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CS 756 1

Basic Queueing TheoryM/M/* Queues

These slides are created by Dr. Yih Huang of George Mason University. Students registered in Dr. Huang's courses at GMU can make a single

machine-readable copy and print a single copy of each slide for their own reference, so long as each slide contains the copyright statement, and GMU facilities are not used to produce paper copies.

Permission for any other use, either in machine-readable or printed form, must be obtained from

the author in writing.

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Introduction

�Queueing theory provides a mathematical basis for understanding and predicting the behavior of communication networks.

�Basic Model

ServerQueue

Arrivals Departures

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� Major parameters:– interarrival-time distribution– service-time distribution– number of servers– queueing discipline (how customers are taken

from the queue, for example, FCFS)– number of buffers, which customers use to wait

for service� A common notation: A/B/m, where m is the

number of servers and A and B are chosen from – M : Markov (exponential distribution)– D: Deterministic – G: General (arbitrary distribution)

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M/M/1 Queueing Systems

� Interarrival times are exponentially distributed, with average arrival rate λλλλ.

�Service times are exponentially distributed, with average service rate µµµµ.

�There is only one server.

�The buffer is assumed to be infinite.

�The queuing discipline is first-come-first-serve (FCFS).

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System State

�Due to the memoryless property of the exponential distribution, the entire state of the system, as far as the concern of probabilistic analysis, can be summarized by the number of customers in the system, i.

– the past/history (how we get here) does not matter

�When a customer arrives or departs, the system moves to an adjacent state (either i+1 or i-1).

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� In equilibrium,

�Let

�We have

State Transition Diagram

10 2 3 4 5

λ

µ

λ

µ

λ

µ

λ

µ

λ

µ

λ

µ

}{ i state in systemPPi =PP ii 1+= µλ

λ

µi i+1

The rate of movements in both directions should be equal

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� Equations from the state transition diagram:

�Solve

�What is ?

�

PPPPPP

32

21

10

µλµλµλ

=

=

=

PP

PPPP

PPP

k

k 0

0

2

012

001

)(

ρ

ρρρµλ

ρµλ

=

===

==

�

ρ

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�Since

we have

�That is, .�Note that must be less than 1, or else the

system is unstable.

��∞

=

∞

===

00

0

1k

k

kk PP ρ

. 111

100

ρρ

−=�=− PP

ρ)1( ρρ −= k

kP

5

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Average Number of Customers

? )1(00��

∞

=

∞

=

=−==kk

kkkkPN ρρ

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�Average delay per customer (time in queue plus service time):

�Average waiting time in queue:

�Average number of customers in queue:

λµλ −== 1N

T

λµρ

µλµ −=−

−= 11

W

ρρλ−

==1

2

WNQ

6

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Applications

�Consider 24 computer users, each of which produces in average 48 packets per second.

�For every customer, the interarrival times of his packets are exponentially distributed.

�The lengths of packets are also exponentially distributed, with mean 125 bytes.

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Scenar io 1�Users share a T1 line using the standard T1

time-division multiplexing.

�Assume that each user is associated with an infinite buffer (that is, queue).

� In a T1 line, it takes 1/8000 seconds to deliver (or serve) each byte.

�However, due to their variable lengths, the delivery (or service) times of packets are still exponentially distributed.

– The average service rate µ = ?

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� The system can be considered as 24 M/M/1 queues:

We have %7564

48 ==ρ

375.01

75.0 =−

=N

msec4224

1

4864

1 ≈=−

=T

QueueArrivals

Departs to the other endof the T1 line

Server,1/24th ofthe T1 line

Packet

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Scenar io 2

�Users share a 1.544 Mbps line through an IP router.

Packetsfrom24 users

The entireT1 as theserver

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�The (aggregated) arrival rate is

�The service rate is

�We have

This system is 24 times faster than TDM !

. 11524824 =×

. 15366424 =×

msec 4.2384

1

11521536

1

%7564/48

≈=−

=

==

T

ρ

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Discussion�Flaws in the analysis ?

