courtesy of r.martin, rutgers univ introduction to queueing theory comp5416 advanced network...
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Courtesy of R.Martin, Rutgers Univ
Introduction to Queueing TheoryCOMP5416Advanced Network Technologies
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Queuing Theory
• View network as collections of queues– FIFO data-structures
• Queuing theory provides probabilistic analysis of these queues
• Examples:– Average length– Probability queue is at a certain length– Probability a packet will be lost
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Little’s Formula (aka. Little’s Law)
• Little’s Law:– Mean number tasks in system = arrival rate x mean
residence time• Observed before, Little was first to prove• Applies to any system in equilibrium, as long as
nothing in black box is creating or destroying tasks
SystemArrivals Departures
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Example using Little’s Formula
• Consider 40 customers/hour visit Hungry Jack Restaurant– Average customer spend 15 minutes in the restaurant
• What is average number of customers at at given time?
10/25.0/40 custhrhrcustTr r
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Model Queuing System
• Use Little’s formula on complete system and parts to reason about average time in the queue
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Kendal Notation
• Six parameters in shorthand (x/x/x/x/x/x)– First three typically used, unless specified
1. Arrival Distribution2. Service Distribution3. Number of servers4. Total Capacity (infinite if not specified)5. Population Size (infinite)6. Service Discipline (FCFS/FIFO)
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Distributions
• M: Exponential• D: Deterministic (e.g. fixed constant)• Ek: Erlang with parameter k
• Hk: Hyperexponential with param. k• G: General (anything)
• M/M/1 is the simplest ‘realistic’ queue
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Kendal Notation Examples
• M/M/1:– Exponential arrivals and service, 1 server, infinite capacity
and population, FCFS (FIFO)• M/M/n
– Same as above, but n servers• G/G/3/20/1500/SPF
– General arrival and service distributions, 3 servers, 17 queue slots (20-3), 1500 total jobs, Shortest Packet First
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M/M/1 Queue
Single Isolated Link
•Assume messages arrive at the channel according to a Poisson process at a rate messages/sec.
•Assume that the message lengths have a negative exponential distribution with mean 1/ bits/message.
•Channel transmits messages from its buffer at a constant rate c bits/sec
•The buffer associated with the channel can considered to be effectively of infinite length.
=> an M/M/1 queue.
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M/M/1
• The first M says that the interarrival times to the queue are negative-exponential (ie Markovian or memoryless);
• The second M says that the service times are negative-exponential (ie Markovian or memoryless again);
• The 1 says that there is a single server at the queue.
• Shorthand notation for "a queue with Poisson arrivals, negative exponentially distributed message lengths, a single server, and infinite buffer space".
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• Arrival rate messages/sec,
• Message lengths neg exp, mean 1/ bits/message,
• Transmission rate c bits/sec, – ie messages transmitted at rate = c messages/sec.
• Define the state of the system as the total number of messages at the link (waiting + being transmitted).
• The system is therefore a Markov chain, since both the arrival and message length distributions are memoryless.
0 1 2
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M/M/1 queue model
Tw 1/
Tr
r
w
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Solving queuing systems
• Given: : Arrival rate of jobs (packets) : Service rate of the server (output link)
• Solve:– r: average number of items in queuing system– Tr: avg. time an item spends in whole system
– Tw: avg. time an item waits in the queue
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Equilibrium conditions
0 1 2
Boundary for flux balance
Boundary for global balance
,2,11 ipp ii
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• Solving the flux balance equations recursively, we get
02
2
1 ppppi
iii
0pp ii
Define occupancy
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(normalising condition)
(geometric distribution)
)1(
1
00
i
ip
,2,1,0)1( ip ii
Note:Solution valid only for < 1 (stability condition)For 1, there is no steady state!
