04104 DATA COMMUNICATION 2 SAMPLE QUESTIONS & ANSWERS

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Oct 30, 2013 (4 years and 8 months ago)

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04104

DATA

COMMUNICATION 2

SAMPLE QUESTIONS

&

CHAPTER

(7)

MULTIPLEXING

1.

What is multiplexing? Explain its application.

Solution
:

Two communicating stations will not utilize the full

capacity of a data

be possible to share that capacity, sharing is called multiplexing.

Fig shows the multiplexing function in its simplest form. There are n inputs to the multiplexer. The
multiplexer is connected by a single data link to

a demultiplexer. The link is able to carries n separate
channels of data. The multiplexer combines (multiplexes) data from the n input lines and transmits over a
higher
-
The demultiplexer accepts the multiplexed data stream, separate
s (demultiplexes)
the data according to channel, and delivers them to the appropriate output line.

FIGURE
.

Multiplexing

The widespread use of multiplexing in data communications can be explained by the
following:

1.

The higher the da
ta rate, the more cost
-
effective the transmission facility. For a given application

and over a given distance,
the cost per kbps declines with an increase in the data rate of the
transmission facility.

2.

Most individual data
-
communicating devices require re
latively modest data
-
rate support. For
example, for most terminal and personal computer applications, a data rate of between 9600 bps
and 64 kbps

2.

Describe Digital Carrier System.

Solution:

(a)

North American

(b) International (ITU
-
T)

Digital

Number

Data

Level

Number

Data

Signal

of voice

rate

number

of voice

rate

Number

channels

(Mbps)

channels

(Mbps)

DS
-
1

24

1.544

1

30

2.048

DS
-
1c

48

3.152

2

120

8.4
48

DS
-
2

96

6.312

3

480 34.368

DS
-
3

672

44.736

4

1920

139.264

DS
-
4

4032

274.176

5

7680

565.148

MUX

1 li
nk, n channel

DE

MUX

n outputs

n inputs

3.

Describe Analog Carrier System.

Solution:

Number of voice

Bandw
idth

Spectrum

AT&T

1
TU
-
T

channel

12

48 kHz

60
-
108 kHz

Group

Group

60

240 kHz

312
-
552

kHz

Super group

Super
group

300

1.232 MHz

812
-
2044 kHz

Master group

600

2.52 MHz

564
-
3084 kHz

Master group

900

3.872 MHz

8.516
-
12.388 MHz

Super
master

group

N*160

Master group

Multiplex

3,600 16.985 MHz 0.564
-
17.548 MHz Jumbo group

10,800

57.442 MHz 3.142
-
60.566 MHz Jumbo group

multiplex

TABLE

North American and international FDM carrier Standards.

4.

What are the distinct chara
cteristics of FDM use FDM application.

Solution
:

FDM are possible when the useful bandwidth
of the transmission medium
exceeds the required
bandwi
d
th

of signals to be transmitted. A number of signals can be carried simultaneously if each signal
is modula
ted onto a different carrier frequency.

Each modulated signal requires a certain bandwidth centered around its carrier frequency, referred to
as a channel. To prevent interface, the channels are separated by guard bands, whi
ch are unused portions
of the

sp
e
ctrum.

The composite signal transmitted across the medium is analog. The input signals may be either
digital or analog .Digital input
,

the input signal must be passed through modems to be converted to
analog. Each input analog signal must then be m
odulated to the appropriate frequency band.

A number of analog or digital signals [m
i
(t)
,

i = 1
,

N

] are to be multiplexed onto the
same
transmission medium . Each signal m
i

(t
)

i, modulated onto a carrier frequency. The resulting modulated
analog si
gnals are then summed to produce a composite signal m
c

(t
).

The composite signal

has a total bandwidth B. At the receiving end, the composite signal is passed
through N
band pass

filters, each filter centered on frequency and having a bandwidth. The signa
l is
again split into its component parts
.

Each component is then demodulated to recover the original signal.

5.

Explain TDM (Time division Multiplexing) system and show TDM of analog and digital source
is organized?

Solution

Time Divisi
on Multiplexing

A number of signals [m
i
(t) , i=1,n] are to be multiplexed onto the same transmission medium. The
signals carry digital data and are generally digital signals. The incoming data from each source are briefly
buffered. Each buffer is typica
lly one bit or one character in length. The
buffers are

scanned sequentially to
form a composite digital data stream m
c
(t).

The digital signal m
c

(t) may
be transmitted

directly or passed through a modem so that an analog
signal is transmitted.

The data

are organized into frames. Each frame contains a cycle of time slots. The sequence of slots
dedicated to one source, from frame, is called a channel. The slot length equals the transmitter buffer length,
typically a bit or a character.

The character inte
rleaving technique is used with asynchronous sources. The bit
-
interleaving
technique is used with synchronous sources and may also be used with asynchronous sources. Each time slot
contains just one bit.

At the receiver, the interleaved data are demultipl
exed and routed to the appropriate destination
buffer. The output

source which will receive the input data at the same rate.

Figure: Frequency

division multiplexing

(a) Transmitter

f
1

f
2

f
3

f
4

f
5

f
6

c

h

a

n

1

c

h

a

n

2

c

h

a

n

3

c

h

a

n

4

c

h

a

n

5

c

h

a

n

6

Time

frequency

Sub carrier

f
sc 1

Sub carrier

f
sc
2

Sub carrier

f
sc
N

m
1

(t)

m
2

(t)

m
N

(t)

Transmitter

f
c

S(t) = FDM

M
c

(t)

S
scN

(t)

S
sc1

(t)

S
sc2

(t)

(b)
Spectrum of composite signal (positive

f)

(

Figure: Frequency Division Multiplexing

Modem

Buffer

Scan operation

S (t)

m
1

(t)

m
2

(t)

m
n
(t)

m
e

(t)

(a) Transmitter

frequency

Time

M
c

f
(t)

1

0

f

N

B
sc 1

B
sc 2

B
sc N

f
sc N

f
sc 2

f
sc 1

Bandpass filer ,

f
sc 1

Bandpass filer ,

f
sc
2

Bandpass filer ,

f
sc
N

S

(t)

Demodulator

f
sc 1

Demodulator

fsc 2

Demodulator

fsc
N

m
N

(t)

m
2

(t)

m
1

(t)

Figure: TDM of a
nalog and digital sources

m
2
(t)

m
1
(t)

m
n
(t)

m
c
(t)

Buffer

Scan

Operation

Time slot: may be empty o
r occupied

(b) TDM frames

frame

frame

1

2

N

1

2

N

FIGURE

Synchronous time
-
division multiplexing

Pulse
stuffing

Pulse
stuffing

2 bit buffer

2 bit buffer

8 kbps

8 kbps

source 11

7.2 kbps

, digital

source 4

7.2 kbps , digital

Scan

128 kbps

TDM signal

2 bit buffer

4 bit
A/D

64

kb p s

T DM P CM
s i gn a l

f

source 1

source 2

source 3

2kHz, a n a lo g

4
kHz, a n a lo g

2kHz, a n a lo g

f
-

4kHz

TDM P
A
M
signal

16 kHz

6.

