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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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Queueing Theory Short-handNomenclature
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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Queueing Theory Short-handNomenclature
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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Deterministic Queueing Analysis
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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Deterministic Queueing Analysis
(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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Deterministic Queueing Analysis
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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Deterministic Queueing Analysis
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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Deterministic Queueing Analysis
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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Deterministic Queueing Analysis
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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Deterministic Queueing Analysis
(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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Problem 1 - Solution
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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Problem 1 - Solution
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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Problem 1 - Solution
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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Problem 2 - Solution
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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Problem 2 - Solution
(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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Problem 2 - Solution
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:
The corresponding values of the cycle length and the
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:
The average delay per vehicle equals:
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:
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 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:
The average delay per vehicle 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
Intersection with Random Arrivals
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Intersection with Random Arrivals
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
Intersection with Random Arrivals
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Intersection with Random Arrivals
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 ( ) --------------------------------=
Intersection with Random Arrivals
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Intersection with Random Arrivals
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).
---=
Intersection with Random Arrivals
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Intersection with Random Arrivals
Figure 7. Service Rate Definition at a Traffic Intersection.
Red Green
Service rate
[veh/h]
Time
Cycle
0
g
r
Intersection with Random Arrivals
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Intersection with Random Arrivals
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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