topic 6 - traffic flow theory - version 2
DESCRIPTION
Document on Traffic flow basics.TRANSCRIPT
3/19/2013
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Topic 6 (Ch. 5)Traffic Stream Parameters
Mohan Venigalla, Ph.D., P.E.A i t P f CEIE
© Mohan Venigalla
Associate Professor, CEIE
George Mason University
Fairfax, VA 22030‐4444
Progress
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Traffic Streams
• Individual vehicles and drivers make up the t ffi ttraffic stream
• Local characteristics and driver behavior are major factors on its performance
• Drivers and vehicles are not uniform in their make up or behavior
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their make up or behavior
Traffic Streams
• Uninterrupted – freeways, two‐lane rural droads
• Interrupted flow facilities – arterials, local roadways (have external devices that interrupted flow)
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Interrupted Facilities
• Vehicles flow in platoons
A f hi l i t th ith – A group of vehicles moving together with a significant gap between themselves and the next group of vehicles
• Signal timing plans try to take advantage of platoons for continuous flow
Si l l l h il b i d
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• Signals place less than 2 miles apart can be timed to allow for uninterrupted flow between signals
Traffic Stream Parameters
• Macroscopic parameters – describe the traffic stream as a wholestream as a whole
– Traffic flow
– Speed
– Density
• Microscopic parameters ‐ describe the behavior of
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the individual vehicle with respect to each other
– Spacing
– Headway
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Macroscopic Parameters
• Traffic flow – number of vehicles that pass a certain point during a specified time certain point during a specified time interval (vehicles/hour)
• Speed – rate of motion in distance/time (mph)
• Density – number of vehicles occupying a i l th f hi h l ( hi l
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given length of highway or lane (vehicles per mile per lane, vpmpl)
Spacing and Time Headway
• Spacing – the distance between successive hi l i t ffi t th vehicles in a traffic stream as they pass some
common reference point on the vehicles
• Time headway – the time between successive vehicles in a traffic stream as they pass some common reference point on the
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p pvehicles
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Traffic Flow and Time Headway
Traffic Flow given by: nq
tq
n
q= traffic flow in vehicles per unit timen= number of vehicles passing some designated roadway point during time tt= duration of time intervalFlow measurements typically related to generalized period of time; Volume of traffic refers to vehicles per hour
Time Headway given by:
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i
iht1
Time Headway given by:t= duration of time intervalhi=time headway of the ith vehiclen= number of measured vehicle time headways at some designated roadway point
Time Headway and Traffic Flow
• Time headway is defined as the time between the passage of successive vehicles (can be measured from front bumpers or rear bumpers)
Substituting t into the flow equation gives:
orh
nq n
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hq
hi
i
11
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Example Problem
Given the following headways, determine the h d d th flaverage headway and the flow:
4.74s, 3.33s, 4.74s, 8.97s, 11.63s, 3.83s, 14.40s
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Speed and Travel Time• Time mean speed – point measure of speed
• Space mean speed – measure relating to length p p g gof roadway
• Average travel time – total time to traverse a highway
• Average running speed – total time during which vehicle is in motion while traversing a highway segment (no stop time included)
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highway segment (no stop time included)
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Speed and Travel Time
• Operating speed – maximum safe speed a hi l b d i ith t di vehicle can be driven without exceeding
design speed
• 85th percentile speed – speed at which 85% of vehicles are traveling at or below
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Time Mean Speed
• Arithmetic mean of vehicles speeds is given by:given by:
n
uu
n
ii
t
1
ut=time-mean speed in unit distance per unit timeui=spot speed of the ith vehicle
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i p pn=number of measured vehicle spot speeds
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Space Mean Speed
• Time necessary for a vehicle to travel some lvehicle to travel some known length of roadway t
lus
n1
us= space-mean speed in unit distance per unit timel=length of roadway used for travel time measurements of vehiclest(bar)= average vehicle travel time, defined as:
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i
itnt
1
1
ti= time necessary for vehicle i to travel a roadway section of length l
Traffic Density
• Measure using aerial photographs; think of it as the number of vehicles that occupy a pylength of roadway
lnk
k=traffic density in vehicles per unit distancen=number of vehicles occupying some lengthof roadway at some specified timel=length of roadway
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n
iisl
1
l=length of roadway
si=spacing of the ith vehicle (the distance between vehicles i and i-1 measuredfrom front bumper to front bumper
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Spacing and Density
• Substituting the equation for roadway length into the density equation giveslength into the density equation gives
ors
nk n
ii
11
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sk
1
Basic Traffic Stream Modelsukq
Example: average headway is 2.5 s/veh on single lane roadway; averagehi l i i 200’ d i d f ffi
fthk
hrvehq
hrssvehq
svehvehs
q
/00501
/1440
/3600/40.0
/40.0/5.2
1
vehicle spacing is 200’; determine average speed of traffic.
