Performance Analysis of Computer Network

Queueing SystemLecture 12

1

ERF – 2016-2

2

Inside a Router

3

Inside a Router

4

Element of Queue

5

Element of Queue

6

Kendall Notation of Queueing System

7

Kendall Notation of Queueing System

8

Performance Measure if Queueing System

9

Little’s Law

10

Litlle’s Law

Queueing System EquationPerformance Measure of Interests:–Jobs in the Queue–Jobs in the Server–Utilization or Traffic Intensity

–The Response Time:

[ ] [ ]E N E Rl=[ ] [ ]sE N E Sl=

[ ] [ ]

1

, [ ]

sE N E S

and

if X thus X ES

r llr rµ

l r

= =

= <

= =

[ ] [ ] [ ][ ][ ]1

E R E W E SE SE R

r

= +

=-

We further have the performance measure of interests:

2

[ ][ ]1

[ ]1

[ ]1q

E SE W

E N

E N

rr

rrrr

=-

=-

=-

Poisson Process

Poisson Process

Exponential and Posisson

Properties of Poisson

M/M/1 ModelArrival Model: Poisson process

Poisson process is viable model for M/M/1 whenpackets are originated from large population ( such asfrom Internet) of independent users

The superposition of Poisson process yields thePoisson process as depicted in the figures

The M/M/1 Performance MeasuresPoisson arrival with rates λ. Hence the average time of packetarrivals equals to E[A] = 1/ λ

Markovian Service time with rate µ. The average service timeE[S] = 1/ µ

The utilization of M/M/1 queue is obtained from λ/ µ. From Little’s law we also obtain ρ = λ E[S]

The average number of jobs in the system equals to E[N] = λ.E[R]. Thus, E[N] also can be obtained from ρ / 1- ρ

From the Little’s law, the average response time can be obtained as E[R] = E[S] / 1 – ρ

The average waiting time of a job in the queue can be measured by: E[W] = ρ .E[S] / 1- ρ

The DTMC Model of M/M/1

With the generator Matrix Q

M/M/1 SolutionFrom the given performance measure

Find the steady state probability using GBE

From the obtained steady state probability wecan find another performance measure

M/M/1 SolutionThere are three ways to solve the steady-state probabilities

Global Balance Equations

Suppose ‘probability flow balance’ hence:p0λ0 = p1µ1 Þ pi(λi + µi) = pi-1 λi-1 + pi+1µi+1 for I = 1, 2, 3, …; thus we have

1

0 0 10 1

1 0 1

1;1

ik

i ik kk

i k k

p p pllµµ

-

-¥= +

= = +

= =+

ÕåÕ

ip

1ijjp =å

M/M/1 SolutionSince station has infinite buffer, no loss involves, thus:–Overall arrival rate:

–Overall service rate:

If r=λ/µ < 1, the queue is stable and the throughput X = λ.

Then

å¥

=

=0i

iip ll

å¥

=

=0i

iip µµ

0 0[ ] ; [ ] ( 1)i q i

i iE N ip E N i p

¥ ¥

= =

= = -å å

E[R]=E[N]/λ; E[W]=E[Nq]/λ=E[R]-E[S].

The probability of having k jobs in the queuing station:

å¥

=

=ki

ipkB )(

Example: Simple Single Server

M/M/1 Solution: Local Balance

Solving flow equation is easier than GBE: piλi = pi+1µi+1; iÎÀ

The computation is done repeatedly, hence we can find all state probability

M/M/1 with Constant RatesAssume the rate is constant

λ = λi and µ = µi for all states. Hence– pi = p0(λ/µ)i– pi = p0(r)i; i = 0,1,2,…

If r = λ/µ we have by normalisation that– p0 = 1 - r

The steady-state probabilities:– Known then

– pi = (1 - r)ri; i = 0,1,2,…

p0 = 1 - r Þ 1 – p0 = åipi è utilisation.

Throughput X = λ hence r = XE[S].

