Slides 2-tkm

7/27/2019 Slides 2-tkm

1/46

Technical Goals and Requirements

David TipperAssociate ProfessorAssociate Professor

Graduate Telecommunications and

University of PittsburghSlides 2

http://www.sis.pitt.edu/~dtipper/2110.htmlhttp://www.sis.pitt.edu/~dtipper/2110.html

Last Week

Network Design is not a precise science. Many different types of problems

Size: LAN vs. MAN vs WAN.

Lifecycle Stage: greenfield, incremental, etc.

There can be many good answers - no best solution Design involves trade-offs among cost vs. performance

Top Down Design approach useful as a framework

TELCOM 2110 2

Logical Model

Physical Model

Implementation, Testing, Tuning, and Documentation


2/46

Analyzerequirements

Top-Down Network Design Steps

Develop

logical

design

Develop

Monitor and

optimize

network

performance

Im lementphysical

design

Test,optimize, and

document

design

and test

network

Source: P. Oppenheimer

Conceptual Model Network Design

Conceptual Model Design

At end of conceptual model design should havega ere en e

Objectives

Business Goals (e.g., make sales force more responsive tocustomers on sales calls)

Technical Goals (e.g., provide wireless access to corporatedata to sales force)

Requirements

TELCOM 2110 4

Business (e.g.,support XYZ application) Technical (availability, delay, bandwidth, etc.,)

Constraints

Business (organizational, budget, etc.,)

Technical (vendor, technology, sites to connect, security,etc.)


3/46

Technical Requirements & Constraints

From surveys/questionnaires, meetings etc. applicationdata determine technical requirements and constraints

Technical goal is to build a network that meets usersre uirements + some the ma not know the need.

Technical Goals Scalability Availability/reliability Network Performance

Utilization, Throughput, Delay, Delay Jitter, packet loss rate,call/connection blocking rate

Traffic Estimation is needed to estimate performance

TELCOM 2110 5

Manageability/Interoperability

Affordability $$ Need to determine reasonable goal for each category andthe relative importance of each.

Scalability

Scalability

how much growth a network design can support?

can the desi n ada t to chan in network load and QoSrequirements?

Can the network be expanded easily in the future?

Need to examine the network needs out a few years

Key points to understand How many more sites will be added?

TELCOM 2110 6

How many more users will be added? How many more servers, etc will be added? How many and what applications will be added? Technology migration path?


4/46

Scalability

Scalability For logical network design how much additional

investment

For physical design - thought of as expandability andupgrade capability

For example,

Given specific Router

TELCOM 2110 7

Can number of I/O ports be increased?

Can additional software features be added (e.g, VLANcapability, IP Sec, etc.)

Try to set reasonable scalability goals

Availability

Availabi li ty (A) Ability of an item to perform stated function at over time Fraction of the time that an item can be used when needed Value in the 0.0 to 1.0 ran e or 0 - 100%

165 hours uptime in 168 hours/week = 98.21% availability

Mean Time To Repair (MTTR) Average time to restore full functionality to an item

This may include time to travel to item, diagnose, isolate, remove and replace parts

MTTF: Mean-Time To Failure MTBF: Mean-Time Between Failures Variables are related as shown in figure below

limobsT obs

AT

TELCOM 2110 8

Failure Repair Failure Repair

MTBF

MTTRMTTR MTTF MTTFA

MTTF MTTR

MTBF

MTTRA 1


5/46

Component Availability

Consider an IP router MTBF 100,000 Hours

MTTR 6 hours (depends in part on location and type of failure)

= = .

Some representative values of networking equipment

Equipment MTBF Range (hr) MTTR (hr)

Web Server 104 - 106 1

IP Interface Card 104 - 105 2

TELCOM 2110 9

IP Router 105 - 106 6

WLAN AP ~105 2

WDM OXC orOADM

~106 6

Other Metrics

POFOD Probability of failure on demand the likelihood that a systems will fail

when a service request is made.

