02 Vehicle Arrival Bandhan[1]

8/3/2019 02 Vehicle Arrival Bandhan[1]

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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



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(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.



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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.



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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.



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. 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.




some characteristics of the normal distribution to observed one.



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Fig 4- Normal time headway distribution



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.



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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



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some characteristics of the random distribution to observed one.

Fig 6- Pearson type III distribution



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.



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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.



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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.



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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.



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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



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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



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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



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= 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,



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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



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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)



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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.



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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



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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:



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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



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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.



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Table 1: Normal distribution table



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Table 2 : Gamma distribution table

02 Vehicle Arrival Bandhan[1]

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