CS 352 Internet Technology Data Link Layer

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CS 352Internet TechnologyData Link Layer

• Richard Martin
• Dept. of Computer Science
• Rutgers University

2
Data Link Layer
3
Data Link Layer
Host A
Host B
Application Protocol
Application Layer
Application Layer
Presentation Protocol
Presentation Layer
Presentation Layer
Session Protocol
Session Layer
Session Layer
Transport Protocol
Transport Layer
Transport Layer
Network Layer
Network Layer
Network Layer
Network Layer
Data Link Layer
Data Link Layer
Data Link Layer
Data Link Layer
Physical Layer
Physical Layer
Physical Layer
Physical Layer
Router
Router
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Data Link Layer Functionality

• Recall
• Frame creation
• Error detection and/or correction
• Flow control

Creating the illusion of a reliable link
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Modes of Operation

• Simplex
• Half Duplex
• Full Duplex

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Simplex Transmission

• Data one way only

Data Link
One-Way Street
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Half Duplex Transmission

• Data One direction at a time

Sometimes
Data Link
Sometimes
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Full Duplex Transmission

• Data Both ways simultaneously

Forward Direction
Data Link
Reverse Direction
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Framing

• The data unit at the data link layer is called a
  frame
• A frame is a group of bits, typically in sequence
• Issues
• Frame creation
• Frame delineation

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3.1 Frame Creation
Network Layer
one-to-one mapping
Data Link Layer
frame boundaries
control information
network layer data unit
11
Frame Creation (contd)
BAD
Network Layer
Data Link Layer
What would happen If the network address is split
between 2 frames?
12
Frame Delineation

• How to tell when a new frame starts
• Character count
• Frame tags with character stuffing
• Frame tags with bit stuffing

13
Delineation by character count
Control Field containing character count

• Character count lists the number of characters in
  the data field of the frame
• Problem corrupted control fields

14
Frame tagging with character stuffing
Frame tags (special characters)

• Use starting and ending characters (tags) to mark
  boundaries of frame
• Problem What if tag characters occur in the data
  or control portions of the frame?

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Character stuffing

• Insert extra escape characters when a tag appears
  in data field

STX
DLE
Start Tag
ETX
DLE
End Tag
DLE
DLE
Character Stuffed DLE code
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Character Stuffing Example
DLE
DLE
ETX
I am a
jerk trying to
crash your network!
Tagging and Character Stuffing
Untagging and Character Unstuffing
DLE
DLE
ETX
I am a
jerk trying to
crash your network!
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Character StuffingPPP

• PPP Point-to-Point Protocol, used widely in the
  Internet
• PPP uses frame tags with character stuffing

PPP Frame Format
01111110
Control Fields
Payload (up to 1500 bytes)
Checksum
01111110
Character Frame Tags
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Frame tagging with bit stuffing

• Bit strings may be used instead of character
  sequences to delineate frames
• More efficient

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Bit stuffing

• Each frame begins with a start and end bit
  sequence, e.g., 01111110
• When senders data link layer sees five 1s in a
  row, it stuffs a zero bit
• The receiver unstuffs a zero after five
  consecutive 1s.

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Bit Stuffing Example
00110010111111100010100
Tagging and Bit Stuffing
0111111000110010111110110001010001111110
Untagging and Bit Unstuffing
00110010111111100010100
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Bit Stuffing HDLC

• HDLC High-level Data Link Control. Used in
  X.25 networks
• HDLC uses frame tags with bit stuffing

HDLC Frame Format
01111110
Control Fields
Payload (arbitrarily long)
Checksum
01111110
Character Frame Tags
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Error Control

• No physical link is perfect
• Bits will be corrupted
• turn on the microwave
• We can either
• detect errors and request retransmission
• or correct errors without retransmission

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Error Detection

• Parity bits
• Polynomial codes or checksums

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Parity bits

• Append a single parity bit to a sequence of bits
• If using odd parity, the parity bit is chosen
  to make the total number of 1s in the bit
  sequence odd.
• If even parity, the parity bit makes the total
  number of 1s in the bit sequence even

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Parity bits Example
Transmitted Sequence
Parity
Parity Bit
00010101
even
1
01111
even
0
11111111
odd
1
0
10011
odd
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Parity Bits

• Only detects when there are an odd number of bit
  errors
• Does not detect an even number of bit errors
• It also has high overhead it requires 1 extra
  bit for every several bits transmitted

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Polynomial Codes

• Can detect errors on large chunks of data
• Has low overhead
• More robust than parity bit
• Requires the use of a code polynomial
• Example x2 1

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Poly codes

• Data is viewed a single long number,S(x)
• Divide S(X) by an agreed upon divisor, G(X)
• S(x)/G(x) C(x) remainder, R(x)
• Transmit T(x) S(x) R(x)
• Check if ( (T(x)/G(x) ) ! 0) then error!

