Modelling arrival of vehicle at section of road is an important step in traffic flow modelling. It has important application in traffic flow simulation where vehicles are to be generated how vehicles arrive at a section. The vehicle arrival is obviously a random process. This is evident if one observe how vehicles are arriving at a cross section. Some times several vehicles come together, while at other times, they come sparsely. Hence, vehicle arrival at a section need to be characterized statistically. Vehicle arrivals can be modelled in two inter-related ways; namely modelling what is the time interval between the successive arrival of vehicles or modelling how many vehicle arrive in a given interval of time. In the former approach, the random variables the time denoting interval between successive arrival of vehicle can be any positive real values and hence some suitable continuous distribution can be used to model the vehicle arrival. In the later approach, the random variables represent the number of vehicles arrived in a given interval of time and hence takes some integer values. Here in this approach, a discrete distribution can be used to model the process. This chapter presents how some continuous distributions can be used to model the vehicle arrival process.
An important parameter to characterize the traffic is to model the inter-arrival time of vehicle at a section on the road. The inter-arrival time or the time headway is not constant due to the stochastic nature of vehicle arrival. A common way of modeling to treat the inter-arrival time or the time headway as a random variable and use some mathematical distributions to model them. The behavior of vehicle arrival is different at different flow condition. It may be possible that different distributions may work better at different flow conditions. Suppose the vehicle arrive at a point at time t1,t2,…. Then the time difference between two consecutive arrivals is defined as headway. This is shown as a time-distance diagram in figure 1.
In fact the headway consist of two components, the occupancy time which is the duration required for the vehicle to pass the observation point and the time gap between the rear of the lead vehicle and front of the following vehicle. Hence, the headways h1 = t2 - t1, h2 = t3 - t2, … It may be noted that the headways h1, h2, … will not be constant, but follows some random distribution. Further, under various traffic states, different distribution may best explain the arrival pattern. A brief discussion of the various traffic states and suitable distributions are discussed next. Generally, traffic state can be divided into three; namely low, medium and high flow conditions. Salient features of each of the flow state is presented below after a brief discussion of the probability distribution.
The low flow traffic can be modeled using the negative exponential distribution. First, some basics of negative exponential distribution is presented. The probability density function f(t) of any distribution has the following two important properties: First,
![]() | (2) |
This gives an expression for the probability that the random variable t takes a value with in an interval, which is essentially the area under the probability density function curve. The probability density function of negative exponential distribution is given as:
![]() | (3) |
where λ is a parameter that determines the shape of the distribution often called as the shape parameter. The shape of the negative exponential distribution for various values of λ (0.5, 1, 1.5) is shown in figure 2.
The probability that the random variable t is greater than or equal to zero can be derived as follow,
The negative exponential distribution is closely related to the Poisson distribution which is a discrete distribution. The probability density function of Poisson distribution is given as:
![]() | (7) |
where, p(x) is the probability of x events (vehicle arrivals) in some time interval (t), and λ is
the expected (mean) arrival rate in that interval. If the mean flow rate is q vehicles per hour,
then λ = vehicles per second. Now, the probability that zero vehicle arrive in an interval t,
denoted as p(0), will be same as the probability that the headway (inter arrival time) greater
than or equal to t. Therefore,
![]() | (8) |
Since mean flow rate is inverse of mean headway, an alternate way of representing the probability density function of negative exponential distribution is given as
![]() | (9) |
where μ = or λ =
. Here, μ is the mean headway in seconds which is again the inverse of
flow rate. Using equation 6 and equation 5 the probability that headway between
any interval and flow rate can be computed. The next example illustrates how a
negative exponential distribution can be fitted to an observed headway frequency
distribution.
An observation from 2434 samples is given table below. Mean headway and the standard deviation observed is 3.5 and 2.6 seconds respectively. Fit a negative exponential distribution.
