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Chapter 4 Data Link Layer

Chapter 4 Data Link Layer. Framing Error control Flow control Multiplexing Link Maintenance Security. Services. Transfers frames across direct connections Directly connected (can be wireless), wire-like Losses & errors, but no out-of-sequence frames More detailed services

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Chapter 4 Data Link Layer

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  1. Chapter 4 Data Link Layer Framing Error control Flow control Multiplexing Link Maintenance Security

  2. Services • Transfers frames across direct connections • Directly connected (can be wireless), wire-like • Losses & errors, but no out-of-sequence frames • More detailed services • Framing (bits ↔ frames) • Error control (protection from impairment) • Flow control • Multiplexing • Link Maintenance • Security: Authentication & Encryption

  3. Examples PPP HDLC Ethernet LAN IEEE 802.11 (WiFi) LAN Packets Packets Data link layer Data link layer Frames A B Physical layer Physical layer Data Link Protocols

  4. Framing(Chapter 5.4 in Leon-Garcia)

  5. Bit stream - frames Frame boundaries can be determined using: Character Counts Control Characters Framing Bits Framing by illegal code received frames transmitted frames Framing 0111110101 0110110111 Framing

  6. Data to be sent A DLE B ETX DLE STX E After stuffing and framing DLE STX A DLE DLE B ETX DLE DLE STX E DLE ETX Control Characters • What about transmission of data (including non-printable characters)? • Introduce DLE (data link escape) = 0x10 • DLE STX (DLE ETX) used to indicate beginning (end) of frame • Insert extra DLE in front of occurrence of DLE STX (DLE ETX) in frame • All DLEs occur in pairs except at frame boundaries. • Transmission of printable characters using ASCII • Octets with HEX value < 0x20 are nonprintable • Use control characters: STX (start of text) = 0x02; ETX (end of text) = 0x03.

  7. HDLC frame any number of bits Flag FCS Control Information Flag Address Bit Stuffing • Frame delineated by flag character • HDLC uses bit stuffing to prevent occurrence of flag 01111110 inside the frame • Transmitter inserts extra 0 after each consecutive five 1s inside the frame • Receiver checks for five consecutive 1s • if next bit = 0, it is removed • if next two bits are 10, then flag is detected • If next two bits are 11, then frame has errors

  8. Data to be sent (a) 0110111111111100 After stuffing and framing 0111111001101111101111100001111110 (b) Data received 01111110000111011111011111011001111110 After destuffing and deframing *000111011111-11111-110* Example: Bit stuffing

  9. Example: Framing in Ethernet • Ethernet complies to standard IEEE 802.3 • An illegal manchester coding is used for framing. • A character count is also included in the header. • All frames have an integral number of bytes. If not, the frame is considered to be received in error.

  10. Error Control Coding(Chapter 3.9 in Leon-Garcia)

  11. c1…cn b1…bk Decoder Channel Encoder Error Control • Two approaches • Forward error correction (FEC) • Error detection & retransmission (ARQ) • Add redundancy (admit only codewords with a certain pattern) • Blindspot: when channel transforms a codeword into another codeword • (n,k) block code • There are capacity-achieving codes • Usually with somewhat high complexity; • When bandwidth is abundant it suffices to use simpler codes.

  12. Information bits: b1, b2, b3, …, bk Check Bit: bk+1= b1+ b2+ b3+ …+ bk modulo 2 Codeword: (b1, b2, b3, …, bk,, bk+!) Single Parity Check • n=k+1 • All codewords have even # of 1s • All error patterns that change an odd number of bits are detectable • Others undetectable • Redundancy: overhead = 1/(k + 1)

  13. Example • Information (7 bits): (0, 1, 0, 1, 1, 0, 0)

  14. n 2 n 4 Probability of Error P[error detection failure] = P[undetectable error pattern] = P[all error patterns with even number of 1s] = p2(1 – p)n-2 + p4(1 – p)n-4 + … • Example: Evaluate above for n = 6, p = 0.01 P[undetectable error] = 0.0014

  15. 1 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 11 row check bits column check bit Two-Dimensional Parity Check

  16. 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 1 Two errors One error 1 0 0 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 1 0 1 0 0 1 1 1 1 0 0 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 Three errors Four errors (undetectable) Arrows indicate failed check bits Error-detecting capability 1, 2, or 3 errors can always be detected; Not all patterns >4 errors can be detected

  17. Hamming Codes • Class of linear block codes • Capable of correcting all single-error patterns • For each m> 2, there is a (2m–1, n-m) Hamming code

  18. m = 3 Hamming Code • Information bits are b1, b2, b3, b4 • Parity checks (binary addition/multiplication) b5 = b1 + b3 + b4 b6 = b1 + b2 + b4 b7 = + b2 + b3 + b4 • Linearity • 24 = 16 codewords