Describe Basic ISDN Interface.

Solution:

Basic ISDN Interface

The basic access structure consist of two 64 kbps B channels and one 16 kbps D cannel, 144 kbps, are
multiplexed over a 192 kbps interface at the S or T

reference point. The remaining capacity use for various
framing and synchronization purposes.

The B channel is the basic user channel. It can be use to carry digital data. The D channel can be
use for a data transmission connection at a lower data rat
e. It is also use to carry control information needed
to setup and terminate the B channel connections.

Each frame of 48 bits includes 16 bits from each of the two B channel, and 4 bits from the D
channel: The remaining bits have the following interpre
tation. In the TE to NT direction. Each frame
begins with a framing bit (F) that is always transmitted as a positive pulse. This is followed by a dc
balancing bit (L) that is set to a negative pulse to balance the voltage. The F
-
L pattern thus acts to
synchronize the receiver on the beginning of the frame.

The next eight bits (B1) are form the first B channel this is followed by another dc balancing bit (L).
Next comes a bit from the D channel, followed by its balancing bit. This is followed by the
auxiliary
framing bit (F
A
), which is set to zero unless it is to be use in a multi
-
frame structure. There follows another
balancing bit (N), eight bits (B2) from the second B channel, and another balancing bit (L); this is follow by
bits from the D channe
l, first B channel, D channel again, second B channel, and the D channel, again.

In the NT to TE direction is similar to the frame structure for transmission in the TE
-
to
-
NT direction.
The following new bits replace some of the dc balancing bits. The
activation bit (A) is use to activate or
deactivate a TE. The N bit is normally set to binary one. The N and M bits may be use for multi
-
framing.
The s bit is reserved. For other future standardization requirements. There are three types of traffic:

B
-
channel traffic: No additional functionality is needed to control
is dedicated to a particular TE at any given time.

D
-

channel traffic: The D
channel
is available for use by all the subscriber devices for both
control signal
ing and packet transmission, so the potential for contention exits, There are two sub
cases:

Incoming traffic:

The LAPD addressing scheme is sufficient to sort out the proper
destination for each data unit.

Outgoing traffic: Access must be regulated so tha
t only one device at a time transmits.
This is the purpose of the contention
-
resolution algorithm.

+

Fig: 7.10

from Page 18

7.

Describe primary ISDN Interface.

Solution:

Primary ISDN Interface

The primary
interface

multiplexes multiple channels acros
s a single transmission medium. In the
case of the primary interface, only a point
-
to
-
point configuration is allowed. Typically, the interface
supports a digital PBX, and providing a synchronous TDM facility for access to ISDN. Two data rate are
defined

for the primary interface: 1.544 Mbps and 2.048 Mbps.

The ISDN interface at 1.544 Mbps is based on the North American DS
-
1 transmission structure,
which is use on the
T1 transmission service. The bit stream is structured into repetitive 193 bit frames.

Each frame consists of 24,

8 bit time slots and a framing bit, which is used for synchronization and other
management purposes. At a data rate of 1.544 Mbps, frames repeat at a rate of one every 125

sec,

or 8000
frames per second. The transmission stru
cture is use to support 23 B channels and 164 kbps D channel.

The line coding for the 1.544Mbps interface is AMI (Alternate Mark Inversion) using B825.

The ISDN interface at 2.048 Mbps is based on the European transmission structure of the same
data ra
te. The bit stream is structure into repetitive 256bit frames. Each frame
consists

of 32,8bit time slots.
The first time slot is use for framing and synchronization purposes:
the remaining 32 time slots support user
channels. At a data rate of 2.048Mbp
s, frames repeat at a rate of one every 125

sec, or 8000 frames per
second. The transmission structure is used to support 30B channels and 1D channel.

The line coding for the 2.048
-
Mbps interface is AMI using HDB3.

(a)

Interfa
ce at 1.54 Mbps

(b) Interface at 2.048 Mbps

Fig: ISDN primary access frame formats

8.

Explain the STS
-
1

Solution:

H1
-
H3: Pointer bytes used in frame alignment and frequency adjust

B2: Bit
-
interleaved parity for line level error monitoring.

K1, K2: Two bytes allocated for signaling between line level automatic protection switching
equipment.

D4
-
D12: 576kbps data communication cannel for alarms, maintena
nce, control, monitoring and

Z1
-
Z2: Reserved for future use.

E2: 64kbps
PCM voice channel for line level orderwire.

J1:64kps channel used to repetitively send a 64
-
octed fixed
-
length string so a
receiving terminal can
continuously verity the intergrity of a path; the contents of the message are user programmable.

B3: Bit
-
interleave parity at the path level.

C2: STS path signal label to designate
equipped versus unequipped STS signals.

G1
: Status byte sent from path terminating equipment back to path originating equipment to convey
status of terminating equipment and path error performance.

F2: 64 kbps channel for path user.

H4: Multiframe indicator for pay load needing frames that are

used when packing lower rate channel.

Z3
-
Z5:Reserved for future use.

1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8

F

Time

slot
1

Time

slot
31

1 2 3 4

5 6 7 8

Time

slot 0

1 frame =193 bits, 125

sec

1 2 3 4 5 6 7 8

Time

slot 0

1 frame =256 bits, 125

sec

1 2 3 4 5 6 7 8

Time

slot 32

1 2 3 4 5 6 7 8

Time

slot 2

1 2 3 4 5 6 7 8

Time

slot 1

framing channel

(9)
Explain Framing and pulse stuffing as used in TDM and note down the lot of TDM carrier standard.

Solution:

Framing:

A link control protocol is not needed to manage the overall

TDM link. A basic requirement for
framing is that it is not providing flag or SYNC characters to bracket TDM frames. It is important to
maintain framing synchronization because if the source and destination are out of step, data on all channels
are lost.

The most common
mechanism for framing is known as added
-
digit framing. To

synchronize, a receiver compares the incoming bits of one frame position to the expected pattern. If the
pattern does not match, successive bit positions are searched. One framing
synchronization is established,
the receiver continues to monitor the framing bit channel. If the pattern breaks down, the receiver must again
enter a
framing search mode.

