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University Of Moratuwa Lecture. 2012. PART 1. TELEPHONE NET WORK. PART 2. PULSE CODE MODULATION. Exercise 1: Convert the following denary numbers to binary(Don’t use the method of dividing by 2, use the finger method). (a) 5 (g) 520 (b) 9 (h) 1028 (c) 16 (i) 2050

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part 1
PART 1
  • TELEPHONE NET WORK
part 2
PART 2
  • PULSE CODE MODULATION
slide7
Exercise 1:Convert the following denary numbers to binary(Don’t use the method of dividing by 2, use the finger method)
  • (a) 5 (g) 520
  • (b) 9 (h) 1028
  • (c) 16 (i) 2050
  • (d)33 (j) 4100
  • (e) 67 (k) 8200
  • (f) 120 (l) 16401
answer to exercise 1
Answer to Exercise 1
  • (a) 5=101 (b) 9=1001
  • (c) 16=10000 (d)33=100001
  • (e) 67=1000011 (f) 120=1111000
  • (g) 520=1000001000 (h) 1028=10000000100

(i) 2050=100000000010 (j) 4100=1000000000100

  • (k) 8200=10000000001000 (l) 16401=100000000010001
exercise 2 convert the following from binary to denary using fingers only
Exercise 2Convert the following from binary to Denary(Using fingers only)
  • (a) 101
  • (b) 110
  • (c) 1001
  • (d) 11101
  • (e) 100000
  • (f) 1011010
  • (g) 111000111
answers to exercise 2
Answers to Exercise 2
  • (a) 101 5
  • (b) 110 6
  • (c) 1001 9
  • (d) 11101 29
  • (e) 100000 32
  • (f) 1011010 90
  • (g) 111000111 455
exercise 3 convert the following denary numbers to hexa and then to binary
Exercise 3Convert the following denary numbers to hexa and then to binary
  • (a) 9
  • (b) 20
  • (c) 36
  • (d) 129
  • (e) 518
  • (f) 1030
  • (g) 4095
  • (h) 8200
answers to exercise 3
Answers to Exercise 3
  • Denary Hexa Binary
  • (a) 9 9 1001
  • (b) 20 14 10100
  • (c) 36 24 100100
  • (d) 129 81 10000001
  • (e) 518 206 1000000110
  • (f) 1030 406 10000000110
  • (g) 4095 FFF 111111111111
  • (h) 8200 2008 10000000001000
convert the following samples into encoded format and calculate the signal noise ratio
Convert the following samples into encoded format and calculate the signal /noise ratio
  • 700mV -400mV 300mV
  • 100mV 1515mV -95mV
answers
Answers
  • 700mV -400mV 300mV
  • 11011101 01010001 11001001

175 50 ∞

  • 100mV 1515mV -95mV

10110001 11110000 0011000

25 72 295

part 3
PART 3
  • HIGHER ORDER PCM
part 4
PART 4
  • BASICS OF OPTICAL FIBRE
what is snell s law
What is Snell’s Law?
  • This describes the bending of light rays when it travels from one medium to another.

Air

Glass

Water

Air

slide46

Snell\'s law states that the ratio of the sines (Sin) of the angles of incidence and refraction is equivalent to the ratio of velocities in the two media, or equivalent to the opposite ratio of the indices of refraction.

Sin Ө1 n 2

=

Sin Ө2 n 1

Sin Ө1 n 2

=

Sin Ө2 n 1

n 1 Sin Ө1 = n 2 Sin Ө 2

PO - Ray of Incidence n 1 - RI for medium 1

OQ - Ray of Refraction n 2 - RI for medium 2

Ө 1 - Angle of Incidence

Ө 2 - Angle of Refraction

slide47

TOTAL INTERNAL REFLECTION

n 1 Sin Ө1 = n 2 Sin Ө 2

With the increase of the angle of incidence, the angle of

refraction increases accordingly.