h | h + dh | pio |
0.0 | 1.0 | 0.012 |
1.0 | 2.0 | 0.178 |
2.0 | 3.0 | 0.316 |
3.0 | 4.0 | 0.218 |
4.0 | 5.0 | 0.108 |
5.0 | 6.0 | 0.055 |
6.0 | 7.0 | 0.033 |
7.0 | 8.0 | 0.022 |
8.0 | 9.0 | 0.013 |
9.0 | > | 0.045 |
Total | 1.00 | |
Solution: The solution is shown in Table 2. The headway range and the observed probability (or proportion) is given in column (2), (3) and (4). The observed frequency for the first interval (0 to 1) can be computed as the product of observed frequency pi and the number of observation (N). That is, fio = pi × N = 0.012 × 2434 = 29.21 and is shown in column (5). The probability that the headway greater than t = 0 is computed as p(t ≥ 0) = e-0 = 1 (refer equation 5) and is given in column (6). These steps are repeated for the second interval, that is fio = 0.178 × 2434 = 433.25, and p(t ≥ 1) = e-1 = 0.751. Now, the probability of headway lies between 0 and 1 for the first interval is given by the probability that headway greater than zero from the first interval minus probability that headway greater than one from second interval. That is pi(0 ≤ t ≤ 1) = pi(t > 0) - pi(t > 1) = 1.00 - 0.751 = 0.249 and is given in column (7). Now the computed frequency fic is pi ×N = 0.249 × 2434 = 604.904 and is given in column (8). This procedure is repeated for all the subsequent items. It may be noted that probability of headway > 9.0 is computed by 1-probability of headway less than 9.0 = 1 - (0.249 + 0.187 + …) = 0.076.
No | h | h + dh | pio | fio | p(t >= h) | pic | fic |
(1) | (2) | (3) | (4) | (5) | (6) | (7) | (8) |
1 | 0.0 | 1.0 | 0.012 | 29.21 | 1.000 | 0.249 | 604.904 |
2 | 1.0 | 2.0 | 0.178 | 433.25 | 0.751 | 0.187 | 454.572 |
3 | 2.0 | 3.0 | 0.316 | 769.14 | 0.565 | 0.140 | 341.600 |
4 | 3.0 | 4.0 | 0.218 | 530.61 | 0.424 | 0.105 | 256.705 |
5 | 4.0 | 5.0 | 0.108 | 262.87 | 0.319 | 0.079 | 192.908 |
6 | 5.0 | 6.0 | 0.055 | 133.87 | 0.240 | 0.060 | 144.966 |
7 | 6.0 | 7.0 | 0.033 | 80.32 | 0.180 | 0.045 | 108.939 |
8 | 7.0 | 8.0 | 0.022 | 53.55 | 0.135 | 0.034 | 81.865 |
9 | 8.0 | 9.0 | 0.013 | 31.64 | 0.102 | 0.025 | 61.520 |
10 | 9.0 | > | 0.045 | 109.53 | 0.076 | 0.076 | 186.022 |
Total | 2434 | 1.000 | 2434 | ||||
The probability density function of the normal distribution is given by:
![]() | (10) |
where μ is the mean of the headway and σ is the standard deviation of the headways. The shape of the probability density function is shown in figure 4.
The probability that the time headway (t) less than a given time headway (h) is given by
and the value of this is shown as the area under the curve in figure 5 (a) and the probability of time headway (t) less than a given time headway (h + δh) is given by
Although the probability for headway for an interval can be computed easily using equation 13, there is no closed form solution to the equation 11. Even though it is possible to solve the above equation by numerical integration, the computations are time consuming for regular applications. One way to overcome this difficulty is to use the standard normal distribution table which gives the solution to the equation 11 for a standard normal distribution. A standard normal distribution is normal distribution of a random variable whose mean is zero and standard deviation is one. The probability for any random variable, having a mean (μ) and standard deviation (σ) can be computed by normalizing that random variable with respect to its mean and standard deviation and then use the standard normal distribution table. This is based on the concept of normalizing any normal distribution based on the assumption that if t follows normal distribution with mean μ and standard deviation σ, then (t - μ)∕σ follows a standard normal distribution having zero mean and unit standard deviation. The normalization steps shown below.
The first and second term in this equation be obtained from standard normal distribution table. The following example illustrates this procedure.If the mean and standard deviation of certain observed set of headways is 2.25 and 0.875 respectively, then compute the probability that the headway lies in an interval of 1.5 to 2.0 seconds.
Solution: The probability that headway lies between 1.5 and 2.0 can be obtained using equation 14, given that μ = 2.25 and σ = 0.85 as:
Suppose α is the minimum possible headway and if we set α = μ - σ than about 60% of headway will be greater than α. Alternatively, if we set α = μ - 2σ, than about 90% of the headway will be greater than α. Further, if we set α = μ- 3σ, than about 99% of the headway will be greater than α. To generalize,
Given that observed mean headway is 3.5 seconds and standard distribution is 2.6 seconds, then compute the probability that the headway lies between 0 and 0.5. Assume that the minimum expected headway is 0.5 seconds.
Solution: First, compute the standard deviation to be used in calculation using equation 15, given that μ = 3.5, σ = 2.6, and α = 0.5. Then:
![]() | (16) |
Second, compute the probability that headway less than zero.
