1. Univ. of TehranIntroduction to Computer Network1 An Introduction to Computer Networks University of Tehran Dept. of EE and Computer Engineering By: Dr. Nasser Yazdani Point-to-Point link Lecture 6: Point-to-Point link

2. Univ. of TehranIntroduction to Computer Network2 Concepts: Data signal Links Link functions Encoding Framing Error Detection Error correction or reliable transmission Outline

3. Univ. of TehranIntroduction to Computer Network3 Link Functions Functions 1. Construct Frame with Error Detection Code 2. Encode bit sequence into analog signal 3. Transmit bit sequence on a physical medium (Modulation) 4. Receive analog signal 5. Convert Analog Signal to Bit Sequence 6. Recover errors through error correction and/or ARQ Adaptor Signal Adaptor: convert bits into physical signal and physical signal back into bits

4. Univ. of TehranIntroduction to Computer Network4 Encoding Goal: send bits from one node to another node on the same physical media Encode binary data onto signals This service is provided by the physical layer Bits flow is between network adaptors. Problem: specify a robust and efficient encoding scheme to achieve this goal How to send clock information? Extract from changes in signals.

5. Univ. of TehranIntroduction to Computer Network5 Encoding (crieria) Criteria for encoding: Bit rate (in limited BW) Recovering time information (in LAN) Error detecting Immunity to noise and interference Complexity and cost of implementation. Bit rates V.s Baud Rates Baud rates is the number of pulses sent over the link or changing in signals.

6. Winter 2008CS244a Handout 126 Encoding for clock recovery Problem : Different hosts use locally-generated clocks of nominally the same frequency, but slightly different. E.g. 10MHz +/- 100ppm (“parts per million”) 1. The receiver needs to “recover” the senders clock from the data stream, for example : 10MHz clock +/- 100ppm Flip- Flop Flip- Flop Sender Sender’s Clock Flip- Flop Clock Recovery Unit 10MHz clock +/- 100ppm Receiver Network Link 1) One part per million equals 10 -4 %. Elasticity buffer

7. Univ. of TehranIntroduction to Computer Network7 Assumptions We use two discrete signals, high and low, to encode 0 and 1 The transmission is synchronous, i.e., there is a clock used to sample the signal In general, the duration of one bit is equal to one or two clock ticks If the amplitude and duration of the signals is large enough, the receiver can do a reasonable job of looking at the distorted signal and estimating what was sent.

8. Univ. of TehranIntroduction to Computer Network8 Non-Return to Zero (NRZ) 1  high signal; 0  low signal 0 01010110 NRZ (non-return to zero) Clock

9. Univ. of TehranIntroduction to Computer Network9 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

10. Univ. of TehranIntroduction to Computer Network10 Non-Return to Zero Inverted (NRZI) 1  make transition; 0  stay at the same level Solve previous problems for long sequences of 1’s, but not for 0’s 0 01010110 Clock NRZI (non-return to zero intverted)

11. Univ. of TehranIntroduction to Computer Network11 Manchester 1  high-to-low transition; 0  low-to-high transition Addresses clock recovery and baseline wander problems Disadvantage: needs a clock that is twice as fast as the transmission rate Efficiency of 50% 0 01010110 Clock Manchester

12. Univ. of TehranIntroduction to Computer Network12 Encodings (cont) Bits NRZ Clock Manchester NRZI 0010111101000010

13. Univ. of TehranIntroduction to Computer Network13 4-bit/5-bit (100Mb/s Ethernet) Goal: address inefficiency of Manchester encoding, while avoiding long periods of low/high signals Solution: Use 5 bits to encode every sequence of four bits such that no 5 bit code has more than one leading 0 and two trailing 0’s Use NRZI to encode the 5 bit codes Efficiency is 80% 0000 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 0111 01111 4-bit 5-bit 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101 4-bit 5-bit  Unused code are used for control. I.e. 11111 is for line idle or 00000 for the line is dead.  FDDI is using this scheme.