When reaches φ290°, there is no refraction and φ1reaches a critical angle (φc)

Beyond the critical angle, light ray becomes totally internally reflected

slide48

Attenuation in Fibreoptical fibre behaves differently for different wavelength of light. The following diagram shows that. The three windows of wavelengths where the attenuation is lower is given below. Hence these 3 windows are mostly used for practical purposes.

1 general observation on attenuation and the present day technology
1. General Observation on Attenuation and the Present Day Technology
  • Attenuation is low between 1500nm-1700nm in wavelength.
  • This gives rise to operate 24Tbps speed
  • How?

C=fλ where C=3*108

  • And f1-f2=[c/(1500nm)]-[c/1700nm]=24Tbps
  • The present day technology goes up to 10Gbps or 40Gbps.
  • STM1 STM4STM16STM64…… STM256

155.52Mbps 620Mbps2.5Gbps10Gbps40Gbps

6.4ns 1.6ns400ps100ps25ps

present day technology adapting to the optical fibre
Present day technology adapting to the optical fibre

The following 2 major factors play a vital role in designing the maximum capacity of an optical fibre

  • How far the digital multiplexing can be achieved
    • As at present , 488ns micro information of a bit pertaining to 2Mbps PCM stream will be reduced to 25ps when it goes through STM64 (10Gbps). If the technology improves to shrink less than 25ps , then the number of bits in the higher order PCM will be more than 10Gbps.
  • To transmit 10Gbps, the optical fibre requires a bandwidth of around 0.078ns = 78ps ( for 1 wavelength)
  • If the available bandwidth in the optical fibre is 200ns , the number of wavelengths that can be produced is around 2400 , which will result in producing a total of 24Tbps.
  • Hence both Time Division Multiplexing and Dense Wave Division Multiplexing can further improve the traffic carrying capacity of an optical fibre up to a total of 24Tbps.
number of wavelengths 24 10 3 gb 10 gb 2400 wavelengths
Number of wavelengths = ( 24 * 103 Gb ) / 10 Gb = 2400 wavelengths

Future Scenarios

Theoretical Maximum of an Optical Fibre Cable

Transponders

λ1

488ns

100 ps

1

TDM

Optical Fibre

2

λ2

2Mbps

10Gbps

λ2399

2399

Only 1 core is needed

2400

λ2400

part 5

PART 5

COMMON CHANNEL SIGNALLING

message types
MESSAGE TYPES
  • BASIC MESSAGE
  • HOMOGENIOUS MESSAGE
  • NON HOMOGENEOUS MESSAGE
basic message

Instruction

Instruction

Data

Data

Data

Variable

Fixed

Instruction

Label

Label

OPC

DPC

CIC

14bits

14bits

12bits

Basic Message

= K1

Message for Homogenous Network

= K2

OPC – Originating Point Cord

DPC – Destination Point Cord

CIC – Circuit Identification Cord

why not

SIO

K2

SIO

K2

Data

Instruction

Label

4bits

4bits

NO

WHY NOT?

Message for Non-Homogenous Network

SIO - Service Information Octel

K2 - Message for Homogenous Network

National or International Message

User

Part

Now we are ready with the complete message, can we transmit it just as it is?

slide62

IAM

ACM

ANC

CBK

slide63

H1

H0

Spare reserved for national use

Spare reserved for international and basic national use

Spare reserved for national use

basic concept of message transmission to establish a call
Basic concept of message transmission to establish a call

IAM

B

ACM

A

ANC

Ringing current to subscriber “B” n ringback tone to subscriber “A”

Speech

CBK

Variable

Node X

Node Y

H0

H1

IAM

(Initial address message)

20 Bits

4 Bits

0001

0001

Dial Number

H0

H1

ACM

(Answer complete mesaage )

0100

0001

Fixed (8 Bits)

