 2419 9.0 9.5 17
20 9.5 10.0 253
Solution:
(ii) Negative exponential distribution
First observed probability of headway= 20/57=0.35
By this procedure we can find out all other observed probabilities.
As we know P (t≤ h) = e-t/μ
So, P (t+0.5≥h)= e-t+2/μ
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 15/25
15
When t=0 sec, t=0.5 sec, P (t≤ h) = e-t/μ=e0=1.
(Given, µ=mean headway=5 sec) P (t+2.0≥h)= e-t+2/μ
= e-0.5/5
=0.9
Therefore, P(t≤h≤t+2.0)= P (t≤ h)-P(t+2.0≥h)=1-0.9=0.095
Now, the no. of headways estimated by negative exponential distribution=0.095*1320=126
By this method we can find the estimated no. of headways by negative exponential
distribution which is solved in a tabular manner.
(ii)Normal distribution
Given, α=0.5, mean headway= 5 sec
Standard deviation( for minimum headway)=(5-0.5)/2=2.25 sec
As we know,. P(t≤ h≤ t+2) = P(h≤ (t+0.5)-µ/ σ)- P(h≤ (t- µ)/ σ)
no
h h+0.5
Obs.headway
obs
prob.
P (t≤ h) P(t≤h≤t+0.5) Actual
no of
h/w
1 0.0 0.5 81 0.061 1.00 0.095 1262 0.5 1.0 87 0.066 0.90 0.086 114
3 1.0 1.5 90 0.068 0.82 0.078 1034 1.5 2.0 102 0.077 0.74 0.070 93
5 2.0 2.5 87 0.066 0.67 0.064 846 2.5 3.0 102 0.077 0.61 0.058 76
7 3.0 3.5 83 0.063 0.55 0.052 698 3.5 4.0 81 0.061 0.50 0.047 629 4.0 4.5 65 0.049 0.45 0.043 56
10 4.5 5.0 36 0.027 0.41 0.039 51
11 5.0 5.5 40 0.030 0.37 0.035 4612 5.5 6.0 41 0.031 0.33 0.032 4213 6.0 6.5 33 0.025 0.30 0.029 38
14 6.5 7.0 32 0.024 0.27 0.026 3415 7.0 7.5 26 0.020 0.25 0.023 31
16 7.5 8.0 20 0.015 0.22 0.021 2817 8.0 8.5 22 0.017 0.20 0.019 25
18 8.5 9.0 24 0.018 0.18 0.017 2319 9.0 9.5 17 0.013 0.17 0.016 2120 9.5 10.0 253 0.192 0.15 0.150 197
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 16/25
16
= P( h≤0.5)-P(h≤0)
=P(h≤(0.5-5)/2.25)- P(h≤(0-5)/2.25)
(from the normal table) = 0.010
Actual no. of headways observed during the first interval is= 0.010*1320=13
By the above procedure we can find all the estimated no. of headways. The complete
calculation is done in the following tabular form.
No. h h+0.5 Observed
no. of
headways
Observed
prob.
P(t≤h) P(t≤h≤t+0.5) Actual
no. of
headways
1 0.0 0.5 81 0.061 0.013 0.010 132 0.5 1.0 87 0.066 0.023 0.015 20
3 1.0 1.5 90 0.068 0.038 0.022 294 1.5 2.0 102 0.077 0.060 0.031 415 2.0 2.5 87 0.066 0.091 0.042 56
6 2.5 3.0 102 0.077 0.133 0.054 717 3.0 3.5 83 0.063 0.187 0.065 86
8 3.5 4.0 81 0.061 0.252 0.076 1009 4.0 4.5 65 0.049 0.328 0.084 110
10 4.5 5.0 36 0.027 0.412 0.088 11611 5.0 5.5 40 0.030 0.500 0.088 116
12 5.5 6.0 41 0.031 0.588 0.084 11013 6.0 6.5 33 0.025 0.672 0.076 10014 6.5 7.0 32 0.024 0.748 0.065 86
15 7.0 7.5 26 0.020 0.813 0.054 7116 7.5 8.0 20 0.015 0.867 0.042 56
17 8.0 8.5 22 0.017 0.909 0.031 4118 8.5 9.0 24 0.018 0.940 0.022 29
19 9.0 9.5 17 0.013 0.962 0.015 2020 9.5 10.0 253 0.192 0.977 0.036 47
(iii)Pearson type III distribution
As we know from the pdf of Pearson type III distribution that,
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 17/25
17
f(t)= λ/ Г(k)[(t-α)λ]k-1
e-λ(t-α)
Now, k=(µ-α)/σ =(5-0.5)/3.9=1.154
,λ=k/(µ-α)=0.256
Г(k)= 0.933 (from gamma distribution table)
Putting respective values in the given equation we get,
f(t=2)=0.256/0.933((2-0.5)0.256) 1.154-1e-0.256(2-0.5)=0.161
f(t=4)=(putting t=t+2 in the above eqn)=0.109
P(2≤h≤2.5)= P(t≤h≤t+δt)={[f(t)+f(t+δt)]/2}dt
=(0.161+0.148)/2*2
= 0.71
Actual no. of headways observed in this interval=0.071*1320=94
Thus by applying the above method we can find out all the actual no. of headways found
by this distribution technique. The complete solution is given in the following tabular
man
No. h h+0.5 Observed
no. of
headways
Observed
prob.
