TRAFFIC ENGINEERING COURSE

(PWE 8322)

Instructor: Usama Elrawy Shahdah, PhDLecture # 02

Contact Information2

Email: usama.shahdah@mans.edu.eg,

usama.elrawy@gmail.com

Office Location: 2nd floor next to the Production

Engineering Block

Office hours: 10:00 AM – 3:00 PM

On Tuesdays by appointment

Course Website3

Google Group

https://groups.google.com/d/forum/traffic2015_2016

“Subscribe to this group”

You will need a Google (i.e., Gmail) Account. You

can create one via:

https://accounts.google.com/SignUp?service=mail&co

ntinue=https%3A%2F%2Fmail.google.com%2Fmail%

2F&ltmpl=default

Two vehicles A and B

y-axis is distance or space, and the x-axis is time.

Afront represents the position of the front bumper of vehicle A as a function of time

Lines Arear represent the position of the rear bumper of vehicle A

Traffic Flow Characteristics:

Time-Space Diagram4

Headway and Gap5

Headway and Gap6

Headway (t): the elapsed time between the front of avehicle (vehicle A ) passing a location and the front of thefollowing vehicle (vehicle B ) passing the same location

t = tB,f – tA,f

Time headway is usually measured in seconds.

Separation Time or Gap (tg): the elapsed time between the rear of a lead vehicle passing a location and the front of the following vehicle passing the same location

tg = tB,f – tA,r

Separation time is usually measured in seconds.

Separation time and time headway are different by an amount equal to the time taken for the lead vehicle’s length to pass a fixed location

t = tg + LA/sA where sA is the velocity of vehicle A

Spacing and clearance7

Distance headway (or Spacing): the distance

between the front of the lead vehicle and the front

of the following vehicle measured at a specific time.

Distance headway is usually measured in meters.

Separation distance (or Clearance): the distance

between the rear of the lead vehicle and the front

of the following vehicle measured at a specific time.

Vehicle Arrival Patterns 8

t: time headway

Time headway distribution is linked to the distribution of the vehicle arrivals

Poisson distribution is used to model the vehicles’ arrival

Poisson Process9

The number of events occurring in one segment of timeor space is independent of the number of events in anyprevious segment (i.e. a Poisson process has nomemory).

The mean process rate (e.g. vehicles per hour) is denoted λ and must remain constant for the entire time or space span considered

The shorter the segment of time (or smaller the segment of space), the less likely it is for more than one event to occur during that segment.

Poisson Distribution (Discrete Distribution)

10

Probability Mass Distribution Function (PMF)

A significant property of the Poisson distribution is that:

Mean = Variance

Example: Using the Poisson distribution

11

Assume that the average traffic flow rate on a single

lane one-way street is 300 vph. Compute the

probability of observing 0, 1, and ≥ 2 vehicles in any

30 second period.

Solution12

30 Sec

Solution: Probability of X vehicles arriving

during a 30 second period 13

Solution14

What if we did not convert the units of t and ?

This solution is incorrect and quite different from the correct solution.

Distribution of Time Headways 15

In the gap acceptance process we are interested in

the availability of time headways, not the number

of vehicles arriving during a fixed time period.

Fortunately, when the Poisson distribution describes

the arrival of vehicles, then the distribution of time

headways between vehicles can be described by

the Exponential distribution.

Probability of headways being larger than some

value (Exponential dist.)

Exponential Distribution 16

Continuous distribution

Probability density function (PDF):

The distribution of time headways:

( ) xf x e

the average arrival rate:

then the exponential distribution can also be written as

Difficulties with using exponential Dist.17

Very small time headways are possible

For a traffic stream with an average arrival rate of 300 vph:

there is a 4% probability that time headways will be less than 0.5 seconds and

an 8% probability that headways will be less than 1.0 seconds.

In reality, very small headways are not possible as the headway must be at least as long as the time required for the length of the lead vehicle to pass the observation location.