H0

H1

ANC

0110

0001

No Data

H0

H1

CBK

No Data

0110

0011

how the common channel signalling works
HOW THE COMMON CHANNEL SIGNALLING WORKS
  • ASSUME A CALL IS ESTABLISHED IN A NETWORK WHERE THERE ARE TWO EXCHANGES(EX X & EX Y) ARE CONNECTED WITH 16 PCM SYSTEMS.
  • THE CALL IS CONNECTED VIA CIRCUIT NUMBER 305. ASSUME P(0) TS16 & P1(1) IS USED FOR COMMON CHANNEL SIGNALLING.
  • DRAW HOW THE SIGNALS ARE ESTABLISHED BETWEEN THE EXCHANGES(assume the call is establised, and after the call, A keeps the receiver first)
  • Calculate the total times taken for forward & backward signalling
slide66

Need to transfer message between A to B

X

exchange

P0f

Y

exchange

P1f

P15f

P0b

P1b

P15b

Customer A

Customer B

helicopter view
Helicopter View

Exchange X

Exchange X

IAM

( P0f TS16 )

ACM

( P0b TS16 )

( P9 TS28)

RBT

( P0b TS16 )

ANS

speaking

( P9f TS28)

( P9b TS28)

( P0f TS16 )

CBR

error control
ERROR CONTROL
  • FORWARD ERROR CORRECTION
  • Detect and correct the error
  • In unidirectional transmission
  • BACKWARD ERROR CORRECTION
  • Detect the error and request for retransmission
  • In bydirectional transmission
slide75

= (31 * 14) + (30 *2)

494

Number of voice channel for voice

communication between X and Y

Channel number that we use

If we numbered voice channel from 1 to 494 : -

Select related TS

30 + 30 =60

305 – 60 = 245

245 / 31 = 7 mod 28

P9 TS28 (PCM no = 9 , TS no = 28)

= 305

7 + 2 = 9

number that we dial 15904607
Number that we dial = 15904607

IAM

K=56 bits

Message

8 bits

8 bits

Total bits = 150

slide77

ACM

K=16 bits

Message

8

bits

8

bits

8

bits

Total bits = 110

slide78

ANC

K=8 bits

Message

8

bits

8

bits

8

bits

Total bits = 102

slide79

CBR

K=8 bits

Message

8

bits

8

bits

8

bits

Total bits = 102

conclusion
Conclusion
  • time for forward message = 2.34 ms
  • time for forward message = 4.906 ms
phases of a call
Phases of a call

Dial Tone

Dialing

Signaling

Answer

Speak

Release

Ring back Tone

error control1
ERROR CONTROL
  • FORWARD ERROR CORRECTION
  • Detect and correct the error
  • In unidirectional transmission
  • BACKWARD ERROR CORRECTION
  • Detect the error and request for retransmission
  • In bydirectional transmission
cyclic reduncy code or frame chech sequance
CYCLIC REDUNCY CODE OR FRAME CHECH SEQUANCE
  • DESIGNED TO DETECT NOISE BURST
  • ACCORDING TO THE NOISE CHARACTERISTICS A POLYNOMIAL IS IDENTIFIED(N+1 BITS)
  • SHIFT THE MESSAGE BY N BITS
  • THEN DIVIDE BY MOULO 2 THE SHIFTED MESSAGE BY THE POLYNOMIAL
  • GET THE RESIDUAL OF N BITS & SHIFT THE MESSAGE BY THESE BITS AS CRC
  • AT THE RECEIVER IF THERE ARE NO ERRORS, YOU WILL NOT GET ANY RESIDUAL WHEN YOU DIVIDE THE RECIEVED MESSAGE BY THE SAME POLYNOMIAL
hint to answer
Hint to answer
  • Write the polynomial in x
  • Draw the 1 bit shift registers and the circuit diagram
  • Write the timing equations for n+1 th step for each output
  • Sketh the output map– no of columns=no of outputs+steps+input(pl add to the message the no of zeros or crc depending upon the situation, no of rows has to be input+2
  • Carryout the timing equation for each step, the last step will give you the output
slide91
CRC

