Lecture 11

C O tliCourse Outline• Introduction to Transportation • Highway Users and their Performance• Geometric Design• Pavement Design• Speed Studies - Project• Speed Studies - Project• Traffic• Queuing• Intersections

Lecture # 11Department of Civil and Environmental Engineering 2

• Level of Service in Highways and Intersections

Previous classPrevious class

• Basic Conceptsa. Flow Rateb S ib. Spacingc. Headwayd Speed – 2 typesd. Speed – 2 typese. Density

• RelationshipsRelationships• Graphs


Relationships…Relationships…

nn

∑n

h

nq = t = hii

n

1∑

tnq =

∑i =

ih1

i = 1t

qnk ==⎤⎡=

nsu11

1q = uk

ulk ==

( )∑=

⎥⎦

⎤⎢⎣

⎡

i itln 1

11 q uk


Consider a linear relationship between speed and density:p p y

⎟⎟⎞

⎜⎜⎛

=kuu 1 ⎟⎟⎠

⎜⎜⎝−=

jf k

uu 1

⎟⎟⎞

⎜⎜⎛

−=uukq

2

⎟⎟⎠

⎜⎜⎝

=f

j uukq


MODELS OF TRAFFIC FLOWMODELS OF TRAFFIC FLOW

T ffi i l if l di t ib t d• Traffic is rarely uniformly distributed– equal time between arriving vehicles or headways?

• Must make some assumption for arrival patterns (distribution)(distribution)


Poisson ModelPoisson Model

• Approximation of non-uniform flow

( ) etnPtn λλ −

=)(Where:• P(n) = probability of having n vehicles arrive in time t

n!nP )(

• P(n) = probability of having n vehicles arrive in time t,• t = duration of the time interval over which vehicles

are counted,• λ = average vehicle flow or arrival rate in vehicles

per unit time, andb f h l l i h ( 2 718)


• e = base of the natural logarithm (e = 2.718).

Poisson Distribution ExamplePoisson Distribution Example

Assume:mean = variance (sd^2)λ = 360 veh/h = 0.1 veh/st = 20 sec

)( et tn λλ −

!)()(n

etnP λ=


Poisson IdeasPoisson Ideas

• Probability of exactly 4 vehicles arriving– P(n=4)

P b bilit f l th 4 hi l i i• Probability of less than 4 vehicles arriving– P(n<4) = P(0) + P(1) + P(2) + P(3)

• Probability of 4 or more vehicles arrivingProbability of 4 or more vehicles arriving– P(n≥4) = 1 – P(n<4) = 1 - P(0) + P(1) + P(2) + P(3)

A t f ti b t i l f i hi l• Amount of time between arrival of successive vehicles

( ) ( ) ( ) 36000

0 qtttetthPP −−

−

≥ λλλ


( ) ( ) ( ) 3600

!00 qtt eethPP ===≥= λ

Poisson ModelPoisson Model• The assumption of Poisson vehicle arrivals also

implies a distribution of the time intervals betweenimplies a distribution of the time intervals between the arrivals of successive vehicles (time headway).

• To show this note that the average arrival rate as:• To show this, note that the average arrival rate as:

q3600

qλ =

Where:• λ = average vehicle arrival rate in veh/s,


• q = flow in veh/h, and 3600 = number of seconds per hour.

Poisson ModelPoisson Model• Substituting into P(n) equation:

( ) ( ) !

3600 3600

neqt = nP

qt-n

• The probability of having no vehicles arrive in a time interval of length t (P(0)) is equivalent to the

!n

time interval of length t (P(0)) is equivalent to the probability of a vehicle headway, h, being greater than or equal to the time interval tthan or equal to the time interval t.