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Cyclic Redundancy Check

• CRC Example of a polynomial code
• Procedure
• 1. Let r be the degree of the code polynomial.
  Append r zero bits to the end of the transmitted
  bit string. Call the entire bit string S(x)
• 2. Divide S(x) by the code polynomial using
  modulo 2 division.
• 3. Subtract the remainder from S(x) using modulo
  2 subtraction.
• The result is the checksummed message
• Easy to build in hardware

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Why polynomial?

• Treat numbers as polynomials, not as integers,
  Why?
• Polynomial arithmetic has no carries
• Arithmetic same a XOR gate
• 0-0 0
• 1-0 1
• 0-1 1 (wraparound)
• 1-1 0

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Generating a CRCExample
Message 1011 1 x x3 0 x x2 1 x x 1
x3 x 1 Code Polynomial x2 1
(101)
Step 1 Compute S(x) r 2 S(x) 101100
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Generating a CRCExample (contd)
Step 2 Modulo 2 divide
Remainder
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Generating a CRCExample (contd)
Step 3 Modulo 2 subtract the remainder from S(x)
101100
- 01
101101
Checksummed Message
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Decoding a CRC

• Procedure
• 1. Let n be the length of the checksummed message
  in bits
• 2. Divide the checksummed message by the code
  polynomial using modulo 2 division. If the
  remainer is zero, there is no error detected.

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Decoding a CRCExample
101101
Checksummed message (n 6)
1011
Original message (if there are
no errors)
Remainder 0 (No error detected)
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Decoding a CRCAnother Example

• When a bit error occurs, there is a large
  probability that it will produce a polynomial
  that is not an even multiple of the code
  polynomial, and thus errors can usually be
  detected.

Remainder 1 (Error detected)
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Choosing a CRC polynomial

• The longer the polynomial, the smaller the
  probability of undetected error
• Common standard polynomials
• (1) CRC-12 x12 x11 x3 x2 x1 1
• (2) CRC-16 x16 x15 x2 1
• (3) CRC-CCITT x16 x12 x5 1

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Putting it all together
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Error Correction

• Parity bits and polynomial codes catch errors,
  but can we correct them without retransmitting
  information?
• Yes, using Hamming Codes
• Example of a Forward Error Correction (FEC) Scheme

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Hamming Codes

• Hamming codes, like polynomial codes, are
  appended to the transmitted message
• Hamming codes, unlike polynomial codes, contain
  the information necessary to locate a single bit
  error

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Calculating a Hamming Code

• Procedure
• Place message bits in their non-power-of-two
  Hamming positions
• Build a table listing the binary representation
  of each of the message bit positions
• Calculate the check bits

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Hamming Code Example
1 0 1 1
Message to be sent
1 0 1 1 1 2 3 4 5 6
7 20 21 22
Position
2n check bits
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Hamming Code Example
1 0 1 1
Message to be sent
1 0 1 1 1 2 3 4 5
6 7 20 21 22
Position
2n check bits
Calculate check bits
3 21 20 0 1 1 5 22 20
1 0 1 6 22 21 1 1 0 7
22 21 20 1 1 1
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Hamming Code Example
1 0 1 1
Message to be sent
1 0 1 1 1 2 3 4 5
6 7 20 21 22
1
Starting with the 20 position Look at positions
with 1s in them Count the number of 1s in
the corresponding message bits If even, place a
1 in the 20 check bit, i.e., use odd
parity Otherwise, place a 0
Position
2n check bits
Calculate check bits
3 21 20 0 1 1 5 22 20
1 0 1 6 22 21 1 1 0 7
22 21 20 1 1 1
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Hamming Code Example
1 0 1 1
Message to be sent
1 0 1 1 1 2 3 4 5
6 7 20 21 22
1
0
Repeat with the 21 position Look at positions
those positions with 1s in them Count the
number of 1s in the corresponding message
bits If even, place a 1 in the 21 check
bit Otherwise, place a 0
Position
2n check bits
Calculate check bits
3 21 20 0 1 1 5 22 20
1 0 1 6 22 21 1 1 0 7
22 21 20 1 1 1
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Hamming Code Example
1 0 1 1
Message to be sent
1 0 1 1 1 2 3 4 5
6 7 20 21 22
1
0
1
Repeat with the 22 position Look at positions
those positions with 1s in them Count the
number of 1s in the corresponding message
bits If even, place a 1 in the 22 check
bit Otherwise, place a 0
Position
2n check bits
Calculate check bits
3 21 20 0 1 1 5 22 20
1 0 1 6 22 21 1 1 0 7
22 21 20 1 1 1
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Hamming Code Example

• Original message 1011
• Sent message 1011011
• Now, how do we check for a single-bit error in
  the sent message using the Hamming code?