Input Data

  • Polynomial:P=11001,P(x)=x4+x3+x0
  • X4 X3 X2 X1 X0
  • Timing equations
  • An + In = Dn+1
  • Dn = Cn+1
  • Cn = Bn+1
  • An + Bn = An+1

I

+

+

A

B

C

D

slide92

An + Bn = An+1

Cn = Bn+1

Cn+1= Dn+1

An + In = Dn+1

slide93

An + Bn = An+1

Cn = Bn+1

Cn+1= Dn+1

An + In = Dn+1

question
QUESTION
  • SHOW THE FOLLOWING RECEIVED MESSAGE IS IN ERROR,FOR THE SAME TRASMITTED MESSAGE ie 10010101010100
  • Received message:10110100010100
  • WRITE THE ERROR MESSAGE EQUATION
error equation
ERROR EQUATION
  • TRANSMITTED MESSAGE + RECIEVED= ERROR MESSAGE
  • 10010101010100
  • 10110100010100

00100001000000 = ERROR MESSAGE

E(X)=X6 + X11

instances where the crc is failed to answer
INSTANCES WHERE THE CRC IS FAILED TO ANSWER?
  • THER ARE INSTANCES WHERE THE CRC WILL FAILED TO ANSWER, ONE SUCH INSTANCES WILL BE WHEN THERE ARE ERRORS INTRODUCED EQUAL TO THE POLYNOMIAL
when error message is equal to the polynomial example
WHEN ERROR MESSAGE IS EQUAL TO THE POLYNOMIAL (EXAMPLE)
  • ASSUME THE FOLLOWING
  • TRANSMITTED MESSAGE
  • 100101010100
  • RECEIEVED MESSAGE
  • 100101001101
  • POLYNOMIAL
  • 1101
  • SHOW THAT CRC IS FAILED TO IDENTIFY THE ERROR IN THE MESSAGE?
though the residual is 0 there is an error in the receieved message
THOUGH THE RESIDUAL IS 0 THERE IS AN ERROR IN THE RECEIEVED MESSAGE
  • Hint divide the received message by mod 2
  • Then observe that no residuals
  • Write the error message & compare with the polynomial
try a crc sum
TRY A CRC SUM
  • TRANSMIT MESSAGE
  • 11001011101
  • POLYNOMIAL
  • 101101
  • FIND OUT THE CRC
  • DRAW THE CIRCUIT DIAGRAM AND SHOW CLEARLY HOW YOU PRODUCE CRC?
slide108

How CCITT No:7 works- Study about the layered structure

Layer 3

Layer 4

Layer 1

Layer 2

Layer 4

Layer 3

Layer 2

CRC=0

W0

FSN=5,FIB=1

Instructions

DATA

SIO

Label

k1

W5

SCF

K2

SCF

K2

K2

K1

k2

k1

W127

SCF=Sequence control field

DPC=st B

LABEL CONTENTS

BSN=5,BIB=1

SCF

Clear W5

OPC|DPC|CIC

User Part

Signaling Link

Link Control

Error detection and correction

Message type

Message handline

Signal control

Actual message

Station A

Station B

how reroutine is done
How reroutine is done?

Layer 1

Layer 2

Layer 3

Layer 4

Layer 3

Layer 2

CRC=0

W0

FSN=5,FIB=1

DATA

SIO

Label

k1

W5

SCF

K2

SCF

K2

K2

k2

DPC=st C

k1

W127

SCF=Sequence control field

Station A

BSN=5,BIB=1

SCF

Clear W5

W10

SCF

K2

Station B

FSN10 ,BIB0

K1

K2

SCF

K2

Station C

question1
QUESTION
  • SHOW THE FOLLOWING RECEIVED MESSAGE IS IN ERROR,FOR THE SAME TRASMITTED MESSAGE ie 10010101010100
  • Received message:10110100010100
  • WRITE THE ERROR MESSAGE EQUATION
error equation1
ERROR EQUATION
  • TRANSMITTED MESSAGE + RECIEVED= ERROR MESSAGE
  • 10010101010100
  • 10110100010100

