Module 3→ARQ Protocols for

error control
Dr.S.Sivakumar /Associate
Professor/ VIT/ Chennai
 Flow control
• Flow control is a set of procedures that tells the sender how
much data it can transmit before it must wait for an
acknowledgment from the receiver.

• The flow of data must not be allowed to overwhelm the

receiver

• Any receiving device has a limited speed at which it can

process incoming data and a limited amount of memory in
which to store incoming data.
 Flow control
• The receiving device must be able to inform the
sending device before those limits are reached and to
request that the transmitting device send fewer frames
or stop temporarily.

• Incoming data must be checked and processed before

they can be used. For this reason, each receiving
device has a block of memory, called a buffer.

• If the buffer begins to fill up, the receiver must be able

to tell the sender to halt transmission
 Error control
• In the data link layer, the term error control refers
primarily to methods of error detection and
retransmission.

• Error control in the data link layer is often implemented

simply: Any time an error is detected in an exchange,
specified frames are retransmitted. This process is called
automatic repeat request (ARQ).

• Error control in the data link layer is based on automatic

repeat request, which is the retransmission of data.
 Automatic repeat request (ARQ)
• Three main types of the ARQ (for noisy
channel in particular) are:
– Stop-and-wait ARQ
– Go-Back-N ARQ
– Selective Repeat ARQ
Note:- Go-Back-N ARQ and Selective Repeat ARQ
are sub categories of Sliding Window ARQ
 Stop and wait ARQ
• Sender
– Sends one data packet at a time.
– Sends next packet only after receiving
acknowledgement for previous.
• Receiver
– Sends acknowledgement after receiving and
consuming of data packet.
 Causes for Data retransmission

i) Data lost
ii) Acknowledgement lost
iii) Delayed Acknowledgement or data
– RTT (restart a time-out timer)
– Timeout
– Time To Live (TTL)
 Stop and Wait ARQ
Sender Receiver
Frame 0
Timer start Ack 1
Timeout Frame 1
Timer start Ack 0
Timeout Frame 0
Ack 1
 Stop and Wait ARQ (Frame lost)

Sender Receiver
Frame 0
Timer start
Ack 1
Timeout Frame 1
Timer start Frame 1 lost
Timeout
Frame 1
Resend Frame 1 Ack 0
 Stop and Wait ARQ (ACK lost)

Sender Receiver
Frame 0
Timer start
Ack 1
Timeout Frame 1
Timer start Discard
Ack 0 Frame 1
Timeout
Frame 1
Ack 0 lost & so Ack 0
Resend Frame 1
 Stop n Wait Protocol (Numbering Frames)
• We need to assign sequence nos. to data frames and
acknowledgement nos. to the ACK frames
• Sequence nos. are 0,1,0,1,0,1,…. The sequence numbers are
based on modulo-2 arithmetic.
• Acknowledgement nos. are 1,0,1,0,1,0,…
• That is, sequence nos. start with 0 and acknowledgement
nos. start with 1
• An ACK no. also defines the sequence no. of next frame to
receive
• ACK number always announces in modulo-2 arithmetic the
sequence number of the next frame expected
Scenario 1: Stop n Wait
Scenario 2: Stop n Wait
 Scenario in Stop-and-Wait ARQ

• Figure 11.11 below shows an example of Stop-and-Wait

ARQ.
• Frame 0 is sent and acknowledged.
• Frame 1 is lost and resent after the time-out. The resent
frame 1 is acknowledged and the timer stops.
• Frame 0 is sent and acknowledged, but the acknowledgment
is lost.
• The sender has no idea if the frame or the acknowledgment
is lost, so after the time-out, it resends frame 0, which is
acknowledged.
Flow diagram
 Problem in Stop-and-Wait ARQ
• Assume that, in a Stop-and-Wait ARQ system, the bandwidth of
the line is 1 Mbps, and 1 bit takes 20 ms to make a round trip.
What is the bandwidth-delay product? If the system data frames
are 1000 bits in length, what is the utilization percentage of the
link?
• Solution:-
The bandwidth-delay product = (1x 106 )(20 x 10-3 )=20,000 bits
The system can send 20,000 bits during the time it takes for the data
to go from the sender to the receiver and then back again. However,
the system sends only 1000 bits. We can say that the link utilization is
only 1000/20,000, or 5 percent.
• For this reason, for a link with a high bandwidth or long delay, the
use of Stop-and-Wait ARQ wastes the capacity of the link.
 Problem in Stop-and-Wait ARQ

