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Introduction to Queueing Theory
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Motivation
First developed to analyze statistical behavior of phone switches.
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Queueing Systems
model processes in which customers arrive.
wait their turn for service.
are serviced and then leave.
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Examples
supermarket checkouts stands.
world series ticket booths.
doctors waiting rooms etc..
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Five components of a Queueing system:
1. Interarrival-time probability density function (pdf)
2. service-time pdf 3. Number of servers 4. queueing discipline 5. size of queue.
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ASSUME
an infinite number of customers (i.e. long queue does not reduce customer number).
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Assumption is bad in :
a time-sharing model.with finite number of customers.
if half wait for response, input rate will be reduced.
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Interarrival-time pdf
record elapsed time since previous arrival.
list the histogram of inter-arrival times (i.e. 10 0.1 sec, 20 0.2 sec ...).
This is a pdf character.
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Service time
how long in the server? i.e. one customer has a shopping cart full the other a box of cookies.
Need a PDF to analyze this.
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Number of servers
banks have multiserver queueing systems.
food stores have a collection of independent single-server queues.
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Queueing discipline
order of customer process-ing.
i.e. supermarkets are first-come-first served.
Hospital emergency rooms use sickest first.
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Finite Length QueuesSome queues have finite length: when full customers are rejected.
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ASSUME
infinite-buffer. single-server system with first-come.
first-served queues.
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A/B/m notation
A=interarrival-time pdfB=service-time pdfm=number of servers.
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A,B are chosen from the set:
M=exponential pdf (M stands for Markov)
D= all customers have the same value (D is for deterministic)
G=general (i.e. arbitrary pdf)
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Analysibility
M/M/1 is known. G/G/m is not.
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M/M/1 system
For M/M/1 the probability of exactly n customers arriving during an interval of length t is given by the Poisson law.
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Poisson’s Law
Pn (t )(t)n
n!e t (1)
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Poisson’s Law in Physicsradio active decay –P[k alpha particles in t seconds]– = avg # of prtcls per second
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Poisson’s Law in Operations Researchplanning switchboard sizes –P[k calls in t seconds]– =avg number of calls per sec
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Poisson’s Law in Biologywater pollution monitoring –P[k coliform bacteria in 1000 CCs]– =avg # of coliform bacteria per cc
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Poisson’s Law in Transportationplanning size of highway tolls –P[k autos in t minutes]– =avg# of autos per minute
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Poisson’s Law in Opticsin designing an optical recvr –P[k photons per sec over the surface of area A] – =avg# of photons per second per unit area
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Poisson’s Law in Communications in designing a fiber optic xmit-rcvr link –P[k photoelectrons generated at the rcvr in one second] – =avg # of photoelectrons per sec.
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- Rate parameter =event per unit interval (time distance volume...)
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Poisson’s Law and Binomial LawThe Poisson law results from asymptotic behavior of the binomial law in the limit as: –the prob[event]->0 –# of trials -> infinity–the number of events is small compared with the number of trials
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Example:
let
interarrival rate 10 cust. per minn the number of customers
100t interval of finite length
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We have:
P10(t),10P0(t)et
t 0P1(t)tet
a(t)tettet
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a(t)dtt 0
•
1
proof:
eax
dx ea
e tdt e t
e t
e • e 0 1
ax
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a(t)dt e t
dt
a(t)t
prob .V that an interarrival interval is between t and t + t:
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Analysis
Depend on the condition:
we should get 100 custs in 10 minutes (max prob).
= interarrival rate = 10 cust. per min
n = the number of customers = 100
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In maple:P10 (t ), 10
0 5 10 15 200
0.01
0.02
0.03
0.04
t
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To obtain numbers with a Poisson pdf, you can write a program:
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total=length of sequencelambda = avg arrival rateexit x() = array holding numbers with Poisson pdf
local num= Poisson valuer9 = uniform random numbert9= while loop limitk9 = loop index, pointer to x()
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for k9=1 to totalnum=0r9=rndt9=exp(-lambda)
while r9 > t9num = num + 1r9 = r9 * rndwend
x(k9)=numnext k9(returns a discrete Poisson pdf in x())
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Prove:
Poisson arrivals gene-rate an exponential interarrival pdf.
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Prove:
– prob. that an interarrival interval is between t and t +
–This is the prob. of 0 arrivals for time t times the prob. of 1 arrival in Dt:
a(t)t
ta(t)t=P0 (t)P1(t)
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Subst into :
Poisson’s Law:
Pn (t ) =(t)n
n!e−t (1)
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Now you get:
P0 (t) =e−t
P1(t)=te−t
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We already knew:
In the limit as the exponential factor in:
t → 0
P1(t)=te−t
approaches 1.
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So by subset:
a(t)t=P0 (t)P1(t)
a(t)t=e−tte−t=and
a(t)dt =e−tdt (2)
with a (t )dtt 0
• 1
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e axdx eax
a
e tdt e t
e t
e • e 0 1
a (t )dtt 0
• 1
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We have made several assumptions:exponential interarrival pdf.
exponential service times (i.e. long service times become less likely)
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The M/M/1 queue in equilibrium
queue
server
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State of the system:
There are 4 people in the system.