( ) ( ) 36000 qt-= eth= PP ≥“negative exponential” or simply called


( ) ( )0 q= eth = PP ≥ simply called“exponential” distribution

Change of mean and distribution shape0.25

Mean = 0 2 vehicles/minute

Change of mean and distribution shape

0.20

ranc

e

Mean = 0.2 vehicles/minute


0.15

ty o

f Occ

ur

0.10

Prob

abili

t

0.00

0.05


0.000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Arrivals in 15 minutes

Change of mean and inter-arrival timesChange of mean and inter arrival times

1.0

0 7

0.8

0.9

ance



0.5

0.6

0.7

of E

xced

a

0.3

0.4

Prob

abili

ty

0.1

0.2

P


0.00 2 4 6 8 10 12 14 16 18 20

Time Between Arrivals (minutes)

Limitations of the Poisson ModelLimitations of the Poisson Model

M ( b f ti i d)• Mean (average number of cars per time period) must be equal to the variance (variance over all time period)time period)

• Otherwise use an alternative model (negative• Otherwise use an alternative model (negative binomial, etc.).

Lecture #Department of Civil and Environmental Engineering 14

Poisson DistributionPoisson DistributionLet’s grab 6 hrs of 5 min aggregated counts from a station in the Portland freeway network (I-5) for one laneDoes it match what we expect?Does it match what we expect?

mean = 84.57 veh/5 min/ln (1,014 veh/hr/ln)standard deviation = 9.35 vehvariance = 87 44 veh^2

Poisson Model

0.04

0.05

0.05

Observed Data

6

7

variance = 87.44 veh 2

0.02

0.02

0.03

0.03

0.04

P(n)

in t=

5 m

in

3

4

5

Freq

uenc

y

0.00

0.01

0.01

0 7 14 21 28 35 42 49 56 63 70 77 84 91 98 105

112

119

N vehicles

0

1

2

0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102

108

114

120


How do we know if a distribution is a good representation of reality?Any objective test?

Problem 5.8.Problem 5.8.• An observer has determined that the time

headways between successive vehicles on aheadways between successive vehicles, on a section of highway, are exponentially distributed and that 60% of the headways between vehicles a d a 60% o e ead ays be ee e c esare 13 seconds or greater. If the observer decides to count traffic in 30-second time intervals, estimate the probability of the observer counting exactly four vehicles in an interval.


Queuing delays significanceQueuing delays significance

• Queuing delays can account for up to 90% or more of a driver’s total trip travel time.

E l f i• Examples of queuing:• Traffic Signals• Toll booths• Traffic incidents (accidents and vehicle

disablements)


Queueing TheoryQueueing Theory

Objects passing through point with restriction onObjects passing through point with restriction on maximum rate of passageInput + storage area (queue) + restriction +Input storage area (queue) restriction outputCustomers, arrivals, arrival process, server, , , p , ,service mechanism, departures, discipline (FIFO)

Input Storage Output


Restriction

Queueing Theory: Study of Congestion Phenomena

Applications: Airplane waiting for takeoff tollApplications: Airplane waiting for takeoff, toll gate, wait for elevator, taxi stand, ships at a port, grocery store, telecommunications, circuits…g yInterested in: maximum queue length, typical queueing times …. SERVICE LEVEL !

Input Storage Output


Restriction

Queueing Theory: common assumptionsQueueing Theory: common assumptions• Arrival patterns (λ, in vehicles per unit time):

equal time intervals (derived from the assumption of– equal time intervals (derived from the assumption of uniform, deterministic arrivals) and

– exponentially distributed time intervals (derived from the p y (assumption of Poisson-distributed arrivals).

• Departure patterns (μ in vehicles per unit time)• Departure patterns (μ, in vehicles per unit time), – equal time intervals (derived from the assumption of

uniform, deterministic arrivals) and )– exponentially distributed time intervals (derived from the

assumption of Poisson-distributed arrivals).


Queueing Theory: Some definitionsQueueing Theory: Some definitionsD/D/1

Deterministic arrivals

G/G/mGeneral arrivalsG l d tDeterministic departures

1 channel departuresGraphical solution easiest

General departuresMulti-channel departuresSIMULATION

M/D/1Exponential arrivalsDeterministic departures1 channel departuresMathematical solution

M/M/1Exponential arrivalsExponential departures1 channel departures


pMathematical solution

Queuing disciplineQueuing discipline• first-in-first-out (FIFO), indicating that the first

vehicle to arrive is the first vehicle to depart andvehicle to arrive is the first vehicle to depart, and

• last in first out (LIFO) indicating that the last• last-in-first-out (LIFO), indicating that the last vehicle to arrive is the first to depart.