What is the utilization percentage of the link in above question,

if we have a protocol that can send up to 15 frames before
stopping and worrying about the acknowledgments?
The bandwidth-delay product is still 20,000 bits. The system can
send up to 15 frames or 15,000 bits during a round trip. This
means the utilization is 15,000/20,000, or 75 percent. Of
course, if there are damaged frames, the utilization percentage is
much less because frames have to be resent.
Note:-There is no pipelining in Stop-and-Wait ARQ because
we need to wait for a frame to reach the destination and be
acknowledged before the next frame can be sent
 Go-Back-N ARQ Protocol
• Pipelining improves the efficiency of the transmission if the number
of bits in transition is large with respect to the bandwidth-delay
product.
• To improve the efficiency of transmission, multiple frames must be
in transition while waiting for acknowledgment. The first is called
Go-Back-N Automatic Repeat Request
• In this protocol, we can send several frames before receiving
acknowledgments; sender keep a copy of these frames until the
acknowledgments arrive.
• In the Go-Back-N Protocol, the sequence numbers are modulo 𝟐𝒎 ,
where m is the size of the sequence number field in bits.
• The sequence numbers range from 0 to 2𝑚 - 1.
• For example, if m is 4, the only sequence numbers are 0 through
15 inclusive. The sequence numbers are
0, 1,2,3,4,5,6, 7,8,9, 10, 11, 12, 13, 14, 15,0, 1,2,3,4,5,6,7,8,9,10, 11, ...
→ In other words, the sequence numbers are modulo 𝟐𝒎
 Go-Back-N Protocol

• In Go-Back-N, the ACK no. is cumulative and defines the seq. no of the
next packet expected
• Send Window: It is an imaginary box covering the sequence nos of the
data packets that can be in transit or can be sent.
• In each window position, some of these sequence nos define the packets
that have been sent. Others define those that can be sent
• The max. size of the window is 2𝑚 - 1
• Let the size be fixed and set to the max. value
Note: Some protocols may have a variable window size
• Let us consider a sliding window of size 7 (m=3) for the Go-Back-N
protocol
 Go back N ARQ (ACK lost)

Timer start 6 5 4 3 2 1 1
2
3
4
1
Timeout 2
Ack lost 3, 4
6 5 5
6 Discarded
6 5 4 3

Retransmit
 Go_back_N ARQ (Frame Received out of order)

Transmitter Receiver
Timer start 6 5 4 3 2 1 1
2
3
4 Discarded
1
Timeout 2
Ack lost 3, 4
6 5 5
6 Discarded
6 5 4 3

Retransmit
 Sliding Window of size 7 (m=3)
Send Window for Go-Back-N ARQ

The window itself is an abstraction; three variables define its size and location at any
time. We call these variables Sf (send window, the first outstanding frame),
Sn (send window, the next frame to be sent), and Ssize (send window, size).
The variable Sf defines the sequence number of the first (oldest) outstanding frame.
The variable Sn holds the sequence number that will be assigned to the next frame to
be sent.
Finally, the variable Ssize defines the size of the window, which is fixed in our protocol.
 Sliding Window of size 15 (m=4)
Send Window before and after sliding

Send window is an abstract concept defining an imaginary box of size 2𝑚 - 1 with

three variables: Sf Sn and Ssize’. Frames 0, I, and 2 are acknowledged, so the
window has slide to the right three slots. Note that the value of Sf is 3 because
frame 3 is now the first outstanding frame.
 Receive window for Go-Back-N ARQ

The sequence numbers to the left of the window belong to the frames already received and
acknowledged; the sequence numbers to the right of this window define the frames that cannot be
received. The receive window also slides, but only one slot at a time.
When a correct frame is received (and a frame is received only one at a time), the window slides.
 Send Window Size

• We can now show why the size of the send window must be less
than 2m ?
• As an example, we choose m =2, which means the size of the
window can be 2m - 1, or 3.
• Fig 11.15 compares a window size of 3 against a window size of 4.
• If the size of the window is 3 (less than 22) and all three
acknowledgments are lost, the frame timer expires and all three
frames are resent. The receiver is now expecting frame 3, not
frame 0, so the duplicate frame is correctly discarded.
• If the size of the window is 4 (equal to 22) and all acknowledgments
are lost, the sender will send a duplicate of frame 0. However, this
time the window of the receiver expects to receive frame 0, so it
accepts frame 0, not as a duplicate, but as the first frame in the
next cycle. This is an error
 Selective Repeat ARQ
Frame 0
0 1 2 3 4 5
Frame 1 0 1 2 3 4 5
0 1 2 3 4 5
ACK2 0 1 2 3 4 5