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hrmimiveh
hrveh
k
qu
mivehmiftftvehk
ftvehvehft
k
/5.54/4.26
/1440
/4.26/5280/005.0
/005.0/200
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Speed‐Density Model
)1(f k
kuu
jf k
u=space mean speed in mi/hruf= free-flow speed in mi/hrk=density in veh/mikj=jam density in veh/hr
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Flow‐Density Model
)(2
jf k
kkuq
jk
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Speed‐Flow Model
)(2
j
uukq )(
fj u
ukq
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Example Problem
• Given an estimate of density of 16.05 vpmpl t d f 6 h d t i th j at a speed of 60mph; determine the jam density and flow rate at 60mph. Assume car length is 15’ and at jam density spacing between vehicles is 15’.
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Questions
• If a vehicle is traveling at a cruising speed of 55 h i th t55 mph, is that:
– Time mean speed, or
– Space mean speed?
• Is free‐flow speed, what is the space mean speed?
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speed?
– Ans: uf
Problem 5.5 (p 169)
• On a specific westbound section of a highway, studies show that the speed density relationship isstudies show that the speed‐density relationship is
• It is known that the capacity is 4200 veh/h and the
)1(
5.3
jf k
kuu
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jam density is 210 veh/mi. what is the space‐mean speed of the traffic at capacity and what is the free‐flow speed?
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Solution to Problem 5.5
• Given: kj = 210; qcap = 4200; )1(
5.3
jf k
kuu
• We have, q = u.k or
• At capacity, dq/dk = 0
)(
5.3
jf k
kkkuq
)5.41(0
5.3
cap
f k
ku
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•
jk
Volume
• Planning (non‐directional) volume measures
– Average annual daily traffic (AADT)
– Average annual weekday traffic (AAWT)
– Average daily traffic (ADT), average 24 hour volume that can be measured by season,
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ymonth, week, day, etc.
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Volume
• Hourly volumes – used for design and ti l l ioperational analysis
– Peak hour volume – single highest hourly volume
– Directional design hour volume –
• AADT x K x D = DDHV (K = proportion of daily ff d k h f k
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traffic during peak hour, D = proportion of peak traffic traveling in peak direction)
Volume
• Peak hour factor – describes the relationship between hourly volume and maximum rate of flow within the hourly volume and maximum rate of flow within the hour
– PHF = hourly volume/maximum rate of flow OR
– PHF = V/(4 x V15)
• PHF range –
1 0 (each 15 minute period equal) to
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1.0 (each 15 minute period equal) to
0.25 (one 15 min period contains all traffic)
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Peak Hour Factor Example
15 min period Vehicle Count Flow Rate (vph)
7:20AM 389 1556
7:35AM 495 1980
7:50AM 376 1504
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8:05AM 363 1452
7:20-8:20AM 1623 1623
Peak Hour Factor Example
• Determine the Peak Hour Factor
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Chapter 5
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QUEUING
Introduction
• Macroscopic relationships and analyses are l bl b tvery valuable, but
• A considerable amount of traffic analysis occurs at the microscopic level
• In particular, we often are interested in the elapsed time between the arrival of successive
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elapsed time between the arrival of successive vehicles (i.e., time headway)
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Introduction
• The simplest approach to modeling vehicle arrivals is to assume a uniform spacingarrivals is to assume a uniform spacing
• This results in a deterministic, uniform arrival pattern—in other words, there is a constant time headway between all vehicles
• However, this assumption is usually unrealistic, as vehicle arrivals typically follow a random process
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vehicle arrivals typically follow a random process
• Thus, a model that represents a random arrival process is usually needed
Introduction
• First, to clarify what is meant by ‘random’:
F f t t b id d t l• For a sequence of events to be considered truly random, two conditions must be met:
1. Any point in time is as likely as any other for an event to occur (e.g., vehicle arrival)
2. The occurrence of an event does not affect the probability of the occurrence of another event (e.g., the
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p y ( g ,arrival of one vehicle at a point in time does not make the arrival of the next vehicle within a certain time period any more or less likely)
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Introduction
• One such model that fits this description is the P i di t ib tiPoisson distribution
• The Poisson distribution:
– Is a discrete (as opposed to continuous) distribution
– Is commonly referred to as a ‘counting
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y gdistribution’
– Represents the count distribution of random events
Poisson Distribution
)()(
etnP
tn
!