Pulse Stuffing

If each source has a separate clock, any variation among clock
s could cause loss of synchronization.
The data rates of the input data streams are not related by a simple rational number. For both these problems,
a technique known as pulse stuffing. With pulse stuffing, the outgoing data rate of the multiplexer
,
excl
uding framing bits, is higher than the sum of the maximum instantaneous incoming rate. The stuffed
pulses are inserted at fixed locations in multiplexer frame format, so that they may be identified and
removed at the demultiplrxer.

(a) North American

(
b) International (1TU
-
T)

Digital
signal
number

Number of
voice
channels

Data rate
(Mbps)

Level
number

Number of
voice
channels

Data rate
(Mbps)

DS
-
1

DS
-
1c

DS
-
2

DS
-
3

DS
-
4

24

48

96

672

4032

1.544

3.152

6.312

44.736

274.176

1

2

3

4

5

30

120

480

1920

7680

2.
048

8.448

34.368

139.264

565.148

Table: North American and International
JDM carrier standards

9.

Draw the frame
-
format for SONET/SDH and show in the table from overhead octet of STS
-
1.

Solution:

SONET is intended to provided a specification for taking ad
vantage of the high
-
speed digital
transmission capacity of optical fiber.

FIGURE STS1 Frame Format

87 octets

90 octets

3 octets

8 octets

6 octe
ts

The basic SONET building block is the STS
-
1 frame, which consists of
810 octets and is transmitted
once every 125

, for an overall data

rate of 51.84 Mbps. The frame can logically be viewed as a matrix of
T rows of 90 octets each, with transmission being one row at a time, from left to right and top to bottom.

The first three columns (3 octets*9 rows=27 octets) of the frame are devot
Nine octets are related to section
-

The remainder of the frame is payload, which is provided by the path layer. The payload includes a
column of path overhead. The lin
e

over

contains a pointer that indicates where the path overhead starts
.

Section

Framing
A1

Framing
A2

STS
-
ID
C1

BIP
-
8
B2

Order
Wire E1

User F1

Data
Com D1

Data
Com D2

Data
Com D3

Pointer
H1

Pointer
H2

Pointer
Action
H3

BIP
-
8
B2

APS K1

APS K2

Data
Com D4

Data
Com D5

Data
Com D6

Data
Com D7

Data
Com D8

Data
Com D9

Data
Com
D10

Data
Com
D11

Data
Com
D12

Growth
Z1

Growth
Z2

Order
wire E2

(a)

T
race

J1

BIP
-
8

B3

Signal level
C2

Path

status

G1

User

F2

Multiframe
H4

Growth

Z3

Growth

Z3

Growth

Z5

Figure:

SONET/STS
-

CHAPTER

(8)

CIRCUIT SWITCHING

10.

Write short notes on circuit
-
switching network.

Solution:

Figure 8.1+

Communication via circuit switching implies that there is a dedicated communication path between two
stations. That path is a connected sequence of links between netwo
rk nodes. On each physical link,

a
logical channel is dedicated to

the connection.

Communication via circuit
-
switching involves three

phases:

1.
Circuit Establishment:

Before any signals can be transmitted, on end
-
to
-
end circuit must be
established. F
or example, station A sends a request to node 4 requesting a connection to station E. so,
the link from A to 4 is a dedicated line. Node 4 must find the next node leading to node 6. Node 4
selects the link to node 5, and sends a message requesting
conne
ction to E. Therefore a dedicated path
has been established from a through 4 to 5 and similarly, 5 to 6. Node 6 completes the connection to E.
In completing the connection, a test is made to determine it E is busy or is prepared to accept the
connection
.

2.
Data Transfer:

Information can now be transmitted from A through network to E. The data may be
analog or digital, depending on the nature of the network. The path is A
-
4

through 5,5
-
6 channel, and internal switching t
hrough 6, 6
-
E link, Generally, the connection is full
-
duplex.

3.
Circuit Disconnect:

After some period of data transfer, the connection is terminated, usually by the
action of one of the two stations. Signals must be propagated to nodes 4,5 and 6 to de
-
allocate the
dedicated resources.

11.

Describe and explain for generic architectural components for public telecommunication network.

Solution:

Four generic architectural components

1.

Subscribers: The devices that attach to the network. (e g. telephone)

2.

L
ocal loop: The link between the subseriber and the network, also referred to as the subscriber loop.
Almost

all local loop is in a range a few kilometers to a few tens of kilometers.

3.

Exchanges: The switching canters in the network, which is directly suppo
rts subscribers is known as
an end office. An end office supports many thousands of subscribers in localized area.

4.

Trunks: The branches between exchanges. Trunks carry multiple voice frequency circuits using
either FDM or synchronous TDM. Earlier, these
we
re referred to as carrier systems.

Figure:

Public
-

Circuit
-

Switching network

12.

What are the signaling functions and describe any two.

Solution:

Signaling Functions

1.

Audible communication with the subsc
riber, including dial tone, ringing tone, busy signal and so
on.

2.

Transmission of the number dialed to switching office, that will attempt to complete a connection.

3.

Transmission of information between switches indicating that a call cannot be completed.

4.

Transmission of information between switches indicating that a call has ended and that the path
can be disconnected.

5.

A signal to make a telephone ring.

6.

Transmission of information used for billing purposes.

7.

Transmission of information giving the status of

equipment or trunks in the network. This
information may be used for routing and maintenance purposes.

8.

Transmission of information used in diagnosing and isolating system failures.

9.

Control of special equipment such as satellite channel equipment.

Digital PBX

Telephone

o

o
o o

o o

o o

o o

o o
o o

o o

o o

o o

o o

o o

o o

o o

o o

o o
o o

o o

o o

o o

o o

o o

o o

o o

o o

o o
o o

o o

o o

o o

Telephone

End Office

Long
-
Distance
Office

Long
-
Distance
Office

End
Office

Su
bscriber
Loop

Connecting Trunk

Intercity Trunk

Signali
ng can also be classified functionally as supervisory, address, call
-
information, and
network management.

(1)
Supervisory control signal

The term supervisory is generally used to refer to control function, that have a binary
character C true/false; on
deal with the availability of the called subscriber and of the need network resources.

(2)

Address signals identify a subscriber, and it is generat
ed by a calling subscriber when dialing
a telephone number. The resulting address may be propagated through the network to support the
routing function and to locate and ring the called subscriber’s phone.

(3)
Call
-
information control signals

-

provide

information to the subscriber about the status of a call.