0 1 2 3 4 5 Frame 2 lost
Frame 3
0 1 2 3 4 5
0 1 2 3 4 5
NACK
Frame2
0 1 2 3 4 5
0 1 2 3 4 5
 Selective-Repeat (SR) ARQ

• Selective-Repeat ARQ →Advantage in this error-control

method is that only the frame which is in error is resent.
• SR ARQ is preferable for Noisy channel and Go-Back-N ARQ
preferred for noiseless channel
• First, the size of the send window is much smaller; it is 2m-1
• Second, the receive window is the same size as the send
window .
• For example, if m = 4, the sequence numbers go from 0 to
15, but the size of the window is just 2m-1 = 8 (wheras size of
send window is 15 in Go-Back-N Protocol since 2m - 1).
Send Window of SR ARQ
 Receive Window of SR ARQ

• The Selective Repeat Protocol allows as many frames as the size of the receive window to arrive
out of order and be kept until there is a set of in-order frames to be delivered to the network
layer.
• Because the sizes of the send window and receive window are the same, all the frames in the
send frame can arrive out of order and be stored until they can be delivered.
• Also note that the receiver never delivers packets out of order to the network layer
 The two well-known implementations of sliding window protocol
are:-
1. Go back N Protocol
2. Selective Repeat Protocol
Go back N Protocol
Go back N protocol is an implementation of a sliding window
protocol.
The features and working of this protocol are explained in the
following points-
Point-01:

In Go back N, sender window size is N and receiver window

size is always 1.

In Go back N,
• Sender window size = N. Example in Go back 10, sender
window size will be 10.
• Receiver window size is always 1 for any value of N.
Point-02:

Go back N uses cumulative acknowledgements.

In Go back N,
• Receiver maintains an acknowledgement timer.
• Each time the receiver receives a new frame, it starts a new
acknowledgement timer.
• After the timer expires, receiver sends the cumulative
acknowledgement for all the frames that are unacknowledged
at that moment.
NOTE
A new acknowledgement timer does not start after the expiry
of old acknowledgement timer. It starts after a new frame is
received.
Point-03:
Go back N may use independent acknowledgements too.

• The above point does not mean that Go back N cannot use
independent acknowledgements.
• Go back N may use independent acknowledgements too if
required.
• The kind of acknowledgement used depends on the expiry of
acknowledgement timer.
Example-
Consider after the expiry of acknowledgement timer, there is
only one frame left to be acknowledged.
• Then, Go back N sends the independent acknowledgement
for that frame.
Point-04:

Go back N does not accept the corrupted frames and silently

discards them.

In Go back N,
• If receiver receives a frame that is corrupted, then it silently
discards that frame.
• The correct frame is retransmitted by the sender after the time
out timer expires.
• Silently discarding a frame means-“Simply rejecting the frame
and not taking any action” (like not sending a NACK to the
sender to send the correct frame)
Point-05:

Go back N does not accept out of order frames and silently

discards them.

In Go back N,
• If receiver receives a frame whose sequence number is not
what the receiver expects, then it silently discards that frame.
• All the following frames are also discarded.
• This is because receiver window size is 1 and therefore
receiver cannot accept out of order frames.
Point-06:

Go back N leads to retransmission of entire window if for any

frame, no ACK is received by the sender.

In Go back N,
• Receiver silently discards the frame if it founds the frame to
be either corrupted or out of order.
• It does not send any acknowledgement for such frame.

• It silently discards the following frames too.

Thus, if for any particular frame, sender does not receive any
acknowledgement, then it understands that along with that
frame, all the following frames must also have been discarded
by the receiver.
• So, sender has to retransmit all the following frames too along
with that particular frame.
• Thus, it leads to the retransmission of entire window.
• That is why, the protocol has been named as “Go back N“.
Point-07:
 Go back N leads to retransmission of lost frames after expiry of time out
timer.

