CSS CSS432 Point-to-Point Links Textbook Ch2.1.2 – 2.5 Professor: Munehiro Fukuda

CSS Basic Knowledge about Links Full-duplex versus Half-duplex  Full  Half Local cable types  Category 5 twisted pair  Coax  Multimode fiber LED-based, 2km  Single-mode fiber laser-diode-based, 40km Carrier-supported cables/fibers  DS1 (T1): 1.544Mbps, 24-digital voice circuits of 64 Kbps each  DS3 (T3): Mbps, 28 DS1 links  STS-1 (OC-1): Synchronous Transport signal, the base link speed= Mbps  STS-3, 12, 48, 192 (OC-3, OC-12, OC-48, OC-192) at time t at time u

CSS Basic Knowledge (Cont’d) Last-Mile Links  POTS: Plain Old Telephone Service  ISDN: Integrated Services Digital Network  ADSL: Asynchronous Digital Subscriber Line  VDSL: Very high data rate Digital Subscriber Line Wireless Links  AMPS, PCS, GSM: Wide-area analog/digital-based mobile phone services  IEEE : wireless LAN with up to 54Mbps transfer rate  BlueTooth radio interface: 1Mpbs piconet in 10m CODEC Voice line 28.8~56Kbps Digital line Kbps Mbps Kbps STS-N 12.96~ 55.2Mbps

CSS Focusing on ADSL To telephone switch Residence A Residence B 255 frequencies for downstream 31 use for upsreams 2 taken for control Long distance Preparing 286 frequencies and agreeing at the available frequencies between two modems Fitting to most usages where a user tends to retrieve Internet information.

CSS Encoding Signals propagate over a physical medium  modulate electromagnetic waves  e.g., vary voltage Encode binary data onto signals  e.g., 0 as low signal and 1 as high signal  known as Non-Return to zero (NRZ) Consecutive 1s and 0s  Baseline wander: if the receiver averages signals to decide a threshold  Clock recovery: both sides must be strictly synchronized Bits NRZ

CSS Alternative Encodings Non-return to Zero Inverted (NRZI)  Encode 1: make a transition from current signal  Encode 0: stay at current signal to encode a zero  solves the problem of consecutive 1s Signal flips every 1 is encountered. Manchester  Transmit XOR of the NRZ encoded data and the clock  Only 50% efficient.  Used by Ethernet/Token Ring 0 1

CSS B/5B Encoding Every 4 bits of data encoded in a 5-bit code 5-bit codes selected to have  no more than one leading 0  no more than two trailing 0s  See textbook Table 2.4 on page 79 Thus, never get more than three consecutive 0s Resulting 5-bit codes are transmitted using NRZI Achieves 80% efficiency  Because 5 th bit is redundant. Used by FDDI 01??? but not 001?? ??100 but not ?1000

CSS Encoding Example Bits NRZ Clock Manchester NRZI

CSS Framing Break sequence of bits into a frame Typically implemented by a network adaptor Frame Bits Adaptor Node BNode A

CSS Approaches HeaderBody CRC Beginning sequence Ending sequence Sentinel-based  delineate frame with special pattern:  e.g., HDLC, SDLC, PPP  problem: ending sequence appears in the payload  solution: bit stuffing sender: insert 0 after five consecutive 1s receiver: delete 0 that follows five consecutive 1s Counter-based  include payload length in header  e.g., DDCMP Frame error  problem: count field corrupted (frame error)  solution: Wait for the next beginning sequence Catch when CRC fails

CSS Approaches (cont) Clock-based  each frame is 125us long  e.g., SONET: Synchronous Optical Network  STS-n (STS-1 = Mbps ) Interleaved every byte: keep 51Mbps for each STS-1

CSS Error Detections Add k bits of redundant data to an n-bit message  want k << n (i.e. k is extremely small as compared to n)  e.g., k = 32 and n = 12,000 (1500 bytes) in CRC-32 Parity Bit Internet Checksum Algorithm Cyclic Redundancy Check

CSS Parity Odd parity Even parity Two-dimensional parity Parity bits Parity byte Data

CSS Internet Checksum Algorithm View message as a sequence of 16-bit integers; Convert each 16-bit integer into a ones-complement arithmetic Sum up each integer Increment the result upon a carryout Example: 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); } 5: : : : (-3): w/ carry: This corresponds to -8 Ones-complement message What if 5 and 3 were changed In 6 and 2 ?

CSS Cyclic Redundancy Check Represent n-bit message as n-1 degree polynomial  e.g., MSG= 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 sender receive r M(x)2 k – M(x) 2 k % C(x) = C(x)-divisible message: P(x) If P(x) % C(x) == 0 P(x) was sent successfully

CSS CRC (cont) Sender polynomial M(x)  M(x) = , C(x) = 1101, thus k = 3  M(x)2 3 = (<< 3)  M(x)2 3 % C(x) = 101  M(x)2 3 – M(x)2 3 % C(x) = = P(x) Noise polynomial E(x) Receiver polynomial P(x) + E(x)  E(x) = 0 implies no errors (P(x) + E(x)) % C(x) == 0 if:  E(x) was zero (no error), or  E(x) is exactly divisible by C(x) To retrieve M(x): P(x) / 2 3, (truncate the last 3 bits) Just appended The last 3 bits of P(x) eor eor eor eor eor eor

CSS CRC Calculation Using Shift Register sender receive r M(x) CRC CRC + M(x) M(x) CRC’ If CRC == CRC’, no errors x0x0 x1x1 x2x2 x0x0 x1x1 x2x Example: M(x) = C(x) = will be transferred.

CSS Acknowledgements & Timeouts Receiver can’t distinguish if the 1 st and 2 nd frames are identical

CSS Stop-and-Wait Stop a transmission and wait for ack to return or timeout Add 1-bit sequence number to detect a duplication of the same frame Problem: Can’t utilize the link’s capacity Example  1.5Mbps link x 45ms RTT = 67.5Kb (8KB)  If the frame size is 1KB and the sender waits for 45ms RTT 1KB for 45ms RTT 1/8th link utilization SenderReceiver Frame 0 Frame 1 Frame 0 Ack 0 Ack 1

CSS Sliding Window Allow multiple outstanding (un-ACKed) frames Upper bound on un-ACKed frames, called window SenderReceiver T ime … …

CSS 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/Back up to SWS frames for future retransmission  SWS LARLFS ……

CSS 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 LFRLFRLFA …… Send a comulativeAck discard

CSS Sequence Number Space SeqNum field is finite; sequence numbers wrap around Sequence number space must be larger than # 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 a correct rule Intuitively, SeqNum “slides” between two halves of sequence number space

CSS Reviews  Encoding (NRZ, NRZI, Manchester, and 4B/5B)  Framing (Sentinel and counter-based)  Error Detections (Parity, Internet checksum, and CRC)  Stop-and-wait  Sliding window Exercises in Chapter 2  Ex. 2 and 5 (4B/5B, NRZI, and bit stuffing)  Ex. 16 (Internet Checksum)  Ex. 18 (CRC)  Ex. 24 (Sliding Window: Apply Ex. 25’s cases (a) and (b) to Ex. 24)