Intersection Analysis

7/31/2019 Intersection Analysis

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Introduction to Transportation Engineering

Applications of Queueing Theory to IntersectionAnalysis Level of Service

Dusan Teodorovic and Antonio A. Trani

Civil and Environmental EngineeringVirginia Polytechnic Institute and State University

Blacksburg, Virginia

Spring 2005


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Material Covered

Application of deterministic queueing models to studyintersection level of service

Study various types of intersection controls schemes usedin transportation engineering

Most of the material applies to ground transportationmodes (highways)


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Basic Ideas

Traffic control represents a surveillance of the motion ofvehicles and pedestrians in order to secure maximumefficiency and safety of conflicting traffic movements.

Traffic lights or traffic signals are the basic devices usedin traffic control of vehicles on roads.

They are located at road intersections and/or pedestrian

crossings.

The first traffic light was installed even before there wasautomobile traffic (London on December 10, 1868). The

current traffic lights were invented in USA (Salt LakeCity, (1912), Cleveland (1914), New York and Detroit(1920)).


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Basic Definitions

Drivers move toward the intersection from differentapproaches

Every intersection is composed of a number ofapproaches and the crossing area (see Figure)

Each approach can have one, or more lanes. The trafficstream is composed of all drivers who cross the

intersection from the same approach

During green time, vehicles from the observed approachcan leave the stop line and cross the intersection

The corresponding average flow rate of vehicles thatcross the stop line is known as a saturation flow

.


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Intersection Geometry

Figure 1. Typical Road Intersection.

Crossing area

Approach


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Flow Conditions at an Intersection

In most cases, queues of vehicles are establishedexclusively during the red phases, and are terminatedduring the green phases.

Such traffic conditions are known as a undersaturatedtraffic conditions

.

An intersection is considered an unsaturated intersection

when all of the approaches are undersaturated.

Traffic conditions in which queue of vehicles can arrive atthe upstream intersection are known as a oversaturated

traffic conditions

.


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Traffic Signal Control Strategies

Many isolated intersections operate under thefixed-timecontrol strategies

.

These strategies assume existence of the signal cycle

thatrepresents one execution of the basic sequence of signalcombinations at an intersection.

A phase represents part of the signal cycle, during which

one set of traffic streams has right of way. Figure 2shows two-phase traffic operations for the intersection.

The cycle contains only two phases. Phase 1 is related to

the movement of the north-southbound vehicles throughthe intersection. Phase 2 represents the movement of theeast-westbound vehicles.


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Traffic Control Strategies

The cycle length represents the duration of the cyclemeasured in seconds. The sum of the phase lengthsrepresents the cycle length.

For example, in the case shown in Figure 2., the cyclelength could be 90 seconds, length of the Phase 1 couldbe 50 seconds, while the length of the Phase 2 could be

equal to 40 seconds. The cycle length is a design parameter of the intersection

as well as the green times allocated to each phase. Trafficengineers can modify the settings of intersection

controllers based on demand needs at the intersection.

c


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Control Strategies

Figure 3. Intersection with Three Phases.

Phase 1 Phase 2 Phase 3

Cycle


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Control Strategies

Higher number of phases is usually caused by trafficengineers wish to protect some movements (usually left-turning vehicles)

Protection assumes avoiding potential conflicts withthe opposing traffic movement, and/or pedestrians

There is always a certain amount of lost time (few

seconds) during phase change. For example, when thegreen light changes to red there is am amber light periodto warn drivers of an impending change

Obviously, the higher the number of phases, the better theprotection, and the higher the value of the lost timeassociated with a phase change.


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Control Strategies

Traffic signals are control devices. The typical sequenceof lights at the intersection approach could be: Red, RedAll, Green, Amber, Red, Red All,....

Figure 4. Definition of Green, Amber and Red Times.

Flow [veh/h]

Time

0

Green

Effective green

Saturation

flow

RedRed All

Amber


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Control Strategies

Green time, effective green,red time, and effectivered are linguistic expressions frequently used by trafficengineers

In theory, all drivers should cross the intersection duringthe green light. In reality, no one driver starts his/her carexactly in a moment of the green light appearance

Similarly, at the end of a green light, some drivers speedup, and cross the intersection during the amber light

Green Time represents the time interval within thecycle when observed approach has green indication

. Onthe other hand, Effective Green represents the timeinterval during which observed vehicles are crossing

theintersection.