An observation from 2434 samples is given table below. Mean headway observed was 3.5 seconds and the standard deviation observed was 2.6 seconds. Fit a normal distribution, if we assume minimum expected headway is 0.5.
h | h + dh | pio |
0.0 | 1.0 | 0.012 |
1.0 | 2.0 | 0.178 |
2.0 | 3.0 | 0.316 |
3.0 | 4.0 | 0.218 |
4.0 | 5.0 | 0.108 |
5.0 | 6.0 | 0.055 |
6.0 | 7.0 | 0.033 |
7.0 | 8.0 | 0.022 |
8.0 | 9.0 | 0.013 |
9.0 | > | 0.045 |
Total | 1.00 | |
Solutions The given headway range and the observed probability is given in column (2), (3) and (4). The observed frequency for the first interval (0 to 1) can be computed as the product of observed frequency pi and the number of observation (N) i.e. pio = pi × N = 0.012 × 2434 = 29.21 as shown in column (5). Compute the standard deviation to be used in calculation, given that μ = 3.5, σ = 2.6, and α = 0.5 as:
No | h | h + δ h | pio | fio = pio × N | p(t ≤ h) | p(t < h < t + δ h) | fic = pic × N |
(1) | (2) | (3) | (4) | (5) | (6) | (7) | (8) |
1 | 0.0 | 1.0 | 0.012 | 29.21 | 0.010 | 0.038 | 92.431 |
2 | 1.0 | 2.0 | 0.178 | 433.25 | 0.048 | 0.111 | 269.845 |
3 | 2.0 | 3.0 | 0.316 | 769.14 | 0.159 | 0.211 | 513.053 |
4 | 3.0 | 4.0 | 0.218 | 530.61 | 0.369 | 0.261 | 635.560 |
5 | 4.0 | 5.0 | 0.108 | 262.87 | 0.631 | 0.211 | 513.053 |
6 | 5.0 | 6.0 | 0.055 | 133.87 | 0.841 | 0.111 | 269.845 |
7 | 6.0 | 7.0 | 0.033 | 80.32 | 0.952 | 0.038 | 92.431 |
8 | 7.0 | 8.0 | 0.022 | 53.55 | 0.990 | 0.008 | 20.605 |
9 | 8.0 | 9.0 | 0.013 | 31.64 | 0.999 | 0.001 | 2.987 |
10 | 9.0 | > | 0.045 | 109.53 | 1.000 | 0.010 | 24.190 |
Total | 2434 | ||||||
As noted earlier, the intermediate flow is more complex since certain vehicles will have interaction with the other vehicles and certain may not. Here, Pearson Type III distribution can be used for modelling intermediate flow. The probability density function for the Pearson Type III distribution is given as
where λ is a parameter which is a function of μ, K and α, and determine the shape of the distribution. The term μ is the mean of the observed headways, K is a user specified parameter greater than 0 and is called as a shift parameter. The Γ() is the gamma function and given as
Step wise procedure to fit a Pearson Type III distribution
![]() | (23) |
Although the closed form solution of Γ(K) is available, it is difficult to compute. Hence, it can be obtained from gamma table. For, example:
The Gamma function is defined as:
![]() | (24) |
The Gamma function can be evaluated by the following approximate expression also:
![]() | (25) |
An observation from 2434 samples is given table below. Mean headway observed was 3.5 seconds and the standard deviation 2.6 seconds. Fit a Person Type III Distribution.
h | h + dh | pio |
0.0 | 1.0 | 0.012 |
1.0 | 2.0 | 0.178 |
2.0 | 3.0 | 0.316 |
3.0 | 4.0 | 0.218 |
4.0 | 5.0 | 0.108 |
5.0 | 6.0 | 0.055 |
6.0 | 7.0 | 0.033 |
7.0 | 8.0 | 0.022 |
8.0 | 9.0 | 0.013 |
9.0 | > | 0.045 |
Total | 1.00 | |
No | h | h + δ h | pio | fio | f(t) | pic | fic |
(1) | (2) | (3) | (4) | (5) | (6) | (7) | (8) |
1 | 0 | 1 | 0.012 | 29.2 | 0.000 | 0.132 | 321.2 |
2 | 1 | 2 | 0.178 | 433.3 | 0.264 | 0.238 | 580.1 |
3 | 2 | 3 | 0.316 | 769.1 | 0.213 | 0.185 | 449.5 |
4 | 3 | 4 | 0.218 | 530.6 | 0.157 | 0.134 | 327.3 |
5 | 4 | 5 | 0.108 | 262.9 | 0.112 | 0.096 | 233.4 |
6 | 5 | 6 | 0.055 | 133.9 | 0.079 | 0.068 | 164.6 |
7 | 6 | 7 | 0.033 | 80.3 | 0.056 | 0.047 | 115.3 |
8 | 7 | 8 | 0.022 | 53.6 | 0.039 | 0.033 | 80.4 |
9 | 8 | 9 | 0.013 | 31.6 | 0.027 | 0.023 | 55.9 |
10 | >9 | 0.045 | 109.5 | 0.019 | 0.044 | 106.4 | |
Total | 1.0 | 2434 | 1.0 | 2434 | |||
Solutions Given that mean headway (μ) is 3.5 and the standard deviation (σ) is 2.6. Assuming the expected minimum headway (α) is 0.5, K can be computed as
This chapter covers how the vehicle arrival can be modelled using various distributions. The negative exponential distribution is used when the traffic is low and is most simplest of the distributions in terms of computation effort. The normal distribution on the other hand is used for highly congested traffic and its evaluation require standard normal distribution tables. The Pearson Type III distribution is a most general kind of distribution and can be used intermediate or normal traffic conditions.