In Go back N,
• Consider a frame being sent to the receiver is lost on the way.
• Then, it is retransmitted only after time out timer expires for that frame at sender’s
side.
Efficiency of Go back N Efficiency of any flow control protocol is given

Efficiency = Sender Window Size (N) in Protocol / (1 + 2a)

Problem-01:
Using stop and wait protocol, the sender wants to transmit
10 data packets to the receiver. Out of these 10 data packets,
every 4th data packet is lost. How many packets will the
sender have to send in total?

Solution
1, 2, 3, 4, 4, 5, 6, 7, 7, 8, 9, 10, 10

• The lost packets are- 4, 7 and 10.

• Thus, sender will have to send 13 data packets in total.

Problem-02:
Station A needs to send a message consisting of 9
packets to station B using a sliding window (window size
3) and Go Back N error control strategy. All packets are
ready and immediately available for transmission.
If every 5th packet that A transmits gets lost (but no ACKs
from B ever get lost), then what is the number of packets
that A will transmit for sending the message to B ?
Solution:

• Total number of packets to be sent = 9

• Go back N is used where N = 3
• Every 5th packet gets lost

Total 16 packet Transmissions.

Go back N strategy: Remember these 2 points below
and try the solution:-

1. The window should slide ahead as we go on receiving

ack and we will miss ack only for frame which is lost i.e
frame 5 according to question

2. We should go back and send frame starting from

frame no which was lost till the end of current window
size.

Suppose if frame 5th was lost we should resent 5 6 7

because currently these 3 frames are in window.

Step-by-step analysis

Step-01: Since sender window size is 3, so sender sends 3 packets

(1, 2, 3)

Total packets sent till now from sender side = 3

Step-02: After receiving the acknowledgement for packet-1, sender
slides its window and sends packet-4.

Total packets sent till now from sender side = 4

Step-03: After receiving the acknowledgement for packet-2, sender
slides its window and sends packet-5.
 Total packets sent till now from sender = 5
Step-04: After receiving the acknowledgement for packet-3, sender
slides its window and sends packet-6.

Total packets sent till now from sender = 6

Step-05: After receiving the acknowledgement for packet-4, sender
slides its window and sends packet-7.

Total packets sent till now from sender side = 7

Step-06:
According to question, every 5th packet gets lost.
• So, packet-5 gets lost and when time out occurs, sender retransmits
packet-5.
• In Go back N, all the following packets are also discarded by the
receiver.
• So, packet-6 and packet-7 are discarded by the receiver and they
are also retransmitted.
• Thus, the entire window (5, 6, and 7) is retransmitted.

So, we have Total packets sent till now from

sender side = 10
Now, the next 5th packet that will be lost will be packet-7. (6, 7, 5, 6,
7)
Step-07: After receiving the acknowledgement for packet-5, sender
slides its window and sends packet-8.

Total packets sent till now from sender side = 11

Step-08: After receiving the acknowledgement for packet-6, sender
slides its window and sends packet-9.

Total packets sent till now from sender side = 12

Step-09: According to question, every 5th packet gets lost.
• So, packet-7 gets lost and when time out occurs, sender
retransmits packet-7 and the following packets namely packets 8
& 9.
• Thus, the entire window (7, 8 and 9) is retransmitted.

So, we have Total packets sent till now from sender

side = 15
Now, the next 5th packet that will be lost will be packet-9. (8, 9, 7, 8,
9)
Step-10: After receiving the acknowledgement for packet-7, sender
slides its window.

Total packets sent till now from sender side = 15

Step-11: After receiving the acknowledgement for packet-8, sender

slides its window.

Total packets sent till now from sender side = 15

Step-12:
According to question, every 5th packet gets lost. So, packet-9 gets
lost and When time out occurs, sender retransmits packet-9.
So, we have

Total packets sent till now from sender side = 16

Finally, all the 9 packets got transmitted which took total 16 number of
transmissions.

Problem-03:

In Go back 4, if every 6th packet that is being transmitted is lost

and if total number of packets to be sent is 10, then how many
transmissions will be required?
 Solution
• Solve in the same way as we have solved Problem-02.
• Total number of transmissions required will be 17.

SELECTIVE REPEAT (SR) ARQ

Problem-04

Answer is: 3405126

Problem-0 5:
A 1 Mbps satellite link connects two ground stations. The
altitude of the satellite is 36,504 km and speed of the signal is 3
x 108 m/sec. What should be the packet size for a channel
utilization of 25% for a satellite link using go-back-127 sliding
window protocol?
Solution
Given
• Bandwidth = 1 Mbps
• Distance = 2 x 36504 km = 73008 km
• Propagation speed = 3 x 108 m/sec
• Efficiency = 25% = 1/4
• Go back N is used where N = 127

Let the packet size be L bits.