Used in systems where services requested infrequently (e.g., shut down of

ROCOF Rate of fault occurrence the frequency of occurrence of failures failure

intensity rate Used in systems with frequent requests for services (e.g., transaction

processing of credit cards)

Unavailability (U) The fraction of the time that an item cannot be used when needed

U= 1 A

Other expressions for unavailability Downtime per year

Downtime in units of minutes per year Obtained by multiplying U by minutes in a year

0.99999 availability (5 `9s availability), 0.00001 unavailability, 5.256 downtime per year (in minutes)


6/46

Availability Goals

Availability level Downtime per year Downtime per Week

90% 36.5 days 16.8 hours

95% 18.25 days 8.4 hours

99% 3.65 days 1.68 hours

99.9% 8.76 hours 10.1 hours

99.99% 52.6 min 1.01 min

99.999% 5.25 min 6.05 seconds

99.9999% 31.5 seconds 0.605 seconds

Telecom equipment traditionally five 9 availability carrier class equipment

Availability

Availability Goals depend on application and userrequirements may vary with location Highly available voice service at customer support call center

Five 9s at call center

Lower available voice over IP (VoIP) service in engineering dept. Three 9s availability for engineering dept.

Challenge how to provide higher availability for onlycertain services/applications

Network/System availability is the amount of time anetwork/system is available to users

TELCOM 2110 12

Need to work with users to set reasonable goals

Higher availability goal more costly design

Given component availability how to find network/systemavailability?


7/46

System Availability

System availability calculated from componentavailability Ai, and unavailability Ui,

If devices in parallel

1 2 n

1

n

series i

i

A A

n

1

n

s i

i

U U

TELCOM 2110 13

2

n

1

1 (1 )n

par all el i

i

A A

1pa ra ll el i

i

U U

System Availability - Example

DM

System

OA

WD

LineSy

A single bidirectional line in WDM optical network

OA

The availability of thebidirectional linesystem = ?

W

Line

Mstem80km 100km 80km

Equipment MTBF (hrs) MTTR (hrs)

Bidirectional OA 5*105 24

Bidirectional 5*105 6

TELCOM 2110 14

WDM LineSystem

Equipment CC (km) MTTR (hrs)

Terrestrial FiberOptic Cable

450 24


8/46

MTBF Physical Cable

Physical cables MTBF can be specified using the Cable Cut (CC) metric

Avera e cable len th that results in a sin le cable cut er ear

CC = 450 km means that per 450 km cable, there will be onaverage on cable cut each year

Example, given CC = 450km and cable length = 260 km,

( 365 24)( )

length of the cable (km)

CCMTBF hours

TELCOM 2110 15

450 365 24 15161.5260

cable km hMTBF hkm

System Availability - Example

Devices in series

Availability of bidirectional line (Aline)

2 2

2 2(1 ) (1 ) (1 )

24 24 6

line cable OA line system

line systemcable OA

cable OA lline system

A A A A

MTTRMTTR MTTR

MTBF MTBF MTBF

h h h

TELCOM 2110 16

5 5 15161.5 5 10 5 10

0.998297

h h h

450 365 24Note. 15161.5

260cable

km hMTBF h

km


9/46

Series-Parallel Reduction

For complex systems need to apply series parallelreduction to determine overall availablity

TELCOM 2110 17

+ series|| parallel

Availability Analysis

General Methodology:1) Get unavailability values of all components and sub-

systems.

2) Draw parallel and series availability relationships

3) Reduce the system availability model by repeatedapplications of the parallel/series availabilitysimplifications.

4) If not completely reduced

Use approximation methods to estimate availability.

Typically take a conservative approach and go with alower bound


10/46

Availability Analysis

Lower bound on unavailability The contributions of parallel elements to the unavailability isnot taken into account

AB

C

D

E

F

G H

Lower bound of Us: UA+UH

that the system does not meet the availability requirements

Several software packages for calculation of availability have approximation methods and simulation support for complicated

network availability analysis

Availability and Design

How do availability goals affect network design?

Basic Techni ues to increase availabilit1. Increased component/system availability

Use components with larger MTTF and/or shorter MTTR In general increased component reliability increased cost

2. Redundancy Duplicate system components and services

For example Space Shuttle has 3 fully redundant flightcomputers (+ manual operation!)

