Queueing Theory
2008
Queueing theory definitions
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(Bose) “the basic phenomenon of queueing arises
whenever a shared facility needs to be accessed for
service by a large number of jobs or customers.”
(Kleinrock) “We study the phenomena of standing,
waiting, and serving, and we call this study Queueing
Theory." "Any system in which arrivals place demands
upon a finite capacity resource may be termed a
queueing system.”
(Mathworld) “The study of the waiting times, lengths,
and other properties of queues.”
排队论是专门研究带有随机因素，产生拥挤现象的优化理
论。也称为随机服务系统。
Applications of Queueing Theory
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Telecommunications
Determining the sequence of computer operations
Predicting computer performance
One of the key modeling techniques for computer
systems / networks in general
 Vast literature on queuing theory
 Nicely suited for network analysis
Traffic control
Airport traffic, airline ticket sales
Layout of manufacturing systems
Health services (eg. control of hospital bed
assignments)
Queuing theory for studying networks
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View network as collections of queues
 FIFO data-structures
Queuing theory provides probabilistic
analysis of these queues
Examples:
 Average length (buffer)
 Average waiting time
 Probability queue is at a certain length
 Probability a packet will be lost
Model Queuing System
Customers
Queue
Server
Queuing System

Use Queuing models to
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Describe the behavior of queuing systems
Evaluate system performance
Customer n
Arrival
event
Delay
Begin
service
Activity
End
service
Time
Interarrival
Arrival
event
Begin
service
Delay
End
service
Activity
Time
Customer n+1
Characteristics of queuing systems
Kendall Notation 1/2/3(/4/5/6)
1.
2.
3.
4.
5.
6.
Arrival Distribution
Service Distribution
Number of servers
Total storage (including servers)
(infinite if not specified)
Population Size
(infinite if not specified)
Service Discipline
(FCFS/FIFO)
Distributions
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M: stands for "Markovian / Poisson" ,
implying exponential distribution for
service times or inter-arrival times.
D: Deterministic (e.g. fixed constant)
Ek: Erlang with parameter k
Hk: Hyperexponential with param. k
G: General (anything)
Poisson process & exponential distribution
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Inter-arrival time t (time between arrivals) in
a Poisson process follows exponential
distribution with parameter  (mean)
无后效性 － 不管多长时间(t)已经过去，逗留
时间的概率分布与下一个事件的相同
fT(t)
Pr(t )  e
E (t ) 
 t

1

E (T ) 
1

t
Examples
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M/M/1:
 Poisson arrivals and exponential service,
1 server, infinite capacity and population,
FCFS (FIFO)
 the simplest ‘realistic’ queue
M/M/m/m
 Same, but
 m servers,
 m storage (including servers)
 Ex: telephone
Analysis of M/M/1 queue

Given:
•
: Arrival rate (mean) of customers (jobs)
(packets on input link)
•
m: Service rate (mean) of the server
(output link)
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Solve:
 L: average number in queuing system
 Lq average number in the queue
~ “1”
 W: average waiting time in whole system
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Wq average waiting time in the queue ~ “1/m”
M/M/1 queue model
L
Lq

m
1
m
Wq
W
Derivation
Po
2
1
0
 P k -1
 P1
m P1
...
 P1  mP 2
k
k-1
mP k
mP 2
 P 0  mP 1
P k
k+1
m P k +1
 P k  mP k+1


m
k
 P0
k
Pk 
k   P0
m
since all probability sum to one

k

 P0  1
k0
Solving W, Wq and Lq

Lq
Wq
Pn


m
2


, L
1 
1 

1

, W 
m (1   )
m (1   )
 (1   )  n
For stability, mean arrival rate must be less
than mean service rate

Utility factor  
<1
m
Response Time vs. Arrivals
Waiting vs. Utilization
0.25
W(sec)
0.2
0.15
0.1
0.05
0
0
0.2
0.4
0.6
 (%)
W 
1
m 
0.8
1
1.2
Example
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On a network router, measurements show
 the packets arrive at a mean rate of 125
packets per second (pps)
 the router takes about 2 millisecs to
forward a packet
Assuming an M/M/1 model
What is the probability of buffer overflow if
the router had only 13 buffers
How many buffers are needed to keep packet
loss below one packet per million?
Example
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Arrival rate λ = 125 pps
Service rate μ = 1/0.002 = 500 pps
Router utilization ρ = λ/μ = 0.25
Prob. of n packets in router =
(1  ρ)ρ  0.75(0.25)
n

n
Mean number of packets in router =
ρ
0.25

 0.33
1  ρ 0.57
Example
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Probability of buffer overflow:
= P(more than 13 packets in router)
= ρ13 = 0.2513 = 1.49x10-8
= 15 packets per billion packets
To limit the probability of loss to less than
10-6:
n
6
ρ  10

(
n  log 10
6
)/ log (0.25)
= 9.96