Calculating Transmission Delay(Tt)
Transmission delay (Tt)
= Packet size / Bandwidth
= L bits / 1 Mbps
= L μsec

Calculating Propagation Delay(Tp)

Propagation delay (Tp)
= Distance / Speed
= (73008 x 103 m) / (3 x 108 m/sec)
= 24336 x 10-5 sec
= 2,43,360 μsec
Calculating Value of ‘a’
a= Propagation Delay / Transmission Delay
a = Tp / Tt
a = 243360 μsec / L μsec
a = 243360 / L
Calculating Packet Size
Efficiency (η) = N / (1+2a)
Substituting the values, we get-
1/4 = 127 / (1 + 2 x 243360 / L)
1/4 = 127 x L / (L + 486720)
L + 486720 = 508 x L
507 x L = 486720
L = 960 bits = 960/8 bytes

That is the packet size = 960 bits or 120 bytes.

Module 3 Framing: Byte Stuffing
& Bit Stuffing
Dr.S.Sivakumar
Associate Professor Senior
SENSE - VIT Chennai
 FRAMING
• Data transmission in the physical layer means moving bits in the
form of a signal from the source to the destination
• The physical layer provides bit synchronization to ensure that the
sender and receiver use the same bit durations and timing
• The data link layer, on the other hand, needs to pack bits into
frames, so that each frame is distinguishable from another. Our
postal system practices a type of framing.
• The simple act of inserting a letter into an envelope separates one
piece of information from another; the envelope serves as the
delimiter.
• Also each envelope defines the sender and receiver addresses
since the postal system is a many-to-many carrier facility
 FRAMING
• Framing in the data link layer separates a message from one
source to a destination, or from other messages to other
destinations, by adding a sender address and a destination
address.
• The destination address defines where the frame is to go; the
sender address helps the recipient acknowledge the receipt.
• When a message is carried in one very large frame, even a
single-bit error would require the retransmission of the whole
message
• When a message is divided into smaller frames, a single-bit
error affects only that small frame.
 Fixed-Size and Variable-Size Framing
Frames can be of fixed size or variable size.
• In fixed-size framing, there is no need for defining the
boundaries of the frames; the size itself can be used as a
delimiter. An example of this type of framing is the ATM WAN,
which uses frames of fixed size called cells
• Variable-Size Framing is prevalent in LANs. In variable-size
framing, we need a way to define the end of the frame and
the beginning of the next.
• Historically, two approaches were used for this purpose: a
character-oriented approach and a bit-oriented approach.
 Character-Oriented Protocols
• In a character-oriented protocol, data to be carried are 8-bit
characters from a coding system such as ASCII
• The header, which normally carries the source and destination
addresses and other control information, and the trailer, which
carries error detection or error correction redundant bits, are also
multiples of 8 bits.
• To separate one frame from the next, an 8-bit (I-byte) flag is added
at the beginning and the end of a frame. The flag, composed of
protocol-dependent special characters, signals the start or end of a
frame.
• Character-oriented framing was popular when only text was
exchanged by the data link layers. The flag could be selected to be
any character not used for text communication.
 Character-Oriented Protocols
• we send other types of information such as graphs, audio, and
video. Any pattern used for the flag could also be part of the
information. If this happens, the receiver, when it encounters this
pattern in the middle of the data, thinks it has reached the end of
the frame.
• To fix this problem, a byte-stuffing strategy was added to
character-oriented framing. In byte stuffing (or character stuffing),
a special byte is added to the data section of the frame when there
is a character with the same pattern as the flag. The data section is
stuffed with an extra byte
• This byte is usually called the escape character (ESC), which has a
predefined bit pattern. Whenever the receiver encounters the ESC
character, it removes it from the data section and treats the next
character as data, not a delimiting flag
 Byte stuffing
• Byte stuffing by the escape character allows the presence of
the flag in the data section of the frame, but it creates
another problem.
• What happens if the text contains one or more escape
characters followed by a flag? The receiver removes the
escape character, but keeps the flag, which is incorrectly
interpreted as the end of the frame.
• To solve this problem, the escape characters that are part of
the text must also be marked by another escape character.
In other words, if the escape character is part of the text, an
extra one is added to show that the second one is part of the
text.
 FF
Process of wrapping data with certain info before
sending out