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Using Hamming Codes to Correct Single-Bit Errors
1 0 1 1 0 0 1
Received message
1 0 1 1 0 0 1 1 2 3 4 5 6
7 20 21 22
Position
2n check bits
Calculate check bits
3 21 20 0 1 1 5 22 20
1 0 1 6 22 21 1 1 0 7
22 21 20 1 1 1
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Using Hamming Codes to Correct Single-Bit Errors
1 0 1 1 0 0 1
Received message
1 0 1 1 0 0 1 1 2 3 4 5 6
7 20 21 22
Starting with the 20 position Look at positions
with 1s in them Count the number of 1s in
both the corresponding message bits and the 20
check bit and compute the parity. If even
parity, there is an error in one of the four bits
that were checked.
Position
2n check bits
Calculate check bits
3 21 20 0 1 1 5 22 20
1 0 1 6 22 21 1 1 0 7
22 21 20 1 1 1
Odd parity No error in bits 1, 3, 5, 7
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Using Hamming Codes to Correct Single-Bit Errors
1 0 1 1 0 0 1
Received message
1 0 1 1 0 0 1 1 2 3 4 5 6
7 20 21 22
Repeat with the 21 position Look at positions
with 1s in them Count the number of 1s in
both the corresponding message bits and the 21
check bit and compute the parity. If even
parity, there is an error in one of the four bits
that were checked.
Position
2n check bits
Calculate check bits
3 21 20 0 1 1 5 22 20
1 0 1 6 22 21 1 1 0 7
22 21 20 1 1 1
Even parity ERROR in bit 2, 3, 6 or 7!
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Using Hamming Codes to Correct Single-Bit Errors
1 0 1 1 0 0 1
Received message
1 0 1 1 0 0 1 1 2 3 4 5 6
7 20 21 22
Repeat with the 22 position Look at positions
with 1s in them Count the number of 1s in
both the corresponding message bits and the 22
check bit and compute the parity. If even
parity, there is an error in one of the four bits
that were checked.
Position
2n check bits
Calculate check bits
3 21 20 0 1 1 5 22 20
1 0 1 6 22 21 1 1 0 7
22 21 20 1 1 1
Even parity ERROR in bit 4, 5, 6 or 7!
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Finding the errors location
1 0 1 1 0 0 1 1 2 3 4 5 6
7
Position
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Finding the errors location
1 0 1 1 0 0 1 1 2 3 4 5 6
7
Position
No error in bits 1, 3, 5, 7
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Finding the errors location
erroneous bit, change to 1
1 0 1 1 0 0 1 1 2 3 4 5 6
7
Position
Error must be in bit 6 because bits 3, 5, 7 are
correct, and all the remaining information agrees
on bit 6
ERROR in bit 2, 3, 6 or 7
ERROR in bit 4, 5, 6 or 7
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Finding the errors locationAn Easier
Alternative to the Last Slide
Check bits
22 21 20
3 21 20 0 1 1 5 22 20
1 0 1 6 22 21 1 1 0 7
22 21 20 1 1 1
E
E
NE
6
1
1
0
E error in column NE no error in column
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Hamming Codes

• Hamming codes can be used to locate and correct a
  single-bit error
• If more than one bit is in error, then a Hamming
  code cannot correct it
• Hamming codes, like parity bits, are only useful
  on short messages

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Flow Control

• What happens if the sender tries to transmit
  faster than the receiver can accept?
• Data will be lost unless flow control is
  implemented

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Controlling the Flow of Data
Busy Client
Slow Server
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Some Flow Control Algorithms

• Flow control for the ideal network
• Stop and Wait for noiseless channels
• Stop and Wait for noisy channels
• Sliding window protocols
• Sliding window with error control
• Go Back N
• Selective Repeat

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Flow control in the ideal network