f(t) P(t≤h≤t+0.5) Actual
no. of
headways
1 0.0 0.5 81 0.061 0.000 0
2 0.5 1.0 87 0.066 0.000 0.044 583 1.0 1.5 90 0.068 0.176 0.087 115
4 1.5 2.0 102 0.077 0.172 0.083 1105 2.0 2.5 87 0.066 0.161 0.077 1026 2.5 3.0 102 0.077 0.148 0.071 94
7 3.0 3.5 83 0.063 0.135 0.064 85
8 3.5 4.0 81 0.061 0.122 0.058 779 4.0 4.5 65 0.049 0.110 0.052 69
10 4.5 5.0 36 0.027 0.099 0.047 62
11 5.0 5.5 40 0.030 0.089 0.042 5512 5.5 6.0 41 0.031 0.079 0.037 49
13 6.0 6.5 33 0.025 0.071 0.033 4414 6.5 7.0 32 0.024 0.063 0.030 39
15 7.0 7.5 26 0.020 0.056 0.027 3516 7.5 8.0 20 0.015 0.050 0.024 3117 8.0 8.5 22 0.017 0.044 0.021 28
18 8.5 9.0 24 0.018 0.039 0.019 2519 9.0 9.5 17 0.013 0.035 0.017 22
20 9.5 10.0 253 0.192 0.031 0.166 219
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 18/25
18
3. Vehicle generation
3.1. Background
Until now throughout the study continuous distribution or headway modeling is being
studied with different distribution techniques. Now we will try to study the vehicle
generation . We will try to generate vehicles as per a given distribution so that
whenever a random number is generated a vehicle arrives. Basically here sequence
of outcomes is generated for a continuous random variable. We will use Monte-Carlo
method for this purpose. By this method we can successfully simulate any repetitive,
random event, such as random pattern of vehicle arrival ,repeated application of it
leads to an approximation of the outcome of a system. For a sequence of independent
and uniformly distributed numbers in the 0 to 1 range we can generate vehicles
following certain distribution. Here vehicles are generated assuming their headways
are following some distribution techniques. Here we have generated only such
vehicles follow exponential distribution. Some previously generated random
numbers are used for this purpose.
3.2. Simulation outcomes of continuous random variables
Now if we have a set of random numbers in the 0 to 1 region, say RN. Then we can
transform them to a particular outcome. Let us consider some vehicles with headways
following exponential distribution. For simplicity we are discussing only considering thatthe headways follow negative exponential distribution. Headways can also follow all
other distribution (e.g.: normal distribution).
F( x )=( 1- e-ax
) (18)
In order to transform RN to the given outcome we need to equate RN with eqn 3.1. Using
inverse transformation technique we get,.
RN=( 1- e-ax )
so, e-ax=1- RN
or, -ax=ln(1- RN )
or, x=-1/a ln(1- RN ) (19)
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 19/25
19
3.3. Numerical example
Using the following random numbers generate vehicle arrival for a period of 20 sec.
Assume headways to follow exponential distribution with mean time headway 6 sec.
[0.59, 0.45, 0.26, 0.70,0.14, 0.28, 0.93, 0.76,0.54, 0.45,0.65]
Solution:
a=1/mean headway=1/6
from eqn (3.2), x=-1/a l n(1- RN )
for the first arrival , h1=1/6ln(1-0.59)=5.43 sec
similarly for the second arrival, , h2=1/6ln(1-0.45)=3.57 sec
therefore, cumulative headway=5.43+3.57=9sec
similarly we can find other headways. Calculation is done in the following table.