Difficulties with using exponential Dist.18

Difficulties with using exponential Dist.19

A tractor-trailer truck configuration having:

a length of 25m and

travelling at 60 km/h

requires 1.5 seconds for the vehicle length to pass a fixed location.

Naturally, the time for the vehicle length to pass a location decreases with shorter vehicles and faster travel speed.

For example, a car having a length of 6m and travelling at 60 km/h requires only 0.36 seconds to pass a location.

Drivers require some reaction time and therefore travel at a separation distance that provides sufficient time to respond to the actions of the lead vehicle.

Notes20

This constrain (i.e., very short headway) is only truewhen examining time headways for a single lane.

If data are collected for traffic traveling in morethan one lane and time headways are measuredbetween consecutive vehicles passing a pointregardless of the lane they are in,

then very small headways are possible as theheadway is computed between two vehicles thatare not physically acting as a lead and followingvehicle.

Shifted Exponential Distribution of Time

Headways 21

We can modify the exponential distribution to prevent very small

(unrealistic) headways by introducing another parameter, α.

The Mean and the Variance:

Example: Using the shifted exponential distribution

22

Consider a traffic stream with a mean arrival rate of

300 vph. If the time headway distribution can be

modeled using a shifted exponential distribution with

α= 2 seconds, then determine

(a) the probability of a time headway being greater

than or equal to 2 seconds; and

(b) the probability of a time headway being greater

than or equal to 3 seconds.

Solution23

Shifted Exponential Distribution:

α = 0s, α = 2s, α = 4s24

Hypothesis on an underlying Distribution

25

How do we confirm that vehicle arrivals follow Poisson distribution?

How do we confirm that headway time follow exponential distribution?

Solution: use Chi‐Square test

Null hypothesis: H0

Use Chi-Square test to accept/reject H0

Example:

H0:headways can be assumed to follow Exponential distribution

Type I and Type II Errors26

In hypothesis testing, there are four outcomes possible, two of which lead to incorrect decisions.

Decision

True Situation Do not reject Ho Reject Ho

Ho is true Correct decision Wrong decision

(No error) (Type I error)

Ho is false Wrong decision Correct decision

(Type II error) (No error)

α = P (Type I error) = P (reject H0 / H0 is true)

β = P (Type II error) = P (Do not reject H0 / H0 is false)

How to Perform Chi-Square Test27

Compute the theoretical frequencies (fi, t) for each

category

Compute the Chi-Square statistics for all categories

( )

Refer to table of Chi-‐Square distribution

is Chi-Squared distributed

We expect low values for if our hypothesis is

correct

2

22

Chi-square test28

The Chi Square statistic is computed as follows,

Chi-Square test29

When using the Chi Square test, the intervals

for the distributions must be chosen so that fi,t

≥ 5.

There is no requirement that each interval

represent the same range of values for the time

headways.

30

The table value for the Chi Square statistic is

dependent on two parameters;

the selected level of significance (usually 5%) and

the degrees of freedom

Example: Chi Square test

31

Time headway data from one direction of a two

lane road have been collected over a period of 1

hour and are summarized in table 1 in terms of the

frequency distribution.

A uniform bin size of 2 seconds has been arbitrarily

chosen.

The average flow rate during the period of

observation was 300 vph.

32

Chi-square example33

We assume that the observed headways can bemodelled using an exponential distribution with λ =300 vph.

Consequently, we can use the exponentialdistribution to estimate the theoretical frequencydistribution (column 4 of Table 1) for each headwayinterval.

Note that the theoretical frequency for the last row(interval 20) reflects the frequency of headwaygreater than 38 seconds.