�Still such a drastic difference in results convincingly reveals the inefficiency of TDM.

�This partly explains the momentum toward using the Internet as the universal information infrastructure.

� In general, allowing customers to share a pool of resources is far more efficient than allocating a fixed portion to each customers.

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M/M/m Queueing Systems

All servers are identical, with service rate µ

DepartsQueue

Arrivals

1

2

Servers

m

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State Transition Diagram

Balance equations:

10 2

λ

µ

λ

2µ

λ

3µ µm

m m+1

µm

λ λ λ

µm

���

>≤

=−miPm

miPiP

i

ii

,

,

for

for1 µ

µλ

10

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Solution

Where

Noticing that we have

���

���

�

>

≤=

mim

mP

mii

mpP

Pim

i

i

for ,!

for ,!)(

0

0

ρ

. µ

λρm

=

, 10

=�∞

=i iP

][)1(!

)(1

0

)(

!

1

0 ρρρ−

−

== +�

−

m

m mm

i i

m iP

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The probability that an arriving customer has to wait in queue:

This is known as the Erlang C formula.

)1(!)(

!)(

!

0

0

0

ρρ

ρρ

ρ

−=

=

=

=

�

�

�

∞

=

−

∞

=

∞

=

m

mP

m

mP

mmP

PP

m

mi

mim

mi

im

miiQ

11

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Average number of waiting customers:

ρρ

ρρρ

ρρρ

−=

−=

==

=−=

��

��∞

=

∞

=

+

∞

=+

∞

=

1

)1(!)(

!)(

!

)(

20

0

0

00

0

P

mmP

im

mPm

mPi

PiPmiN

Q

m

i

im

i

mim

imi

miiQ

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Average waiting time in queue:

Average time in the system:

Average number of customers in the system:

)1( ρλρ

λ −== PN

W QQ

λµµρλρ

µµ −+=

−+=+=

mPP

WT QQ 1

)1(

11

ρρ

ρλµ

λµλλ

−+=

−+==

1P

mm

PTN QQ

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M/M/1/K Queueing Systems

�Similar to M/M/1, except that the queue has

a finite capacity of K slots.

�That is, there can be at most K customers in

the system.

� If a customer arrives when the queue is full,

he/she is discarded (leaves the system and

will not return).

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Analysis

Notice its similarity to M/M/1, except that there are no states greater than K . We have

Noticing that we have

λ

µ

λ

µ

λ

µ

λ

µ

λ

µ0 1 2 K- 1 K

���

>≤≤

=Kifor ,0

0for ,0 KiPP

i

i

ρ

10

=� =

K

i iP

ρρ

101

1+−

−= KP

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Poisson ProcessLet random variable N be a “counter” of the number of occurrences of a particular type of events. Clearly, the value of N increases over time. Let be the value of N at time t. Moreover, if , is said to be a counting process.The counting process is said to be a Poisson processhaving rate λ if the number of events in any interval of length t is Poisson distributed with mean λt. That is, for all

)(tN0)0( =N )(tN

)(tN

0, ≥ts,...1,0 ,

!)(

})()({ ===−+ nn

tensNstNP

nt λλ

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Discussion

� The interarrival times of a Poisson process with rate is exponentially distributed with average

� The reverse is also true: if the interarrival times of events are exponentially distributed with average then the event counting process is Poisson with rate .

� Thus, the customer arrival processes of M/M/* queueing systems are Poisson.

� A Poisson customer count and exponentially distributed customer interarrival times are the two sides of the coin.

λ λ/1

λ/1

λ

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Sampling Poisson Arr ivals

�Consider a Poisson customer arrival process of with average rate λ.

�Each customer can be classified as Type I or Type II, with probability p and 1-prespectively.

�Then, the arrival process of Type I customers is also Poisson with average rate pλ.

�Likewise, the arrivals of Type II customers is Poisson with average rate (1− p)λ.

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Application�We know that the customer arrivals at a

barbershop form Poisson process with average rate of 10 customers per hour.