• Assumptions
• Error free transmission link,
• Infinite buffer at the receiver

No acknowledgement of frames necessary Since the
data link is error-free and the receiver can
buffer as many frames as it likes, no frame will
ever be lost
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Flow control in the ideal network (contd)
Busy Client
Slow Server
Infinite Box
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5.2 Stop and Wait with Noiseless Channels

• Assumptions
• Error free transmission link,
• Finite buffer at the receiver

• Problem of Buffer overflow at the receiver
• Buffer overflow may happen at the receiver when
  the sender sends frames at a rate faster than the
  receiver

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Stop and Wait with Noiseless Channels (contd)
Busy Client
Slow Server
Finite Box (once full, Letter is lost)
Bit Bucket
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Stop and Wait with Noiseless Channels (contd)

• Solution Stop-and-Wait
• The receiver sends an acknowledgement frame
  telling the sender to transmit the next data
  frame.
• The sender waits for the ACK, and if the ACK
  comes, it transmits the next data frame.

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Stop and Wait with Noiseless Channels (contd)
Data
Data
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Stop and Wait (contd)

• Note that we assume an error-free transmission
  link and therefore ACK frames will not be lost
• In this flow control protocol, there are two
  types of frames data frames and ACK frames. The
  ACK frames dont contain any particular
  information, since only the arrival of the ACK
  frame at the sender is important.

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5.3 Stop and Wait for Noisy Channels

• Assumptions
• Transmission link may cause errors in frames,
• Finite buffer at the receiver

ACK frames may now be lost
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Problems introduced by a noisy line

• Problem 1 Loss of a data or ACK frame
• Since the transmission link is not error-free, a
  data or ACK frame may be lost, causing the sender
  to wait indefinitely for an ACK

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Loss of an ACK frame
Data
?
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Problems introduced by a noisy line

• Can we solve problem 1 by introducing a timeout
  period for the sender? Yes, but...
• Problem 2 Duplicated frames
• If the ACK frame for a certain data frame is
  lost, the sender will retransmit the same frame
  after a time-out period, and the receiver will
  then have two copies of the same frame

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Duplicated Data Frames
Buy 10 Pikachus!
Wheres the ACK?
Buy 10 Pikachus!
Order 20?
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Stop and Wait for Noisy Channels (contd)

• Solution
• The sender uses a timer to retransmit data frames
  when an ACK has not arrived
• The sender includes a sequence number in each
  frame to distinguish one frame from another.
  This way, the receiver knows when it has received
  duplicate frames.

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Stop-and-Wait with Noisy channels
Wheres the ACK?
Data
Duplicate
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Is Stop and Wait the best we can do?

• Stop and Wait is an effective form of flow
  control, but
• Its not very efficient.
• 1. Only one data frame can be in transit on the
  link at a time
• 2. When waiting for an acknowledgement, the
  sender cannot transmit any frames
• Better solution? Sliding Window

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Sliding Window ProtocolsDefinitions

• Sequence Number Each frame is assigned a
  sequence number that is incremented as each frame
  is transmitted
• Sender Window Keeps track of sequence numbers
  for frames that have been sent but not yet
  acknowledged
• Receiver Window Keeps track of sequence numbers
  for frames the receiver is allowed to accept
• Maximum Sender Window size The maximum number
  of frames the sender may transmit without
  receiving any acknowledgements
• Maximum Receiver Window size The maximum number
  of frames the receiver may receive before
  returning an acknowledgement to the sender

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Simple Sliding Window with Window Size of 1

• A sliding window with a maximum window size of 1
  frame
• Window for a 3-bit sequence number

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Sliding Window example
Sender window
Receiver window
(a)
(b)
(c)
(d)
(a) Initial state, no frames transmitted (b)
Sender transmits frame 0 (c) Receiver receives
frame 0 and ACKs (d) Sender receives ACK
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Simple Sliding Window withWindow size 1 (contd)

• This protocol behaves identically to stop and
  wait for a noisy channel

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Sliding Window ProtocolsGeneral Remarks

• The sending and receiving windows do not
  necessarily have the same maximum size
• Any frame whose sequence number falls outside the
  receiver window is discarded at the receiver
• The sender windows size grows and shrinks as
  frames are transmitted and acknowledged
• Unlike the sender window, the receiver window
  always remains at its maximum size

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Sliding Window ProtocolsPiggybacking
Acknowledgements

• Since we have full duplex transmission, we can
  piggyback an ACK onto the header of an outgoing
  data frame to make better use of the channel