Incurs additional cost due to sparecomponents/capacity


11/46


Increasing system and component availability/reliability

MTTFA

MTTF MTTR

MTBF

MTTRA 1

Increase either MTTF, MTBF or decrease MTTR

Techniques to improve ICT equipment hardware MTBF well known

Adoption of fault tolerant hardware architectures (hotswappable line cards, backup switch cards, redundant cooling,backup power, etc)

Expense is major concern and market who will buy most

WLAN AP with 10 year MTBF?

Software availability a bigger issue increasing lines of code how to make more reliable (increase MTTF or MTBF)?

Micro-reboots, process redundancy, hot upgrades/patchinstallation, model checking code analysis, etc.


MTTR is often a Network Operations andManagement issue some what out of thecontrol of equipment manufacture

MTTR includes Mean time to detect failure

Mean time to diagnose failure

Mean time to fix and return to service

Today typically have fast mean time to detect

Often dont bother with diagnosis > fast replacementrather than repair

oar swap, re oot, reset, etc. Human in the loop and travel is often the bottleneck

Communication links have large MTTR due to need tolocate and physically repair link


12/46


Redundancy/Diversity Techniques seek to increase systemavailabilitynot individual component Combat independent faults by duplicate equipment/services

or examp e prov e our ac up attery supp y to core networequipment in order to combat electrical power outages

Redundancy higher COST

Tradeoff between Availability and COST

Redundancy Effects

PrimaryRouter

Back-upRouter

Scenario Single Router Availability with

1

1 (1 )n

paral lel i

i

A A

24

1 0.90 0.9900

2 0.95 0.9950

3 0.99 0.9999


13/46

Redundancy Example

BP

Ai is an availability of link i

Availability of a connection between S-D:

WPSource (S) Destination (D)

no protection i

i WP

GivenAi= 0.998297 for all links yieldsAno-protection= 0.996597, Aprotection= 0.999983

WPi BPiiiprotection AAA ||

Availability Example

Consider Availability of Internet Connection

TELCOM 2110 26

A = .99999 x (1 - (1-.999)(1 -.999)) x (1 (1-.99) (1-.99)) =0.999889


14/46

Network Performance

Several Performance measures Utilization Throughput Accuracy (BER, Packet Loss) Efficiency Delay and Delay Jitter Call Blocking for circuit switched networks

Typically look at measures during the busy periodof theday set threshold values

Need to know how to estimate values

TELCOM 2110 27

pproac es w en es gn ng ne wor are Queueing Analysis analytical models

Simulation measurements on computer model of the networkdesign

Benchmarking existing network then predict behavior - withempirical model, queueing model or simulation

Network Performance

Typically have a camelback shape to network traffic (bothpacket and circuit switched networks)

Busy time period will vary with network type and. .,

networks)

TELCOM 2110 28


15/46

Network Performance

Busy time period can be defined in several ways Time period (15 min period, 1 hour, 2hours, etc.) Location (system wide, switch, access net, cell tower, link, etc.) , , , .

TELCOM 2110 29

Network Performance - Utilization

Utilization is the percent of total availablecapacity (bandwidth) on a link in use (0-100%)

interval to determine the amount in use (e.g. thebusy time period or some fraction of it)

Link/equipment utilization identifies networkbottleneck points

Data networks usually have utilization < 40 -

TELCOM 2110 30

are o ten t res o s Telephone network utilization much higher

80 90 %

Utilization goals will effect resulting delay


16/46

Network Performance - Throughput

Throughput is defined as the quantity of error-freedata successfully transferred between nodes perunit of time Good ut or La er 2/3 throu h ut

Depends on network access method, the load onthe network and the error rate

Throughput can be expressed in Packets per Second (PPS) than can be sent by a

device with dropping any packets or bps for data

TELCOM 2110 31

Carried load in Erlangs for circuit switched networks

Example IEEE 802.11b wireless LAN channel rate 11Mbps typical max throughput 7 Mbps

Network Performance -Accuracy

Accuracy is a measure to ensure that the datareceived at the destination must be the same as

Data errors are caused by power surges, orspikes, poor physical connections, failing devices,electrical noise, interference, etc.

