1

Technical Goals and Requirements

David TipperAssociate ProfessorAssociate Professor

Graduate Telecommunications and Networking ProgramNetworking Program

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• Technology: wired vs wireless• Technology: wired vs. wireless• 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– Conceptual Model

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Conceptual Model

– Logical Model

– Physical Model

– Implementation, Testing, Tuning, and Documentation

2

Analyze requirements

Top-Down Network Design Steps

Develop logical design

Develop

Monitor and optimize network

performance

Implementphysical design

Test, optimize, and

document design

Implement and test network

Source: P. Oppenheimer

Conceptual Model Network Design

• Conceptual Model Design

• At end of conceptual model design should have gathered/identifiedgathered/identified

• Objectives– Business Goals (e.g., make sales force more responsive to

customers on sales calls)

– Technical Goals (e.g., provide wireless access to corporate data to sales force)

• Requirements

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– Business (e.g.,support XYZ application)

– Technical (availability, delay, bandwidth, etc.,)

• Constraints – Business (organizational, budget, etc.,)

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

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Technical Requirements & Constraints

• From surveys/questionnaires, meetings etc. application data determine technical requirements and constraints

• Technical goal is to build a network that meets user’s requirements + some they may not know they need.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

• Traffic Estimation is needed to estimate performanceSecurity

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– Security– Manageability/Interoperability– Affordability $$

• Need to determine reasonable goal for each category and the relative importance of each.

Scalability

• Scalability – how much growth a network design can support?

• can the design adapt to changing network load and QoS ca t e des g adapt to c a g g et o oad a d 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?– How extensive will networks be at each site?

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How extensive will networks be at each site?– 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?

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Scalability

• Scalability – For logical network design – how much additional

traffic can be added – without substantial additionaltraffic can be added without substantial additional investment

– For physical design - thought of as expandability and upgrade capability

– For example,• Given specific Router

• Can interface bit rate be upgraded

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• Can interface bit rate be upgraded

• Can number of I/O ports be increased?

• Can additional software features be added (e.g, VLAN capability, IP Sec, etc.)

– Try to set reasonable scalability goals

Availability

• Availability (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 range or 0 - 100% Uptime g

• 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

UptimeA

T

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Failure Repair Failure Repair

MTBF

MTTRMTTR MTTF MTTFA

MTTF MTTR

MTBF

MTTRA 1

5

Component Availability

• Consider an IP router– MTBF 100,000 Hours

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

– A = 1 – MTTR/MTBF = 0 99994A 1 MTTR/MTBF 0.99994

• Some representative values of networking equipment

Equipment MTBF Range (hr) MTTR (hr)

Web Server 104 - 106 1

IP Interface Card 104 - 105 2

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IP Router 105 - 106 6

WLAN AP ~105 2

WDM OXC or OADM

~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

chemical plant)chemical plant)• 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 – AU– 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)

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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 user requirements – may vary with location– Highly available voice service at customer support call center

• Five 9’s at call center

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

– Challenge – how to provide higher availability for only certain services/applications

• Network/System availability is the amount of time a network/system is available to users– Can be expressed as percent uptime or downtime

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– Can be expressed as percent uptime or downtime• Need to work with users to set reasonable goals

• Higher availability goal more costly design

• Given component availability – how to find network/system availability?

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System Availability

• System availability calculated from component availability Ai, and unavailability Ui,

• If devices in series• If devices in series

• If devices in parallel

1 2 n

…

1

1

n

series ii

A A

n

1

n

s i

i

U U

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1

2

n

…

1

1 (1 )n

parallel ii

A A

1parallel i

i

U U

System Availability - Example

WD

MS

yste

m

OA

WD

ML

ine Sy

A single bidirectional line in WDM optical network

OA

• The availability of the bidirectional line system = ?

WL

ine

O Mystem80km 100km 80kmO

Equipment MTBF (hrs) MTTR (hrs)

Bidirectional OA 5*105 24

Bidirectional 5*105 6

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WDM Line System

Equipment CC (km) MTTR (hrs)

Terrestrial Fiber Optic Cable

450 24

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MTBF – Physical Cable

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

• Average cable length that results in a single cable cut per yearAverage cable length that results in a single cable cut per year

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

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

( 365 24)( )

length of the cable (km)

CCMTBF hours

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450 365 2415161.5

260cable

km hMTBF h

km

System Availability - Example

• Devices in series

• Availability of bidirectional line (Aline)

2 2

2 2

2 2

(1 ) (1 ) (1 )

24 24 6(1 ) (1 ) (1 )

line cable OA line system

line systemcable OA

cable OA lline system

A A A A

MTTRMTTR MTTR

MTBF MTBF MTBF

h h h

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2 25 5

(1 ) (1 ) (1 )15161.5 5 10 5 10

0.998297h h h

450 365 24Note. 15161.5

260cable

km hMTBF h

km

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Series-Parallel Reduction

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

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+ series|| parallel

Availability Analysis

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

systems.2) Draw parallel and series availability relationships3) Reduce the system availability model by repeated

applications of the parallel/series availability simplifications.