14. Univ. of TehranIntroduction to Computer Network14 Framing How to distinguish between data and garbage. Break sequence of bits into a frame Typically implemented by network adaptor Frames Bits AdaptorAdaptor Node B Node A

15. Univ. of TehranIntroduction to Computer Network15 Byte-oriented Approaches Sentinel-based delineate frame with special pattern: 01111110 or STX, ETX, etc characters. e.g., HDLC, SDLC, PPP Problem: special pattern or characters appear in the payload. Character escaping or stuffing: in BISYNC (from IBM) with DLE character. Body 8 16 8 CRC Beginning sequenc e Ending Header

16. Univ. of TehranIntroduction to Computer Network16 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

17. Univ. of TehranIntroduction to Computer Network17 Bit-oriented Approaches Frame is a collection of bits, SDLC (IMB) (synch. Data Link Control), later changed to HDLC (high) by OSI Beginning and end with 01111110 problem: How if the sequence appear in the body. solution: Bit stuffing: add after 5 consecutive 1 a zero. Beginning SequenceHeader Body …….CRCEnding Sequence 8 16 8

18. Univ. of TehranIntroduction to Computer Network18 Clock-Based Framing e.g., SONET: Synchronous Optical Network By Bellcore Use NRZ, but scramble it for enough transition Multiplex multiple low-speed links into a high sp STS-n (STS-1 = 51.84 Mbps), each frame 125 us Overhead Payload 90 columns 9 rows

19. Univ. of TehranIntroduction to Computer Network19 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. Two dimensional parity.

20. Univ. of TehranIntroduction to Computer Network20 Error Detecting (goals) Goals: Reduce overhead, i.e., reduce the number of redundancy bits Increase the number and the type of bit error patterns that can be detected

21. Winter 2008CS244a Handout 1321 Bit error rate An example Assume an N-bit packet, with known BER and independent errors: Packet Error Rate = PER = 1 – (1 – BER) N PER ~= N (BER) if N (BER) << 1 e.g. N = 10 4, BER = 10 -7 = PER = 10 -3 In practice, bit errors occur in bursts: Perhaps caused by mechanical switches that switch slowly relative to a bit-time. If a bit is in error, it is likely that the next bit is in error too. Therefore, bit errors are not independent. So for a given BER, PER < N (BER)

22. Univ. of TehranIntroduction to Computer Network22 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); }

23. Univ. of TehranIntroduction to Computer Network23 Even Parity Check EDC field has 1 bit Sender: If number of 1’s in payload is odd, the check bit is 1, else it is 0 Receiver: Accept the packet if payload number of 1’s match the value of the check bit. Can detect an odd number of bit errors

24. Univ. of TehranIntroduction to Computer Network24 Two-dimensional Parity Add one extra bit to a 7-bit code such that the number of 1’s in the resulting 8 bits is even (for even parity, and odd for odd parity) Add a parity byte for the packet Example: five 7-bit character packet, even parity 0110100 1011010 0010110 1110101 1001011 1 0 1 1 0 10001101

25. Univ. of TehranIntroduction to Computer Network25 How Many Errors Can you Detect? All 1-bit errors Example: 0110100 1011010 0000110 1110101 1001011 1 0 1 1 0 10001101 error bit odd number of 1’s Can detect AND correct 1-bit errors

26. Univ. of TehranIntroduction to Computer Network26 Cyclic Redundancy Code (CRC) T M m R MSB i.e. T = M.2 r + R Modulo-2 addition (XOR) Add r bits of redundant data to an n-bit message Where r << n e.g., r= 32 and n = 12,000 (1500 bytes) r

27. Univ. of TehranIntroduction to Computer Network27 CRC(cont) Represent n-bit message as n-1 degree polynomial e.g., MSG=10011010 as M(x) = x 7 + x 4 + x 3 + x 1 Let k be the degree of some divisor polynomial e.g., C(x) = x 3 + x 2 + 1 Transmit polynomial T (x) that is evenly divisible by C(x) shift left k bits, i.e., M(x)x k subtract remainder of M(x)x k / C(x) from M(x)x k

28. Univ. of TehranIntroduction to Computer Network28 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.

29. Univ. of TehranIntroduction to Computer Network29 CRC(cont) Example: M(x)= 110101, C(x) = 1001 1 0 0 11 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 0 1 0 1 1 R The final transmitted message is: T(x) = 1 1 0 1 0 1 0 1 1

30. Univ. of TehranIntroduction to Computer Network30 Selecting C(x) All single-bit errors, as long as the x k and x 0 terms have non-zero coefficients. All double-bit errors, as long as C(x) contains a factor with at least three terms Any odd number of errors, as long as C(x) contains the factor (x + 1) Any ‘burst’ error (i.e., sequence of consecutive error bits) for which the length of the burst is less than k bits. Most burst errors of larger than k bits can also be detected See Table 2.6 on page 102 for common C(x)