3 in the queue. 1 in the server.
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Memory of M/M/1:
The amount of time the person in the server has already spent being served is independent of the probability of the remaining service time.
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Memoryless
M/M/1 queues are memoryless (a popular item with queueing theorists, and a feature unique to exponential pdfs).P kequilibrium prob .
that there are k in system
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Birth-death system
In a birth-death system once serviced a customer moves to the next state.
This is like a nondeterminis-tic finite-state machine.
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State-transition Diagram
The following state-transition diagram is called a Markov chain model.
Directed branches represent transitions between the states.
Exponential pdf parameters appear on the branch label.
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Single-server queueing system
0 1 2 k-1 k k+1...
Po P1 Pk-1 Pk
μPk μPk+1μP1 μP2
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Symbles:
=mean arrival rate (cust. /sec)
μ =mean service rate (cust./ sec)
P0mean number of transitions/ secfrom state 0 to 1
μP1mean number of transitions/ sec
from state 1 to 0
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States
State 0 = system empty
State 1 = cust. in server
State 2 = cust in server, 1 cust in queue etc...
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Probalility of Given State
Prob. of a given state is invariant wrt time if system is in equilibrium.
The prob. of k cust’s in system is constant.
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Similar to AC
This is like AC current entering a node
is called detailed balancing
the number leaving a node must equal the number entering
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Derivation
P0 = μP1
P1 =P0
μ
P1 = μP2
P 2 =P1μ
3
3a
4
4a
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by 3a
P 2 = P0
μμ
P2 =2P0
μ 2
P k = μP k+1
=4
since
5
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then:
P k =kP0
μ k =ρkP06
where = traffic intensity < 1ρ =μ
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since all prob. sum to one
ρ kP0k = 0
∞
∑ =1=P0 ρ k
k =0
∞
∑ = 16a
Note: the sum of a geometric series is
ρ kk = 0
∞
∑ =1
1− ρ7
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Suppose that it is right, cross multiply and simplify:
ρ k
k 0
•
1
1 ρ
ρ kk=0
∞
∑ − ρ ρ k
k=0
∞
∑ = 1
ρ kk=0
∞
∑ − ρ k
k=1
∞
∑ = ρ 0 = 1SoQ.E.D.
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subst 7 into 6a
P0
1−ρ=1
P0 ρkk0
•
16a
7a and
P0 =1−ρ=prob server is empty
7b
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subst into
P k =kP0
μ k =ρkP06
yields:
P k =(1−ρ)ρk8
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Mean value:
let N=mean number of cust’s in the system
To compute the average (mean) value use:
E[k ]= kPkk=0
∞
∑8a
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Subst (8) into (8a)
P k =(1−ρ)ρk
E[k ]= kPkk=0
∞
∑
E[k ]= k(1−ρ)ρkk=0
∞
∑ =(1−ρ) kρkk=0
∞
∑
8
8a
8b
we obtain
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differentiate (7) wrt k
ρ kk = 0
∞
∑ =1
1− ρ
Dk ρkk=0
∞
∑ =Dk1
1−ρ= kρk−1k=0
∞
∑ =1
(1−ρ)2
7
we get
8c
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multiply both sides of (8c) by ρ
kρkk=0
∞
∑ =ρ
(1−ρ)2
E[k ]=N =(1−ρ)ρ
(1−ρ)2=
ρ(1−ρ)
8d
9
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Relationship of , N
ρ
0 0.2 0.4 0.6 0.8 10
20
40
60
80
rho
as r approaches 1, N grows quickly.
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T and
T=mean interval between cust. arrival and departure, including service.
=mean arrival rate (cust. /sec)
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Little’s result:
In 1961 D.C. Little gave us Little’s result:
T =N
=ρ / 1−ρ
=1 / μ1−ρ
=1
μ −10
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For example:
A public bird bath has a mean arrival rate of 3 birds/min in Poisson distribution.
Bath-time is exponentially distributed, the mean bath time being 10 sec/bird.
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Compute how long a bird waits in the Queue (on average):
=0.05 cust / sec = 3 birds / min * 1 min / 60 sec
= mean arrival rate
μ = 0.1 bird / sec = 1 bird
10 sec
= mean service rate
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Result:
So the mean service-time is 10 seconds/bird =(1/ service rate)T =
1μ −
=1
0.1−0.05=20sec
for wait + service
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Mean Queueing Time
The mean queueing time is the waiting time in the system minus the time being served, 20-10=10 seconds.
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M/G/1 Queueing System
Tannenbaum says that the mean number of customers in the system for an M/G/1 queueing system is:
N =ρ +ρ2 1+Cb2
2(1−ρ)11
This is known as the Pollaczek-Khinchine equation.
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What is Cb
Cb = standard deviation mean
of the service time.
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Note:
M/G/1 means that it is valid for any service-time distribution.
For identical service time means, the large standard deviation will give a longer service time.
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Introduction to Queueing Theory