The document discusses the functions of the data link layer, including framing data, error detection and correction, and flow control. Specifically, it describes how the data link layer: 1. Organizes the physical layer bit stream into frames and applies error detection techniques like checksums. 2. Uses flow control methods like acknowledgments and windowing to prevent fast senders from overwhelming slow receivers and ensure reliable transmission. 3. Provides an interface between the physical layer and network layer, allowing network layer packets to be reliably transmitted over a communication channel.

1. Prepared by:
anil shrestha
Tribhuvan university

2. a)Services Provided to the Network Layer
b)Framing
c)Error Control
d)Flow Control

3. DLL purpose? The goal of the data link layer is to provide reliable, efficient
communication between adjacent machines connected by a single
communication channel. Specifically:
1. Group the physical layer bit stream into units called frames. Note that
frames are nothing more than ``packets'' or ``messages''. By convention,
we'll use the term ``frames'' when discussing DLL packets.
2. Sender checksums the frame and sends checksum together with data. The
checksum allows the receiver to determine when a frame has been damaged
in transit.
3. Receiver recomputes the checksum and compares it with the received value.
If they differ, an error has occurred and the frame is discarded.
4. Perhaps return a positive or negative acknowledgment to the sender. A
positive acknowledgment indicate the frame was received without errors,
while a negative acknowledgment indicates the opposite.
5. Flow control. Prevent a fast sender from overwhelming a slower receiver.
For example, a supercomputer can easily generate data faster than a PC can
consume it.
6. In general, provide service to the network layer. The network layer wants to
be able to send packets to its neighbors without worrying about the details
of getting it there in one piece.

4. Functions of the Data Link Layer:
a)Provide service interface to the network
layer
b)Dealing with transmission errors
c)Regulating data flow
1.Slow receivers not swamped by fast senders

5. Relationship between packets and frames.

6. (a) Virtual communication.
(b) Actual communication.

7. Placement of the data link protocol.

8.  Framing by character count.
A character stream. (a) Without errors. (b) With one error.
Problem: Even if the error is detected, the receiver cannot figure out
where the next frame starts ... its cannot resynchronize.

9. (a) A frame delimited by flag bytes.
(b) Four examples of byte sequences before and after stuffing.
Problem: Too tied to the 8-bit per character format ... UNICODE uses 16-bits/char

10. Frames that need to be send in a bit stream:
FFlalagg
The sender sends the following bit stream:
Flag Esc Flag
The receiver will ignore this flag.
Frames that need to be send in a bit stream:
Esc Flag
The sender sends the following bit stream:
Esc Esc Flag Esc Flag
The receiver will ignore this Esc, and accept the flag. The receiver will ignore this flag.

11. The goal is to have 01111110 as a unique bit
pattern.
Bit stuffing
(a) The original data.
(b) The data as they appear on the line.
(c) The data as they are stored in receiver’s memory after destuffing.

12. a)Error-Correcting Codes
b)Error-Detecting Codes

13.  Include enough redundancy to detect and correct errors.
 To understand errors, consider the following:
 Messages (frames) consist of m data (message) bits
and r redundancy bits, yielding an n = (m+r)-bit codeword.
 Hamming Distance. Given any two codewords, we can
determine how many of the bits differ. Simply exclusive or
(XOR) the two words, and count the number of 1 bits in the
result.
 Significance? If two codewords are d bits apart, d errors are
required to convert one to the other.
 A code's Hamming Distance is defined as the minimum
Hamming Distance between any two of its legal codewords
(from all possible codewords).
 In general, all possible data words are legal. However, by
choosing check bits carefully, the resulting codewords will
have a large Hamming Distance. The larger the Hamming
distance, the better able the code can detect errors.

14. Use of a Hamming code to correct burst errors.

15.  Error-correcting codes are widely used on wireless links that
are noisy.
 However, they generate too large transmission overhead for
reliable links such as copper wire or fiber. Therefore, here
error-detection codes are used.
 When error is detected, the data is retransmitted.
 The goal for error correcting codes it to add redundancy to
the data so that the errors are not only detected but can be at
the same time corrected (without retransmission).
 For error-detecting codes the goal is to only detect the errors
with the minimal transmission overhead. They are based on
polynomial code also known as CRC (Cyclic Redundancy
Check)
 A k-bit frame is regarded as polynomial with coefficients 0
and 1 with terms from xk-1 to x0
 For example: 110001 -> x5 + x4 + x0

16. Polynomial arithmetic is done modulo 2 using the rules of algebraic field theory.
Both addition and subtraction are identical to exclusive OR. For exampe:
10011011 11110000
+11001010 -10100110
-------------- -------------
01010001 01010110
The sender and receiver must agree on a generator polynomial G(x).
G(x) must have the first and last bit equal to 1.
For a given frame, we consider its polynomial M(x) (longer than G(x)).
The checksum is the reminder from the division M(x)*xr / G(x),
where r is the degree of G(x).
Polynomial T(x) obtained as M(x)*xr - checksum
represents the check-summed frame that is divisible by G(x).
An example division is shown on the next page, where the frame is
1101011011 (corresponds to M(x))
and the generator polynomial G(x) = x4 + x + x0 -> 10011.
M(x)*xr -> 11010110110000 (we added 4 zeros at the end)