• For virtually all traffic-oriented queues, the FIFO queuing discipline is the more appropriate of thequeuing discipline is the more appropriate of the two.


D/D/1 QueuingD/D/1 Queuing

d t i i ti i l d d t ith• deterministic arrivals and departures with one departure channel (D/D/1 queue)

• D/D/1 queue lends itself to a graphical or mathematical solutionmathematical solution.


Queueing Theory : Conservation PrincipleQueueing Theory : Conservation PrincipleCustomers don’t disappearArrival times of customers completely characterizes p yarrival process.Time/accumulation axes

N(x,t)Uniform arrivals/departures

2

3 A=l(t)D=m(t)

1

2


t1 t2 t3Time, t @ x

Queueing Theory: Departure ProcessQueueing Theory: Departure ProcessObserver records times of departure for corresponding objects to construct D(t).p g j ( )

N(x,t)

4

A(t)

1234

D(t)


Time, t @ xt1 t2 t3 t4t1′ t2′ t3′ t4′

Queueing Theory: AnalysisQueueing Theory: AnalysisIf system empty at t=0:

Vertical distance is queue length at time t: Q(t)=A(t)-D(t)For FIFO horizontal distance is waiting time for jth customer.

Uniform arrivals/departuresN(x,t)

Uniform arrivals/departures

Q(t) 2

3 A=l(t)D=m(t)

Wj 1


t1 t2 t3Time, t @ x

Queueing Theory: AnalysisQueueing Theory: Analysis

Horizontal strip of unit height width WjHorizontal strip of unit height, width Wj

N(x,t)

4

A(t)W2

1234

D(t)


Time, t @ xt1 t2 t3 t4t1′ t2′ t3′ t4′

Queueing Theory: AnalysisQueueing Theory: AnalysisAdd up horizontal strips total delayTotal time spent in system by some numberTotal time spent in system by some number of vehicles (horizontal strips)

N(x,t)

Total Delay=Area

4

A(t)Total Delay=Area

1234

D(t)


Time, t @ xt1 t2 t3 t4t1′ t2′ t3′ t4′

Queueing TheoryQueueing TheoryTotal delay = WAverage time in queue: w = W/nAverage number in queue: Q = W/T

N(x,t)

4

A(t)

1234

D(t)


Time, t @ xt1 t2 t3 t4t1′ t2′ t3′ t4′

EXAMPLE 5.7EXAMPLE 5.7• Vehicles arrive at an entrance to a recreational

park There is a single gate (at which all vehiclespark. There is a single gate (at which all vehicles must stop), where a park attendant distributes a free brochure. The park opens at 8:00 A.M., at ee b oc u e e pa ope s a 8 00 , awhich time vehicles begin to arrive at a rate of 480 veh/h. After 20 minutes, the arrival flow rate declines to 120 veh/h and continues at that level for the remainder of the day. If the time required to distribute the brochure is 15 seconds anddistribute the brochure is 15 seconds, and assuming D/D/1 queuing, describe the operational characteristics of the queue


characteristics of the queue.

EXAMPLE 5 7 - SOLUTIONEXAMPLE 5.7 - SOLUTION

• Begin by putting arrival and departure rates into common units of vehicles per minute.

veh/h480 t λ

h/h120

min 20for veh/min 8 = min/h 60 veh/h480 ≤=

t λ

/ i

min 20 for veh/min 2 = min/h 60 veh/h120

>=

tμ allfor veh/min 4 = s/veh 15 s/min 60

=


EXAMPLE 5 7 - SOLUTIONEXAMPLE 5.7 - SOLUTION

• Begin by putting arrival and departure rates into common units of vehicles per minute.

