Download 2532524-queuing v1

Introduction to Queueing Theory
First developed to analyze statistical behavior of phone
switches.
Queueing Systems
used to model processes in which customers arrive, wait
their turn for service, are serviced and then leave. Eg:
supermarket checkouts stands, world series ticket booths,
doctors waiting rooms etc..
Five components of a Queueing system:
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Chap. 4 handout
1. Interarrival-time probability density function (pdf)
2. service-time pdf
3. Number of servers
4. queueing discipline
5. size of queue.
ASSUME an infinite number of customers (i.e. long queue
does not reduce customer number). Bad assumption in a
time-sharing model, with finite number of customers, if
half wait for response, input rate will be reduced.
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Chap. 4 handout
Interarrival-time pdf – e.g. record elapsed time since
previous arrival, list the histogram of interarrival times (i.e.
10 0.1 sec, 20 0.2 sec ...) This is a pdf character.
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.
Number of servers - banks have multiserver queueing
systems, food stores have a collection of independent
single-server queues.
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Chap. 4 handout
Queueing discipline - order of customer processing i.e.
supermarkets are first-come-first served. Hospital
emergency rooms use sickest first.
Some queues have finite length: when full customers are
rejected.
ASSUME infinite-buffer, single-server system with firstcome, first-served queues.
A/B/m notation
A=interarrival-time pdf
B=service-time pdf
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Chap. 4 handout
m=number of servers.
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)
M/M/1 is known, G/G/m is not.
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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Chap. 4 handout
(1)
(  t) n   t
Pn (t ) 
e
n!
This law appears in:
physics (radio active decay (P[k alpha particles in t
seconds] with  = avg # of prtcls per sec
operations research planning switchboard sizes P[k calls
in t seconds] with  =avg number of calls per sec
biology water pollution monitoring P[k coliform bacteria in
1000 CCs]  =avg # of coliform bacteria per cc
Transportation planning size of highway tolls P[k autos in
t minutes]  =avg# of autos per minute
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Chap. 4 handout
optics in designing an optical recvr P[k photons per sec
over the surface of area A]  =avg# of photons per second
per unit area
Communications in designing a fiber optic xmit-rcvr link
P[k photoelectrons generated at the rcvr in one second]
with  =avg # of photoelectrons per sec.
 =rate parameter=event per unit interval (time distance
volume...)
The Poisson law results from asymptotic behavior of the
binomial law (in the limit as the prob[event]->0, # of trials
-> infinity and the number of events is small compared
with the number of trials)
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Chap. 4 handout
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Chap. 4 handout
let
  interarrival rate  10 cust. per min
n  the number of customers = 100
t  interval of finite length
P10 (t),   10
P0 (t )  e t
t  0
P1 (t)   te  t
a(t)t  e t  te  t
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Chap. 4 handout

 a(t)dt  1
t0
proof:

e ax
e dx 
a
ax
 e t
 t
 
0
  e dt    e  e  e  1
a(t)dt  e  t  dt
 t
a(t)t  prob. that an interarrival interval is between t and t + t
e.g.
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Chap. 4 handout
  interarrival rate  10 cust. per min
n  the number of customers = 100
we should get 100 custs in 10 minutes (max prob). In
maple:
P10 (t),   10
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Chap. 4 handout
0.04
0.03
0.02
0.01
0
5
10
15
t
20
0
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Chap. 4 handout
To obtain numbers with a Poisson pdf, you can write a
program:
total=length of sequence
lambda = avg arrival rate
exit x() = array holding numbers with Poisson pdf
local num= Poisson value
r9 = uniform random number
t9= while loop limit
k9 = loop index, pointer to x()
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Chap. 4 handout
for k9=1 to total
num=0
r9=rnd
t9=exp(-lambda)
while r9 > t9
num = num + 1
r9 = r9 * rnd
wend
x(k9)=num
next k9
(returns a discrete Poisson pdf in x())
Prove: that Poisson arrivals generate an exponential
interarrival pdf.
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Chap. 4 handout
a(t)t  prob. that an interarrival interval is between t and t + t
This is the prob. of 0 arrivals for time t times the prob. of 1
arrival in t:
a(t)t  P0 (t)P1 (t)
(  t) n   t
e
subst into (1) Pn (t ) 
and you get:
n!
P0 (t)  e   t and P1 ( t )  te   t
In the limit as t  0, the exponential factor in
P1 ( t )  te   t approaches 1 so by subst.:
a(t)t  P0 (t)P1 (t) = a(t)t  e   t  te   t
and
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Chap. 4 handout
a(t)dt  e   t  dt
(2)
with