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Vehicle delays at signalized intersections: Uniform

Vehicle Arrivals

For simplicity, let us assume for the moment thatobserved signalized intersection could be treated as a D/

D/1 (deterministic) queueing system with one server(hence the notation (D/D/1))

We assume uniform arrivals, and uniform departure rate(see Figure 5).


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Queueing Theory Short-handNomenclature

Queues come in different flavors as demonstrated sofar

Kendall developed a simple scheme to designatequeues back in the early 50s. His nomenclature has

been widely adopted

Typically 6 parameters:

a/b/c/d/e/f

a = inter-arrival time distribution (arrivals)

b = service time distribution

c = number of servers

14a


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Typically 6 parameters:

a/b/c/d/e/f

d = service order (i.e., FIFO, LIFO, etc) e = Max. number of customers

f =Size of the arrival population

14b


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Possible outcomes for (a) and (b)

M = Times are neg. exponential (i.e., Poisson arrivals)

D = Deterministic distribution Ek = Erlang distribution

G = general distribution

14c


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Example Queueing Systems we HaveStudied

M/M/1/FIFO/ /

Stochastic queue with neg. exponential timebetween arrivals

Neg. exponential service times

1 server

First in-first out Infinite no. of customers in system

Infinite arrival population

14d


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Example Queueing Systems we HaveStudied

M/M/2/FIFO/15/15

Stochastic queue with neg. exponential timebetween arrivals

Neg. exponential service times

2 servers (2 pavers)

First in-first out Up to 15 no. of trucks in system

15 trucks population14e


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Definition of Queueing Terms for Intersection

Analysis

Figure 5. Arrivals and Departures at an Intersection.

r g

c

Red Green

Cum

ulativenumbero

fvehicles

Cumulative departures

Cumulative arrivals

Time

g0

D t( )

A t( )

A

C

h

B


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Deterministic Queueing Analysis

Let us denote by vehicles arrival rate, and by vehicles

departure rate during the green time period. In the

deterministic case, the cumulative number of arrivals

and the cumulative number of departures are:

(1)

(2)

where:

- the duration of the signal cycle

- effective red

- effective green

A t( )

D t( )

A t( ) t=

D t( ) t=

c

r

g


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The duration of the signal cycle equals:

(3)

The formed queue is the longest at the beginning of

effective green. The queue decreases at the beginning of

effective green.

We denote by the time necessary for queue to dissipate

(Figure 5). The queue must dissipate before the end of

effective green. In the opposite case, the queue would

escalate indefinitely. In other words, queue dissipationwill happen in every cycle if the following relation is

satisfied:

c r g+=

g0


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(4)

The relation (4) will be satisfied if the total number of

vehicle arrivals during cycle length is less than or equalto the total number of vehicle departures during effective

green , i.e.:

(5)

(6)

(7)

g0 g

c

g

td0

c

td

0

g

t0

c t

0

g

c g


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Finally, we get:

(8)

Let us note the triangleABC(Figure 5). Vehicles arrive

during time period . Vehicles depart during time

period . The total number of vehicle arrivals equals the

total number of vehicle departures, i.e.:

(9)

(10)

---

g

c---

r g0+( )

g0

r g0+( ) g0=

( ) g0 r=


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The time period required for queue to dissipate equals:

(11)

We divide both numerator and denominator by . We get:

(12)

Define the utilization factor ( ) (or traffic intensity) of the

intersection as , we can write:

(13)

g0

g0 r

------------=

g0

--- r

1 ---

------------=

---=

g0 r1 ------------=


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The area of the triangleABCrepresents the total

delay of all vehicles arrived during the cycle. This area

equals:

(14)

where is the height of the triangle (ABC).

The ratio represents the slope , i.e.:

(15)

AABC

d

AABC1

2--- r h =

h

h

r g0+( )------------------

h

r g0+( )

------------------=


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The height of the triangle ABC is:

The area of the triangleABCequals:

(16)

The total delay of all vehicles arriving during the cycle

equals:

(17)

h

h r g0+( )=

AABC1

2--- r h

1

2--- r r g0+( )

r2

---------- r g0+( )= = =

d

d r

2

---------- r g0+( ) r

2

---------- r r

1

------------+

r2

2

----------- 1

1

------------+

= = =


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(18)

The average delay per vehicle represents the ratiobetween the total delay and the total number of vehicles

per cycle. The total number of vehicles per cycle equals

. Therefore the average delay per vehicle is:

(19)

or

(20)

d r2

2 1 ( )------------------------=

d

d

c d

dd

c----------=

d

r22 1 ( )------------------------

c------------------------=


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Simplifying the previous expression, average delay per

vehicle is the average:

(21)dr

2

2 c 1 ( ) -------------------------------=


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Example Problem 1

The cycle length at the signalized intersection is 90

seconds. The considered approach has the saturation flow

of 2200 [veh/hr], the green time duration of 27 seconds,

and flow rate of 600 [veh/hr].