00100001000000 = ERROR MESSAGE

E(X)=X6 + X11

slide115

1

17

9

2

8

12

16

3

10

15

11

4

14

13

5

6

7

0

0

0

0

0

1

1

0

0

1

0

0

0

1

0

0

1

0

1

0

1

0

0

0

0

0

1

1

0

1

1

1

0

0

1

0

1

0

0

0

1

1

0

1

0

0

1

0

1

0

0

0

0

0

0

0

1

1

1

0

0

0

0

0

1

0

0

1

0

1

0

0

0

1

0

1

1

1

0

0

0

0

0

0

0

0

0

1

0

0

1

1

0

0

0

0

0

0

0

0

1

1

0

1

0

0

1

1

0

1

0

0

0

1

1

1

0

0

part 6
Part 6
  • Switching network
basic analogue switch
Basic analogue switch

Output

Analogy

C

D

C

D

A

A

Input

B

B

No of points =4

Full available Switch

Non blocking switch

All the voltages generated in the phone can be seen in the points

A

C

D

B

4 4 switching
4 4 switching

16

Yes

Yes

No: of * points =

Is it fully available =

Is it non-blocking =

When the number of inputs and no: of ouptuts increases , we have to think about a alternative solution

quality factors of a switching network
QUALITY FACTORS OF A SWITCHING NETWORK

E

A

B

F

G

C

D

H

To have full availability, we should have at least 1 link from the input small switch to a small output switch as shown above

Blocking:

How to make this non-blocking?

A

E

2

A

E

B

F

2

B

F

G

C

2

C

G

D

H

2

H

D

When A is connected to E, can B be connected to F?

No, therefore this is a blocking network

working of t switch
Working of T switch

Assume a master switch is connected with 2 RSU’s

B

A

slide140

Working of T switch

Assume a master switch is connected with 2 RSU’s as shown in the figure below

Assume 16 PCm systems are connetcted to from RSU to MSU

B

A

Now assume that A is speaking with B

A speaks in p1 TS5 and B in p16 TS10

What will happen at the master switch C ?

Switching Equation

A’s hello !! P1f TS5-------  P16b TS10 B will hear.

B’s hello !! P16f TS10---  P1b TS5 A will hear.

detailed working of t switch timing chart
Detailed working of T-switchTiming chart

A’s hello !! P1f TS5

------ 

P16b TS10 B will hear.

B’s hello !! P16fTS10

---- 

P1b TS5 A will hear.

P1f

TS5

P16b

TS10

P16f

TS10

P1b

TS5

T=0

T=125µs

what really happens at the t switch
What really happens at the T-switch

A’s hello !! P1f TS5

------ 

P16b TS10 B will hear.

B’s hello !! P16fTS10

---- 

P1b TS5 A will hear.

