Chapter 3: The Datalink Layer

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Chapter 3 The Datalink Layer

• CS 455/555 Spring 2003

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Topics to be covered

• Design Issues
• Error detection and correction
• Elementary datalink protocols
• Sliding window protocols
• Example datalink protocols
• SLIP, PPP, DL in ATM

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Design Issues

• Functions Provide a well-defined interface to
  the network layer determining the frame
  boundaries in the bit stream delivered by
  physical layer dealing with transmission errors
  regulating the flows (flow control)
• Services provided Unacknowledged connectionless
  service, acknowledged connectionless service
  acknowledged connection-oriented service
• Framing How can different frames be separated in
  the stream of bits received from the physical
  layer? Certainly, it is not wise to assume
  synchronized clocks between the two datalink
  layers (I.e., sender and receiver).
• Methods of separation (1) Character count (2)
  Starting and ending characters (3) Starting and
  ending flags (sequence of bits) (4) Physical
  layer (5) A combination of the techniuqes

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Design Issues (Contd.)

• Error Control Sequence numbers, timers,
  acknowledgements, retransmissions
• Flow control Rules to throttle a sender from
  sending at a rate higher than what the receiver
  can receive.

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Error Detection and Correction

• Error-detection and error-correction codes
• Codeword datacheck bits
• Hamming distance The number of bits by which two
  code words differ (e.g., 1101 and 1011 have a
  H-distance of 2). For 1101 to become 1011, 2
  single-bit errors would have to occur.
• Suppose we use a special coding in which only 01
  and 10 are recognized as legal where 01
  represents data 0, and 10 means data 1.
  Clearly, in this coding, the patterns 00 and 11
  are illegal. In addition, the distance between 01
  and 10 is 2. So the Hamming distance for the
  coding is 2 (this is the minimum hamming distance
  between any two legal codes).

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Error Detection and Correction (Contd)

• To detect d errors, a coding with a hamming
  distance of d1 is needed. So, the earlier 2-bit
  coding can detect 1 error. Is it true? Suppose we
  send 01 and it becomes either 11 or 00, then the
  receiver knows that these are illegal codes. So
  it can detect them. What if there are two bit
  errors in 01? Then it becomes 10 which cannot
  be detected.

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Error Detection and Correction (Contd)

• To correct d errors, we need a coding with 2d1
  distance. So the earlier example, cannot correct
  any errors? Suppose 01 becomes 00 can this be
  corrected? No. Because it could have been a 01
  or a 10.
• Suppose we use a 3-bit coding where only 000
  and 111 are legal codes. Thus, the distance of
  this code is 3. How many bit errors can this
  detect? 2. How many bit errors can this correct?
  1. Example Suppose 000 becomes 010, since
  the distance between 010 and 000 is 1 while
  between 010 and 111 is 2. So it can correct 010
  as 000.
• Parity bit An example coding where 1 extra bit
  is generated to detect errors. This has a
  distance of 2 and hence can detect single-bit
  errors.
• Odd-parity The number of 1s in any legal code
  word are always odd.
• Even-parity The number of 1s in any legal code
  word are always even.
• Example If 11011001 is a legal 8-bit code, is is
  odd or even parity? How about 11000101??

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Error Detection and Correction (Contd)

• Error-detecting codes Generation and detection
  Use of Polynomial codes or Cyclic redundancy code
  or CRC code
• A generator polynomial G(x) is known to both the
  sender and the receiver. Both the high-end and
  low-end bits of the polynomial must be 1. Ex
  G(x) x4x21. This is represented as 10101
  representing the coefficients of each of the
  exponent terms.
• Given a data frame of m bits and a generator of r
  bits, then the message sent will have mr-1 bits
  where r-1 bits are extra bits generated and
  appended to the data frame.
• How to generate the r-1 bits to be appended?

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Error Detection and Correction (Contd)

• Procedure for generating the code bits
• Step 1 Append r-1 0s to the given data frame.
  (r-1 is the degree of the generator polynomial)
• Step 2 Using modulo 2 division, divide the data
  bit stream from step 1 by the generator
  polynomial.
• Step 3 The remainder is the checksum to be
  appended to the original message which is then
  transmitted.
• Procedure for error-detection The receiver
  divides the received bit stream by the generator
  polynomial using Modulo 2 division. If the
  remainder is 0, then there no errors are
  detected.