A frame typically consists of

▪ Flag: indication for start and end of a frame
▪ Header: source/destination addresses, as well as other control
information
▪ Data from the upper layer
▪ Trailer: error detection/correction code
 CHARACTER / BYTE ORIENTED FRAMING

Flag – Special character that not used for text communication

• When we send other information like graph, audio or video
– special character can also be the part of the information
▪ Byte stuffing strategy
 BYTE STUFFING
Process of adding an extra byte whenever there is an escape or a flag character
in the data
 Bit-Oriented Protocols
Problem with byte oriented framing
Conflict with 16-bit & 32-bit coding have conflict with 8-bit characters
• In a bit-oriented protocol, the data section of a frame is a sequence of bits
to be interpreted by the upper layer as text, graphic, audio, video, and so
on.
• However, in addition to headers (and possible trailers), we still need a
delimiter to separate one frame from the other.
• Most protocols use a special 8-bit pattern flag 01111110 as the delimiter
to define the beginning and the end of the frame
 BIT ORIENTED FRAMING

If a 0 and five consecutive 1s are encountered, an extra bit 0 will

be added regardless of the value of the next bit. This flag can
create the same type of problem we saw in the byte-oriented
protocols.
That is, if the flag pattern appears in the data, we need to somehow inform
the receiver that this is not the end of the frame. We do this by stuffing
1 single bit (instead of I byte) to prevent the pattern from looking like a
flag. The strategy is called bit stuffing
 Bit stuffing
• In bit stuffing, if a 0 and five consecutive I bits are
encountered, an extra 0 is added. This extra stuffed bit is
eventually removed from the data by the receiver.
• Note that the extra bit is added after one 0 followed by five 1s
regardless of the value of the next bit.
• This guarantees that the flag field sequence does not
inadvertently appear in the frame.
• Bit stuffing is the process of adding one extra 0 whenever
five consecutive 1s follow a 0 in the data, so that the
receiver does not mistake the pattern 0111110 for a flag
• Note that even if we have a 0 after five 1s, we still stuff a O.
The 0 will be removed by the receiver
 BIT STUFFING

01111110
 Module 3

HDLC Protocol and PPP

 HDLC
• High-level Data Link Control (HDLC) is a bit-
oriented protocol for communication over
point-to-point and multipoint links.
– Normal Response Mode (NRM)
– Asynchronous Balanced Mode (ABM)
• Normal Response Mode
• A primary station can send commands; a
secondary station can only respond.
• Asynchronous Balanced Mode
The link is point-to-point, and each station can function as
a primary and a secondary.
HDLC defines three types of frames:
HDLC FRAMES
Information frames (I-frames)
Flag Address Control User FCS Flag
Information
Supervisory frames (S-frames)

Flag Address Control FCS Flag

Unnumbered frames (U-frames)

Flag Address Control Management FCS Flag
Information
• Flag field
– It is an 8-bit sequence with the bit pattern 01111110 that identifies both the
beginning and the end of a frame and serves as a synchronization pattern for
the receiver.
• Address field
– The second field of an HDLC frame contains the address of the secondary
station. If a primary station created the frame, it contains a to address. If a
secondary creates the frame, it contains a from address.
• Control field
– The control field is a 1- or 2-byte segment of the frame used for flow and error
control.
• Information field
– The information field contains the user's data from the network layer or
management information
• FCS field
– The frame check sequence (FCS) is the HDLC error detection field. It is
usually a 2 byte or 4 byte standard CRC
 Control Field for I-Frames
• I-frames are designed to carry user data from the network
layer.
• In addition, they can include flow & error control information
(piggybacking)
 Control Field for I-Frames
• I-frames are designed to carry user data from the network
layer.
• The first bit defines the type. If the first bit of the control
field is 0, this means the frame is an I-frame.
• The next 3 bits, called N(S), define the sequence number
of the frame.
• The last 3 bits, called N(R), correspond to the
acknowledgment number when piggybacking is used.
• The P/F field is a single bit with a dual purpose.
• when it is set (bit = 1) and can mean poll or final. It means
poll when the frame is sent by a primary station to a
secondary. It means final when the frame is sent by a
secondary to a primary.
 Control Field for S-Frames
• S-frames do not have information fields. If the first 2
bits of the control field is 10, this means the frame is an S-
frame. The last 3 bits, called N(R), corresponds to the
acknowledgment number (ACK)
• ACK or negative acknowledgment number (NAK)
depending on the type of S-frame.
• The 2 bits called code is used to define the type of
S-frame itself.
➢ Receive ready (RR)→The value of the code subfield is 00,
• This kind of frame acknowledges the receipt of a safe and sound
frame or group of frames. In this case, the value N(R) field defines the
acknowledgment number.
➢ Receive not ready (RNR)→ the value of the code subfield is 10.
• It acknowledges the receipt of a frame or group of frames, and it
announces that the receiver is busy and cannot receive more frames. It
acts as a kind of congestion control mechanism by asking the sender
to slow down.
➢ Reject (REJ) →If the value of the code subfield is 01, it is a REJ
S-frame. This is a NAK frame, but not like the one used for Selective
Repeat ARQ.
➢ Selective reject (SREJ) →If the value of the code subfield is 11, it is
an SREJ S-frame.
• This is a NAK frame used in Selective Repeat ARQ.