Accuracy can be expressed in Bit Error Rate

TELCOM 2110 32

Target values of BER depend on physicalmedium used wireless link 1 in 104 , opticalfiber 1 in 1010


17/46


Packet Loss occurs whenbuffers overflow at routers orgateways in wired networks

In wireless networks acket

originservers

loss due to interference, poorsignal quality, collisions

Packet Loss results inretransmission in applicationsthat require reliability

In real-time applicationsretransmission is not an

publicInternet

1.5 Mbpsaccess link

TELCOM 2110 33

op on a er pac e oss Some low level of packet loss

can be made up by humanbrain from context inaudio/video

institutionalnetwork

10 Mbps LAN


Quality drops quickly with increasing packet loss rate For example for VoIP to have quality comparable to

PSTN need very low loss rate < 0.5%

increase

Host Ain : originaldata

out

TELCOM 2110 34

output link buffers

os


18/46

Network Performance -Efficiency

How much overhead is needed to send traffic across the network

Overhead is due to several factors lets look at some of them: Packetization Overhead Network Protocol Overhead out ng rotoco ver ea s

Remember data is packaged in protocol frames that contain overheaddata, some have more overhead than others Ethernet - 38 bytes per frame IP - 20 bytes per frame TCP - 20 bytes per frame ATM - 5 bytes per cell IP RIP - every 30 seconds sends 532 byte packets

TELCOM 2110 35

Example VoIP (IP/UDP/RTP) Payload efficiency: P/(P+Header)% 20-40%

Header20 Bytes

UDP packetHeader8 Bytes

IP packet

RTP packetHeader12 Bytes Data payload

Network Performance - Delay

Interactive applications demand minimal delaywhen receiving a data stream

Dela must be constant for real-time a licationslike voice/audio and video applications other wiseyou will getjittercausing disruptions in audioquality and jumpiness in video streams

Delay Jitter is the variability in the delay from aconstant

TELCOM 2110 36

data within a network (e.g., router)

For example consider Voice over IP


19/46

IP Telephony Delays

Consider VoIP only network (no gateways or PSTN)

TELCOM 2110 37

Coding

Packetization/SerializationQueueing at RoutersPropagationDejitterDecoding

IP Telephony Delays

Coding Delay Time to gather speech sample compute vocoder

model values for transmission

Value depends on vocoder utilized (0-50ms)

Packetization and Serialization Packetization: Time to gather data from coder for

packet payload, attach headers

Remember the protocol stack for VoIP

TELCOM 2110 38

Output of Vocoder packed in Real Time Protocol (RTP) packets

Which are payload for User Datagram Protocol (UDP) packets

Which are payload for Internet Protocol packets (IP)


20/462

Packetization Delay

VoIP packet (RTP/UDP/IP)

Header20 Bytes


IP packet


TELCOM 2110 39

Delay: N voice samples T ms -> payload P

Payload efficiency: P/(P+Header) %

Net data rate: (P+Header)/T = R Kbps

Packetization and Delay

Data stream(Compressed) Buffer

Accumulationdelay

For example: 10Byte payload from 4-to-1 compression

Header20 Bytes


IP packet


TELCOM 2110 40

ra e voco er Coding Delay: 10Byte 40 samples: 40125s = 5ms

Packet efficiency: 10/(40+10) = 20%

Net data rate: 50Bytes/5ms = 80 Kbps (>64 kbps DSO!)


21/462

Serialization and Transmission

Serialization Delay: time to transmit on access lineboth from caller to network also have this at the otherend of network to called party => G.723a VoIP codec over modem: 64byte packet

/56kbps=11ms 1byte on OC-3 optical fiber to home line (155Mbps) => 0.05

sec Insignificant on high-speed links

Propagation Delay Time to propagate packet down link - depends on distance of

TELCOM 2110 41

Satellite Hop wireless link 250 ms Coast to Coast in North America fiber optic propagation 24 ms

For example fiber optic cable propagates at roughly 2/3 speed oflight (3 x 108 ) meter/sec - so 200km link has propagation delayless than 200/(3 x 108 ) = 0.66 ms

Small enough on short fiber links to ignore

Network Delays

Router delay Time for router to process/transmit packet + delay in router

queues Time to rocess/transmit acket de ends on router switch

speed and link speed for high bandwidth links and corenetwork routers small amount of time 10 20 secs

15

20

25Queueing Delay

Time waiting in router buffers forprocessing and transmission

Value highly dependent on load and

TELCOM 2110 42

0

5

10

00.

