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Link Layer

Link Layer. Dr. Mozafar Bag-Mohammadi Ilam University. Contents. Overview Encoding Framing Error detection Acknowledgements & Timeouts. 1. Overview: Link Layer. Communication service between two physically connected devices: host-router, router-router, host-host

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Link Layer

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  1. Link Layer Dr. Mozafar Bag-Mohammadi Ilam University

  2. Contents • Overview • Encoding • Framing • Error detection • Acknowledgements & Timeouts

  3. 1. Overview: Link Layer

  4. Communication service between two physically connected devices: host-router, router-router, host-host unit of data: frame implemented in“adapter” e.g., PCMCIA card, Ethernet card typically includes: RAM, DSP chips, host bus interface, and link interface M H H H H H H H H H t n l t t t n l n M M application transport network link physical M Link layer network link physical data link protocol M frame phys. link adapter card

  5. Link Layer Services • Framing, link access: • encapsulate datagram into frame, adding header, trailer • implement channel access if shared medium, • ‘physical addresses’ used in frame headers to identify source, dest • different from IP address! • Reliable delivery between two physically connected devices: • seldom used on low bit error link (fiber, some twisted pair) • wireless links: high error rates

  6. Link Layer Services (more) • Flow Control: • pacing between sender and receivers • Error Detection: • errors caused by signal attenuation, noise. • receiver detects presence of errors: • signals sender for retransmission or drops frame • Error Correction: • receiver identifies and corrects bit error(s) without resorting to retransmission

  7. Bits 0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 0 NRZ 2.Encoding • Encode binary data onto signals • Bits flow is between network adaptors. • How to send clock information? • Extract from signals or data. • The simplest way is 0 as low signal and 1 as high signal. This is known as Non-Return to zero (NRZ)

  8. Problem with NRZ • Consecutive 0’s or 1’s may create problems. • Synchronization problem because of difference in the sender or receiver clocks. • The average of signals which is used to distinguish between low or high may move and make the decoding difficult. This is called baseline wander • Unable to recover clock

  9. Alternative Encodings • Non-return to Zero Inverted (NRZI) • make a transition from current signal to encode a 1; stay at current signal to encode a zero • solves the problem of consecutive 1s • Manchester encoding • transmit XOR of the NRZ encoded data and the clock. • 0 is being encoded as a low to high transition and 1 as high to low. • It doubles the rate, then, it is only 50% efficient. • The rate signal changing is called baud rate.

  10. Bits 0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 0 NRZ Clock Manchester NRZI Encodings (cont)

  11. 4B/5B Encoding • 4B/5B: The idea is insert extra bits to break the consecutive bit patterns. • every 4 bits of data encoded in a 5-bit code

  12. 4B/5B Encoding (cont) • 5-bit codes selected to have no more than one leading 0 and no more than two trailing 0s • thus, never get more than three consecutive 0s • resulting 5-bit codes are transmitted using NRZI • achieves 80% efficiency. • Unused code are used for control. I.e. 11111 is for line is idle or 00000 for the line is dead. • FDDI is using this scheme.

  13. 3.Framing • How to distinguish between data and garbage. • Break sequence of bits into a frame • Typically implemented by network adaptor

  14. 8 8 16 16 16 16 8 8 ASCII character 4 eot (end of transmission) ASCII character 1 soh (start of header) 01111110 01111110 Header Header Body Body CRC CRC Byte-oriented Approaches • Sentinel-based • delineate frame with special pattern: 01111110 or STX, ETX, etc characters. • e.g., HDLC, SDLC, PPP

  15. Byte-oriented Approaches • problem: special pattern or characters appear in the payload • solution: • bit stuffing • sender: insert 0 after five consecutive 1s • receiver: delete 0 that follows five consecutive 1s

  16. Byte stuffing

  17. Byte-oriented Approaches (cont) • Counter-based • include payload length in the header • e.g., DDCMP protocol from DEC. • problem: count field corrupted • solution: catch when CRC fails