4

4

2

W0

W5

2

P1f TS5

1

W0

W5

3

Go to W522

1

W522

3

P16f TS10

W1023

W522

Go to W5

8 Bits

W1023

Buffer Memory

10 Bits

Microprocessor-2 actions for each channel in 125µs

Control Memory

types of t switches
TYPES OF T SWITCHES
  • WHAT WE HAHE STUDIED NOW IS OUTPUT CONTROLLED T SWITCH
  • THE INPUT PCM TSS ARE CYLICLY STORE IN THE BUFFER MEMORY
  • THE OUT PUT PCM ARE CYCLICLY ADDRESSING THE CONTROLLED MEMORY, THE CONTENTS WILL TELL YOU WHERE TO READ AND TRANSPORT IT TO THE DESTINATION.
  • THE OUTPUT PCMS ARE RIGIDLY CONNECTED TO THE CONTROLED MEM,ORY THATS WHY IT IS CALLED OUT PUT CONTROLLED T SWITCH
  • SIMILARLY CAN YOU TRY A INPUT CONTROLLED T SWICH AND WRITE ITS CHARACTERISTICS?
characteristics of input controlled t switch
CHARACTERISTICS OF INPUT CONTROLLED T SWITCH
  • INPUT PCMS ARE RIGIDLY CONNECTED TO THE CONNTROLLED MEMORY
  • THE INPUT PCMS ARE CYCLICLY ADDRESSING THE CONTROLLED MEMORY. THE CONTENTS WILL TELL YOU WHER TO STORE IN THE BUFFER MEMORY.
  • THE OUTPUT PCMS ARE CYCLICLY READING THE BUFFER MEMORY.
basic componants of a switching system
BASIC COMPONANTS OF A SWITCHING SYSTEM
  • SWITCHING UNIT
  • CONTROLLED UNIT
  • PERIPARALS
  • SOFTWARE
  • IS IT ANALOGUES TO HUMAN BODY?
  • YES,EXCEPT FREE WILL OF THE HUMAN,THIS SYSTEM WILL HAVE HEART, BRAIN & MIND
comparison of human with animal switching node
Comparison of Human with Animal & SWITCHING NODE

HEART

BRAIN

MIND

TRUTH

FREE WILL/GOOD

HUMAN

TELEPHOE NODE

SWITCHING NETWORK

CONTROL NETWORK

PROGRAMMING

SOFTWARE

ANIMAL

HEART

INSTINC

BRAIN

switching unit
SWITCHING UNIT
  • MAIN FUNCTION—connecting to an input to a output
  • In the case of local node, input will be the customer, and the output will be a route, where the call is destined to
  • Limiting factor: no of connections that can be established simultaneously is the limiting factor
  • MEASURED IN ERLANG
  • Present day technology : analogue & digital switches are now obsolete, now packet switching routers are deployed.
controlled unit
CONTROLLED UNIT
  • FUNCTION: ALL THE MANAGEMENT FUNCTIONS THAT NEED TO CARRYOUT IN ESTABLISHING ACONNECTION WILL BE DONE BY THE CONTROLLED UNIT.
  • THE MANAGEMENT FUNCTIONS ARE /CALL ESTABLISHMENT, SENDING INFORMATION TO THE OTHER NODES, CALL BILLING FUNCTION, CUSTOMER FACCILITY MANAGEMENT ETC.....
  • LIMITATION WILL BE THE OCCUPANCY OF THE PROCESSOR. NORMALLY MORE THAN 80% WILL NOT BE ADVISABLE FOR ANY PROCESSOR. ANOTHER MEASUREMENT WILL BE TLME TAKEN TO ESTABLISH A CONNECTION
  • TECHNOLOGY: CENTRALISED CONTROL FUNCTION HAS BEEN SHIFTED TO DITRIBUTED PROCESSER FUNCTION. MODERN NGN SWITCH WILL HAVE MOST OF THE MANAGEMENT FUNTIONS CENTRALISED TO THE CONTROL PART OF SOFT SWITCH, WHILE ROUTING PART IS DISTRIBUTED TO THE ROUTERS. ANOLGUE CONTROL(WIRED LGIC),IS OBSELETE.
periparals
PERIPARALS
  • THEY ARE THE ANCILIARY EQUIPMENT TO CARRY OUT THE MAJOR FUNTIONALITIES OF THE SYSTEM. THEY BARE REGISTERS, TONE GENERETORS, TIMING DEVICES, ETC...
software
SOFTWARE
  • SOFTWARE WILL PROVIDE ALL DETAILED ACTION PLANS IS BECOMING HIGLY COMPLEX.
  • MODULAR KIND OF SOFTWARE IS NOW ENCOURAGED.
  • WITH NGN TECHNOLOGY THE SOFTWARE HAS BECOME A VITAL ELEMENT FOR THE PROPER FUNCTIONING OF MUTIPLE SERVICE FACILIETIES.
the ip world
THE IP WORLD
  • TODAY WE ARE IN THE IP WORLD
  • ALL THE NETWORKS ARE PUSHING TO THE IP APPLICATIONS
  • MODERN CODING METHODS PUSH IP NETWORKS TO GO FOR REAL TIME APPLICATIONS
  • TDM & ATM NETWORKS ARE REPLACED BY IP
  • LETS STUDY THE PACKET CONCEPTS
major switching types