 a(t )dt  1
t 0
proof :

e ax
e dx 
a
ax
 e
 t
e t
dt 
 e  t  e     e   0  1

We have made several assumptions:
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Chap. 4 handout
1. exponential interarrival pdf.
2. exponential service times (i.e. long service times become
less likely)
The M/M/1 queue in equilibrium
queue
server
There are 4 people in the system (and this describes the
state of the system).
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Chap. 4 handout
3 in the queue, 1 in the server.
The amount of time the person in the server has already
spent being served is independent of the probability of the
remaining service time. M/M/1 queues are memoryless (a
popular item with queueing theorists, and a feature unique
to exponential pdfs).
Pk  equilibrium prob . that there are k in system
In a birth-death system once serviced a customer moves
to the next state. This is like a non-deterministic finite-state
machine. 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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Chap. 4 handout
These are the states for a single-server queueing
system.
Po
2
1
0
 P1
 P k -1
 P1
...
k
k-1
P k
P 2
P k
k+1
P k +1
  mean arrival rate (cust. /sec)
 P0  mean number of transitions / sec from state 0 to 1
  mean service rate (cust./sec)
P1  mean number of transitions / sec from state 1 to 0
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State 0 = system empty
State 1 = cust. in server
State 2 = cust in server, 1 cust in queue etc...
Prob. of a given state is invariant wrt time if system is in
equilibrium. The prob. of k cust’s in system is constant.
This is like AC current entering a node and is called
detailed balancing thus, the number leaving a node must
equal the number entering:
3
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 P 0  P 1
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Chap. 4 handout
P0

3a
P1 
4
 P1  P 2
4a
P2 
 P1

by 3a

4b
P2 
P0

2 P 0
= P2 

2
since
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Chap. 4 handout
 P k  P k+1
5
then
k P0
P k  k   k P0

6
where  

= traffic intensity < 1

since all prob. sum to one,

6a

k0
k
P0  1 P0

 k  1
k 0
and since the sum of a geometric series is
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Chap. 4 handout

k 
7
k0
1
1 
subst 7 into 6a :
7a
P0
1
1 
and
7b
P0  1  =prob server is empty
subst into(6)
8
P k  (1  ) k
let N=mean number of cust’s in the system
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Chap. 4 handout
To compute the average (mean) value use:
E[k ] 
8a

 kPk
k 0
subst (8) into (8a) to obtain 8b
8
P k  (1  ) k
8b
E[k ] 

 k(1   )
k 0
k
 (1   )  k k
differentiate (7) wrt k


k0

1
1
8c
Dk    Dk
  k k 1 
1   k 0
(1   ) 2
k0
multiply both sides of (8c) by 
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k
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Chap. 4 handout

8d
9
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
(1   )2
k0


E[k ]  N  (1   )

(1   ) 2 (1   )
 k k 
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Chap. 4 handout
80
60
40
20
0
0.2
0.4
0.6
0.8
rho
1
0
as  approaches 1, N grows quickly.
T=mean interval between cust. arrival and departure,
including service.
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Chap. 4 handout
  mean arrival rate (cust. /sec)
In 1961 D.C. Little gave us Little’s result:
10
T
N


 /  1/
1


1  1    
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.
Compute how long a bird waits in the Queue (on average).
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Chap. 4 handout
  0.05 cust / sec = 3 birds / min * 1 min / 60 sec
= mean arrival rate
 = 0.1 bird / sec =
1 bird
=mean service rate
10 sec
So the mean service-time is 10 seconds/bird
=(1/ service rate)
T
1
1

 20 sec for wait + service
   0.1  0.05
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Chap. 4 handout
The mean queueing time is the waiting time in the system
minus the time being served, 20-10=10 seconds.
Tannenbaum says that the mean number of customers in
the system for an M/G/1 queueing system is:
1  Cb2
N  
2(1  )
(11)
Cb 
2
standard deviation
mean
of the service time.
This is known as the Pollaczek-Khinchine equation.
Note: M/G/1 means that it is valid for any service-time
distribution.
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Chap. 4 handout
For identical service time means, the large standard
deviation will give a longer service time.
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Chap. 4 handout