-

are audible tones that can be heard by the caller or operator with the proper

Phone set.

(4)
Network management signal

-

is used for maintenance, troubleshooting, and overall operat
or of the network.

-
may be in the form of message, such as a list of preplanned routes being sent to a

station to
update its routing tables.

13.

Explain Time Division Switching Techniques of circuit switching with diagram.

Solution:

Time
-
division swit
ching

All modem circuit switches are digital time
-
division techniques for establishing and maintaining
“circuits”. Time
-
division switching involves the partition of a lower
-
speed bit stream into pieces, or slots,
are manipulated by control logic to rout
e data from input to output.

In figure, each device attaches to the switch through a full
-
duplex line. These lines are connected
through controlled gates to a high
-
speed digital bus. Each line is assigned a time slot for providing input.
For the dura
tion of the slot, that line’s gate is enabled, allowing a small burst of data onto the bus. For that
same time slot, one of the other line gates is enabled for output. Daring successive time slots, different
input/output parings are enabled, allowing a n
umber of connections to be carried over the shared bus. An
attached device achieves full
-
duplex operation by transmitting during one assigned time slot and receiving
occurring time slots. One iteration for all time slots is referred to as a frame. The i
nput assignment may be
fixed, the output assignments vary to allow various connections. When a time slot begins, the designated
input line may insert a burst of data onto the line.

The designated output line, during that time, copies the data. The tim
e slot must equal the
transmission time of the input plus the propagation delay between input lines, the data rate on the bus must
be high enough that the slots recur sufficiently frequently. Input data on each line are buffered at the gate.
Each buffer m
ust be cleared, by enabling the gate quickly enough to avoid overrun. The actual data rate
must be high enough to also account for the wasted time due to propagation delay.

In blocking switch, is no fixed assignment of input lines to time slots: they are

allocated on demand.
The data rate on the bus dictates how many connections can be made at a time.

TDM bus switching scheme, can accommodate lines of varying data rates.

Control of TDM bus switch

Figure Implements how the control for a TDM bus

switch. A control memory indicates which gates
are to be enabled during each time slot. Six words memory are needed. A controller cycles through the
memory. During the first time slot of each cycle, the input gate from device 1 and the output gate to
device
3 are enabled, allowing data to pass from device 1 to device 3 are the bus. The remaining words are
accessed in successing time slots and treated.

FIGU
RE

IDM bus switching

FIGURE

C
ontrol of TDM bus switch

Control unit

Control memory

1

2

3

4

5

6

Ne
twork
Interface

Control logic

Ful
l
-
du
ple
x
lin
es
to
atta
che
d
dev
ice
s

1 3

2 5

3 6

4 1

5 2

6 4

14.

Draw the block diagram of in
-
channel and common
-
channel signaling and explain common
-
channel.

LEGEND

SIG= Per
-
trunk signaling equipment

CC
-
SIG=

Common
-
channel signaling equipment

FIRURE

Inchannel and

common
-
channel signaling

Common Channel Signaling

Control signals are carried over paths completely independent of the voice channels. One
independent control signal path can carry the signals for a number of subscriber channels and is a common
cont
rol channel for these subscriber channels.

In figure (b), the signal path is separated from the path for voice or other subscriber signals. The
control signals are message that are passed between a switch and network management center.

Two modes of ope
ration are used.

(1)

Associated mode: The common channel closely tracks along its entire lengte the interswitch
trunk groups that are served between endpoints. The control signals are an different channels
from the subscriber signal, and, inside the switch,
the control signals are routed directly to a
control signal processor.

(2)

Non
-

The network is augmented by additional nodes, known as signal
transfer points. There is no close or simple assignment of control channel to trunk groups.
Therefo
re, there are now two separates networks with links between them so that the control
portion of the network can exercise control over the switching nodes that are serving the
SIG

SIG

SIG

SIG

SIG

SIG

SIG

Office B

Trunks

(a) Inchannel

Office D

CC
-
SIG

CC
-
SIG

Processor
Office D

SIG

Processor
Office D

Office D

Office A

subscriber calls. Network management is more easily exerted in one or more centr
al control
points can be established. The non
-
associated mode is the mode used in ISDN.

The control signals are transferred directly from one control processor to another,
without being tied to a voice signal.

It is less susceptible

to accidential or intentional interface between subscriber and
control signals, call
-
setup time is reduced and it is more attractive because computer hardware
costs are dropped.

With common
-
channel signaling, forwarding of control information can
-
overlap

the
circuit
-
setup process.

Both the signaling protocol and the network architecture are more complex than
inchannel signaling.

(a) Associated

Speech

LEGEND

Speech network

Signaling network

Swi
tching points
(speech)

LEGEND

(b) Non
-
associated

FIGURE COMMON
-
CHANNEL SIGNALING MODES

15.

How many types of routing techniques in circuit
-
switched

network? Explain them.

Solution:

Two types: (1) Alternate Routing

(1)
Alternate Routing

The essence of alternate
-
routing scheme is that possible routes to be used between two end
offices are predefine
d. It is the responsibility of the originating switch to select the appropriate route for each
call. Each switch is given a set of pre
-
planned routes for each destination, in order of preference C i.e, a
direct link connection between two switches. If th
is trunk is unavailable, then the second choice is to be
tired, and so on. The routing sequence are based on historical traffic patterns and designed to optimize the
use of network resources.

There are two types of Alternate
-
Routing Scheme.

(a)

Fixed
-
alterna
te Routing

If there is only one routing scheme defined for each source
-
destination pair, the scheme is known as a fixed
-
alternate
-
routing scheme.

(b)

Dynamic Alternate
-
routing

A different set of pre
-
planned routes is used for different time periods. The rout
ing decision is based both
on current traffic status and historical traffic patterns.

(2)

-
routing scheme is designed to enable switches to react to changing traffic
patterns on the network. Such schemes require greater m
anagement overhead, as the switches must
exchange information to learn of network conditions.

Dynamic traffic Management (DTM) is a routing capability. DTM uses a central controller to find
the best alternate route choices depending on congestion in the
network. The central controller collects the
status data from each switch in the network every 10 seconds to determine preferred alternate routes. Each
call is first attempted on the direct path, if any exits, between source and destination switches. If

the call is
blocked, it is attempted on a two
-
-
path.

16.

Explain the elements of a circuit
-
switched node.

Elements of a circuit
-
switch node are:

(1)

digital switch

(2)

network interface element

(3)

control unit

(1)

Digital switch

The function of the d
igital switch is to provide a transparent signal path between any pair of
attached devices, and to connect directly between them by using full
-
duplex transmission.