Analyze traffic conditions in the vicinity of the

intersection. Calculate average delay per vehicle. Assume

that the D/D/1 queueing system adequately describesconsidered intersection approach.


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Problem 1 - Solution

The corresponding values of the cycle length and the

green time are:

;

The flow rate ( ) and the service rate ( ) are:

Traffic intensity equals:

c 90 s[ ]= g 27 s[ ]=

600

veh

hr---------

600

3600------------

veh

s--------- 0.167

veh

s---------= = =

2200veh

hr---------

2200

3600------------

veh

s--------- 0.611

veh

s---------= = =


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The duration of the red light for the considered approach

is:

The number of arriving vehicles per cycle is:

---

0.167veh

s---------

0.611veh

s

---------

---------------------------- 0.273= = =

r c g 90 27 63 s[ ]= = =

c 0.167 vehs

--------- 90 s[ ] 15.03 veh[ ]= =


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The number of departing vehicles during green light is:

We conclude that the following relation is satisfied:

This means that the traffic conditions in the vicinity of the

intersection are undersaturated traffic conditions.

The average delay per vehicle is estimated using:

g 0.611veh

s--------- 27 s[ ] 16.497 veh[ ]= =

c g

dr

2

2 c 1 ( ) -------------------------------=


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d63

2

2 90 1 0.273( ) --------------------------------------------- 30.33 s[ ]= =


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Example Problem 2

The cycle length at the signalized intersection is 60

seconds. The considered approach has the saturation flow

of 2200 [veh/hr], the green time duration of 15 seconds,

and flow rate of 400 [veh/hr]. Analyze traffic conditions inthe vicinity of the intersection. Assume that the D/D/1

queueing system adequately describes the intersection

approach considered.

Calculate: (a) the average delay per vehicle; (b) the

longest queue length; (c) percentage of stopped vehicles.


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Problem 2 - Solution:

(a) The corresponding values of the cycle length and the

green time are:

;

The red time is:

The flow rate and the service rate are:

c 60 s[ ]= g 20 s[ ]=

r c g 60 20 40 s[ ]= = =

400veh

hr---------

400

3600------------

veh

s--------- 0.111

veh

s---------= = =


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The utilization factor for the queue is:

The average delay per vehicle equals:

2200veh

hr---------

2200

3600------------

veh

s--------- 0.611

veh

s---------= = =

---

0.111veh

s---------

0.611

veh

s---------

---------------------------- 0.182= = =

d r

2

2 c 1 ( ) -------------------------------=


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(b) The longest queue length happens at the end of a

red light (Figure 5). The quantity is calculated as

follows:

(c) Vehicles arrive all the time during the cycle. The total

number vehicles arrived during the cycle equals:

d40

2

2 60 1 0.182( ) --------------------------------------------- 16.3 s[ ]= =

Lmax

Lmax

Lmax r 0.111veh

s--------- 40 s[ ] 4.44 vehicles[ ]= = =

A c 0.111veh

s--------- 60 s[ ] 6.66 vehicles[ ]= = =


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All vehicles that arrive during time interval are

stopped. The total number of stopped vehicles equal:

The time period required for queue to dissipate is

estimated using equation:

We get:

r g0+( )

S

S r g0+( )=

g0

g0 r

------------=

S r g0+( ) r r

------------+

0.111 40 0.111 40

0.611 0.111---------------------------------+

= = =


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The percentage of stopped vehicles equal:

[%]

S 5.43 vehicles[ ]=

PS

A--- 100

5.43

6.66---------- 100 81.53= = =


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Example Problem 3

A simple T intersection is signalized. There are two

approaches indicated in the figure. The cycle length at the

signalized intersection (Figure) is 50 seconds.

Phase 1 Phase 2 Approach 1

Cycle

Approach 2


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Example Problem 3

Approach 1 has the saturation flow of 2200 [veh/hr], the

effective green time duration of 35 seconds, and the flow

rate of 600 [veh/hr]. Approach 2 has the saturation flow of

2000 [veh/hr], the effective green time duration of 15seconds, and the flow rate of 550 [veh/hr]. Assume that

the D/D/1 queueing system adequately describes

considered intersection approach.