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Error Detection and Correction (Contd)

• Example Given an 8-bit word 11001100, generate
  checksum using the generator polynomial of 10101.

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Error Detection and Correction (Contd)

• What bit pattern is transmitted? 110011001100
• What happens at the receiver end? Receiver
  divides (Modulo 2 division) 110011001100 by the
  generator polynomial 10101. If the remainder is
  0, then there is no detected error.

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Elementary Data Link Protocols

• Unrestricted Simplex Protocol Assume (I)
  Negligible processing time (ii) Unrestricted
  buffering (iii) No errors in the channel (Fig
  3.9)
• A Simplex Stop-and-Wait Protocol Relax the
  assumption of infinite buffers and negligible
  processing time Sender sends a frame and waits
  until it gets an ACK before sending the next one.

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Elementary Data Link Protocols (Contd.)

• A Simplex Protocol for a Noisy Channel Here, a
  data frame could be in error when received. A
  receiver simply discards frames in error. The
  sender sets a timer, and when it times out,
  resends the same frame. In case, the original Ack
  was lost, a retransmission occurs. To avoid the
  problem of duplicate frames being delivered to
  the network layer at the receiver, a 1-bit
  sequence number is now introduced.

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Elementary Data Link Protocols (Contd.)

• ARQ (Automatic Repeat Request) or PAR (Positive
  Acknowledgement with Retransmission) The last
  simplex protocol with 1-bit sequence number is an
  example of this type where the sender does waits
  for a ve ACK before sending the next.

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Sliding Window Protocols

• Duplex, more than one frame sent in each
  direction
• Piggybacking ACK for the other station is
  piggybacked on the data frame being sent to it.
• Each outbound frame contains a sequence number,
  ranging from 0 to up to a maximum of 2n-1 where n
  denotes the number of bits used for the sequence
  number field. If 3 bits are used, then the seq
  is from 0 to 7.

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Sliding Window Protocols (Contd.)

• Sender maintains a sending window and receiver
  maintains a receiving window.
• Sending window indicates frames sent but no ACK
  received yet
• At the sender, whenever a new packet is received
  from the NW-layer, it is assigned a sequence
  number which is outside the sending window. The
  upper edge of the sending window is then shifted
  up to include it. Similarly, when an ACK is
  received for the frame at the lower end of the
  sending window, the lower end is moved up.
• Receiving window indicates frames that are
  permitted to be accepted at the receiver. A frame
  falling outside the window is discarded. When a
  frame whose seq equal to the lower edge of the
  window is received, it is delivered to the
  network layer and the window is rotated up by one
  unit.

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Sliding Window Protocols (Contd.)

• A One-bit sliding window protocol Here, only
  1-bit is used for the sequence number. So it acts
  like the stop-and-wait protocol.
• Go back n Protocol Let us assume sender window
  size n gt 1 and receiver window size 1. While
  sender may send multiple frames before receiving
  an ACK, the receiver only receives the next
  expected one and send ACK for that one. So when a
  frame that is ahead of what is expected arrives,
  the receiver simply discards it. Similarly, when
  a sender timeout, it retransmits all
  unacknowledged frames (not just the one that
  timed out). It can waste a lot of bandwidth.
• Example of Go back n Suppose a sender has frames
  2,3,,7 outstanding. If frames 2 and 3 were in
  error, but 4 was received correctly at the
  receiver. Since it was expecting 2 and not 4, the
  receiver simply discards 4 and all subsequent
  ones. When the sender times out on 2, it resends
  2,3,7, which are still unacknowledged.

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Go-back-N Example

• Assume 3-bit sequence number
• Station A sends frames 0..7.
• A piggybacked ACK for frame 7 eventually comes to
  A.
• Station A sends another eight frames 0..7.
• Another piggybacked ACK for frame 7 comes to A.
• Is the new ACK for the old set or the new set of
  7?
• Solution Window-size max_seq_number

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Sliding Window Protocols (Contd.)

• Selective repeat To reduce the number of
  retransmissions, here window size of receiver gt
  1. Suppose the receiver window size is 5. If, as
  in previous example, frames 2 and 3 are lost, but
  4,5,6,7 are received correctly, then the receiver
  would store 4,5, and 6 (since it expects 2 and 3,
  it reserves 2 buffers for them) and discards 7.
• The trade-off here is between network bandwidth
  and buffer space at the datalink layer.