(2)

Network
-
Interface Element

The network
-
interface elements the functions and hardware need
ed to connect digital device,
such as data processing devices and digital telephone, to the network. Analog telephone can also be
attached if the network interface contains the logic for converting to digital signals. It provide the links
for constructin
g multiple
-
node networks.

(3)

Control unit

The control unit performs three general tasks. First, it establishes connections. To establish
the connection, the control unit must handle and acknowledge the request, determine it the
switch. Second, the control

unit must maintain the connection. Third, the control unit must
tear down the connection.

18.

State and explain cross
-
bar switch limitation.

Output Lines

FIGURE Space
-
division switch

Input Lines

Control unit

0

0

0

0

0 Digital

0 Switch

0

0

0

0

0

0

0

0

0

Network

Interface

Full
-
duplex lines

To attached devices

FIGURE Elements of a circuit switch note

Figure shows a simple crossbar matrix with 20 full
-
duplex I/O l
ines. The matrix has 10 inputs
and 10 outputs, each station attaches to the matrix via one input and one output line. Interconnection is
possible between any two lines by enabling the appropriate crosspoint.

Crossbar switch limitations

The number of cro
sspoints grows with the square of the number of attached stations. This
is costly for a large switch.

The lost of crosspoint prevents connection between the two devices whose lines intersect
at that crosspoint.

The crosspoints are inefficiently utilized
, even when all of the attached devices are active,
only a small fraction of the crosspoints are engaged.

CHAPTER (9)

PACKET SWITCHING

19.

Explain two shortcomings of circuit
-
switching and the advantages of packet

switc
hing.

Solution:

Two shortcomings of circuit
-
switching

In a typical user/host data connection, much of the time the line is idle. Thus, with data connections,
a circuit
-
switching is inefficient.

In a circuit
-
switching network, the connection provides fo
r trans mission at constant data rate.
Thus, each of the two devices that are connected must transmit and receive at the same data rate as the other;
this limits the utility of the network in interconnecting a variety of host computers and terminals.

-
switching

Line efficiency is greater, as a single node
-
to
-
node link, can be dynamically shared by many packets
over time. The packets are queued up an transmitted as rapidly as possible over the link.

A packet sw3itching netwo
rk can perform data
-
rate conversion. Two stations of different data rates
can exchange packets because each connects to its node at its proper data rate.

When traffic becomes heavy on a packet
-
switching network, packets are still accepted, but delivery
de
lay increases.

Priorities can be used. Thus, if a node has a number of packets queued for transmission

, it can
transmit the higher
-
priority packets first. Because these packets will less delay than lower
-
priority packets.

(20)

Explain briefly the comparison

in timing for circuit and packet switching with diagram.

FIGURE

Event timing for circuit
-
switching and packet switching

User Data

3

3

Propagation
delay

Pro
cessing
delay

Call request
signal

Call accept
signal

Call request
packet

Acknow
-
ledgement
signal

Call accept
packet

Acknow
-
ledgement
packet

pkt 1

pkt 2

pkt
3

pkt 1

pkt 2

pkt 3

pkt 1

pkt 2

pkt 3

pkt 1

pkt 2

pkt 3

pkt 1

pkt 2

pkt 3

pkt 1

pkt 2

pkt 3

In figure , there types of delay are occurred;

(1)

Propagation delay
:

The time it takes a signal to propa
gate from one node to the next. This time is
generally negligible.

(2)

Transmission Time
: The time it takes for a transmitter to send out a block of data.

(3)

Node delay
: The time it takes for a node to perform the necessary processing as it switches data.

For circuit
-
switching, There is a processing delay at each node during the call request but is not
needed the call
-
accept because the connection is already set
-
up.

For Virtual
-
circuit packet switching, the processing delay at each node in both call r
equest and call
accept packet.
Because this packet is queued at each node and must wait its turn for retransmission, although
the route is established.

Datagram packet switching does not require a call setup. It will be faster than the above two
swit
ching techniques in short messages but the processing time may be longer than virtual
-
circuit packets in
long messages.

(21)

-
switching.

Solution:

In adaptive routing, the routing decisions are made change as conditions i
n the network change. The
principle conditions that influence routing decisions are

Failure: When a node or trunk fails, it can no longer be used as a part of a route.

Congestion: When a particular portion of the network is having congestion, it is d
esirable to roué
packets around, rather than through, the area of congestion.

There are several drawbacks in the use of adaptive routing:

The routing decision is more complex, therefore, the processing burden on network nodes increases

ive strategies depend on status information that is collected at one place but used
at another, therefore, the traffic burden on the network increases.

An adaptive strategy may react too quickly, causing congestion
-
producing oscillation; if it reacts too
s
lowly, the strategy will be irrelevant.

Although this strategy has these problems, it is used in widespread because of two reasons:

It can improve performance as seen by the network user.

It can aid in congestion control.

A convenient way to classify adapt
ive routing strategies is on the basis of information source: local,
adjacent nodes, all nodes. Adaptive strategies can be either distributed, or centralized.

In distributed case, each node exchanges delay information with other nodes. Based on incoming
information, a node tries to estimate the delay situation through the network, and applies a least
-
cast routing
algorithm.

In centralized case, each node reports its link delay status to central node, which designs routes based
on this incoming information

and sends the routing information back to the nodes.

(22)
Co
mpare the different communication switching techniques by using tables.

Solution:

Comparison of communication switching techniques

Circuit switching

Datagram packet

switching

Virt
ual
-
circuit packet
switching

Dedicated transmission
path

Continuous transmission
of data

Fast enough for interactive

Messages are not stored

The path is established for
entire conversation

Call setup delay;
negligible transmission
delay

Busy signal if c
alled party
busy

setup; no delay for
established calls

Electromechanical or
computerized switching
nodes

User responsible for
message loss protection

Usually no speed for or
code conversion

Fixed bandwidth
transmission

bits after call
setup

No dedicated path

Transmission of packets

Fast enough
for interactive

Packets may be stored until
delivered

Route established for each
packet

Packet transmission delay

Sender may be notified if
packet not delivered

creases packet
delay:

Small switching nodes

Network may be responsible
for individual packets

Speed and code conversion

Dynamic use of bandwidth

message

No dedicated path

Transmission of packets

Fast enough for interactive

Pack
ets stored until
delivered

Route established for entire
conversation

Call setup
delay; packet
transmission delay

Sender notified of
connection denia
l

setup; increases packet
delay

Small switching nodes

Network may be
responsible

for packet
sequences

Speed and code conversion

Dynamic use of bandwidth

packet

23.
Draw the diagrams of External and Internal virtual circuit and diagram.

Solution:

A

1. 3

1. 2

1. 1

2. 3

2. 2

2. 1

Packet
-
switched
network

C

1. 3

1. 2

1. 1

B

2. 3

2. 2

2. 1

(a
) External vertical circuit.