Calculate: (a) the average delay per vehicle for every

approach; (b) Allocate effective red and green time among

approaches in such a way to minimize the total delay of

the T intersection.


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Problem 3 -Solution

(a) Approach 1:


green time are:

;

The red time equals:

The flow rate and the service rate are respectively equal:

c 50 s[ ]= g1 35 s[ ]=

r1 c g1 50 35 15 s[ ]= = =

1 600veh

hr---------

600

3600------------

veh

s--------- 0.167

veh

s---------= = =


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Problem 3 -Solution

The utilization factor for the queue in approach 1 is:


1 2200veh

hr---------

2200

3600------------

veh

s--------- 0.611

veh

s---------= = =

1

111-----

0.167veh

s---------

0.611veh

s---------

---------------------------- 0.273= = =

d1 r12

2 c 1 1( ) --------------------------------- 15

2

2 50 1 0.273( ) --------------------------------------------- 3.09 s[ ]= = =


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Problem 3 -Solution

Approach 2:


green time are:;

The red time is:

The flow rate and the service rate are:

c 50 s[ ]= g2 15 s[ ]=

r2 c g2 50 15 35 s[ ]= = =

2 550veh

hr---------

550

3600------------

veh

s--------- 0.153

veh

s---------= = =


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Problem 3 -Solution

The utilization factor for the queue equals:


2 2000veh

hr---------

2000

3600------------

veh

s--------- 0.555

veh

s---------= = =

2

222-----

0.153veh

s---------

0.555veh

s---------

---------------------------- 0.276= = =

d2r

2

2

2 c 1 2( ) --------------------------------- 35

2

2 50 1 0.276( ) --------------------------------------------- 16.92 s[ ]= = =


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Problem 3 -Solution

(b) The total delay per cycle of all vehicles on both

approaches is the sum of the delays of every approach:

substituting the definitions of and (see equation 21)

Since:

after substitution, we get:

TD 1 d1 2 d2+=

d1 d2

TD 1r1

2

2 c 1 1( ) --------------------------------- 2

r22

2 c 1 2( ) ---------------------------------+=

r1 r2+ c=


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The total delay is minimal when:

After substitution, we get:

TD 1 r12

2 c 1 1( ) --------------------------------- 2 c r1( )

2

2 c 1 2( ) ---------------------------------+=

TD[ ]dr1d

--------------- 0=

1r1

2

2 c 1 1( ) --------------------------------- 2

c r1( )2

2 c 1 2( ) ---------------------------------+d

r1d----------------------------------------------------------------------------------------------------- 0=

1r1

c 1 1( )

-------------------------- 2c r1( )

c 1 2( )

-------------------------- 0=

S i


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Problem 3 -Solution

After solving the equation, we get:

These are the optimal values of green and red times tominimize the intersection delay.

0.167r1

50 1 0.273( )------------------------------------- 0.153

50 r1( )

50 1 0.276( )------------------------------------- 0=

r1 24 s[ ]=

g1 c r1 50 24 26 s[ ]= = =

r2 26 s[ ]=

g2 24 s[ ]=

P bl 3 S l ti


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Problem 3 -Solution

We can recalculate the average delays per vehicle:

Approach 1

Approach 2:

These delays compare favorably with those obtained

before (3.09 and 16.92 seconds, respectively for

approaches 1 and 2).

d1r

1

2

2 c 1 1( ) ---------------------------------24

2

2 50 1 0.273( ) --------------------------------------------- 7.92 s[ ]= = =

d2r2

2

2 c 1 2( ) ---------------------------------

262

2 50 1 0.276( ) --------------------------------------------- 9.34 s[ ]= = =

V hi l D l t Si li d I t ti R d


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Vehicle Delays at Signalized Intersections: Random

Vehicle Arrivals

Traffic flows are characterized by random fluctuations

The delay that a specific vehicle experiences depends on

the probability density function of the interarrival times,as well as on signal timings and the time of a day whenthe vehicle shows up

Obviously, individual vehicles experience at a signalizedapproach various delay values.

I t ti ith R d A i l


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Intersection with Random Arrivals

Figure 6. Intersection with Random Arrivals.

Red Green

Cumulative

number of

vehicles

Cumulative arrivals

Time

Uniform delay

Overflow Delay



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Let us calculate the delay for the vehicle arriving attime (Figure 6). The overall delay is composed of theuniform delay and the overflow delay , i.e.:

(22)

The uniform delay represents delay that would beexp rienced by a vehicle when all vehicle arriveuniformly and when traffic conditions are unsaturated(see Equations in previous sections).