10.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

10s msec to 10s secs

Queueing Delay nonlinear withincreases of network load


22/462

Network Delays

Delay Jitter defined as the variation of the delay for twoconsecutive packets

Due to variation of

Router delay (processing time + queueing time)

TELCOM 2110 43

Network Delays

Jitter buffer Jitter buffer to smooth out playout of packets to destination

Allows packets to arrive out of order

Note 30 ms holds one G.723 packet, typical values 30-100 msec

CO

Receive Buffer

TELCOM 2110 44

EC

Jitter eliminated if

buffer is sufficiently large


23/462

Example of End-to-End Delay Budget

Often design on basis of a Target Delay Budget Sender

Coding Delay 5

Packetization delay 30

Serialization delay 11

Network Routers 5 @ 7ms each 35

Propagation 25

Receiver

If no congestion.

TELCOM 2110 45

Serialization, de-packet, decode 46

Total 182 ms ITU recommend max of 400ms for VoIP and target

ideal of 150ms

Network Performance - Response Time

Response time is a network performance goal thatusers care about

from the networked system

Users begin to notice when response time is 100ms

(.1 seconds) or greater

Get interaction delay when have to wait on the networkedsystem

TELCOM 2110 46


24/462

Queueing Theory

Queueing theory : Mathematical analysis of waitinglines

Queueing Theory is the primary analytical frameworkfor evaluating performance in the initial stage of systemdesign.

Analytical Model of the system based on stochasticprocesses

Approximates real system by focusing on contention at

TELCOM 2110 47

shared resources.

Examples: shared medium router, window flow controlled session,time shared computer system

Model of Router

TELCOM 2110 48


25/462

Nomenclature of a Queueing System

The input process how customers arrive

The system structure waiting space number of servers, etc.

The service process

Kendalls Notation 1/2/3/4/5/6

A Shorthand notation to describe a queueing system containing aqueueing system.

1 : Customer arriving pattern (Interarrival time distribution).

TELCOM 2110 49

2 : Service pattern (Service time distribution).

3 : Number of parallel servers.

4 : System capacity. 5 : Queueing discipline.

6: Customer Population

Characteristics of the Input Process (1)

1. Arrival pattern or Arrival Process

Customers may arrive at a queueing system either in some regular

.

When customers arrive regularly at a fixed interval, the arrival

pattern can be easily described by a single number the rateof arrival

When customers arrive according to some random fashion,

the arrival pattern is described by a probability distribution.

Arrival process characterized by interarrival distribution

50


26/462

Characteristics of the Input Process

Probability distributions that are commonly usedto describe the arrival process are:

: ar ov an or memory ess , mp es e o ssonprocess for arrivals means the number of arrivalsover a time interval has a Poisson distribution this isequivalent to the time between customers arrivingbeing exponentially distributed.

D : Deterministic, fixed interarrival times

TELCOM 2110 51

Ek: Erlang distribution of order k

G : General probability distribution GI: General and independent (inter-arrival time)

distribution.

Characteristics of the Input Process

Behavior of the arriving customers

Customer arriving at a queueing system may behave

differently when the system is full (due to finite waiting queue)

or when all servers are busy.

Blocking System :

The arriving customers when system is full are

considered lost dropped from systems

Non-Blocking System :

52

The arriving customers are placed in queues of infinitesize.

Balking or Discouraged arrivals : customers refuse to join

queue when line too long or the arrival rate decreases with

line length


27/462

Characteristics of the Service Process

2. Service distribution describe the time take by a server to

process a customer. Can be deterministic or probabilitistic

In fashion similar to interarrival process use abbreviations to

describe common cases

M : Markovian (or memoryless), implies the exponentially

distributed service times.