4) If not completely reduced Use approximation methods to estimate availability. Typically take a conservative approach and go with a lower bound

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Availability Analysis

• Lower bound on unavailability– The contributions of parallel elements to the unavailability is

not taken into account

AB

C

D

E

F

G H

Lower bound of Us: UA+UH

– Quick evaluation of a lower bound on U can be enough to conclude– Quick evaluation of a lower bound on U can be enough to conclude 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 Techniques to increase availabilityBasic Techniques to increase availability 1. 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

Can be partial/fully redundant systems• Can be partial/fully redundant systems• For example Space Shuttle has 3 fully redundant flight

computers (+ manual operation!)• Incurs additional cost due to spare

components/capacity

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Availability and Design

• 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 (hot swappable line cards, backup switch cards, redundant cooling, backup power, etc)

• Expense is major concern and market – who will buy most reliable system when replacing often? Who wants to pay for areliable system when replacing often? Who wants to pay for a 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/patch installation, model checking code analysis, etc.

Availability and Design

• MTTR is often a Network Operations and Management issue – some what out of the control 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 don’t bother with diagnosis –> fast replacement rather than repair

B d b• Board swap, reboot, reset, etc.

• Human in the loop and travel is often the bottleneck

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

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Availability and Design

• Redundancy/Diversity Techniques seek to increase system availability not individual component– Combat independent faults by duplicate equipment/services

F l id 24 h b k b tt l t t k– For example provide 24 hour backup battery supply to core network equipment in order to combat electrical power outages

– Redundancy higher COST

– Tradeoff between Availability and COST

Redundancy Effects

PrimaryRouter

Back-upRouter

Scenario Single RouterAvailability

Availability with Redundancy

1

1 (1 )n

parallel ii

A A

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Availability Redundancy

1 0.90 0.9900

2 0.95 0.9950

3 0.99 0.9999

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Redundancy Example

BP

• Ai is an availability of link i

• Availability of a connection between S-D:

WPSource (S) Destination (D)

A Ano protection ii WP

A A

• Given Ai = 0.998297 for all links yieldsAno-protection = 0.996597, Aprotection= 0.999983

WPi BPi

iiprotection AAA ||

Availability Example

Consider Availability of Internet Connection

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

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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 period of the day set threshold values

• Need to know how to estimate valuesA h h d i i t k

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• Approaches when designing network are – Queueing Analysis – analytical models– Simulation – measurements on computer model of the network

design– Benchmarking existing network – then predict behavior - with

empirical model, queueing model or simulation

Network Performance

• Typically have a camelback shape to network traffic (both packet and circuit switched networks)

• Busy time period will vary with network type and application (e g is commuting time in urban cellularapplication (e.g., is commuting time in urban cellular networks)

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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.)– Service (data voice SMS etc )Service (data, voice, SMS, etc.)

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Network Performance - Utilization

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

• Bandwidth utilization is measured over a timeBandwidth utilization is measured over a time interval to determine the amount in use (e.g. the busy time period or some fraction of it)

• Link/equipment utilization identifies network bottleneck points– Data networks usually have utilization < 40 -

60% ft th h ld

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60% are often thresholds– Telephone network utilization much higher

80 – 90 % • Utilization goals will effect resulting delay

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Network Performance - Throughput

• Throughput is defined as the quantity of error-free data successfully transferred between nodes per unit of time (Goodput or Layer 2/3 throughput)( p y g p )

• Depends on network access method, the load on the 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 networks

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networks – Carried load in Erlangs for circuit switched networks– Example IEEE 802.11b wireless LAN – channel rate 11

Mbps – typical max throughput 7 Mbps

Network Performance -Accuracy

• Accuracy is a measure to ensure that the data received at the destination must be the same as the data sent by the sourcethe data sent by the source

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

• Accuracy can be expressed in Bit Error Rate (BER) or packet error rate (PER)

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(BER) or packet error rate (PER)

• Target values of BER depend on physical medium used wireless link – 1 in 104 , optical fiber 1 in 1010

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Network Performance -Accuracy