8 20 t t for ≤ min

( )( ) min20for202160 t t + >−

• the number of vehicle departures is:

4 t tf llLecture # 11Department of Civil and Environmental Engineering 32

4 t t for all

EXAMPLE 5.7 - SOLUTION

A=l(t)=2(t)A l(t) 2(t)Equation of line =

160 + 2(t-20)

80 veh

A=l(t)=8(t)D=m(t)=4(t)


EXAMPLE 5.7 - SOLUTIONEXAMPLE 5.7 SOLUTION• When the arrival curve is above the departure

curve a queue condition will existcurve, a queue condition will exist. • The point at which the arrival curve meets the

departure curve is the moment when the queuedeparture curve is the moment when the queue dissipates (no more queue exists).

• The point of queue dissipation can be determinedThe point of queue dissipation can be determined by equating appropriate arrival and departure equations, that is

( ) t = t + 4202160 −


• Solving for t gives t = 60 minutes.

EXAMPLE 5.7 - SOLUTIONEXAMPLE 5.7 SOLUTION• Thus the queue that began to form at 8:00 A.M. will

dissipate 60 minutes later (9:00 A M ) at whichdissipate 60 minutes later (9:00 A.M.), at which time 240 vehicles will have arrived and departed (4 veh/min 60 min).e / 60 )

• Individual vehicle delay: – Under FIFO queuing discipline, the delay of an individual q g p , y

vehicle is given by the horizontal distance between arrival and departure curves. S b i ti f Fi 5 7 th 160th hi l t i– So, by inspection of Fig. 5.7, the 160th vehicle to arrive will have the longest delay of 20 minutes (the longest horizontal distance between arrival and departure


curves)

• The total length of the queue is given by the vertical distance between arrival and departurevertical distance between arrival and departure curves at that time.

• The longest queue (longest vertical distance• The longest queue (longest vertical distance between arrival and departure curves) will occur at t = 20 minutes and is 80 vehicles longg

• Total vehicle delay, defined as the summation of the delays of each individual vehicle, is given by the total area between arrival and departure curves


• In this example, the areas between arrival and departure curves can be determined by summingdeparture curves can be determined by summing triangular areas, giving total delay, Dt, as

D = + t12

80 20 12

80 40( ) ( )

i

× ×

h

• Because 240 vehicles encounter queuing-delay

= -2400 minveh

(as previously determined), the average delay per vehicle is 10 minutes (2400 veh-min/240 veh),

d th l th i 40 hi l


and the average queue length is 40 vehicles(2400 veh-min/60 min).

Problem 5.14.Problem 5.14.• Vehicles begin to arrive at a parking lot at 6:00

A M at a rate 8 per minute Due to an accidentA.M. at a rate 8 per minute. Due to an accident on the access highway, no vehicles arrive from 6:20 to 6:30 A.M. From 6:30 A.M. on, vehicles 6 0 to 6 30 o 6 30 o , e c esarrive at a rate of 2 per minute. The parking lot attendant processes incoming vehicles (collects parking fees) at a rate of 4 per minute throughout the day. Assuming D/D/1 queuing, determine total vehicle delaydetermine total vehicle delay.


EXAMPLE 5.8EXAMPLE 5.8• After observing arrivals and departures at a highway

toll booth over a 60-minute time period thetoll booth over a 60-minute time period, the observer notes that the arrival and departure rates (or service rates) are deterministic but, instead of ( ) ,being uniform, they change over time according to a known function. The arrival rate is given by the f ( ) 2function (t) = 2.2 + 0.17t − 0.0032t2 and the departure rate is given by (t) = 1.2 + 0.07t, where tis in minutes after the beginning of the observationis in minutes after the beginning of the observation period and (t) and (t) are in vehicles per minute. Determine the total vehicle delay at the toll booth


Determine the total vehicle delay at the toll booth and the longest queue assuming D/D/1 queuing.

M/D/1 QueuingM/D/1 Queuing• exponentially distributed times between the

arrivals of successive vehicles (Poisson arrivals)arrivals of successive vehicles (Poisson arrivals)

• deterministic departures and• deterministic departures, and

d t h l• one departure channel

O• Obvious example– Traffic Signals


M/D/1M/D/1• Basic relationship:

λμλρ =

Where:• ρ = traffic intensity, and is unit-less,• λ = average arrival rate in vehicles per unit time,

and• μ = average departure rate in vehicles per unit

ti


time.