D : Deterministic, constant service times

Ek: Erlang distribution of order k service time distribution

PH hase t e distribution

53

G : General service time distribution

Characteristics of the System Structure

3. Physical number and layout of servers

Default assumption of parallel and identical servers

Integer number of servers

A customer at the head of the queue can go to any server who

is free, and leave the system after receiving service from that

server.

4. The system capacity

The s stem ca acit is the maximum number of customers that a

54

queueing system can accommodate, inclusive of those customers

at the service facility

Finite (integer value) - maximum number of customers in the

systems

Infinite (default value)


28/462

Characteristics of the Service Process

5. Queueing discipline how customers are selected for service

from the line

rs - ome- rs - erve

Last-Come-First-Served (LCFS)

Priority

Process sharing

Random

Longest Queue First

Etc.

55

6. The size of the customer population

Infinite : the number of potential customers from external sources isvery large as compared to those in the system.

Finite : the arrival process (rate) is affected by the number of

customers already in the system.

Example of Notation

M/D/2/50/FIFO/

1. Ex onentiall distributed interarrival times.

2. Deterministic service times

3. Two parallel servers

4. Waiting space for 48 customers + 2 in service

5. First in First Out processing from the queue

.

What one can say generally about queueing

systems?

56


29/462

Nomenclature

Standard notation

mean arrival rate of customers/time unit

mean serv ce rate n customers t me un t

n(t) number of customers in the system at time t

i= limtP{n(t) = i}

is server utilization rememberfor stability

L Average number of customers in systems

Lq - Average number of customers in the queues

know L = Lq +

W Average delay in system (includes server + queue)

Wq Average delay in queue

know W = Wq + 1/

Littles Law

L = W

57

Nomenclature

Standard notation - relationships


30/463

Basic Queue Analysis

Consider single queue case focus on basic models widely

used in network performance analysis

Data networks and database systems

M/M/1

M/M/1/K

Telephony

M/M/C Erlang C

59

M/M/C/C Erlang B

All are Markovian queues, study using Birth Death process CTMC

Markovian Queues Analysis

Develop state transition diagram

System state is indicated by the number ofcustomers in the system at time t {n(t), t0}

Flow Balance Equations

Derive steady state probability i = P{n(t) = i} outflowinflow

TELCOM 2110 60

Apply Littles theorem to obtain meanperformance metrics. L = W

i i


31/463

Single Queue Analysis (M/M/1)

Most basic Markovian queue is the M/M/1//FIFO/ queue

Customers arrive according to a Poisson process with exponentially

distributed interarrival times (IAT)

P{ IAT t} = 1 e-t , mean interarrival time = 1/

Customers are served by a single server with exponential service time

distribution P(service time < t ) = 1 e-t

61

mean service time = 1/

The arrival rate () and service rate () do not depend upon the number of

customers in the system or time

Consider behavior of n(t) number of customers in the system at time t

forms a Markov Process

M/M/1 Queue From state transition diagram flow balance get the

equations to solve for the steady state probabilities

0j

flow out statej= flow in statej

62

0j11)( jjj

i

i 1Also use Normalization equation


32/463

M/M/1 Continued)1( nn Geometric distribution

1

where

Mean Number in System L Mean Delay W = L//11

W iiL

Stability Condition

TELCOM 2110 63

i

Variance of number in system L

2)1(

L

Variance of Delay W

22 )1(

1

W

M/M/1 Queue - Mean Behavior

At heavy load smallchanges in rho resultin large change in L

64

1/


33/463

M/M/1 Example

Consider a concentrator that receives messagesfrom a group of terminals and transmits them

.

The packets arrive according to a Poissonprocess with one packet every 2.5 ms and thepacket transmission times are exponentiallydistributed with a mean of 2 ms. That is thearrival rate = 1 packet/2.5 ms = 400 packets/sec

TELCOM 2110 65

Service rate = 1packet/2ms = 500 packets/sec

Find the average delay through the system Utilization = = 400/500 = .8

Delay W = 1/(500 400) = .01 secs = 10 msecs

M/M/1/K

The system has a finite capacity of size K.