A
logical connection is setup between two stations. Packets are labeled with a
vertical circuit number and a sequence number. Packets arrive in sequence.

(b)

External datagram.

Each packet is transmitted independently. Packets are

labeled with th
e destination
address and may arrive out of sequence.

(c
)Internal virtual circuit:

A route for packets between two stations is defined and labeled. All packet for the
virtual circuit follow the same route and arrive in sequence.

(d
)

Int
ernal

datagram:

Each packet is treated independly by the network. Packets are labeled with the
destination address and may arrive at the destination node out of sequence.

FIGURE

External and Internal virtual circuits and diagram.

24. Describe cont
rol mechanisms for congestion control in packet
-
switched network.

Solution:

Control mechanisms for congestion control

1.

Send a control packet from a congested node to source or all source nodes. This choke p
acket will
have the effect of stopping or slowing the rate of transmission from source and limit the total
number of packets in the network. This approach requires additional

traffic on the network during
a period of congestion.

2.

Rely on routing information
, Routing Algorithms, such as ARPANETs, provide link delay
information to other nodes, which influences routing decisions. This information could also be
used to

influence the rate at which new packets are produced. Because these delays are being
influence
d by the routing decision, they may vary too rapidly to be used effectively for congestion
control.

A

B. 3

B. 2

B. 1

C. 3

C. 2

C.

1

Packet
-
switched
network

C

B. 1

B. 3

B. 2

B

C
.
2

C
.
1

C
.
3

A

C

B

VC # 1

1

2

3

4

5

6

VC # 2

A

C

B

1

2

3

4

5

6

3

3

2

1

2

1

3.

Make use of an end
-
to
-
end probe packet. Such a packet could be time
-
stamped to

measure the
delay between two particular endpoints. This procedure has the di
to the network.

4.

Allow packet
-
switching nodes to add congestion inf
ormation to packets.

There are two possible approaches:

-
A node could add such information to packets going in the direction opposite of the
congestion.This inf
ormation quickly reaches the source node, which can reduce the reduce the
flow of packet into the network.

-
A nod
e could add such information to packets going in the same direction as the congestion. The
d or returns the signal back to the source in the
packets (or acknowledgements) going in the reverse direction.

25.

Draw X.25 packet format.

Solutions:

Q

1

0

Group #

Channel #

P(R)

M

P(S)

0

(a)

Data packet with 3bit
sequence numbers

0

0

0

1

Group #

Channel #

Packet type

1

(b)

Control packet for virtual
call with

3bit sequence numbers

0

0

0

1

Group #

Channel #

P ( R )
Packet type

1

( c )

RR, RNR, and REJ packets with 3bit sequence numbers

Q

D

1

0

Group #

Channel #

P(S)

0

P(R)

M

(d) Data packet with 7bit

sequence numbers

User Data

User Data

0

0

1

0

Group #

Channel #

Packet type

1

( e)

Control packet for virtual cells with 7bit sequence numbers

0

0

1

0

Group #

Channel #

Packet Type

1

P(R)

0

(f)

RR, RNQ and REJ packe
ts with 7bit sequence numbers

L
EGEND

P(S) = send sequence number

FIGURE
: X.25 Packet Formats

26.

Explain fixed
R
outing
and build a centralized routing directory for a given packet
-
switched networks.

2

5

3
4

6

1

1
4

2

3

8

8

1

3

2

3

1

7

4

2

1

5

2

1

3

4

5

6

FIGURE

Example packed
-
switched network

Fixed Routing

For Fixed Routing, a route is selected for each source
-
destination pair of nodes in the network. The
routes are fixed, with the exception that they might change if there is movement in the topology of the
n
etwork. Thus, the link costs used in designing routes could be based on expected traffic or capacity.

Figure shows how fixed routing might be implemented. A certain routing matrix is created at a
network control center. The matrix shows, for each sourc
e
-
destination pair of nodes, the identity of the next
node on the route.

At each node along a route, it is only necessary to know the identity of the next node, not the entire
route. Ea

With fixed routing, there is no difference between routing for data
-
grams and virtual circuits.

Advantage: It is simplicity and it work well in a reliable network with a stable

Disadvantage: It has lack of flexibility and it does not react to network

congestion or failures.

CENTRAL ROUTING

DIRETORY

From Node

-

1

5

1

4

5

2

-

2

2

4

5

4

3

-

5

3

5

4

4

5

-

4

5

4

4

5

5

-

5

4

4

5

5

6

-

Node 1 Directory

Node 2 Directory

Node 3 Directory

Destination

Next Node

Destination

Next Node

Destination

Next Node

2

2

1

1

1

5

3

4

2

3

2

2

4

4

4

4

4

5

5

4

5

4

5

5

6

4

6

4

6

5

Node 1 Directory

Node 2 Directory

Node 3 Directory

Destination

Next Node

Destination

Next Node

Destination

Next Node

1

1

1

4

1

5

2

2

2

4

2

5

3

5

3

3

3

5

5

5

4

4

4

5

6

5

6

6

5

5

27.

Explain the flooding technique of packet
-
switching.

Solution:

Flooding

1 2 3 4 5 6

1

2

3

4

5

6

To Node

Flooding require no network information. A packet is send by a source node to every one of its
neighbors. At each node, an inco
ming packet is retransmitted on all outgoing links expect for the link on
which it arri
ved. The packet must have some u
nique identifier so that the node which is equal to the

To stop the incessa
nt retransmission of packet, each node remembers the identity of those packets
it has already retransmitted. When duplicate copies of the packet arrive, they are discarded. A simpler
technique is to include a hop count field with each packet. The count can

originally be set to some maximum
value, such as the diameter of the network. Each time a node passes on a packet, it decrements the count by
one. When the count reaches zero, the packet is discarded.

The flooding technique has three remarkable properties
:

All possible routes between source and destination are tried. Thus, no matter what link or
node outages have occurred, a packet will always get through of, at least one path between
source and destination exists.

Because all routes are tried, at least on
e copy of the packet to arrive at the destination will
have used a minimum
-
hop route.

All nodes that are directly or indirectly connected to the source node are visited.

As the first property, this technique is highly
robust

and could be used to send emerg
ency
messages. The second property, it might be used to initially set up the route
for a virtual circuit. The third
property, it can be useful for the dissemination of important information to all nodes.