17. Calculation of the
polynomial code checksum.

18. Upon receiving the check-summed frame, the receiver divides it by G(x):
[T(x) + E(x)] / G(x)
Since T(x) / G(x) is always zero, the result is always E(x) / G(x).
The errors containing G(x) as a factor will slip by, all other errors will be caught.
Single bit errors will be detected:
We have E(x)=xi for a single bit error,
E(x) / G(x) will not be zero, since G(x) must have the first and last bit equal to 1.
All errors consisting of an odd number of inverted bits will be detected
if G(x) is divisible by (x + 1).
E(x) consists of odd number of terms, e.g., x5 + x2 + x0
and therefore, cannot be divisible by (x+1).
Since E(x) has an odd number of terms E(1)=1.
If E(x) = (x + 1) Q(x), then E(1) = (1 + 1) Q(1) = 0, a contradiction.
The polynomial G(x) used in IEEE 802 standard is
x32 + x26 + x23 + x22 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x1 + 1

20.  In parity check, a parity bit is added to every
data unit so that the total number of 1s is even
(or odd for odd-parity).

21.  Suppose the sender wants to send the word world.
In ASCII the five characters are coded as
 1110111 1101111 1110010 1101100 1100100
 The following shows the actual bits sent
 11101110 11011110 11100100 11011000 11001001

26.  The sender follows these steps:
• The unit is divided into k sections, each of n
bits.
• All sections are added using one’s complement
to get the sum.
• The sum is complemented and becomes the
checksum.
• The checksum is sent with the data.

27.  The receiver follows these steps:
• The unit is divided into k sections, each of n
bits.
• All sections are added using one’s complement
to get the sum.
• The sum is complemented.
• If the result is zero, the data are accepted:
otherwise, rejected.

28.  Suppose the following block of 16 bits is to be sent
using a checksum of 8 bits.
 10101001 00111001
 The numbers are added using one’s complement
 10101001
 00111001
------------
Sum 11100010
 Checksum 00011101
 The pattern sent is 10101001 00111001
00011101

29.  Now suppose the receiver receives the pattern
sent in Example 7 and there is no error.
 10101001 00111001 00011101
 When the receiver adds the three sections, it will
get all 1s, which, after complementing, is all 0s and
shows that there is no error.
 10101001
 00111001
 00011101
 Sum 11111111
 Complement 00000000 means that the
pattern is OK.

30.  Ensuring the sending entity does not
overwhelm the receiving entity
 Preventing buffer overflow
 Transmission time
 Time taken to emit all bits into medium
 Propagation time
 Time for a bit to traverse the link

32. a)A Simplex Stop-and-Wait Protocol.
b)Sliding window protocol.
- A One Bit sliding window protocol.
- A protocol using Go Back N.
- A protocol using selective Repeat.

33.  Source transmits frame
 Destination receives frame and replies with
acknowledgement
 Source waits for ACK before sending next
frame
 Destination can stop flow by not send ACK
 Works well for a few large frames

34.  Large block of data may be split into small
frames
 Limited buffer size
 Errors detected sooner (when whole frame received)
 On error, retransmission of smaller frames is needed
 Prevents one station occupying medium for long
periods
 Stop and wait becomes inadequate

36.  Allow multiple frames to be in transit
 Receiver has buffer W long
 Transmitter can send up to W frames without
ACK
 Each frame is numbered
 ACK includes number of next frame expected
 Sequence number bounded by size of field (k)
 Frames are numbered modulo 2k

39.  Receiver can acknowledge frames without
permitting further transmission (Receive Not
Ready)
 Must send a normal acknowledge to resume
 If duplex, use piggybacking
 If no data to send, use acknowledgement frame
 If data but no acknowledgement to send, send last
acknowledgement number again, or have ACK valid
flag (TCP)

40.  Source transmits single frame
 Wait for ACK
 If received frame damaged, discard it
 Transmitter has timeout
 If no ACK within timeout, retransmit
 If ACK damaged,transmitter will not recognize
it
 Transmitter will retransmit
 Receive gets two copies of frame
 Use ACK0 and ACK1

42.  Simple
 Inefficient

43.  Based on sliding window
 If no error, ACK as usual with next frame
expected
 Use window to control number of outstanding
frames
 If error, reply with rejection
 Discard that frame and all future frames until error
frame received correctly
 Transmitter must go back and retransmit that frame
and all subsequent frames

44.  Receiver detects error in frame i
 Receiver sends rejection-i
 Transmitter gets rejection-i
 Transmitter retransmits frame i and all
subsequent

45.  Frame i lost
 Transmitter sends i+1
 Receiver gets frame i+1 out of sequence
 Receiver send reject i
 Transmitter goes back to frame i and
retransmits

46.  Frame i lost and no additional frame sent
 Receiver gets nothing and returns neither
acknowledgement nor rejection
 Transmitter times out and sends
acknowledgement frame with P bit set to 1
 Receiver interprets this as command which it
acknowledges with the number of the next
frame it expects (frame i )
 Transmitter then retransmits frame i

47.  Receiver gets frame i and send
acknowledgement (i+1) which is lost
 Acknowledgements are cumulative, so next
acknowledgement (i+n) may arrive before
transmitter times out on frame i
 If transmitter times out, it sends
acknowledgement with P bit set as before
 This can be repeated a number of times before
a reset procedure is initiated

49.  Also called selective retransmission
 Only rejected frames are retransmitted
 Subsequent frames are accepted by the receiver
and buffered
 Minimizes retransmission
 Receiver must maintain large enough buffer
 More complex login in transmitter