When a data frame arrives at a router, instead of
immediately sending a separate ACK frame, the
router waits until it is passed the next data
frame to send. The acknowledgement is attached
to the outgoing data frame.
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Sliding Window with Maximum Sender Window Size WS

• With a maximum window size of 1, the sender waits
  for an ACK before sending another frame
• With a maximum window size of WS, the sender can
  transmit up to WS frames before being blocked
• This allows the sender to transmit several frames
  before waiting for an acknowledgement

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Sender-Side Window with WS2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(a) Initial window state (b) Send frame 0 (c)
Send frame 1 (d) ACK for frame 0 arrives
(e) Send frame 2 (f ) ACK for frame 1 arrives (g)
ACK for frame 2 arrives, send frame 3 (h) ACK for
frame 3 arrives
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Pipelining
packet 5
packet 4
packet 6
Ack 0
Ack 2
Ack 1
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Pipelining

• Sliding window with WS gt 1 is also called
  pipelined communication

data stream
99
51
50
A
B
49
1
0
ACK stream
By allowing several frames onto the link before
receiving an acknowledgement, pipelining keeps
the link from being idle
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5.4.3 Sliding Window with Maximum Receiver
Window Size WR

• With a maximum window size of 1, the receiver
  must receive and process every frame in sequence
• With a maximum window size of WR, the receiver
  can receive and process up to WR frames before
  acknowledging them
• This is useful when frames are lost the receiver
  can still accept and buffer frames after the
  missing frame

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Receiver-Side Window with WR2
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(a) Initial window state (b) Nothing happens (c)
Frame 0 arrives, ACK frame 0 (d) Nothing happens
(e) Frame 1 arrives, ACK frame 1 (f) Frame 2
arrives, ACK frame 2 (g) Nothing happens (h)
Frame 3 arrives, ACK frame 3
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(No Transcript)
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What about Errors?

• What if a data or acknowledgement frame is lost
  when using a sliding window protocol?
• Two Solutions
• Go Back N
• Selective Repeat

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5.5 Sliding Window with Go Back N

• When the receiver notices a missing or erroneous
  frame, it simply discards all frames with greater
  sequence numbers and sends no ACK
• The sender will eventually time out and
  retransmit all the frames in its sending window

90
Go Back N
Timeout interval
Sender
0
1
2
3
4
2
3
4
5
6
Maximum window size 8
ACK 0
ACK 1
ACK 2
ACK 3
ACK 4
ACK 5
ACK 6
Receiver
0
1
E
D
D
2
3
4
5
6
Maximum window size 8
Discarded by receiver
Frame with error
Time
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Go Back N (contd)

• Go Back N can recover from erroneous or missing
  frames
• But
• It is wasteful. If there are errors, the sender
  will spend time retransmitting frames the
  receiver has already seen

92
Sliding Window with Selective Repeat

• The sender retransmits only the frame with errors
• The receiver stores all the correct frames that
  arrive following the bad one. (Note that the
  receiver requires a frame buffer for each
  sequence number in its receiver window.)
• When the receiver notices a skipped sequence
  number, it keeps acknowledging the last good
  sequence number
• When the sender times out waiting for an
  acknowledgement, it just retransmits the one
  unacknowledged frame, not all its successors.

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Selective Repeat
Timeout interval
Sender
0
1
2
3
4
2
5
6
Maximum window size 8
ACK 0
ACK 1
ACK 4
ACK 5
ACK 1
ACK 1
ACK 6
Receiver
0
1
E
2
5
6
3
4
Maximum window size 8
Buffered by receiver
Frame with error
Time
94
Recap
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Layer 2 - Data Link Layer
DATA LINK LAYER
1
1
0
001100
1 0 1 0
00110
111001
101010
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Data Link Layer Design Issues

• A Well Defined Service Interface to the Network
  Layer.
• How are the bits grouped into frames.
• Handling Transmission Errors
• Regulating the flow of Frames to prevent swamping

97
OSI Architecture
98
OSI Operations
Data
Data
Bits
Bits
99
Data Link Layer Design Issues

• Virtual Communication (a) and Actual
  Communication (b)

• the Data Link Layer provides services to the
  Network Layer
• Services Provided to the Network Layer
• Unacknowledged connectionless service
• Acknowledged connectionless service
• Acknowledged connection-oriented service