• Packet Loss occurs when buffers overflow at routers or gateways in wired networks

• In wireless networks packet

originservers

In wireless networks packet loss due to interference, poor signal quality, collisions

• Packet Loss results in retransmission in applications that require reliability

• In real-time applications retransmission is not an

ti ft k t l

publicInternet

1.5 Mbps access link

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option after packet loss • Some low level of packet loss

can be made up by human brain from context in audio/video

institutionalnetwork

10 Mbps LAN

Network Performance -Accuracy

• 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%• Packet Loss increase is highly nonlinear with load• Packet Loss increase is highly nonlinear with load

increase

limited shared

Host Ain : original data

H t B

out

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limited shared output link buffers

Host B

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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

R ti P t l O h d– Routing Protocol Overheads• Remember data is packaged in protocol frames that contain overhead

data, 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

• Overhead effects delay and link sizing

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• Overhead effects delay and link sizing• 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 delay when receiving a data stream

• Delay must be constant for real-time applications y pplike voice/audio and video applications other wise you will get jitter causing disruptions in audio quality and jumpiness in video streams

• Delay Jitter is the variability in the delay from a constant Delay caused by network devices that move the

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• Delay caused by network devices that move the data within a network (e.g., router)

• For example consider Voice over IP

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IP Telephony Delays

•Consider VoIP only network (no gateways or PSTN)•Major Delays in IP Telephony Systems

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•Major Delays in IP Telephony Systems •Coding•Packetization/Serialization•Queueing at Routers•Propagation•Dejitter•Decoding

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

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pOutput 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)

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Packetization Delay

• VoIP packet (RTP/UDP/IP)

total header = 40 Bytes

• Assume

total header = 40 Bytes

Header20 Bytes

UDP packetHeader8 Bytes

IP packet

RTP packetHeader12 Bytes Data payload

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Assume – 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 t d

Header20 Bytes

UDP packetHeader8 Bytes

IP packet

RTP packetHeader12 Bytes Data payload

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rate vocoder

– Coding Delay: 10Byte 40 samples: 40×125s = 5ms

– Packet efficiency: 10/(40+10) = 20%

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

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Serialization and Transmission

• Serialization Delay: time to transmit on access line both from caller to network also have this at the other end of network to called party– 1 byte on 64kbps line => 125 sec1 byte on 64kbps line 125 sec– 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

link and medium

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link and medium • 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 of

light (3 x 108 ) meter/sec - so 200km link has propagation delay less 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 process/transmit packet depends on router switch p p p

speed and link speed – for high bandwidth links and core network routers small amount of time 10 – 20 secs

15

20

25Queueing Delay–Time waiting in router buffers for processing and transmission

–Value highly dependent on load and QoS mechanisms deployed in router

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0

5

10

00.

10.

20.

30.

40.

50.

60.

70.

80.

9

QoS mechanisms deployed in router 10’s msec to 10’s secs

•Queueing Delay nonlinear with increases of network load

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Network Delays• Delay Jitter defined as the variation of the delay for two

consecutive packets• Due to variation of

– Routes of packets– Routes of packets

– Router delay (processing time + queueing time)

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Network Delays

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

• Allows packet delivery times to vary• Allows packet delivery times to vary

• Allows packets to arrive out of order

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

CO

D

Receive Buffer

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DE

C

Jitter eliminated ifbuffer is sufficiently large

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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

• ReceiverJitter buffer 30

If no congestion.

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– Jitter buffer 30– 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 that users care about– Users recognize the amount of time to receive a responseUsers recognize the amount of time to receive a response

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 networked system

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Queueing Theory

• Queueing theory : Mathematical analysis of waiting lines

• Queueing Theory is the primary analytical framework for evaluating performance in the initial stage of system design.

• Analytical Model of the system – based on stochastic processes

• Approximates real system by focusing on contention at

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shared resources.• Examples: shared medium router, window flow controlled session,

time shared computer system

Model of Router

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Nomenclature of a Queueing System

• The input process – how customers arrive

• The system structure – waiting space – number of servers, etc.

• The service process

• Kendall’s Notation 1/2/3/4/5/6– A Shorthand notation to describe a queueing system containing a

queueing system.

– 1 : Customer arriving pattern (Interarrival time distribution).

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– 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 pattern or in a random fashionpattern or in a random fashion.

When customers arrive regularly at a fixed interval, the arrival pattern can be easily described by a single number the rate of 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

Characteristics of the Input Process

• Probability distributions that are commonly used to describe the arrival process are:

M M k i ( l ) i li th P i– M : Markovian (or memoryless), implies the Poisson process for arrivals – means the number of arrivals over a time interval has a Poisson distribution – this is equivalent to the time between customers arriving being exponentially distributed.