�Among the customers, 40% are males and 60% are females.

�Then the interarrival times of male customers are exponentially distributed with an average rate of 4 per hour.

�The interarrival times of female customers are exponentially distributed with an average rate of 6 per hour.

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Exercise� Consider the router configuration below.

� The lengths of arriving packets are exponentially distributed with an average of 1000 bits.

2000PacketsPer sec.

1Mbps

2Mbps

Port 1

Port 2

40%

60%

Port 0

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Questions

�Argue that queues A and B are independent M/M/1 systems.

�Compute the average length of queue A in bits.

�For a packet destined for port 2, compute its expected time at the router (including transmission time).

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�Compute the average time a packet spent at the router (including transmission time).

�Compute the average number of packets at the router(including the ones in transmission).

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Merging Poisson Arr ivals

�Given two exponential variables and with rates and , the random variable

is also exponential, with rate .

�Consider two Poisson arrivals, with average rates and .

�The merged arrival process will also be a Poisson process, with the average rate

X 1 X 2

λ1 λ 2

},min{ 21 XXX =λλ 21+

λ1 λ 2

λλ 21+

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Application

�Consider the router configuration below.

Router

Port 1

Port 2

Port 0

Packetarrival

Packetarrival

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�The lengths of arriving packets are exponentially distributed with an average of 1000 bits.

– Why do we care about packet lengths ?

�Packet arrivals at ports 1 and 2 are exponentially distributed with average rates of 2000 and 3000 packets per second, respectively.

�The transmission rate of port 0 is 10Mbps.

�The whole system can be modeled as a single M/M/1 queueing system, with an arrival rate of 5000 and service rate of 10,000.

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Burke's TheoremIn its steady state, an M/M/m queueing system with arrival rate λ and per-server service rate µproduces exponentially distributed inter-departure times with average rate .

Application: Two cascaded, independently operating M/M/m systems can be analyzed separately.

Server 1

M/M/1 system 1

Server 2

Departs

M/M/1 system 2

µ1

µ2λ

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Pitfall�Consider the system below where the

servers are transmission lines.

�Packets lengths are exponentially distributed with an average of 1000 bits.

�Can the two queues be analyzed separately ? Why ?

1Mbps 2Mbps

500PacketsPer sec.

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Discussion

In general: any “ feedforward” network of independently-operating M/M/m systems can be analyzed in this system-by-system decomposition.

λ

p

1- p

µ1

µ 2

µ 3

µ4

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Question: How about networks that do contain “ feedbacks” ?

Answer : the interarrival times of some systems may not be exponentially distributed and thus cannot be analyzed as independent M/M/m queues

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Jackson's TheoremFor an arbitrary network of k M/M/1 queueing

systems,

where

That is, in terms of the number of customers in each system, individual systems act as if they are independent M/M/1 queues (they may not).

),()...()(),...,,( 221121 nPnPnPnnnP kkk =

. )1()( ρρ jjjjn jnP −=

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ApplicationConsider the network below. The arrival rate

and probabilities and are known.

We first compute the arrival rates and :

λ p1

p2

CPU µ 1

I / O µ2

λ λ 1 λ1 λ 1p1

λ1p2

λ2λ2

λ1 λ 2

ppp

p

12211

12221

/ ,/

, λλλλ

λλλλλ==�

=+=

21

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Let and . By Jackson's theorem,

And

Total number of customers in system is

Average time in system is

µλρ 111 /= µλρ 222 /=

)1()1(),( 2211 ρρρρ −−= jijiP

ρρ

ρρ

2

22

1

11

1 ,

1 −=

−= NN

.21 NNN +=

.)1()1( 2

2

1

1

ρλρ

ρλρ

λ −+

−== N

T

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DiscussionConsider the packet switching network below.

Can we cite the Jackson's theorem, model the transmission lines as servers, and analyze them as separate M/M/1 queues ? Why ?

Router 1

Router 2

Router 3

Router 4

Packets