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Sliding Window Protocols (Contd.)

• Go back n Given that n-bit sequence numbers are
  used, what is the maximum window size at the
  sender? Answer 2n-1, which is the maximum
  sequence number. That is, if we use 3-bit seq
  (0,1,,7) then, the senders window can at most
  be 7.
• In other words, while the allowed seq range is
  8, at most 7 frames can be outstanding for ACK at
  any time at the sender. It also needs multiple
  timers.

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Sliding Window Protocols (Contd.)

• Selective Repeat Protocol Both sender and
  receiver maintain a window of acceptable sequence
  numbers. Senders window size starts from 0 and
  grows until a predefined maximum window size
  ((MAX_SEQ1)/2). The receivers window is always
  ((MAX_SEQ1)/2).

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Selective Repeat (example)

• Scenario 3-bit sequence number If the window
  size is MAX_SEQ_NUMBER?
• Station A sends frames 06 to station B.
• Station B receives all seven frames and
  cumulatively acknowledges through RR 7.
• Because of a noise burst, RR 7 is lost.
• A times out and retransmits frame 0.
• B has already advanced its receive window to
  accept frames 7,0,1,2,3,4,5. Thus, it assumes
  that frame 7 has been lost and that this is a new
  frame 0 which it accepts.
• Solution Window size (max_seq_number1)/2

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Sliding Window Protocols (Contd.)

• Performance Issues
• Stop-and-wait Flow control Time to receive one
  frame and receive an acknowledgement (TF)
  2Propagation time Transmission time (per
  frame) 2Processing timeTransmission time
  (ACK)
• Utilization U Transmission time (per frame)/
  TF
• If ACK transmission time and processing time are
  assumed to be negligible, then
• UFrame Trans. time/(2propagation time Frame
  Trans. Time)
• If a Prop. Time/ Trans. Time (frame)
• U 1/(2a1)

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Sliding Window Protocols (Contd.)

• With errors P --- Prob. That a single frame is
  in error N is the window size

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Sliding Window Protocols (Contd.)

• Assume frame size 1000 bytes
• Prob of error in a frame 10-2
• Link length 10 km
• Speed of propagation 200 km/msec
• Speed of transmission 1 Mbps
• Number of bits in sequence 3
• Determine the utilization of the channel in the
  case of stop-and-wait, the Go-back-N, and
  selective reject protocols.

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• Transmission delay 8000/106 8 msec
• Propagation delay 10/200 0.1 msec
• a0.1/8 0.0125
• P0.01
• Stop-and-wait U (1-P)/(12a)0.09/1.0250.088
  or 8.8
• Go-back-N N23-17 Ngt(12a) U
  0.99/(1.00025) or 99
• Selective repeat N22 4 Ngt(12a) U 0.99

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Data Link Layer in ATM

• Transmission convergence sublayer (TC) Actually
  part of ATM Physical layer
• Each cell has 5-byte header and 48-byte data
• 5-byte header 4-byte (VCcontrol info)1-byte
  checksum
• 1-byte checksum(called HEC) is computed only on
  the 4-byte header field and uses x8x2x1 as the
  generating polynomial
• OAM cells Operation and maintenance cells used
  by ATM to exchange control information among the
  switches
• Idle cells When using a synchronous transmission
  medium (physical layer), if ATM switch has
  nothing to send, it generates idle cells

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Data Link Layer in ATM

• At the sending end, TC sublayer takes a sequence
  of cells, adds a HEC to each one, converts the
  result to a bit stream, and matches the bit
  stream to the speed of the underlying physical
  medium.
• At the receiving end, the TC layer takes the
  incoming bit stream, locates the cell boundaries,
  verifies the headers, discards invalid cells,
  processes OAM cells, and passes the data cells to
  the ATM layer.

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Data Link Layer in ATM

• How does TC sublayer recognize cell boundaries in
  a bit stream?
• It uses the HEC.

Cell-by-cell check
Correct HEC detected
Bit-by-bit check
HUNT
PRESYNCH
Incorrect HEC detected
? Consecutive correct HECs
SYNCH
? Consecutive incorrect HECs