introduction to queueing theory - computer science...

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Introduction to Queueing Theory

Based on the slides of Prof. Hiroyuki OhsakiGraduate School of Information Science & Technology, Osaka University, Japan

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Contents

Introduction to Queueing TheoryLittle’s TheoremStandard Notation of Queueing SystemsPoisson Process and its PropertiesM/M/1 Queueing SystemM/M/m Queueing SystemM/M/m/m Queueing SystemM/G/1 Queueing System

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Introduction to Queueing Theory

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What is Queueing Theory?

Primary methodological framework for analyzing network delayOften requires simplifying assumptions since realistic assumptions make meaningful analysis extremely difficultProvide a basis for adequate delay approximation

queue

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Packet Delay

Packet delay is the sum of delays on each subnet link traversed by the packetLink delay consists of:

Processing delayQueueing delayTransmission delayPropagation delay

node

node

node

packet delay

link delay

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Link Delay Components (1)

Processing delayDelay between the time the packet is correctly received at the head node of the link and the time the packet is assigned to an outgoing link queue for transmission

head node tail node

outgoing link queue

processing delay

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Queueing delayDelay between the time the packet is assigned to a queue for transmission and the time it starts being transmitted

head node tail node

outgoing link queue

queueing delay

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Transmission delayDelay between the times that the first and last bits of the packet are transmitted

head node tail node

outgoing link queue

transmission delay

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Propagation delayDelay between the time the last bit is transmitted at the head node of the link and the time the last bit is received at the tail node

head node tail node

outgoing link queue

propagation delay

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Queueing System (1)

Customers (= packets) arrive at random times to obtain serviceService time (= transmission delay) is L/C

L: Packet length in bitsC: Link transmission capacity in bits/sec

queue

customer (= packet)service (= packet transmission)

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Queueing System (2)

Assume that we already know:Customer arrival rateCustomer service rate

We want to know:Average number of customers in the systemAverage delay per customer

customer arrival rate

customer service rate

average delay

average # of customers

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Little’s Theorem

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Definition of Symbols (1)

pn = Steady-state probability of having ncustomers in the systemλ = Arrival rate (inverse of average interarrivaltime)µ = Service rate (inverse of average service time)N = Average number of customers in the system

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Definition of Symbols (2)

NQ = Average number of customers waiting in queueT = Average customer time in the systemW = Average customer waiting time in queue (does not include service time)S = Average service time

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Little’s Theorem

N = Average number of customersλ = Arrival rateT = Average customer time

N = λTHold for almost every queueing system that reaches a steady-stateExpress the natural idea that crowded systems (large N) are associated with long customer delays (large T) and reversely

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Illustration of Little’s Theorem

Assumption:The system is initially emptyCustomers depart from the system in the order they arrive

delay T1

delay T2

α(τ)

β(τ)

N(τ)

t

# o

f cu

stom

erar

rival

s/de

part

ures

0

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Application of Little’s Theorem (1)

NQ = Average # of customers waiting in queueW = Average customer waiting time in queue

NQ = λWX = Average service timeρ = Average # of packets under transmission

ρ = λXρ is called the utilization factor (= the proportion of time that the line is busy transmitting a packet)

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Application of Little’s Theorem (2)

λi = Packet arrival rate at node iN = Average total # of packets in the network

∑ =

= n

ii

NT

1λ

node1

noden

nodei

λ1

λi

λn

iii TN λ=

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Little’s Theorem: Problem

Customers arrive at a fast-food restaurant as a Poisson process with an arrival rate of 5 per minCustomers wait at a cash register to receive their order for an average of 5 minCustomers eat in the restaurant with probability 0.5 and carry out their order without eating with probability 0.5A meal requires an average of 20 minWhat is the average number of customers in the restaurant? (Answer: 75)

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Standard Notation of QueueingSystems

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Standard Notation of Queueing Systems (1)

X/Y/Z/KX indicates the nature of the arrival process

M: Memoryless (= Poisson process, exponentially distributed interarrival times)G: General distribution of interarrival timesD: Deterministic interarrival times

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X/Y/Z/KY indicates the probability distribution of the service times

M: Exponential distribution of service timesG: General distribution of service timesD: Deterministic distribution of service times

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X/Y/Z/KZ indicates the number of serversK (optional) indicates the limit on the number of customers in the systemExamples:

M/M/1, M/M/m, M/M/∞, M/M/m/mM/G/1, G/G/1M/D/1, M/D/1/m

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Poisson Process and its Properties

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Poisson Process

A stochastic process A(t) (t > 0, A(t) >=0) is said to be a Poisson process with rate λ if

1. A(t) is a counting process that represents the total number of arrivals in [0, t]

2. The numbers of arrivals that occur in disjoint intervals are independent

3. The number of arrivals in any [t, t + τ] is Poisson distributed with parameter λτ

K,1,0,!