– D : Deterministic, fixed interarrival times

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– 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 g q g y y

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 infinite size.

Balking or Discouraged arrivals : customers refuse to join queue when line too long or the arrival rate decreases with line length

27

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 – phase type distribution

53

PH phase type distribution

G : General service time distribution

Characteristics of the System Structure

3. Physical number and layout of servers♦ Default assumption of parallel and identical serversp p

♦ 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 system capacity is the maximum number of customers that a

54

♦ The system capacity is the maximum number of customers that a 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)

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Characteristics of the Service Process

5. Queueing discipline – how customers are selected for service from the line

Fi t C Fi t S d (FCFS/FIFO)♦ First-Come-First-Served (FCFS/FIFO)

♦ 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 is very 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. Exponentially distributed interarrival times1. Exponentially 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

6 Infinite population of customers6. Infinite population of customers

♦ What one can say generally about queueingsystems?

56

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Nomenclature

♦ Standard notation mean arrival rate of customers/time unit

i t i t /ti it mean service rate in customers/time unit

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

πi = limt ∞ P{n(t) = i}

is server utilization remember for 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/♦ Little’s Law

L = W

57

Nomenclature

♦ Standard notation - relationships

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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 of

i h i { ( ) 0}customers in the system at time t {n(t), t0}

• Flow Balance Equations– Derive steady state probability i = P{n(t) = i}

outflowinflow

1

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• Apply Little’s theorem to obtain mean performance metrics. L = W

i

i 1

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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

10 0j

flow out state j = flow in state j

62

0j11)( jjj 10 j

i

i 1Also use Normalization equation

32

M/M/1 Continued

)1( nn Geometric distribution

1

where

Mean Number in System L Mean Delay W = L/

)1(

/1

)(

1

W

)1( iiL

Stability Condition

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)1()( )1( 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 small changes in rho result in large change in L

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1/

33

M/M/1 Example

• Consider a concentrator that receives messages from a group of terminals and transmits them over a single transmission lineover a single transmission line.

• The packets arrive according to a Poisson process with one packet every 2.5 ms and the packet transmission times are exponentially distributed with a mean of 2 ms. That is the arrival rate = 1 packet/2.5 ms = 400 packets/sec

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• 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

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• The state space will be truncated at state K.

34

M/M/1/K (2)

Kj 111)( jjj

10 0j

Flow Balance Equations

67

1 jj Kj

Also use Normalization equation

i

i 1

M/M/1/K

Kn 0 nn

h

Solving equations yields

0

)1(

n

n

where

Normalized offered load

Solving normalization equation one gets

1

68

11 Kn

1

1

Kn

1

1From L’Hopital’ rule get

35

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

0 largest K largest i discrete uniform

69

M/M/1/K Probability of blocking (Pb) = Loss Rate

)1( KP

111

)(

KkbP

1

1

KP kb

1

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

M/M/1/K

ebe

P

)1(Effective server utilization : (actual utilization of the system)

e

1

1

0 1

)1(

1

K

KK

ii

KiL

1

1

00 KiiL

k

i

K

ii

Average number in the system

1

1

71

2

1

1

0

K

iK

k

i

M/M/1/K

Other performance measures

q

e

WW

LW

1

Mean Delay

Mean Queueing Delay

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eq

q

LL

Mean Number in Queue

37

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 link is a T1 line (1.544Mbps), packets arrive according to a Poisson process with mean rate 659.67 packets/sec, the packet lengths are p p , p gexponentially 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

0.875

Thus the packet loss rate = blocking probability

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0165.0875.1

875).875.1(

1

)1(17

16

1

K

K

bP

38

Teletraffic Modeling♦ Historically in the telephone system traffic measured or

specified in Erlangs (in honor of A.K..Erlang) –♦ Denote Erlangs by E or Erl and the offered load in erlangs by a

ll t h ldi ti ta = average call rate x average holding time = 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 ½ min => 20 x 3x .5/60 = .5E60 k 1 ll/h f 1 i > 60 1 1/60 1E

Telcom 2700 75

60 make 1 call/hour for 1 min => 60 x 1 x 1/60 = 1E100 users produce a = 3.5 E load or au = 35mE per user

♦ Given T traffic channels - what is GoS? or How many users can be supported 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

• C identical servers process customers in parallel.