Due to the random nature of vehicle arrivals, the arrivalrate during some time periods can go over the capacity,

causing overflow queues.

D

D

d dR

D d dR+=

d

Considering Random Arrivals


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Considering Random Arrivals

The overflow delay represents the delay that is causedby short-term overflow queues. This delay can be easilycalculated using queueing theory techniques.

dR

Crossing area

Queueing System



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We assume that vehicle interarrival times are

exponentially distributed. The service rate is deterministic

(we denote by departure rate from the artificial queue

into the signal), and there is only one server.

This means that the artificial Queueing System is M/D/1

queueing system (single server with Poisson arrivals and

deterministic service times).

The average delay per customer in the M/D/1 queueing

system equals:

(23)

dR 2

2 1 ( ) --------------------------------=



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where:

- utilization ratio in the M/D/1 queueing system

The utilization ratio in the M/D/1 queueing systemequals:

(24)

The departure rate from the artificial queue into the signal

can be expressed in terms of departure rates from the

traffic signal . The departure rate equals during green

time. During red time, departure rate equals zero (seeFigure 6).

---=



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Figure 7. Service Rate Definition at a Traffic Intersection.

Red Green

Service rate

[veh/h]

Time

Cycle

0

g

r



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Departure rate during whole cycle :

(25)

(26)

The utilization ratio in the M/D/1 queueing system

:

(27)

i.e.:

0 r g+

c---------------------------=

g

c---=

---

g

c---

-----------= =


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(28)

The quantity is known as a volume to capacity ratio.

The average vehicle delay is:

(29)

(30)

It has been shown by simulation that Equation (30)overestimate the average vehicle delay. The following two

formulas for average vehicle delay calculation were

proposed as a corrections of the equation (30):

c g

----------=

D d dR+=

Dr

2

2 c 1 ( ) -------------------------------

2

2 1 ( ) --------------------------------+=

Websters formula:


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(31)

Allsops formula:

(32)

Dr2

2 c 1 ( ) -------------------------------

2

2 1 ( ) -------------------------------- 0.65

c

2-----

1

3---

2 5 g

c----------+

+=

D9

10------

r2

2 c 1 ( ) -------------------------------

2

2 1 ( ) --------------------------------+=

Example Problem 4


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p

Using data given in the Example Problem 1, calculate: (a)

average delay per vehicle using Allsops formula. (b)

Calculate duration of the green time necessary to achieve

average delay per vehicle of 40 seconds.

Solution:

(a) The cycle length, green time, arrival rate, departure

rate, traffic intensity, volume to capacity ratio, and red

time duration are:

c 90 s[ ]=

g 27 s[ ]=

600veh 600 veh

0 167veh

= = =


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600hr

---------3600------------

s--------- 0.167

s---------= = =

2200veh

hr---------

2200

3600------------

veh

s--------- 0.611

veh

s---------= = =

---

0.167 vehs

---------

0.611veh

s---------

---------------------------- 0.273= = =

---

gc------

0.167veh

s---------

0.611veh

s---------

----------------------------

27 s[ ]90 s[ ]-----------------------------------------

0.273

0.3------------- 0.91= = = =

Solution - Problem 4


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The average delay per vehicle based on Allsops formula

equals:

(b) The average delay per vehicle is:

r c g 90 27 63 s[ ]= = =

D9

10------

r2

2 c 1 ( ) -------------------------------

2

2 1 ( ) --------------------------------+=

D 910------ 63

2

2 90 1 0.273( ) --------------------------------------------- 0.91

2

2 0.167 1 0.91( ) -------------------------------------------------+=

D 52.083 s[ ]=

D9 r

2 2+=


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Green time to achieve a delay of 40 seconds per vehicle.

D10------

2 c 1 ( ) -------------------------------

2 1 ( ) --------------------------------+=

r2

2 c 1 ( ) -------------------------------

10

9------ D

2

2 1 ( ) --------------------------------=

r 2 c 1 ( ) [ ] 109------ D

2

2 1 ( ) --------------------------------=

r 2 90 1 0.273( ) [ ]10

9------ 40

0.912

2 0.167 1 0.91( ) -------------------------------------------------=

r 47 s[ ]=

g c r 90 47= =

g 43 s[ ]=

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Intersection Analysis