)(})()({ ===−+ − n

nentAtAP

nλττ λτ

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Properties of Poisson Process (1)

Interarrival times τn are independent and exponentially distributed with parameter λ

The mean and variance of interarrival times τn

are 1/λ and 1/λ^2, respectively

0,1}{ ≥−=≤ − sesP sn

λτ

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If two or more independent Poisson process A1, ..., Ak are merged into a single process A = A1 + A2 + ... + Ak, the process A is Poissonwith a rate equal to the sum of the rates of its components

A1

Ai

Ak

A

independent Poisson processes

Poisson process

mergeλ1

λi

λk

∑ ==

k

ii

1λλ

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If a Poisson process A is split into two other processes A1 and A2 by randomly assigning each arrival to A1 or A2, processes A1 and A2

are Poisson

A1

A2

A

Poisson processes

Poisson process

split randomlyλ1

λ2

with probability p

with probability (1-p)

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M/M/1 Queueing System

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M/M/1 Queueing System

A single queue with a single serverCustomers arrive according to a Poisson processwith rate λThe probability distribution of the service time is exponential with mean 1/µ

Poisson arrival with arrival rate λ

Exponentially distributed service timewith service rate µ

single server

infinite buffer

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M/M/1 Queueing System: Results (1)

Utilization factor (proportion of time the server is busy)

Probability of n customers in the system

Average number of customers in the system

µλρ =

)1( ρρ −= nnp

ρρ−

=1

N

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M/M/1 Queueing System: Results (2)

Average customer time in the system

Average number of customers in queue

Average waiting time in queue

λµλ −==

1NT

ρρλ−

==1

2

WNQ

λµρ

µ −=−=

1TW

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M/M/1 Queueing System: Problem

Customers arrive at a fast-food restaurant as a Poisson process with an arrival rate of 5 per minCustomers wait at a cash register to receive their order for an average of 5 minutesService times to customers are independent and exponentially distributedWhat is the average service rate at the cash register? (Answer: 5.2)If the cash register serves 10% faster, what is the average waiting time of customers? (Answer: 1.39min)

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M/M/m Queueing System

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M/M/m Queueing System

A single queue with m serversCustomers arrive according to a Poisson processwith rate λThe probability distribution of the service time is exponential with mean 1/µ

Poisson arrival with arrival rate λ Exponentially

distributedservice time with rate µ

m servers

infinite buffer

1

m

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M/M/m Queueing System: Results (1)

Ratio of arrival rate to maximal system service rate

Probability of n customers in the systemµ

λρm

=

⎪⎪⎩

⎪⎪⎨

⎧

>

≤=

−

−+=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡−

=∑

mnm

mp

mnn

mp

p

m

mm

k

kmp

mm

n

n

m

k

!

!

)(

1

)1(!

)(

!

)(

0

0

1

00

ρ

ρ

ρρρ

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Probability that an arriving customer has to wait in queue (m customers or more in the system)

Average waiting time in queue of a customer

Average number of customers in queue

)1(!

)(0

ρρ−

=m

mpP

m

Q

)1( ρλρ

λ −==

QQ PNW

ρρ−

== ∑∞

=+

10

Q

nnmQ

PnpN

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Average customer time in the system

Average number of customers in the systemλµµµ −

+=+=m

PWT

Q11

ρρρλ−

+==1

QPmTN

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M/M/m Queueing System: Problem

A mail-order company receives calls at a Poisson rate of 1 per 2 minThe duration of the calls is exponentially distributed with mean 2 minA caller who finds all telephone operators busy patiently waits until one becomes availableThe number of operators is 2 on weekdays or 3 on weekendWhat is the average waiting time of customers in queue? (Answer: 0.67min and 0.09min)

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M/M/m/m Queueing System

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M/M/m/m Queueing System

A single queue with m servers (buffer size m)Customers arrive according to a Poisson processwith rate λThe probability distribution of the service time is exponential with mean 1/µ

Poisson arrival with arrival rate λ Exponentially

distributedservice time with rate µ

m servers

buffer size m

1

m

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M/M/m/m Queueing System: Results

Probability of m customers in the system

Probability that an arriving customer is lost(Erlang B Formula)

mnn

pp

np

n

n

m

n

n

,,2,1,!

1

1

!

1

0

00

K=⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=∑

µλ

µλ

∑ =

= m

n

n

m

m

n

mp

0!/)/(

!/)/(

µλµλ

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M/M/m/m Queueing System: Problem

A telephone company establishes a direct connection between two cities expecting Poisson traffic with rate 0.5 calls/minThe durations of calls are independent and exponentially distributed with mean 2 minInterarrival times are independent of call durationsHow many circuits should the company provide to ensure that an attempted call is blocked with probability less than 0.1? (Answer: 3)

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M/G/1 Queueing System

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M/G/1 Queueing System

A single queue with a single serverCustomers arrive according to a Poisson processwith rate λThe mean and second moment of the service time are 1/µ and X2

Poisson arrival with arrival rate λ

Generally distributed service timewith service rate µ

single server

infinite buffer

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M/G/1 Queueing System: Results (1)

Utilization factor

Mean residual service timeµλρ =

2

2XR

λ=

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M/G/1 Queueing System: Results

Pollaczek-Khinchin formula

WT

XRW

+=

−=

−=

µ

ρλ

ρ1

)1(21

2

)1(2

)1(2

22

22

ρλρλ

ρλλ

−+==

−==

XTN

XWNQ

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Conclusion

Queueing models provide qualitative insights on the performance of computer networks, and quantitative predictions of average packet delayTo obtain tractable queueing models for computer networks, it is frequently necessary to make simplifying assumptionsA more accurate alternative is simulation, which, however, can be slow, expensive, and lacking in insight

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