• Customers arrive according to a Poisson process with rate • Customer service times exponentially distributed with mean 1/• The system has a finite capacity of size C customer arriving when• The system has a finite capacity of size C - customer arriving when

all C servers are busy is dropped

• Called Blocked Calls Cleared (BCC) model

76

)1( be P

bP

e

39

M/M/C/C

• Analysis parallels M/M/1/K. Consider n(t) behavior get birth-death state transition diagram below

32 C)1( C

Cj 111 )1()( jjj jj

10 0j

flow out state j = flow in state j

77

Cj 111 )1()( jjj jj

Cj 1)( ccC

10

jjNormalization condition

M/M/C/C

Solving the equations for i , one gets

Ciai

i 10

Where is the offered loadin Erlangs

a

timeholdingcallxratearrivalcalla ii ! 0

10

jjPlugging into the normalization condition

One gets

timeholdingcallxratearrivalcalla

78

ci

n

ai

a

i

ac

n

n

i

i

i ,...2,1

!

!!

0

0

c

n

n

n

a

0

0

!

1

40

ca

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

Erlang B Formula

c

n

nc

na

ca

acB

0 !

!),(

B(c,a) Erlang’s B formula, Erlang’s blocking formulaErlang B formula can be computed from the recursive formula

Valid for M/G/c/c queue

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Erlang B formula can be computed from the recursive formula

),1(

),1(),(

acBac

acBaacB

Usually determined from table or charts

Traffic Engineering Erlang B table

Telcom 2700 80

41

Erlang B Charts

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M/M/C/C

))(1( acB The carried load

• Other metrics

)),(1( acBe

)),(1( acBc

ae

)),(1( acBa

L

Effective throughput of the system

Mean server utilization

Mean number in the system

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Average delay in the system

Distribution of delay is just the exponential service time

1

W

42

Traffic Engineering Example

• A T1 line can support 24 circuit switched phone calls? What is the maximum load a T1 link can support while providing 0.5% call blocking?F th E l B t bl ith 24 h l d 0 5% ll bl kiFrom the Erlang B table with c = 24 channels and 0.5% call blocking

the maximum load = 11.56 Erlangs

• A company has a PBX that connects to the local phone company. During the busy hour PBX handles 410 calls with 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 Erlangs(b) Given that the local phone sells bandwidth only in units of ½ T1

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( ) p ylines (i.e. 12 DS0s), how many ½ 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 is greater than 33 is 36 which is 1½ T1 lines

Security

• Security design is becoming one of the most important aspects of network designmost important aspects of network design

• Network design must ensure against loss of business data or disruption of business activity

• Need to understand the risk of data loss

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• Security Concepts– COMMSEC: security at communications level

– INFOSEC: security at information level

43

Security Threats

• System Intrusion– Improper access to network and hosts resources

• Denial of service• Denial of service– Disable network and hosts

• Snooping• Spoofing• Data manipulation• Physical damage

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• Physical damage• Information Assurance – info security + info

availability

Security Impact on Network

• Security Mechanisms must be put in place to provide security– Physical Security Measures

• Servers/cabling in locked rooms

• Backup power and storage, etc

• Impacts physical design

– Electronic Security Measures

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• Authentication, packet filters, encryption

• Firewalls

– Impacts network performance => greater delays and requires more capacity

44

Typical Security Topologies

Internet

Firewall

Enterprise NetworkDMZ

Proxy Web Server DNS, Mail Servers, IDS

Manageability

• There are various ways to manage a network and different 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 and

security mechanisms needed

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y– Number of platforms supported, tools needed etc.

• Also need to consider interoperability with existing infrastructure and management

45

Affordability

• Affordability is sometimes called cost-effectiveness

• Want to carry the maximum amount of traffic for a given financial cost

• Financial costs include non-recurring equipment costs and recurring network

i

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operating costs

• Campus, Metro and WAN costs are areas where a good design can save $

Ranking

• Useful to have users/management rank performance 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 distributed among the categories of interest and users must allocate the points among the performance categories

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allocate the points among the performance categories• (scalability, availability,delay, security, etc.)

• Provides Guidance to optimizing network design

46

Making Tradeoffs

• Scalability 20

• Availability 30Availability 30

• Network performance 15

• Security 5

• Manageability 5

• Usability 5Usability 5

• Adaptability 5

• Affordability 15

Total 100

Summary

• From surveys/questionnaires, meetings etc. application data determine technical requirements and constraints

• Technical goal is to build a network that meets user’s• Technical goal is to build a network that meets user s 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 rateT ffi E ti ti b d d

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• Traffic Estimation may be needed– Security– Manageability/Interoperability– Affordability $$

• Need to determine reasonable goal for each category and the importance of each.