31. Winter 2008CS244a Handout 1331 Circuit for Calculating CRC +++ g2g2 g1g1 g0g0 M T R G(x) = g 0 + g 1 x + … + g L-1 x L-1 Multiplier

32. Univ. of TehranIntroduction to Computer Network32 Hamming Distance Given codewords A and B, the Hamming distance between them is the number of bits in A that need to be flipped to turn it into B Example H(011101,000000)=4 If all codewords are at least d Hamming distance apart, then up to d-1 bit errors can be detected and (d-1)/2 error can be corrected.

33. Univ. of TehranIntroduction to Computer Network33 Reliable Transmission Error detection code Very costly Acknowledgment and timeout Send Acks in opposite direction. Retransmit if sender did not receive Ack.

34. Univ. of TehranIntroduction to Computer Network34 Acknowledgements & Timeouts Sender Receiver Frame ACK T Timeout Time Sender Receiver Frame ACK T Timeout Frame ACK T Timeout Sender Receiver Frame ACK T imeout Frame ACK T imeout Sender Receiver Frame T imeout Frame ACK T imeout (a) (c) (b)(d)

35. Univ. of TehranIntroduction to Computer Network35 Stop-and-Wait Problem 1: keeping the pipe full Example 1.5Mbps link x 45ms RTT = 67.5Kb (8KB) 1KB frames imples 1/8th link utilization Problem 2: How about repeated frames? Add a sequence number to frames. Sender Receiver

36. Univ. of TehranComputer Network36 Performance of Stop & Wait? Without Error T (total time) = 2t p + t f + 2t proc + t ack T proc and t ack are usually negligible. Take a= t p /t f, then utilization η= t f /( 2t p + t f ) =1/(2a+1) Then, utilization depends on a frame Sender Receiver t p = propagate time tftf

37. Univ. of TehranComputer Network37 Performance of Stop & Wait? a = (D/V)/(L/R) where D= distanceV= Velocity of light L= Length of packet R= Rate a= RD/VL Constant Then, L determines the link utilization, the longer the packet the more link utilization. frame Sender Receiver t p = propagate time tftf

38. Univ. of TehranComputer Network38 Performance of Stop & Wait? With Error N r is the expected value of sending a frame. η= 1/ N r (2a+1), Then what is N r ? ρ= prob. Of a frame in error N r = i ρ i-1 (1-ρ) =1/ (1-ρ) η= (1-ρ)/(2a+1), frame Sender Receiver t p = propagate time tftf

39. Univ. of TehranIntroduction to Computer Network39 Sliding Window Allow multiple outstanding (un-ACKed) frames Upper bound on un-ACKed frames, called window SenderReceiver T ime … …

40. Univ. of TehranIntroduction to Computer Network40 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  SWS LARLFS ……

41. Univ. of TehranIntroduction to Computer Network41 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 LFA discarded Send cumulative ACKs  RWS LFR LFA ……

42. Univ. of TehranIntroduction to Computer Network42 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

43. Univ. of TehranComputer Network43 Performance sliding window? Without Error, the same assumption as stop and wait. t p = a, t f =1 W ≥ 2a+1 => η= 1 W η= W/(2a+1) It is the same for selective repeat/reject. frame Sender Receiver 2a +1 for 1 st ack. tftf

44. Univ. of TehranComputer Network44 Performance of sliding window? With Error, W ≥ 2a+1 => η= 1/ N r, Then what is N r ? W η= W/ N r (2a+1) ρ= prob. Of a frame in error N r = i ρ i-1 (1-ρ) =1/ (1-ρ) W ≥ 2a+1 => η= 1- ρ, W η= W(1- ρ)/ (2a+1) frame Sender Receiver 2a +1 for 1 st ack. tftf

45. Univ. of TehranIntroduction to Computer Network45 Concurrent Logical Channels Multiplex 8 logical channels over a single link Run stop-and-wait on each logical channel Maintain three state bits per channel channel busy current sequence number out next sequence number in Header: 3-bit channel num, 1-bit sequence num 4-bits total same as sliding window protocol Separates reliability from order