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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
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Logical Model
Physical Model
Implementation, Testing, Tuning, and Documentation
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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
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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. 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
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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?
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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
investment
For physical design - thought of as expandability andupgrade capability
For example,
Given specific Router
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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
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Failure Repair Failure Repair
MTBF
MTTRMTTR MTTF MTTFA
MTTF MTTR
MTBF
MTTRA 1
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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
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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)
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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 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
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Need to work with users to set reasonable goals
Higher availability goal more costly design
Given component availability how to find network/systemavailability?
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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
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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
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WDM LineSystem
Equipment CC (km) MTTR (hrs)
Terrestrial FiberOptic Cable
450 24
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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
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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
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5 5 15161.5 5 10 5 10
0.998297
h 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 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
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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
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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 (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.
Availability and Design
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
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Availability and Design
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
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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)
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
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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 periodof theday set threshold values
Need to know how to estimate values
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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)
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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.) , , , .
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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 -
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are o ten t res o s 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-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
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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
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Target values of BER depend on physicalmedium used wireless link 1 in 104 , opticalfiber 1 in 1010
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Network Performance -Accuracy
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
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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
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%
increase
Host Ain : originaldata
out
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output link buffers
os
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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 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
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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
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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)
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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
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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)
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Packetization Delay
VoIP packet (RTP/UDP/IP)
Header20 Bytes
UDP packetHeader8 Bytes
IP packet
RTP packetHeader12 Bytes Data payload
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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
UDP packetHeader8 Bytes
IP packet
RTP packetHeader12 Bytes Data payload
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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!)
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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
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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
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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
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Network Delays
Delay Jitter defined as the variation of the delay for twoconsecutive packets
Due to variation of
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 packets to arrive out of order
Note 30 ms holds one G.723 packet, typical values 30-100 msec
CO
Receive Buffer
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EC
Jitter eliminated if
buffer 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
Receiver
If no congestion.
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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
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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
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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
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).
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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
.
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
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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
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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
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 :
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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
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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)
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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?
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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
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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 ofcustomers in the system at time t {n(t), t0}
Flow Balance Equations
Derive steady state probability i = P{n(t) = i} outflowinflow
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Apply Littles theorem to obtain meanperformance metrics. L = W
i i
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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
0j
flow out statej= flow in statej
62
0j11)( jjj
i
i 1Also use Normalization equation
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M/M/1 Continued)1( nn Geometric distribution
1
where
Mean Number in System L Mean Delay W = L//11
W iiL
Stability Condition
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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
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1/
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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
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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.
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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
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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
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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
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eq LL Mean Number in Queue
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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
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0165.0875.1
875).875.1(
1
)1(17
16
1
K
K
bP
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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
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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
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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
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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
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),1(
),1(),( acBac
acBaacB
Usually determined from table or charts
Traffic Engineering Erlang B table
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Erlang B Charts
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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
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Average delay in the system
Distribution of delay is just the exponential service time
1W
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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
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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
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Security Concepts
COMMSEC: security at communications level
INFOSEC: security at information level
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Security Threats
System Intrusion Improper access to network and hosts resources
Disable network and hosts
Snooping
Spoofing
Data manipulation
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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
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Authentication, packet filters, encryption Firewalls
Impacts network performance => greaterdelays and requires more capacity
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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
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Number of platforms supported, tools needed etc.
Also need to consider interoperability with existinginfrastructure and management
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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
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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
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(scalability, availability,delay, security, etc.)
Provides Guidance to optimizing network design
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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.
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