  19. Hamming (7,4) code

  20. Rearrange parity check equations: All codewords must satisfy these equations Note: each nonzero 3-tuple appears once as a column in check matrix H b1 b2 0 = 1 0 1 1 1 0 0 b3 0 = 1 1 0 1 0 1 0 b4 = Hbt = 0 0 = 0 1 1 1 0 0 1 b5 b6 b7 Parity Check Equations 0 = b1 + b3 + b4 + b5 0 = b1 + b2 + b4 + b6 0 = + b2 + b3 + b4 + b7 • In matrix form:

  21. 0 0 1 0 0 0 0 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 Single error detected 1 0 1 s = H e= = 0 1 0 0 1 0 0 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 1 1 0 0 Double error detected s = H e= = + = 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 1 0 1 0 1 Triple error not detected s = H e= = + + = 0 Hamming Code: Error Detection

  22. o o o b1 b2 o o o o o Minimum distance • Undetectable error pattern must have 3 or more bits • At least 3 bits must be changed to convert one codeword into another codeword Set of n-tuples within distance 1 of b2 Set of n-tuples within distance 1 of b1 Distance 3 • Spheres of distance 1 around each codeword do not overlap • If a single error occurs, the resulting n-tuple will be in a unique sphere around the original codeword

  23. General Hamming Codes • For m> 2, the Hamming code is obtained through the check matrix H: • Each nonzero m-tuple appears once as a column • The resulting code corrects all single errors • P[undetectable error] = P[ error is a codeword] ≈ (# of codewords with dmin) x pdmin • Animated example http://www.systems.caltech.edu/EE/Faculty/rjm/SAMPLE_20040708.html

  24. R b (Receiver) (Transmitter) + e Error pattern Hamming Codes: Error-correction • Compute syndrome: s = HR = H (b + e) = Hb + He = He • If s = 0, then the receiver accepts R as the transmitted codeword, find the corresponding k-bit message • If s is nonzero, then an error is detected • Hamming decoder assumes a single error has occurred • Each single-bit error pattern has a unique syndrome • The receiver matches the syndrome to a single-bit error pattern and corrects the appropriate bit

  25. s = HR = He 7p s = 0 s = 0 1–3p 3p No errors in transmission Undetectable errors Correctable errors Uncorrectable errors (1–p)7 7p(1–3p) 7p3 21p2 Hamming Codes: Performance • Assume bit errors occur independent of each other and with probability p

  26. Other Error Control Codes • Good practical codes for error detection: • Internet Check Sums • CRC Polynomial Codes They can detect the vast majority of errors. • Good codes for error “correction”: • Turbo codes • Low-density parity-check (LDPC) codes

  27. Internet Checksum • Several Internet protocols (e.g. IP, TCP, UDP) use check bits to detect errors in the header • A checksum is calculated for header contents and included in a special field. • Treating each 16-bit word in data as an integer, find x = b0 + b1 + b2+ ...+ bL-1 modulo 216-1 • The checksum is then given by: bL = - x modulo 216-1 Thus, the headers satisfy the following pattern: 0 = b0 + b1 + b2+ ...+ bL-1 + bL modulo 216-1

  28. Encoder for g(x) = x3 + x + 1 g0 = 1 g1 = 1 g3 = 1 0,0,0,i0,i1,i2,i3 Reg 2 Reg 1 Reg 0 + + Polynomial Codes • Convenient mathematical formulation of coding • Polynomials as codewords • Implemented using shift-register circuits • Called cyclic redundancy check (CRC) codes • Excellent for detecting burst errors

  29. Automatic Repeat Request (ARQ)(Chapter 5 in Leon-Garcia)

  30. Each layer provides a service to the layer above. It does so by executing a peer-to-peer protocol. The protocol uses the the services of the layer below. Peer-to-Peer Protocols n + 1 n + 1 n n n – 1 n – 1

  31. Service Models • The service model specifies the manner in which information is transferred. • Connection-oriented • Connectionless

  32. n + 1 peer process send n + 1 peer process receive Layer n connection-oriented service SDU SDU Connection-Oriented • Connection setup • Message transfer • Connection release • Example: TCP, PPP

  33. Connectionless Transfer Service • No setup • Each message sent independently • Must provide all address information per message • Simple & quick • Example: UDP, IP n + 1 peer process send n + 1 peer process receive Layer n connectionless service SDU

  34. Automatic Repeat Request (ARQ) • Purpose: To pass to the receiver every frame correctly, only once, in order. • Bad things can happen: Error, arbitrary delay, out-of-order arrival, or loss. Aim at very high reliability. • Assume if frames arrive, they arrive in-order for now. We save the out-of-order problem for later. • Basic elements: • Error-detecting code • ACKs (positive acknowledgments) • NAKs (negative acknowledgments) • Timeout mechanism

  35. Stop-and-Wait ARQ Transmit a frame, wait for ACK Error-free packet Packet Information frame Transmitter Receiver Timer set after each frame transmission Control frame

  36. (a) Frame 1 lost Time-out Time A Frame 0 Frame 1 Frame 1 Frame 2 ACK ACK B (b) ACK lost Time-out Time A Frame 0 Frame 1 Frame 1 Frame 2 ACK ACK ACK B Need for Sequence Numbers • In cases (a) & (b) the transmitting station A acts the same way • But in case (b) the receiving station B accepts frame 1 twice • Question: How is the receiver to know the second frame is also frame 1? • Need a sequence number: Slast=SN of most recent transmitted frame.