The principal disadvantage of flooding is the

high traffic load that is directly proportional to the
connectively of the network.

28.

Using diagram, show the effect of packet size on transmission time.

Y

(a ) 1 packet message

(b ) 2 packet message

(c ) 5 packet message

Data

Data

Data

Data

1

Data

2

Data

1

Data

2

Data

1

Data

2

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

X

a

b

Y

X

a

b

Y

X

a

b

( d ) 10
-

packet messag
e

Figure : Effect of packet size on transmission time

1

2

3

4

5

6

7

8

9

10

1

2

3

4

5

6

7

8

9

10

1

2

3

4

5

6

7

8

9

10

X

a

b

Y

29.

Draw the diagram of sequence of events: X.25 protocol.

30.

Draw an

Call Request

Call Connected

Incoming Call

Call Accepted

Data R=0 , S=0

Data R=2, S=0

Data R= 0, S =2

Data R=2, S =0

Data R=1, S=3

Data R=2 , S=4

Data R=5, S=1

Clear Request

Data
R=0, S=0

Data R=0 , S =1

Data R=0 , S =1

Data R=1, S =3

Network

User Network

Interface

User System B

User Network

Interface

User System A

Data R=0, S=2

Data R=5, S =1

Clear Comfirmation

Clear Indication

Clear Confirmation

Data R=1 , S

=4

A initiates
clearing

Of the virtual
call

When A is
informed

That the call is
connected , it can
begin to sent data
packet

A initiate a virtual
call to B

B accepts the call

The packets are
delivered in sequence

B has no data packet

With which to
acknowledege
packets S= 2 . it
se
nds a control
packet

Figure

Sequence Of events : X .25 protocol

30.Draw an
d explain X.25 Interface.

Solution:

User

Process

Pack
et

Line

Access

Physical

DEF

DCE

FIGURE
:

X.25 interface

X
-
25 specif
ies an interface between a host system and a packet
-
switching network. This
standard is almost univer
sally used for interfacing to packet
-
switching and is employed in ISDN.

The standard specifically calls for three layers of functionality

Physical layer

Packet layer

The physical layer deals with the

physical interface between an attached station
(computer,
terminal) and the link that attaches that station to the packet
-
switching
node. The

standard refers to user
machines as data terminal equip
ment (DTE) and to a packet
-
switching node to with a DTE is attached as
data circuit
-
terminating equipment (DCE).

The link layer provides for the reliab
le transfer of data.
transmitting the data as a sequence of frames. The lin
k layer standard is referred to as LAPB ( Link Access
Protocol Balanced ).

The packet layer provides an external
-
virtual
-
circuit service.

Packet

Access

Physical

X.21 physical interface

To remote user process

Multi
-
channel logical interface

CHAPTER (10)

FRAME RELAY

31.

Describe the key difference between Frame relaying and X.25 pa
cket
-
switching service and draw

the
operation of these network
.

Solution:

The key different between frame relaying and X.25 packet switching service

Call control signaling is carried on a separate logical connection from user data. Thus, intermediate
no
des need not maintain state tables or process messages relating to call control on an individual per
-
connection basis

Multiplexing and switching of logical connections take place at layer 2 instead of layer 3, eliminating
one entire layer of processing.

Th
ere is no hop
-
by
-
hop flow control and error control. End
-
to
-
end flow control and error control, if
they are employed at all, are the responsibility of a higher layer.

(a) Packet
-
switching network

(b) Frame relay network

Figure

Packet switching versus frame relay: source sending, destination responding

32.

Explain Frame Relay Protocol Architecture.

Solution:

Frame relay protocol architecture

Two separate planes of operation are needed in the frame
-
mode bearer servi
ce.

Control

Plane
(C
)

The control plane for frame
-
mode bearer services is similar to that for common
-
channel signaling in
circuit
-
switching services, in that a separate logical channel is used for control information. In the case
of ISDN, control signal
ing is done over the D
-
channel, to control the establishment and termination of
frame
-
mode virtual calls on the D, B and H channels.

User Plane (U)

For the actual transfer of information between end users, the user
-
plane

Access P
rocedure for Frame
-
Mode Barrier Service,

Only the core functions of LAPP are used for frame relay:

Frame delimiting, alignment, and transparency.

Frame multiplexing, demultiplexing using the address field.

Source

Destination

Source

Destination

Intermediate node

Intermediate node

16

1

2

15

9

8

7

10

14

3

4

13

12

5

6

11

1

8

5

4

2

7

3

6

Inspection of the frame to ensure that it consist
s of an integral number of octets prior to zero
-
bit
extraction.

Inspection of the frame to ensure that it is neither

too long not too short.

Detection of transmission errors

Congestion control functions

Control Plane User Plane

Q
.931/Q.933

User
-
selectable TE
function

LAPD (Q.922)

LAPF core
(Q.922)

User (TE)

User Plane

Control Plane

Q.931/Q.933

LAPF control

(Q.922)

LAPD

(Q.921)

LAPF core

(Q.922)

Network (NT)

FIGURE

User
-
network interface protocol architecture.

33.

Discuss

Solution:

User Data Transfer

The operation of frame relay for user data transfer is best explained by the frame format, shown in
figure. This is the format defined for the minimum
-
function LAPF protocol (known as LAPF co
re
protocol).
There is no control field. This has following implications:

There is only one frame type, used for carrying user data. There are no control frames.

It is not possible to use in

band signaling: a logical connection can only carry user data.

It

is not possible to use perform flow control and error control, as there are no sequence number.

The flag and frame check sequence (FCS)
field function as in LAPD and LAPB. The information
field carries higher
-
layer data. If the user selects to imp
-
to
-
end, then a data link frame can be carried in this field. A common selection will be use the full LAPF
protocol (known as LAPF control protocol) in order to perform functions above the LAPF core function
s.

The address field has a default length of 2 octets and may be extended to 3 or 4 octets. It carries a
data link connection identifier (DLC1) of 10, 17, or 24 bits. DLC1 serve the same function as the virtual
circuit number in X.25. It allows mul
tiple logical frame relay connections to be multiplexed over a single
channel.

The length of address field is determined by the address field extension (EA) bits. The CIR bit is
application specific and is not used by standard frame relay protocol
. The remaining bits have to do with
congestion control.