å¥

=

=0

0 1i

i prå

=i

pr

10

Other Performance Measure

E[N] = ρ / 1 – ρ

E[R] = 1 / µ - λ

In case we have E[S] , thus E[R]= E[S]/1 - ρ

A Performance of Router

Router A sends 8 pps, on the average, to Router B. The mean size of a packet is 400 bytes with exponentially distributed. They are connected with 64 kbps line speed. –What is the utilization of the router ?

1864 / / (400 8 / ) 20 /

0.4

skbps s x bit packet packtes s

lµ

lrµ

-== =

= =

A Performance of Router–What is the E[N] ?

–What is the probability of the number of packet is 10 or more ?

[ ] 0.671

E N rr

= =-

10 410Nr r -= =

M/M/1 AnalysisWe start with simple Queuing M/M/1

Model as a birth death process

Generate Markov Chain

Birth death is a Markov Process in which transition are only allowed between neighboring states

Steady state solution of a birth death process–Steady state probability of being in state n is expressed as:

and is probability being in state 0

np

0 1 10

0 1... 2

... ; 0,1,2,...,nnp p wherenl l l

µ µ µ-= = ¥

0p

For the proof of

where previously we have

Because M/M/1 is a special case for birth death process where

1

0 0 10 1

1 0 1

1;1

ik

i ik kk

i k k

p p pllµµ

-

-¥= +

= = +

= =+

ÕåÕ

01 0

1

p plµ

=

i j

i j

l l

µ µ

=

=

Hence by simplification

–In general or traffic intensity. Thus, the probability of being in state i is implicitly determined by traffic intensity –And

–Hence

0

i

ip plµ

æ ö= ç ÷è ø

l rµ

æ ö=ç ÷

è ø

0i

ip pr=

0 2

1 11 ...

p rr r r¥= = -

+ + + +

(1 ) iip r r= -

Example of M/M/1

Consider the following Queue

–The utilization can be calculated:

–E[N] in the system can be calculated as well:

–E[R] can also be calculated

0.30.5

lµ==

0.3 0.60.5

r = =

0.6[ ] 1.51 0.6

E N = =-

1[ ] 5E Rµ l

= =-

[ ] 1.5[ ] 50.3

E NE Rl

= = =

Probability of being in state 5

n Probability of the are number 10 IP Packets or more in the system:

5 55 (1 ) 0.4 0.6 0.03p xr r= - = =

10 0.06r =

M/M/1/m Single Server Queue

Suitable to model M/M/1 in practice

Total of jobs in the queuing station is limited by m capacity

Whenever m has been reached, there are no jobs arrive

Consequently, we have the following flows

Steady state solution for M/M/1/m

1

0 00

0

0 1

1

; 1,2,..., ;

;

1

0,1,2,...,11

;

1

ii

m

ii m

ii

j

p p

p

i m

p

p p

i m

lµ

r

rrr

l lµ

r

rr

µ

-

=

+

+

æ ö= = = =ç

-=

---

÷ø

==

è

=

Õ

Example: M/M/1/5The probability of packet lost equals to p5 –

– See the table below.

55 6

11

p r rr-

=-

The blocking probability pn is a function of m, l,and µ

Cases –(1) a single M|M|1 with arrival λ and service rate Kµ; –(2a) M|M|K with arrival λ and service rate µ; and –(2b) K number of M|M|1 each with arrival λ/K and service rate µ.

Three queuing Cases

Solution(1) E[R] = E[S]/(K-r) where r=λ/µ.

(2)

(3) E[R] = KE[S]/ (K-r) = K. E[R] in Solution (1)

20

)1(!][][][

K

K pKK

SESEREr

r-

+=

Example: Performance measure of HTTP 1.1

An ISP involves two HTTP server A and B which incorporatesversion 1.0 and 1.1, respectively . The average responsetime of Server A < Server B.

Is Server A better than server ?– A is better than B. Since λ is constant, the smallest service rate

E[S] makes the smallest E[R]. Hence, System A has smallestservice time that reflects better service. Service time is timeneeded to serve a packet arrives at a server.

To fulfill real time requirement, what are the performancemeasures ?– E[R] and smallest E[W]

To fulfill reliability service, what are the performancemeasures?– By ignoring the Network delay, the retransmission attempt of

TCP is interested measure.

Simulation

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