  37. The transmitting station misinterprets duplicate ACKs Question: How is the receiver to know second ACK is for frame 0? Time-out Time A Frame 0 Frame 0 Frame 2 Frame 1 ACK ACK B Sequence Numbers (c) Premature Time-out • Need SN in ACK: Rnext=SN of next frame expected by the receiver. • Implicitly acknowledges receipt of all prior frames. • What if ACK only ifSlast=Rnext?

  38. 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Rnext Slast Timer Slast Receiver B Transmitter A Rnext How many bits for SN? 1-Bit SN Suffices

  39. Slast Receiver B Transmitter A Frame 0 lost/error ACK 0 lost/error Rnext Error-free frame 0 arrives (0,0) (0,1) ACK 0 arrives ACK 1 arrives Global State: (Slast, Rnext) (1,0) (1,1) Error-free frame 1 arrives Frame 1 lost/error ACK 1 lost/error Finite State Machine

  40. Last frame bit enters channel ACK arrives First frame bit enters channel Transmitter waits for ACK t A B t Receiver processes frame and prepares ACK First frame bit arrives at receiver Last frame bit arrives at receiver S/W Efficiency

  41. t0 = total time to transmit 1 frame if no error A tproc B tprop tprop tproc tack S/W Transmission Time frame tftime bits/info frame bits/ACK frame channel transmission rate

  42. Efficiency on Error-free channel Overhead bits (header, CRC) Effective transmission rate: Transmission efficiency: Effect of frame overhead Effect of Delay-Bandwidth Product Effect of ACK frame

  43. Delay-Bandwidth Product nf=10,000 bits, na=no=200 bits S/W inefficient for very high speeds or long delays

  44. Average Transmission Number Proposition: Let Pf be the frame error probability. Then the average number of transmissions per successful frame is 1/ (1–Pf ). Proof: The number of transmissions to first correct arrival has geometric distribution. E.g., if 1-in-10 gets through, then in average 10 tries to success.

  45. Efficiency in Channel with Errors • Assuming time-out is equal to t0 (it should be larger) Effect of frame loss • If bit-error-rate is p, then

  46. Go-Back-N • A sliding-window protocol. • Keep channel busy by continuing to send frames • Allow a window of up to Ws outstanding frames • If ACK for oldest frame arrives before window is exhausted, we can continue transmitting • If window is exhausted, pull back and retransmit all outstanding frames

  47. 4 frames are outstanding; so go back 4 Go-Back-4: Time fr 0 fr 1 fr 2 fr 3 fr 4 fr 5 fr 3 fr 4 fr 5 fr 6 fr 6 fr 7 fr 8 fr 9 A B out of sequence frames ACK1 ACK2 ACK4 ACK5 ACK3 ACK7 ACK6 ACK9 ACK8 Rnext 0 1 2 3 3 4 5 6 7 8 9 Go-Back-N ARQ • Frame transmission are pipelined to keep the channel busy • Frame with errors and subsequent out-of-sequence frames are ignored

  48. Time-out Stop-and-Wait ARQ Time fr 1 fr 0 fr 0 A B ACK1 Receiver is looking for Rnext=0 If window exhausted, go back N Go-Back-N ARQ fr 0 fr 1 fr 2 fr 3 fr 0 fr 1 fr 2 fr 3 fr 4 fr 5 fr 6 Time A B ACK1 ACK5 ACK2 ACK6 ACK4 ACK3 Receiver is looking for Rnext=0 Out-of-sequence frames Choose Window Size > RTT

  49. Go-Back-N with Timeout • Problem with Go-Back-N as presented: • If frame is lost and source does not have frame to send, then window will not be exhausted and recovery will not commence • Use a timeout with each frame • When timeout expires, resend all outstanding frames

  50. Receiver Transmitter Send Window (size Ws) Receive Window (size 1) ... Frames transmitted and ACKed Slast Srecent Slast+Ws-1 Frames received Buffers Rnext oldest un-ACKed frame Slast Timer Slast+1 Timer ... most recent transmission Srecent Timer ... max SN allowed Slast+Ws-1 Go-Back-N Transmitter & Receiver Receiver will only accept error-free frame with SN Rnext. When the frame arrives Rnext is incremented by 1, so the receive window slides forward by 1.

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