Estimating the theoretical frequency 34

t t/average t P(T>=t) P(t ≤ T ≤ t+Δt)F(t ≤ T ≤ t+Δt) =

P(t ≤ T ≤ t+Δt) * N

0 0.00 1

2 0.17 0.846482 0.153518275 46.06

4 0.33 0.716531 0.129950414 38.99

6 0.50 0.606531 0.110000651 33.00

8 0.67 0.513417 0.093113541 27.93

10 0.83 0.434598 0.078818911 23.65

12 1.00 0.367879 0.066718767 20.02

14 1.17 0.311403 0.056476217 16.94

16 1.33 0.263597 0.047806086 14.34

18 1.50 0.22313 0.040466978 12.14

20 1.67 0.188876 0.034254557 10.28

22 1.83 0.15988 0.028995857 8.70

24 2.00 0.135335 0.024544463 7.36

26 2.17 0.114559 0.020776439 6.23

28 2.33 0.096972 0.017586876 5.28

30 2.50 0.082085 0.014886969 4.47

32 2.67 0.069483 0.012601547 3.78

34 2.83 0.058816 0.01066698 3.20

36 3.00 0.049787 0.009029403 2.71

38 3.17 0.042144 0.007643225 2.29

>38 0 0.042143844 12.64

N = the total

number of

observed

headways

Chi-square example35

theoretical frequency shows that in intervals 15 to

19, the estimated frequency is < 5

therefore we must aggregate some of the intervals

combine intervals 15 through 19 into a single

interval (representing headways between 28 and

38 seconds)

The observed frequency for this interval is 18 and

the estimated frequency is 16.4.

Observed versus expected frequencies36

Chi-square example37

Chi-square table value38

Degrees of freedom

n = (I - 1) – p = (16 – 1) – 1 = 14

Assume level of significance of 5%

Chi-square critical value = 23.685

Chi-Square table39

Recall acceptance/rejection regions for

Hypothesis test40

Reject Ho

1-

Do not

reject HoReject Ho

Two-tailed

1-

Do not

reject HoReject Ho

Left-tailed

1-

Do not

reject Ho

Reject Ho

Right-tailed

Null Hypothesis: H0 : μ = μo

Alternative Hypothesis1. Two-tailed test:

H1 : μ μo

2. Left-tailed test:

H1: μ < μo

3. Right-tailed test:

H1: μ > μo

Conclusion41

The calculated Chi Square value is less than the

table value

implying that there is no evidence to reject the null

hypothesis and

we can safely use the exponential distribution with λ= 300 vph to model the observed data.

Notes on Chi-Square Test42

It is possible to show that different distributions could

present the same data

The test does not prove a distribution

It does not oppose the desire to assume a distribution

The test is not directly on the hypothesized distribution

The test is on the expected versus observed number of

samples

The actual test is between the histograms

The size of each category (interval) has a significant role

Generating Exponential Headways 43

exponential distribution

We want to solve for the headway t. set

Then and

U1, U2, ..., UN are independent uniform random variables between 0 and 1.0.

Use RAND() function in Microsoft Excel

To generate exponentially distributed time headways

Generating Shifted Exponential

Headways 44

Use the same procedure

Macroscopic Measures of a Traffic Stream

45

Speed

Unit: km/h

Variable: S

Flow

Unit: Veh/h

Variable: V

Density

Unit: Veh/km

Variable: D

Speed46

Rate of motion expressed as distance

unit per unit of time

Space mean speed (harmonic average)

Time mean speed (arithmetic average)

Time mean speed >= Space mean

speed

The relationship …

Rate of Flow / Volume47

Volume: Total number of vehicles passing a given

point during a given time interval

Interval may be an hour, day, week, or even a year

Rate of Flow: Equivalent hourly rate at which

vehicles pass a given point during a given time

interval less than one hour

Peak-Hour Factor48

Density49

Number of vehicles occupying a given

length of roadway, averaged over

time

Fundamental Traffic Flow Relationship50

Flow= Density × Speed(SMS)

51

Macroscopic Speed-Flow-Density Relationship

52

There is a maximum speed at which vehicles will travel

(free flow speed, Sf)

The minimum speed vehicles can travel at is zero

There is a maximum number of vehicles that can

occupy a given segment of roadway (jam density, Dj).

There must be a condition at which the maximum flow

occurs (Capacity, Vc)

If vehicles are very close together, people tend to drive

more slowly (as density increases, speeds decrease)

Home Reading53

Chapter #5

Traffic engineering book 3rd edition by Roess et. al

Questions54

Thanks for your time