Major switching types

Circuit ,Message and Packet switching

chracteristics of signalling voice
CHRACTERISTICS OF SIGNALLING & VOICE
  • CHANNEL ASSOCIATED SIGNALLING

SIGNALLING & VOICE IN ONE TUNNEL-WASTAGE IN SIGNALLING, AND LESS THAN 50% EFFICIENT IN VOICE MEDIA DUE TO THE CHARACTERISTICS OF VOICE.

COMMON CHANNEL SIGNALLING

SIGNALLING IS COMMONLY USED FOR MANY VOICE CHANNELS, HENCE SIGNALLING CHANNEL IS EFFICIENT. VOIE CHANNELS CARRIES THE SAME INEFFICIENCY AS IN THE CASE OF CHANNEL ASSOCIATED SIGNALLING

HENCE THE CONCEPTS OF MESSAGE AND PACKET SWITCHING NETWORK TO BE CONSIDERED.

delay analysis of a message switch
DELAY ANALYSIS OF A MESSAGE SWITCH
  • TAKE THE PREVIOUS EXAMPLE, THE FOLLOWING ARE KNOWN
  • MESSAGE SIZE=30, OVERHEAD=3, NO OF HOPS=3,THE DELAY FOR ONE HOP/OCTET=1 MSEC.
  • THE TOTAL DELAY FOR 3 HOPS=33*3 MSEC
  • TOTAL DELAY=NO OF HOPS*NO OF OCTET IN MESSAGE+NO OF OVERHEAD OCTETS IN THE MESSAGE(THIS RESULT IS TALLYING)
deciding optimum packet size

l

l

l

Packet 1

Packet 2

Packet 3

Packet 1

Deciding Optimum Packet Size
  • Shown above is a message

MESSAGE

.

  • The message will be divided in to equal length packets
  • Each Packet will have a Header
  • The header will consist the following details:
    • Originating Point Code
    • Terminating Point Code
    • Packet Number
slide171

A

B

C

D

T1

T5

T4

T3

T2

Packet 1

Packet 2

Packet 1

Packet 2

Packet 3

Packet 3

Packet 1

Packet 1

Packet 2

Packet 2

Packet 3

Packet 3

T1

T2

1

2

T3

3

T4

2

T5

1

TOTAL DELAY α S + H - 1

Therefore in this case

the total delay time is α 3 + 3 – 1 = 5

  • Where
      • S - Number of Packets
      • H - Number of Hops
delay analysis for packet switching network
DELAY ANALYSIS FOR PACKET SWITCHING NETWORK
  • TAKE TWO EXAMPLES THE 30 OCTET MESSAGE IS (1) MADE TO TWO PACKETS OF 15 OCTET EACH, (2) MADE TO 3 PACKETS OF 10 OCTETS EACH
  • YOU WILL OBSERVE THE FOLLOWING

OVERLAP SENDING OF PACKETS, FOR EXAMPLE, WHEN 1 ST PACKET RCEIVED AT B, THIS PACKET IS SEND FROM B TO C,DURING THIS TIME THE 2ND PACKET IS SEND FROM A TO B. NOTE THAT IN AGIVEN TIME 2 ACTIONS HAVE BEEN MADE