M/D/1M/D/1assuming that ρ <1, for an M/D/1:

• average length of queue in vehicles:( )ρρ = Q−12

2

• average waiting time in the queue (for each hi l )

( )ρ12

vehicle):

ti t i th t ( ti f( )ρμρ−12

= w

• average time spent in the system (summation of average queue waiting time + average departure time (service time)): ρ−2


time (service time)): ( )ρμρ= t−−12

2

M/D/1M/D/1NOTE !

th t th t ffi i t it i l th ( <1) th• that the traffic intensity is less than one (ρ <1), the D/D/1 queue will predict…

NO queue formation– NO queue formation. • Models with random arrivals or departures, such as

the M/D/1 queuing model will predictthe M/D/1 queuing model, will predict– queue formations !

2

( )ρρ = Q−12

2

( )ρμρ−12

= w( )ρμρ = t−−12

2


M/M/1 QueuingM/M/1 Queuing• exponentially distributed times between the arrivals

of successive vehicles (Poisson arrivals)of successive vehicles (Poisson arrivals) • exponentially distributed departure time patterns in

addition to exponentially distributed arrival timesaddition to exponentially distributed arrival times• one departure channel• Traffic applications:• Traffic applications:

– Toll booth where some arriving drivers have the correct toll and can be processed quickly, and others may not p q y, yhave the correct toll, thus producing a distribution of departures about some mean departure rate.


M/M/1M/M/1

μλρ = 1.0ρ <

λ = arrival rateμ = departure (service) rate

– Average length of queue

μ

ρ 2

Q

– Average time waiting in queue

( )ρρ−

=1

Q

Average time waiting in queue

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=λμ

λμ1w

– Average time spent in system

λμ=

1t


λμ −

M/M/N QueuingM/M/N QueuingApplications:

M/M/N i i bl ti t t ll• M/M/N queuing is a reasonable assumption at toll booths on turnpikes or at toll bridges where there is often more than one departure channel availableis often more than one departure channel available (more than one toll booth open).

• M/M/N queuing is also frequently encountered inM/M/N queuing is also frequently encountered in non-traffic but “transportation” applications such as security checks at airports, vessel queueing at ports/airports and so on.

• Other “non-transportation” :


– checkout lines at retail stores, call centers, etc.

M/M/N– Multi-channelM/M/N Multi channel

μλρ = 1.0Nρ <

– Average length of queue⎤⎡+1 1P Nρ

μ

– Average time waiting in queue( ) ⎥⎦

⎤⎢⎣

⎡

−= 2

0

11

! NNNPQ

ρρ

g g q

μλρ 1

−+

=Qw

– Average time spent in system

λρ Qt +

=


λ

M/M/NM/M/N

– Probability of having no vehicles

( )∑−

+= 10

1!!

1N Nnc

NNn

P

ρρρ

– Probability of having n vehicles

( )= −0 1!!n ccNNn ρ

Nnfor !

0 ≤=nPP

n

nρ Nnfor

!0 ≥= − NN

PP Nn

n

nρ

– Probability of being in a queue

( )PP

N

Nρ

=+

>

10


( )NNNP Nn ρ−> 1!

Problem 5.35Problem 5.35• Vehicles leave an airport parking facility (arrive at

parking fee collection booths) at a rate of 500parking fee collection booths) at a rate of 500 veh/h (the time between arrivals is exponentially distributed). The parking facility has a policy that d s bu ed) e pa g ac y as a po cy athe average time a patron spends in a queue while waiting to pay for parking is not to exceed 5 seconds. If the time required to pay for parking is exponentially distributed with a mean of 15 seconds what is the fewest number of paymentseconds, what is the fewest number of payment processing booths that must be open to keep the average time spent in a queue less than 5


average time spent in a queue less than 5 seconds.

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Lecture 11