)(n
P(n) = probability of having n vehicles arrive in time tλ = average vehicle arrival rate in vehicles per unit timet= duration of time interval over which vehicles are countede= base of the natural logarithm
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e= base of the natural logarithm
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Example Application
Given an average arrival rate of 360 veh/hr or 0.1 vehicles per second; with t=20 seconds;vehicles per second; with t=20 seconds; determine the probability that exactly 0, 1, 2, 3, and 4 vehicles will arrive.
p(1) =
p(2) =
p(3) =
!
)()(
n
etnP
tn
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p(3) =
p(4) =
p(>4) = 1 – Sigma p( 1 to 4)
Example Solution
1350)201.0(
)0()20(1.00 e
P
271.0!1
)201.0()1(
135.0!0
)()0(
)20(1.01
eP
P
)5(1)5( nPnP
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053.0
090.0180.0271.0271.0135.01
)()(
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Poisson Example
• Example:
– Consider a 1‐hour traffic volume of 120 vehicles, during which the analyst is interested in obtaining the distribution of 1‐minute volume counts
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Poisson Example #2
(120 h/h ) / (3600 /h ) 0 0333 h/• = (120 veh/hr) / (3600 sec/hr) = 0.0333 veh/s• t = 0.0333 veh/sec 60 sec = 2 veh
= (120 veh/hr) / (60 min/hr) = 2 veh/min
t = 2 veh/min 1 min = 2 veh
OR
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1353.01
1353.01
!0
2)0(
20
e
P
Poisson Example
# of 1‐min intervals with exactly n vehicle arrivals
probability of exactly n vehicles arriving in 1‐min interval
2707.01
1353.02
!1
2)1(
21
e
P
2707.02
1353.04
!2
2)2(
22
e
P
x 60 min = 16.24
x 60 min = 16.24
yg
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1804.06
1353.08
!3
2)3(
23
e
P
And so on…
x 60 min = 10.82
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Poisson Example
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16
18
Poisson Example
4
6
8
10
12
14
16
Fre
qu
en
cy
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0
2
0 1 2 3 4 5 6 7 8 9 10
# of veh arrivals/minute
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Poisson Example
What is the probability of more than 6 cars i i (i 1 i i t l)?arriving (in 1‐min interval)?
6
0
1
616
i
inP
nPnP
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(0.5%)or 005.0
995.01
)012.0036.0090.0180.0271.0271.0135.0(16
nP
Poisson Example
What is the probability of between 1 and 3 i i (i 1 i i t l)?cars arriving (in 1‐min interval)?
32131 nPnPnPnP
%0.18%1.27%1.2731 nP
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%2.72
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Poisson distribution
• The assumption of Poisson distributed hi l i l l i li di t ib tivehicle arrivals also implies a distribution
of the time intervals between the arrivals of successive vehicles (i.e., time headway)
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Negative Exponential
• To demonstrate this, let the average arrival rate, , be in units of vehicles per second, so thatbe in units of vehicles per second, so that
3600
q
Substituting into Poisson equation yields
sec
veh
hsec
hveh
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!3600
)(
3600
n
eqt
nP
qtn
(Eq. 5.25)!
)()(
n
etnP
tn
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Negative Exponential
• Note that the probability of having no hi l i i ti i t l f l thvehicles arrive in a time interval of length
t [i.e., P (0)] is equivalent to the probability of a vehicle headway, h, being greater than or equal to the time interval t.