  18. 4.Error Detection • To send extra information to find error in the frame. • The simplest form is sending two copies, • inefficient. • Sending the sum of values (?) in the frame, checksum. • In Internet, consider 16 bits sequences and then use one-complement to find the result. • Sending parity, odd or even parity. • Odd parity of is

  19. Error Detection • Two dimensional parity, adding one extra bit, parity bit, to code and also find the parity for each bit position for total data

  20. Cyclic Redundancy Code (CRC) • Add k bits of redundant data to an n-bit message • Where k << n • e.g., k = 32 and n = 12,000 (1500 bytes)

  21. CRC(cont) • Represent n-bit message as n-1 degree polynomial • e.g., MSG=10011010 as M(x) = x7 + x4 + x3 + x1 • Let k be the degree of some divisor polynomial • e.g., C(x) = x3 + x2 + 1 • Transmit polynomial T(x) that is evenly divisible by C(x) • shift left k bits, i.e., M(x)xk • subtract remainder of M(x)xk / C(x) from M(x)xk

  22. CRC (cont) • Receiver polynomial T(x) + E(x) • E(x) = 0 implies no errors • Divide (T(x) + E(x)) by C(x); remainder zero if: • E(x) was zero (no error), or • E(x) is exactly divisible by C(x) • All operation is done in modulo 2 in which there is no carry. Then, the operation can be done by XOR only.

  23. CRC(cont) Example: M(x)= 110101, C(x) = 1001 1 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 0 1 0 0 1 0 1 1 The final transmitted message is: T(x) = 1 1 0 1 0 10 1 1 R

  24. Example

  25. Internet Checksum Algorithm • View message as a sequence of 16-bit integers; sum using 16-bit ones-complement arithmetic; take ones-complement of the result. u_short cksum(u_short *buf, int count) { register u_long sum = 0; while (count--) { sum += *buf++; if (sum & 0xFFFF0000) { /* carry occurred, so wrap around */ sum &= 0xFFFF; sum++; } } return ~(sum & 0xFFFF); }

  26. 5.Acknowledgements & Timeouts

  27. Stop-and-Wait Sender Receiver • Problem: keeping the pipe full • Example • 1.5Mbps link x 45ms RTT = 67.5Kb (8KB) • 1KB frames imples 1/8th link utilization

  28. Sender Receiver … ime T … Sliding Window • Allow multiple outstanding (un-ACKed) frames • Upper bound on un-ACKed frames, called window

  29. £ SWS … … LAR LFS SW: Sender • Assign sequence number to each frame (SeqNum) • Maintain three state variables: • send window size (SWS) • last acknowledgment received (LAR) • last frame sent (LFS) • Maintain invariant: LFS - LAR <= SWS • Advance LAR when ACK arrives • Buffer up to SWS frames

  30. £ RWS … … LFR LFA SW: Receiver • Maintain three state variables • receive window size (RWS) • largest frame acceptable (LFA) • last frame received (LFR) • Maintain invariant: LFA - LFR <= RWS • Frame SeqNum arrives: • if LFR =< SeqNum < = LFA accept • if SeqNum < LFR or SeqNum > LFA discarded • Send cumulative ACKs

  31. Sequence Number Space • SeqNum field is finite; sequence numbers wrap around • Sequence number space must be larger then number of outstanding frames • SWS <= MaxSeqNum-1 is not sufficient • suppose 3-bit SeqNum field (0..7) • SWS=RWS=7 • sender transmit frames 0..6 • arrive successfully, but ACKs lost • sender retransmits 0..6 • receiver expecting 7, 0..5, but receives second incarnation of 0..5 • SWS < (MaxSeqNum+1)/2 is correct rule • Intuitively, SeqNum “slides” between two halves of sequence number space

  32. Concurrent Logical Channels • Multiplex 8 logical channels over a single link • Run stop-and-wait on each logical channel • Header: 3-bit channel num, 1-bit sequence num • 4-bits total • same as sliding window protocol • Maintain three state bits per channel • channel busy • current sequence number out • next sequence number in • Separates reliability from order

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