)1( be P

bP

TELCOM 2110 66

The state space will be truncated at state K.


34/463

M/M/1/K (2)

Kj 111)( jjj 10 0j

Flow Balance Equations

67

1jj Kj Also use Normalization equation

i

i 1

M/M/1/K

Kn 0 nn Solving equations yields

0)1( n

w ere Normalized offered load

Solving normalization equation one gets

1

68

11 K1

1

Kn

1

From LHopital rule get


35/463

M/M/1/K Behavior of state probabilities i with

0 largest K largest i discrete uniform

69

M/M/1/KProbability of blocking (Pb) = Loss Rate

1 K 11 KkbP

1

1

KP kb

1

Portion of traffic dropped/rejected = Pb

70

Effective throughput of the system

e= (1-Pb) effective arrival rateExample M/M/1/10

Notice how it is nonlinear


36/463

M/M/1/K ebP )1(

Effective server utilization : (actual utilization of the system)

e

1

1

0 1

)1(

1

KKK

i

iK

iL

1

1

00K

iiLk

i

K

i

i

Average number in the system

1

1

71

2

1

1

0

K

i

K

k

i

M/M/1/K

Other performance measures

q

e

WW

LW

1

Mean Delay

Mean Queueing Delay

TELCOM 2110 72

eq LL Mean Number in Queue


37/463

M/M/1/K Compare with M/M/1/ results

73

M/M/1/K Example

Consider the queue at an output port of router. The transmission linkis a T1 line (1.544Mbps), packets arrive according to a Poissonrocess with mean rate 659.67 ackets/sec, the acket len ths are

exponentially distributed with a mean length of 2048 bits/packet.

If the system size is 16 packets what is the packet loss rate?

model as M/M/1/16 queue with

659.67 , Mbps/2048 bits per packet = 753.9 packets/sec

0.875Thus the packet loss rate = blocking probability

TELCOM 2110 74

0165.0875.1

875).875.1(

1

)1(17

16

1

K

K

bP


38/463

Teletraffic Modeling

Historically in the telephone system traffic measured orspecified in Erlangs (in honor of A.K..Erlang) Denote Erlangs by E or Erl and the offered load in erlangs by a

a = average ca rate x average o ng t me = x th One Erlang = one completely occupied channel Example: a radio channel occupied for 30 min. per hour carries 0.5 Erlangs

Total traffic intensity a = traffic intensity per user x number of users= au x nu

Example 100 subscribers in a cell20 make 1 call/hour for 6 min => 20 x 1 x 6/60 = 2E20 make 3 calls/hour for min => 20 x 3x .5/60 = .5E

Telcom 270075

ma e ca our or m n = x x =100 users produce a = 3.5 E load or au = 35mE per user

Given T traffic channels - what is GoS? or How many users can besupported for a specific GoS? or given load how many channels for GoS?

Basic analysis same for all circuit switched telephony (wired or wireless)

Erlang B model

M/M/C/C Erlang B model

Cidentical servers process customers in parallel.

Customers arrive according to a Poisson process with rate Customer service times exponentially distributed with mean 1/

-all C servers are busy is dropped

Called Blocked Calls Cleared (BCC) model

76

)1( be P

bP

e


39/463

M/M/C/C

Analysis parallels M/M/1/K. Considern(t) behavior get birth-deathstate transition diagram below

32 C)1( C

10 0j

flow out statej= flow in statej

77

Cj 1)( ccC

10

jj

Normalization condition

M/M/C/C

Solving the equations fori , one gets

Cia

i

1

Where is the offered loadin Erlangs

a

i!

10

jjPlugging into the normalization condition

One gets

78

ci

n

a

i

a

i

ac

n

n

i

i

i ,...2,1

!

!!

0

0

c

n

n

n

a

0

0

!

1


40/464

c

Probability of a customer being blocked B(c,a) = i

Erlang B Formula

c

n

nc

n

a

cacB

0 !

