ReviewCh8-Model Queueing System

8/3/2019 ReviewCh8-Model Queueing System

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241-460 Introduction to Queueing

Networks : Engineering Approach

Assoc. Prof. Thossaporn KamolphiwongCentre for Network Research (CNR)

Department of Computer Engineering, Faculty of EngineeringPrince of Songkla University, Thailand

ap er o e ueue ng ys em

Email : [email protected]

Outline

Queueing System

Basic Queueing Systems

Commond Queueing Systems configuration

Kendall Notation

Measuring System Performance

ommon arame ers an e a ons p

Littles result

Chapter 8 : Model Queueing System


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Queueing Systems

Model of systems providing service

customers arrive looking for service and departafter service has been provided


Why analyze Queues

To quantify the operating characteristics of thegiven system

Often need simplifying assumptions for amathematical model

Realistic mathematical model often too complexto solve

Usually requires simulation

Example

Restaurant more employees (better service) vs.fewer employees (less wages)



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QueueingSystem

Basic Queueing Systems

Arrival Departure

Queue Server


Queueing Systems

Server

Queueing System

Arrival Pattern of Customer

Queue Discipline

DepartureArrival

Queues

Service Patterns of Server

System Capacity

Number of Service Channels

Stage of service



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Arrival Pattern of Customer

Server

Queueing System

Identify stochastic processto describe thearrivin stream

DepartureArrival

Queues

Describe the time between successive customerarrivals (interarrival times)


(Continue)

Server

Queueing System

Example of Arrival Process

Poisson arrivals, interarrival times areexponentially distributedM, most commonly used

DepartureArrival Queues

Other distributions e.g.

Deterministic D : interarrival time or service time isconstant

General G



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Arrive Process

2

t0


= 8 /t0 second

Server

Queueing System

Service pattern of Server

DepartureArrivalQueues

The length of time that a customer spends inthe service facility (How long in the server?)

Service Time Distribution

Commonly assumed random variables

Most common distribution exponential



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Server

Queueing System

Queue Discipline

Order of customer processing.

i.e. supermarkets are first-come-first served.

Hos ital emer enc rooms use sickest first.

134


2

First Come, First Served (FCFS)

Last Come, First Served (LCFS)

etc.


System Capacity (Queue)

Server

Queueing System

Where customers who have entered the system waitbefore being served


Can be finite or infinite

We typically assume it is infinite unless the finite numberis small enough that it would seriously affect the model

Finite length: when full customers are rejected.



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Number of Service Channels

Server

Queueing System

Example

banks have multiserver queueing systems.

DepartureArrival

Queues

number of processors in the system

number of I/O channels


Stage of Service

Single stage of service

Multistage

Telecommunications network process messages

through a selected sequence of node



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Common Queueing System

Configurations

Customer Customer

Waiting line Serverarrives leaves

Server 1

Customer

leaves

Customer Customer


Waiting line Server 2arr ves

Server 3

Customer

leaves

Common Queueing SystemConfigurations

Server 1

Customer

leaves

Waiting line Server 2Customer

arrives

Customer

leaves

Customer

leaves

Waiting line


Server 3Waiting line


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Kendall Notation

A/S/m/B/K/SD

S : Service Time distribution

m : Number of servers

B : System capacity

K : Population size (total number of packets)

SD : Service Discipline


Queue Example

M/M/1

exponential service time distribution

single server unlimited number of waiting places

infinite population

FCFS



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Queue Example

M/M/3/20/1500/FCFS

exponential service time distribution

3 server

Buffer size 20

Population 1500

FCFS


Notation for Basic QueueingSystem

Cn : the nth customer to enter the system

N(t) : number of customer in the system at time tn : arrival time for Cntn : interarrival time between Cn-1 and Cn = n -n-1

Queueing systemC9 8 C3 5

wn : waiting time (in queue) for Cnxn : service time for CnTn : system time (queue plus service) for Cn

Tn = wn + xn



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Time-diagram notation for Queues

Cn-1

x x + x +w

Cn+2Cn+1Cn

n+2

Cn

Servicer

tn+1 tn+2

n+1n

Cn+2

Cn+2Cn+1

Cn+1Cn

Queue

Time

wn : wa ng me n queue or nxn : service time for Cntn : interarrival time between Cn-1 and Cnn : arrival time for Cn



Three types of system response of interest

Time a customer spends in the queue

Total time a customer spends in the system(queue + service)

Indication of manner in which customer mayaccumulate

Measure of idle time of servers



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Prob. of the number of requests in the system: Pn

Pn =P[there arenjobs in the system]

escr e e e av our o a queue ng sys em ymeans of the probability vector of the number ofjobs in the system


Utilization or carried load

Single server

u za on s e rac on o e me n w cthe server is busy

In case when the source is infinite and there isno limit on the number of jobs in the singleserver queue

rateservice

ratearrival



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Multiple servers

Utilization or carried load

servers is the mean fraction of active (or busy)servers.