• The HDLC Protocol uses the term selective reject instead of

selective repeat
 Control Field for U-Frames
➢ Unnumbered frames are used to exchange session management
and control information between connected devices.
➢ U-frames contain an information field, but one used for system
management information, not user data.
➢ U-frame codes are divided into two sections: a 2-bit prefix
before the P/F bit and a 3-bit suffix after the P/F bit.
➢ Together these 2 segments (5 bits) can be used to create up to
32 different types of U-frames
 Point to Point Protocol (PPP)
• Today, millions of internet users who need to connect their
home computers to the server of an ISP use PPP
• Point - to - Point Protocol (PPP) is a communication protocol
of the data link layer that is used to transmit multiprotocol
data between two directly connected (point-to-point)
computers.
• It is a byte - oriented protocol that is widely used in
broadband communications having heavy loads and high
speeds.
 Components of PPP
• Point - to - Point Protocol is a layered protocol having three sets of
protocols (LCP, AP & NCP) defined to make PPP powerful
• Encapsulation Component − It encapsulates the datagram so that it
can be transmitted over the specified physical layer.
• Link Control Protocol (LCP) − It is responsible for establishing,
configuring, testing, maintaining and terminating links for
transmission. It also imparts negotiation for set up of options and
use of features by the two endpoints of the links.
• Authentication Protocols (APs) − These protocols authenticate
endpoints for use of services.
• Network Control Protocols (NCPs) − These protocols are used for
negotiating the parameters and facilities for the network layer.
 PPP Frame
• PPP is a character-oriented or byte - oriented protocol
where each field of the frame is composed of one or more
bytes.
Flag Address Control Protocol Payload FCS Flag
• In the above PPP frame format, the Flag, Address, Control are
1 byte fields, Protocol is 1-2 bytes, Payload is variable field &
FCS is 2-4 bytes
• Note: The Flag in PPP is a byte that needs to be escaped
whenever it appears in the data section of the frame.
• The escape byte is 01111101, which means that whenever flag
like pattern appears in data, this extra byte is stuffed (byte
stuffing done) to tell the receiver that next byte is not a flag
• The fields of a PPP frame are −
• Flag − 1 byte that marks the beginning and the end of the
frame. The bit pattern of the flag is 01111110.
• Address − 1 byte which is set to 11111111 in case of
broadcast.
• Control − 1 byte set to a constant value of 11000000.
• Protocol − 1 or 2 bytes that define the type of data
contained in the payload field.
• Payload − This carries the data from the network layer. The
maximum length of the payload field is 1500 bytes.
However, this may be negotiated between the endpoints of
communication.
• FCS − It is a 2 byte or 4 bytes frame check sequence for
error detection. The standard code used is CRC (cyclic
redundancy code)
 Services not provided by PPP
• PPP does not provide flow control
• A sender can send several frames one after another with no
concern about overwhelming the receiver
• PPP uses a simple mechanism for error control. A CRC field is
used to detect errors
• If a frame is corrupted it is silently discarded
• Lack of error control & sequence numbering may cause a
packet to be received out of order
• PPP does not provide sophisticated address mechanism to
handle frames in a multi-point configuration
 Muticast and Broadcast protocol
• Data is transported over a network by three simple methods.
– Unicast
• from one source to one destination i.e. One-to-One
– Broadcast
• from one source to all possible destinations i.e. One-to-
All
– Multicast
• from one source to multiple destinations (group of
members) stating an interest in receiving the traffic i.e.
One-to-Many