!),(

B(c,a) Erlangs B formula, Erlangs blocking formula

Valid for M/G/c/c queue

TELCOM 2110 79

),1(

),1(),( acBac

acBaacB

Usually determined from table or charts

Traffic Engineering Erlang B table

Telcom 270080


41/464

Erlang B Charts

TELCOM 2110 81

M/M/C/C

The carried load

Other metrics

,e )),(1( acB

c

ae

)),(1( acBa

L

Effective throughput of the system

Mean server utilization

Mean number in the system

TELCOM 2110 82

Average delay in the system

Distribution of delay is just the exponential service time

1W


42/464

Traffic Engineering Example

A T1 line can support 24 circuit switched phone calls?What is the maximum load a T1 link can support whileproviding 0.5% call blocking?

rom e r ang a e w c = c anne s an . ca oc ngthe maximum load = 11.56 Erlangs

A company has a PBX that connects to the local phonecompany. During the busy hour PBX handles 410 callswith an average call length of 200 seconds(a) What is the average load in Erlangs?Load = 410 call per hour x 200 sec/call x 1/3600 sec = 22.78 Erlangsb Given that the local hone sells bandwidth onl in units of T1

TELCOM 2110 83

lines (i.e. 12 DS0s), how many T1 lines are needed to achieve 1%call blocking? From the Erlang B table 33 DS0s are needed to

support 22.91 Erlangs of load

the nearest multiple of 12 that isgreater than 33 is 36 which is 1 T1 lines

Security

Security design is becoming one of themost im ortant as ects of network desi n

Network design must ensure against lossof business data or disruption of businessactivity

Need to understand the risk of data loss

TELCOM 2110 84

Security Concepts

COMMSEC: security at communications level

INFOSEC: security at information level


43/464

Security Threats

System Intrusion Improper access to network and hosts resources

Disable network and hosts

Snooping

Spoofing

Data manipulation

TELCOM 2110 85

Information Assurance info security + infoavailability

Security Impact on Network

Security Mechanisms must be put in placeto provide security

Physical Security Measures

Servers/cabling in locked rooms

Backup power and storage, etc

Impacts physical design

Electronic Security Measures

TELCOM 2110 86

Authentication, packet filters, encryption Firewalls

Impacts network performance => greaterdelays and requires more capacity


44/464

Typical Security Topologies

Internet

Firewall

Enterprise NetworkDMZ

Proxy Web Server DNS, Mail Servers, IDS

Manageability

There are various ways to manage a network anddifferent things to manage Fault, accounting, configuration, performance, security

etc.

Management architecture needs to be determined In-band versus out-of-band monitoring/signaling

Centralized vs. distributed monitoring and management

Estimate additional traffic due to management flows andsecurit mechanisms needed

TELCOM 2110 88

Number of platforms supported, tools needed etc.

Also need to consider interoperability with existinginfrastructure and management


45/464

Affordability

Affordability is sometimes called cost-effectiveness

Want to carry the maximum amount oftraffic for a given financial cost

Financial costs include non-recurringequipment costs and recurring network

TELCOM 2110 89

operating costs

Campus, Metro and WAN costs are areaswhere a good design can save $

Ranking

Useful to have users/management rankperformance goals

Low delay more important than availability

Ease of management more important than security

Comparative ranking or absolute

One approach is assume 100 point to be distributedamong the categories of interest and users must

TELCOM 2110 90

(scalability, availability,delay, security, etc.)

Provides Guidance to optimizing network design


46/46

Making Tradeoffs

Scalability 20

Network performance 15

Security 5

Manageability 5

Adaptability 5 Affordability 15

Total 100

Summary

From surveys/questionnaires, meetings etc.application data determine technical requirements andconstraints

requirements + some they may not know they need.

Technical Goals Scalability Availability/reliability Network Performance

Utilization, Throughput, Delay, Delay Jitter, packet loss rate,call/connection blocking rate

ra c s ma on may e nee e Security Manageability/Interoperability Affordability $$

Need to determine reasonable goal for each categoryand the importance of each.

Documents

Slides 2-tkm