100

• Unacknowledged connectionless service
• source machine sends independent frames to the
  destination machine
• the destination machine does not ACK
• no connection is established
• the data link layer does not attempt to recover
  lost frames
• appropriate for low error rate connection
• ok for speech data where late data is worse than
  bad data
• any error recovery is left to higher layers
• Acknowledged connectionless service
• source machine sends independent frames to the
  destination machine
• the destination machine does ACK on indivisual
  frame received
• no connection is established
• if ACK is not received for given time limit, the
  source machine resends
• appropriate for unreliable connection like
  wireless systems

101

• the ACK by the data link layer is not a
  requirement but optimization
• the transport layer (who is responsible for
  end-to-end transport of message) can always send
  a message and wait for a ACK on the message ok
  with reliable fiber connection
• message by message checkup has much bigger
  overhead due to the size
• only a few frames of a message can be in error
  but still the whole message is resent
• much more efficient to do it (ACK on frame) in
  the data link layer esp for wireless systems
• Acknowledged connection-oriented service
• connection is established first
• frames sent over the established route are
  numberd
• the data link layer guarantees the safe delivery
  and orderly delivery of each frame

102

• The Data Link Layer in ATM
• the ATM physical layer covers roughly the OSI
  physical layer and data link layers
• the PMD (physical medium dependent) sublayer acts
  like the OSI physical layer
• the TC (Transmission Convergence) sublayer acts
  like the Data Link layer
• There are no physical layer characteristics
  specific to ATM
• Instead, ATM cells are carried by SONET, FDDI,
  and other transmission systems
• Well look at TC sublayer functionality

103
TCP/IP Protocol Stack vs. OSI
User Processes
Kernel
104
Data Link Layer
Interface Data Unit
Service Access Points
ICI
SDU
Service User
Layer N1
Network Layer
Interface
Service Provider
Layer N
ICI
SDU
Data Link Layer
SDU
Header
ICI Interface Control Information
N-Protocol Data Unit
105
Communication Path
Virtual Data Path
106
Breaking Bits into Frames Character Count Method
Character Count
Error Free
5
1
2
3
4
5
6
7
8
9
8
0
1
2
3
4
5
6
Frame 1 5 Characters
Frame 3 5 Characters
Frame 2 5 Characters
107
Breaking Bits into Frames Starting and Ending
Flags with Bit Stuffing
0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 0
Network Layer
Stuffed Bits
108
Error Control
Start
Timers
PACKET
Stop
109
Flow Control
STOP!!!
STOP!!!
Can Happen When Sender overloads Receiver!
110
Error Correcting

• Error Correcting Codes
• Error Detecting Codes

111
Positive Acknowledgement with Retransmission
Events at Sender Site
Network Messages
Events at Receiver Site
Send Packet 1
Receive Packet 1 Send ACK1
Receive ACK1 Send Packet 2
Receive Packet 2 Send ACK2
Receive ACK2
112
Positive Acknowledgement with Retransmission
(Lost Packet)
Events at Sender Site
Network Messages
Events at Receiver Site
Send Packet 1 Start Timer
Packet should arrive ACK should be sent
ACK would normally arrive at this time
Timer Expires
Retransmit Packet 1 Start Timer
Receive Packet 1 Send ACK1
Receive ACK1 Cancel Timer
113
Sequence Numbers

• Initial Sequence Numbers (ISN)
• A ISN generator is employed which is driven by
  a 32 bit
• clock.
• An ISN generator will cycle every 4.55 Hrs
• Each Connection has a Transmit and Receive
  Sequence
• Number
• MSL is 2 Minutes, But can be changed.

114
Sliding Window Mechanism

• The Sliding Window solves two
• important problems for TCP
• Efficient Transmission
• Flow Control

115
Sliding WindowEfficient Transmission
There are three pointers associated with every
connection
1 2 3 4 5 6 7 8 9 10 11
...
1
2
3
1. One marks left of sliding window 2. One
marks right of sliding window 3. One separates
sent from not sent
Each End of the connection maintains two
windows Transmit Receive
116
Sliding WindowFlow Control
There are two independent Flow problems 1.
Need end to end flow control between source
and Destination 2. Need to allow intermediate
systems (I.e. routers) to control a source
that sends more traffic than the machine
can tolerate
Sliding window will solve 1. The other is
congestion control
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Sliding Windows
Events at Sender Site
Network Messages
Events at Receiver Site
Send Packet 1
Receive Packet 1 Send ACK1
Send Packet 2
Receive Packet 2 Send ACK2
Send Packet n
Receive Packet n Send ACKn
Send Packet 8
Receive ACK1
Receive Packet 8 Send ACK8
Send Packet 9
Receive ACK2
Receive ACKn
Receive ACK8