The condition for stability is < 1


Throughput

Throughput() is defined as the mean numberof jobs whose processing is completed in asingle unit of time,

ThroughputPB

= PB = (1 PB)

With infinite queue, there is no blocking PB = 0



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Throughput

Throughputis average rate of service if the queueis not em t .

Hence

= (1 P0)

(1 P0) is probability that system is NOT empty


Throughput

= (1 P = (1 P

BP

P

1

1intensityTraffic 0



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Total Wait-Time (T)

Total Wait-Time: is mean interval betweencustomer arrival and customer de artureincluding time in the queue and time beingserved.

Customer Aarrive

TCustomer A

leave

time


Total Wait time (T) = Wait time + service time

t0 t0+T

Wait time (Wq)

Wait time(Wq) is the time that a job

spen s n a queue wa ng o e serv ce .

Customer Aarrive

TCustomer A

leave


timet0 t0+T

Wq


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Queue length (Nq

)

Queue lengthNq: is the number of jobs in the

.

timet0 t+T


Nq

Mean number of arrivals in time

)()()( xxPxxPxEProbability Theory

Queueing Theory

Arrival exponential distribution

Mean number of arrivals in time :E(k)!

)(

)( k

e

kP

k


11 !

)()(k

k

k k

ekkPkE


17/24

17


)(kk

ee

kE

k

e

Because

11 kk

1 !

)(k

k

kekE

1k


eekE )(Hence


Mean number of arrivals in timeMean number of arrivals in time == == meanmean interarrivalinterarrival raterate xx interarrivalinterarrival timetime



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Common Parameters and

Relationship

N

Nq

W x


Queue ServerT

Common Parameters andRelationship

: mean inter-arrival rate : mean de arture rate er server : inter-arrival time = 1/

X : service time

Wq : time a customer spends in queue

T : total time a customer spends in the system

= q

Nq : number of customer in queue

N : number of customer in system

N=Nq + customer in service



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Littles Result

Average number of customers in system

= Average arrival rate of customer to that

system x average time spend in that system


Proof for Littles Law

Let

,

(t) is number of departures in (0,t)

N(t) is the number in the system at time t

t is the average arrive rate during (0,t)

= t t



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Littles Result

Arrivals and Departures10 N(t) =(t) - (t)

(t)

(t)mberofcustomer

5

t


Time

Nu

0

t1 t2

Tt3

Littles Result

T710

T2

T3

T4

T5

T6

(t)

(t)N(t)5

M

i

iTArea11


T1

N(t)

2

t0

0

dtttAreaT

0

me

TimeT


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(Continue)

M

i

TT

TdttNdtttArea

Consider TandM , whereMis the number ofarrivals in time (0,T)

Average Time T spent in system

i 100

M

iT

TM

T1

lim


Average number of customer in system

Time average arrive

t

M

tM,lim

T

TdttN

TN

0

1lim

(Continue)

MT

TdttN

Taking the limit of both sides as T

i 10

M

i

iT

T

T

TMT

MdttN

T 10

1lim

1lim


N= T


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Littles Result

Littles Law

=

Average number of customers in system= Avera e arrival rate o customer to that


system x average time spend in that system

Littles Result

Intuition:

time in the system, and the average arrivalrate is , during this time, there can be Wcustomer arriving to the system

Nq = WqN= T = Wq + one being serve in server



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Little result Example

A disk server takes on average, 10 ms to satisfy anI O re uest. If the re uest rate is 100 ersecond, then how many requests are queued atthe server?

Solution

q ,

Nq = Wq = 10x10-3(100) = 1 requests


References

1. Alberto Leon-Garcia, Probability and RandomProcesses for Electrical En ineerin 3rd Ed.Addision-Wesley Publishing, 2008.

2. Robert B. Cooper, Introduction to QueueingTheory, 2nd edition, North Holland,1981.

3. Donald Gross, Carl M. Harris, Fundamentals of-

Interscience Publication, USA, 1998.



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(Continue)

4. Leonard Kleinrock, Queueing Systems VolumnI: Theor A Wile -Interscience PublicationCanada, 1975.

5. Georges Fiche and Gerard Hebuterne,Communicating Systems & Networks: Traffic &Performance, Kogan Page Limited, 2004.

Modeling and Analysis of TelecommunicationsNetworks, John Wiley & Sons, 2004.


Documents

ReviewCh8-Model Queueing System