Vehicle no. RN headway Cumulative
headway
1 0.59 5.43 5.43
2 0.45 3.57 9
3 0.26 1.8 10.8
4 0.70 7.2 18
5 0.28 2 20
4. Conclusion
Today for any type of transportation-roadway related issue proper knowledge on
headway considerations is obvious, which makes our detailed discussion very much
relevant. Classification of headway distribution with three types of traffic scenarios are
discussed . Then the basics of Monte Carlo method and its application to generate vehicle
arrival assuming the headways to follow exponential distribution are discussed. This
Monte-Carlo method forms the heart of the larger simulation models. Vehicle arrival can
also be modeled using Poisson distribution, which is discussed in brief in the next chapter
as an advanced study topic.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 20/25
20
5. References
Kadiyali L.R. “Traffic engineering and transport planning”, Khanna Publishers,
2009, pp. 596-620
Drew.D.R. “Traffic flow theory and control”, McGraw-Hill,1968,pp 124-153
May.A.D. “Traffic flow fundamentals”, Prentice Hall,2002,pp.12-48
Papacostas C.S. “Transportation engineering and planning” ,Prentice , pp-615-620
Ross S.M. “Introduction to probability and statistics” Elsevier press, 2003 ,
pp-171-180
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 21/25
21
6. Advanced study: Poisson distribution of vehicle arrival
6.1. Introduction
The distribution of vehicles in space or time may assume various mathematical forms. In
general it is likely that the spacing in between the vehicles may assume a random and
haphazard manner with gaps of various sizes occurring . Then we can apply Poisson
distribution to illustrate the model. The following traffic conditions need to be satisfied in
order to apply Poisson distribution:
1. An arrival of a vehicle at a given point is completely independent of the
occurrence of any other event. i.e. the number of vehicles arrived during the
previous interval.
2. Equal intervals of time or space are equally likely to contain equal number of
vehicles.
When the above conditions are fulfilled we can use Poisson distribution. It should be kept
in mind that it is applied to the study of number of vehicles because it is a discrete
variable.
The Poisson distribution can be indicated by the following form
P (x) = (mx
e-m
)/ x!
Where ,P (x) = Probability of arrival of x vehicles in any interval of t sec
m= (average rate of arrival) * (time interval)
6.2. Validation of Poisson distribution
In order to check the validity of the Poissonian arrival in a lightly trafficked street free
from the influence of intersections or signals nearby a count of vehicles passing an
observer was taken. The number of vehicle arriving in 20 sec each as counted and
recorded are given below:
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 22/25
22
Number of
Vehicles in time
interval of 20 secs
0 1 2 3 4 5 6
Number of
vehicles intervals
with stated number
of vehicles
- 4 13 23 26 24 14
Total no. of vehicles counted during the study=
(0*1+4*1+2*13+3*23+4*26+5*24+6*14) =408
Total intervals = 4+13+23+26+24+14=104
m= avg. arrival per 20 sec interval=(408/104) =3.92
P (0) = (e-3.92) = 0.02
Intervals with 0 arrival= I (0)=0.02*104=2
P(1)=P(0)m= 0.08 I(1)=8
P(2)=P(1)m/2=.15 I(2)=16
P(3)=P(2)m/3=.2 I(3)=21
P(4)=P(3)m/4=.193 I(4)=20
P(5)=P(4)m/5=.1505 I(5)=16
P(6)=P(5)m/6=.098 I(6)=11
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 23/25
23
No. of
vehicles
0 1 2 3 4 5 6
Observedvalues of
20 sec
interval(O)
- 4 13 23 26 24 14
Expected
values of
20 sec
interval(E)
2 8 16 21 20 16 11
²=∑(O-E)²/E=10.43ּא
C= number of classes =6
Degree of freedom= 4
For a significance level of 5%, χ ²=14.86 (from table 1)
χ ², critical> χ ², observed
We conclude that Poisson distribution represents the observed distribution.
6.3. Limitations of Poisson distribution
• The Poisson distribution only holds good when the traffic is light, not vehicles are
not inhibited by other vehicles, and the flow should be free.
• This is almost impossible to get these conditions right, so we need to find better
way to deal.
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 24/25
24
Table 1: Normal distribution table
8/3/2019 02 Vehicle Arrival Bandhan[1]
http://slidepdf.com/reader/full/02-vehicle-arrival-bandhan1 25/25
25
Table 2 : Gamma distribution table
top related