t | t + δt | Observed Proportion |
0.0 | 1.0 | 191 |
1.0 | 2.0 | 131 |
2.0 | 3.0 | 170 |
3.0 | 4.0 | 98 |
4.0 | 5.0 | 82 |
5.0 | 6.0 | 81 |
6.0 | 7.0 | 44 |
> 7.0 | 2 | |
h | h + dh | piobs |
0.0 | 0.5 | 0.012 |
0.5 | 1.0 | 0.064 |
1.0 | 1.5 | 0.114 |
1.5 | 2.0 | 0.159 |
2.0 | 2.5 | 0.157 |
2.5 | 3.0 | 0.130 |
3.0 | 3.5 | 0.088 |
3.5 | 4.0 | 0.065 |
4.0 | 4.5 | 0.043 |
4.5 | 5.0 | 0.033 |
5.0 | 5.5 | 0.022 |
5.5 | 6.0 | 0.019 |
6.0 | 6.5 | 0.014 |
6.5 | 7.0 | 0.010 |
7.0 | 7.5 | 0.012 |
7.5 | 8.0 | 0.008 |
8.0 | 8.5 | 0.005 |
8.5 | 9.0 | 0.007 |
9.0 | 9.5 | 0.005 |
9.5 | > | 0.033 |
Total | 1.00 | |
h | h+dh | piobs |
0.0 | 0.5 | 0.012 |
0.5 | 1.0 | 0.064 |
1.0 | 1.5 | 0.114 |
1.5 | 2.0 | 0.159 |
2.0 | 2.5 | 0.157 |
2.5 | 3.0 | 0.130 |
3.0 | 3.5 | 0.088 |
3.5 | 4.0 | 0.065 |
4.0 | 4.5 | 0.043 |
4.5 | 5.0 | 0.033 |
5.0 | 5.5 | 0.022 |
5.5 | 6.0 | 0.019 |
6.0 | 6.5 | 0.014 |
6.5 | 7.0 | 0.010 |
7.0 | 7.5 | 0.012 |
7.5 | 8.0 | 0.008 |
8.0 | 8.5 | 0.005 |
8.5 | 9.0 | 0.007 |
9.0 | 9.5 | 0.005 |
9.5 | > | 0.033 |
Total | 1.00 | |
h | h+dh | piobs |
0.0 | 0.5 | 0.012 |
0.5 | 1.0 | 0.064 |
1.0 | 1.5 | 0.114 |
1.5 | 2.0 | 0.159 |
2.0 | 2.5 | 0.157 |
2.5 | 3.0 | 0.130 |
3.0 | 3.5 | 0.088 |
3.5 | 4.0 | 0.065 |
4.0 | 4.5 | 0.043 |
4.5 | 5.0 | 0.033 |
5.0 | 5.5 | 0.022 |
5.5 | 6.0 | 0.019 |
6.0 | 6.5 | 0.014 |
6.5 | 7.0 | 0.010 |
7.0 | 7.5 | 0.012 |
7.5 | 8.0 | 0.008 |
8.0 | 8.5 | 0.005 |
8.5 | 9.0 | 0.007 |
9.0 | 9.5 | 0.005 |
9.5 | > | 0.033 |
Total | 1.00 | |
I wish to thank several of my students and staff of NPTEL for their contribution in this lecture. I also appreciate your constructive feedback which may be sent to tvm@civil.iitb.ac.in
Prof. Tom V. Mathew
Department of Civil Engineering
Indian Institute of Technology Bombay, India
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Thursday 31 August 2023 12:12:44 AM IST