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M/M/1 Queue Length Distribution
0
0.1
0.2
0.3
0.4
0.5
0 1 2 3 4 5 6 7 8 9
0
0.1
0.2
0.3
0.4
0.5
0 1 2 3 5 5 6 7 8 9
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M/M/1 Mean Number in the System
1
)1()1(
)1(}{
2
00
i
ii
i
ipiRE
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M/M/1 Mean Delay
1
11}{
111}{1
}|{}{delayMean
0
0
R
ii
iiRR
TE
REp
i
piTETE
0.00
2.00
4.00
6.00
8.00
10.00
E{T
}
0 0.25 0.5 0.75 1
M/M/1 queue
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M/M/1 Mean Waiting Time
• The waiting time is defined as the time in the queue, excluding service time:
1
11
1
111}{}{ RW TETE
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Summary of M/M/1 (infinite buffer)
1
1}{
1
11}{
1}{
1
W
R
TE
TE
RE
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Little’s Formula
• If messages/sec enter a “system”, and each spends, on average, W seconds there, then the total number in the “system” must be given by W by a simple flow argument.
• Proof for M/M/1:
}{}{
}{1
1
/1
1
11}{
R
R
TERE
RETE
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M/M/1/n – Finite Buffer
0 1
2
n
nipp ii ,,1,00
1
1
00
1
1
n
n
i
ip
nip ini ,,1,0
1
11
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M/M/1/n
• Probability that the buffer will be blocked:
nnnB pP
11
1}blocked is messagePr{
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M/M/1/n exampleExample : = 0.5• Assume that we want
• Choose n so that
• i.e. choose n = 9
310BP
96.91101
105.0
105.01
5.05.0
5.01
5.01
3
31
31
1
1
n
n
n
nn
n
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M/M/1/n – validity of solution
• Solution for the finite buffer queue is valid for all – not just for < 1 as was the case for the infinite
buffer.
• This is because the finite buffer copes with overloads by blocking messages, rather than by creating a backlog of messages.
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M/M/1/n – Little’s formula
and it is only these messages which contribute to the buffer size. This implies that:
)1( BP
}{)1(}{}{ RBR TEPTERE
To use Little's formula for the finite buffer case, note that the rate of non-blocked messages is given by:
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Summary of M/M/1/n
n
i
in
n
ii
n
n
B
RB
iipRE
P
TEPRE
01
0
1
1
)1(}{
1
)1(
}{)1(}{
1) <or 1 be(can
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State dependent Queues
Let i= arrival rate when the system is in state i
i = service rate when the system is in state i
0 1 2
Balance Equations:,2,111 ipp iiii
,2,1021
110 ippi
ii
with solution
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M/M/2
• State dependent with
,3,22
1
,1,0
i
i
i
i
i
0 1 2
0101 2)2(ppp
ii
i
i
i
i
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M/M/1 vs M/M/2
1
1
2
1}{ 2 rate server, 1RTE
)1)(1(
11}{ rateeach servers, 2
RTE
12
1
}{
}{
rateeach servers, 2
2 rate server, 1
R
R
TE
TE
Compare M/M/2 (2 servers – each at rate ) to that of a single link at rate 2 .
(M/M/1 result)
(M/M/2 result)
Conclusion: Single fast server is always more efficient!
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M/M/1 vs M/M/2M/M/1 vs M/M/2
0
1
2
3
4
Occupancy
Mea
n D
elay
(u
nit
s o
f si
ng
le p
acke
t se
rvic
e ti
mes
)
M/M/1: 1 server, rate 2
M/M/2: 2 servers, each rate
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M/M/
(a Poisson distribution)
0 1 2
Ai
Aep
ip
iA
i
i
i with !!
0
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M/M/1 delay distribution
(an exponential distribution)
tT etp
R
)1()1()(
• Probability of delay within certain limit:
tR etT )1(1)Pr(
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Example
Suppose that packets arrive at a link at a Poisson rate of 250 packets/second, and the packet lengths are exponentially distributed with mean length 1000 bits/packet. Assume that the buffer length is sufficiently large that it can be assumed to be effectively infinite.
Find the required capacity of the link in bits/second so that the following criteria are satisfied:
(1) Mean delay 10 msec
(2) Pr{delay 20 msec} 0.10
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Solution
bps000,2502500001000250
:Occupancy
ccc
bps000,350
010.0250000
1000
2500001
11000
1
11}{ :DelayMean
c
c
cc
TE R
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Solution
In order to satisfy both criteria, we need to choose capacity >365,129 bps.
(The next highest standard rate is 384 kbps.)
tR etT )1(}Pr{ :Delay of Percentile
10.0}msec 20Pr{ 1000
20
1000
250000
c
R eT
bps 129,365c