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Negative Exponential
• So from Eq. 5.25,
)()0( thPP
36003600
1
1 qtqt
ee
(Eq. 5.26)
1 !0
10
x
Note:
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This distribution of vehicle headways is known as the negative exponential distribution.
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Negative Exponential Example
• Assume vehicle arrivals are Poisson distributed with an hourly traffic flow of 360 veh/h.an hourly traffic flow of 360 veh/h.
Determine the probability that the headway between successive vehicles will be less than 8 seconds.
Determine the probability that the headway
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Determine the probability that the headway between successive vehicles will be between 8 and 11 seconds.
Negative Exponential Example
• By definition, thPthP 1
818 hPhP
1
18
3600)8(360
3600
ehPqt
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551.0
4493.01
1 3600
e
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Negative Exponential Example
811118 hPhPhP
551.03329.01
551.01
8111
811118
3600)11(360
e
hPhP
hPhPhP
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1161.0
Negative Exponential
1.0
0.2
0.4
0.6
0.8
Pro
b (
h >
= t
)
e^(-qt/3600)
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0.0
0 5 10 15 20 25 30 35
Time (sec)
For q = 360 veh/hr
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Negative Exponential
c.d.f.
0.4
0.6
0.8
1.0P
rob
ab
ility
(h
< t
)
1 - e^(-qt/3600)0.551
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0.0
0.2
0 5 10 15 20 25 30 35
Time (sec)
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Queuing Systems
• Queue – waiting line• Queue – waiting line
• Queuing models – mathematical descriptions of queuing systems
• Examples – airplanes awaiting clearance for takeoff or landing, computer print jobs,
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g, p p j ,patients scheduled for hospital’s operating rooms
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Characteristics of Queuing Systems• Arrival patterns – the way in which items or• Arrival patterns the way in which items or customers arrive to be served in a system (following a Poisson Distribution, Uniform Distribution, etc.)
• Service facility – single or multi‐server
Service pattern the rate at which
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• Service pattern – the rate at which customers are serviced
• Queuing discipline – FIFO, LIFO
D/D/1 Queuing Models
• Deterministic arrivals
• Deterministic departures
• 1 service location (departure channel)
• Best examples maybe factory assembly lines
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Example
Vehicles arrive at a park which has one entry i t ( d ll hi l t t ) P kpoints (and all vehicles must stop). Park
opens at 8am; vehicles arrive at a rate of 480 veh/hr. After 20 min the flow rate decreases to 120 veh/hr and continues at that rate for the remainder of the day. It takes 15 seconds
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to distribute the brochure. Describe the queuing model.
Example Solution
f llths
tvehhr
hrveh
tvehhr
hrveh
i/4min/60
min20min/2min/60
/120
min20min/8min/60
/480
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foralltvehvehs
min/4/15
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Example Solution
Therefore,
Vehicle arrivals can be defined as:
8t for t ≤ 20 min and
160 +2(t ‐ 20) for t > 20 min
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Vehicle departures can be defined as:
4t for all t
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Example Continued
• When will the queue dissipate? q p
160 +2(t‐20) = 4t
t = 60 minutes
• Total vehicle delay is?
The total area between the arrival and departure curves.
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min2400)4080(2
1)2080(
2
1 vehDt
M/D/1 Queuing Model
• M stands for exponentially distributed times s a ds o e po e a y d s bu ed esbetween arrivals of successive vehicles (Poisson arrivals)
• Traffic intensity term is used to define the ratio of average arrival to departure rates:
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M/D/1 Equations
• When traffic intensity term < 1 and constant steady state average arrival and departure rates:
)1(2
)1(2
2
w
Q
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)1(2
2
)1(2
t
M/M/1 Queuing Models
• Exponentially distributed arrival and departure po e a y d s bu ed a a a d depa u etimes and one departure channel
When traffic intensity term < 1
1
2
Q
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1
)(
t
w
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M/M/N Queuing Models
• Exponentially distributed arrival and departure p y ptimes and multiple departure channels (toll plazas for example)
• In this case, the restriction to apply these equations is that the utilization factor must be less than 1.
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0.1N
M/M/N Models
P
N Nnc
1
10
Qw
NNN
PQ
NNnN
n cc
1
)/1(
1
!
)/1(!!
2
10
0
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Q
t
w