THE TOTAL DELAY EXPERIENCED IN (1)72 MSEC (2) 65 MSEC

message vs packet delay
Message vs Packet delay
  • It is easy to calculate message delay rather than packet delay ,why ?
  • When sending packets from one node to another the following process can be adopted as against message transmission
  • (a)Packets can be sent to another node through different paths simultaneously
  • (b)Packetizing at the node and sending packets over the hop can be made in different times to maximize the sending of packets over a hop .
  • (C)Hence , overlap sending & receiving of packets can be achieved in a node . Hence the delay introduced in packet mode is rather complicated (than message) although it is efficient .
contd
Contd….
  • In message mode the delay introduced in (H-1) hops is rather simple and is equal to
  • Message delay = K * (H-1)
  • In packet mode the delay is proportional to addition delay to message length and it is assumed to be
  • D = K*(H-1) + term proportional to message length
  • Delay={ S+ (H-1) } * ZT
calculation of the optimum packet size
Calculation of the optimum packet size
  • Deciding the optimum packet size will depend upon 2 factors
  • Overall delay
  • Overall overhead bits compared to the message
  • Calculate the optimum packet size of 30 octet message, where the overhead for aq packet is 3 octet?
what is the packet network
What is the packet network?
  • A packet is a unit of data that is transmitted across a packet-switched network.
  • A packet-switched network is an interconnected set of networks that are joined by routers or switching routers.
example
Example
  • Packet Switching technology  TCP/IP
  • largest packet-switched network  Internet
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Why ?
  • Data traffic is bursty
    • Logging in to remote machines
    • Exchanging e-mail messages
  • Don’t want to waste reserved bandwidth
    • No traffic exchanged during idle periods
  • Better to allow multiplexing
    • Different transfers share access to same links
goals
Goals
  • To optimize utilization of available link capacity
  • To increase the robustness of communication
concept
Concept
  • A method of transmitting messages through a communication network, in which long messages are subdivided into short packets.
  • The packets are then sent through the network to the destination node.
packet switching techniques
Packet-Switching Techniques

Datagram

Virtual Circuits

Each packet contains addressing information and is routed separately

  • A logical connection is established before any packets are sent; packets follow the same route.
datagram
Datagram
  • Each packet treated independently
  • Packets can take any practical route
  • Packets may arrive out of order
  • Packets may go missing
  • Up to receiver to re-order packets and recover from missing packets
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Need to transmit ‘123’ from computer A to computer B

Computer

A

1

2

3

First data is broken to small pieces (PACKETS)

Computer B

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Packets contain header information that includes a destination address.

Computer A

1

2

3

Routers in the network read this address and forward packets along the most appropriate path to that destination.

Computer B

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Computer

A

2

1

3

Computer B

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Computer

A

2

1

3

Computer B

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Computer

A

2

1

3

Computer B

virtual circuits v datagram
Virtual Circuits v Datagram
  • Virtual circuits
    • Network can provide sequencing and error control
    • Packets are forwarded more quickly
      • No routing decisions to make
    • Less reliable
      • Loss of a node loses all circuits through that node
  • Datagram
    • No call setup phase
      • Better if few packets
    • More flexible
      • Routing can be used to avoid congested parts of the network
advantages
Advantages
  • Line efficiency
    • Single node to node link can be shared by many packets over time
    • Packets queued and transmitted as fast as possible
  • Data rate conversion
    • Each station connects to the local node at its own speed
    • Nodes buffer data if required to equalize rates
  • Packets are accepted even when network is busy
    • Delivery may slow down
  • Priorities can be used
difference between channel associated common channel and packet signaling networks
Difference between channel associated common channel and packet signaling networks

Channel associated signalling(CAS)

Each voice channel will have a supervisory channel(either direct or associate).

Highly inefficient for the signalling channel and less than 50% efficient for the voice channel.

Common Channel signalling – All supervisory

Signals of voice channels are in one time slot

And the voice channels have similar inefficiency as CAS

Voice channel

Supervisor signal

Signalling and voice are going on packets whenever it is needed

Packet network=Signalling and voice are sent in packets ,highly efficient for voice as

well as signalling.The deficiecy experienced for voice channels, in CAS & CCS has overcome

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