Flag

Information

FCS

Flag

1

2.4

var
iab
le 2 1

(a)
Frame Format

8 7 6 5 4 3 2 1

Upper DLC1

CIR

EAO

Lower DLC1

FCEN

BECN

DE

EA 1

8

7 6 5 4 3 2 1

Upper DLC1

C
/
R

EA 0

DLC1

FECN

BECN

DE

EA 0

Lower DLC1 or DL
-
core control

D/
C

EA1

-
3
octets

8 7 6

5 4 3 2 1

Upper DLC1

C/
R

EA 0

0

DLC1

FECN

BECN

DE

EA 0

DL
C1

EA 0

Lower DL
C1 or DL
-
core control

D/
C

EA 1

(d
-
4

octets

L
EGEND

EA = Address field extension bit

C/R = Command/Response bit

D/C = DL
C1 or CORE control indicator

FIGURE
:

LAPF
-

core format.

34.

Draw the effects of congestion control and define the objectives for Frame relay congestion control.

Solution:

`

( a ) Through put

( b) Dela
y

Figure: The effects of congestion

Network Throughput

A

B

A

B

No

congestion

Mild

congestion

Serve

congestion

Serve

congestion

Mild

congestion

No

congestion

The objectives for frame relay congestion control

Maintain, with high probability and minimum variance, an agreed quality of service

Minimize the possible that one end user can monopolize network r
esources at the expense
of
other end users

Be simple to implement, and place little overhead on either end user or network

Distribute network resources fairly among end users

Limit spread of congestion to other net
works and elements within the network

Operate effectively regardless of the traffic flow in either direction between end users

Have minimum intera
ction or impact on other systems in the frame relaying network

Minimize the variance in quality of service del
ivered to individual frame relay connections
during congestion.

35.

Why the congestion
procedures are

used? Explain two bits in the address Field are used

Solution:

Congestion avoidance procedures are used
at the onset of congestion to minimize the effect
on the network.

The two bits are

Notifies the user that congestion
avoidance

procedures should be
initiated where applicable
for traffic in the opposite di
transmitted by the user on this logical connection may encounter congested resource.

Notifies the user that congestion
avoidance

pr
ocedures should be initiated where applicable
for traffic in the same direction of the
this logical connection, has encountered congested resource.

36.

Draw the diagram of frame handler operation

and explain.

Figure:

Frame Handler operation

Frame relay
control point

TE C

TE A

TE B

TE D

TE E

Frame
Handler

DLC1=0

DLC1=0

DLC1=0

DLC1=0

DLC1=0

DLC1=
312

DLC1=
306

DLC1=
306

DLC1=
334

DLC1=
342

D
LC1=
322

Frame Handler Option

Figure shows the operation of a frame handler in which a number of users are
directly connected
to the same frame handler over diffe
rent physical channels. The operation involved relaying a frame through
two or more frame handlers. The decision making is done by a separate module: the frame relay control
point.

Routing is controlled by entries in a connection table based on D
LC1 that map incoming frames
from one channel to another. The frame handler switches a frame from an incoming channel to an outgoing
channel, based on the appropriate entry in the connection table, and translates the DLC1 in the frame before
transmission.
For example, incoming frames from TE B on logical connection 306 are retransmitted to TE D
on 342. Multiple logical connection to TE D are multiplexed over the same physical channel.

All TEs have a logical connection to the frame relay control point w
ith DLC1=0. These
connections are reserved for in
-
channel call control.

37.

Explain the access modes of frame relay by using diagram.

Access modes of Frame Relay

1.

Switched Access:

The user is connected to a switched network, such as ISDN, and the local
exch
ange does not provide the frame
-
handling capability. In this case, switched access must be provided
from the user
’s terminal equipment (TE) to the frame handler elsewhere in the network, this can either be a
demand connection or a semi
-
permanent connection
. In either case, the frame relay service is provided over
a B or H channel.

2.

Integrated Access :
The user is connected to a pure frame
-
relaying network or to a switched
network, in which the local exchange does provide the frame handling capability. In th
is case, the user has

EEE

(a) Switched Access

(b) Integrated Access

LEGEND

TE = Terminal equipment

NE = Network equipment

ET = Exchange Terminal

FH = Frame Handler

Figure
:

Frame Relay Access Modes

ET

ET

FH

NT

TE

Demand (switched) access connection

Semiparmanent access connection

ET

FH

NT

TE

Local ex
change

38.

Explain frame relay connection and access connection for call control.

Solution:

Frame Relay Connection

Frame relay supports multiple connections over a single link which are called data link con
nections,
and each has a unique data link connection identifier (DLC1). Data transfer involves the following
stages:

1.

Establish

a logical connection between two end points, and assign a unique DLC1 to the
connection.

2.

Exchange information in data frames.

Each frame includes a DLC1 field to identify the
connection.

The est
ablishment and release of a logical connection is accomplished by the exchange of messages
over a logical connection. A frame with DLC1 = 0 contains a call control message in the inform
ation
field. Four messages types such as SETUP, CONNECT, RELEASE,

and RELEASE COMPLETE.

SETUP = request message for establish (assign DLC1 with an unused value)

CONNECT = reply message if it accepts the connection (DLC1 value is assigned by accepting

side)

RELEASE
COMPLETE
= req
uest message upon receipt of REL
EASE

message for clear connection.

Access Connection

If the connection is semi
-
permanent, then no call control protocol is required. If the connection

is to be
setup on demand, then th
e user requests such a connection by means of a common channel signaling
protocol between the user and network. First, the calling user must establish a circuit
-
switched connection to
a frame handler that is one of the nodes of the frame relay network with

SETUP, CONNECT, and
CONNECT ACK message,

exchange at the local user network interface and at the interface between the
network and frame handler. The procedures and parameters for this exchange are carried out on the D
channel. The access connection is cr
eated for B channel.

Once the access connection is established, an exchange takes place directly between end user and frame
handling node for each frame mode connection
that is set up. Again, SETUP,

CONNECT, and CONNECT
ACK message are used. The proced
ures and parameters for this exchange are carried out on the same B
channel.

39.

Draw

and

briefly explain the comparison of X.25
and frame relay protocol stacks.

Solution
:

-
User data is transmitted in frames with virtually no processing by the intermed
iate nodes, other
than to check for errors and to route based on connection number.

-

The packed handling functions of
X.25 operate at layer 3 of OSI model. At layer 2, LABB is used.

-

The processing burden on the network for X.25 considerably higher than

for frame relay.

X.25

Packet level

LAPB

Physical

Layer

(a) X.25

LAPF control

LAPF core

Physical layer

(b) Frame relay

Figure

Comparison of X.25 and Frame Relay proto
co
l stacks

Implement by end
system and

network

Implement by end
system and

network

Implement by end
system but not network