Outline Building Blocks Shannon’s Theorem Encoding

1. 11. Physical Layer Outline Building Blocks Shannon’s Theorem Encoding CS 585

2. Building Blocks • Node • Network Adaptor (memory is the speed bottleneck) • Router (input ports, switch fabric, output ports) • Links • C = f λ • Cables • Leased Lines: T1(DS1), T3(DS3), OC-1(STS-1),OC-3(STS-3),OC-48(STS-48,2.48Gbps) • Last mile links: modem, xDSL, ISDN, cable modem CS 585

3. Shannon’s Theorem • C = B log (1+S/N) • C: channel capacity, in bps • B: bandwidth, in hertz • S: average signal power • N: average noise power • dB = 10 * log (S/N) • dB: signal-noise ratio expressed in decibels 2 10 CS 585

4. Shannon’s Theorem • Example: • B : 300 Hz to 3300 Hz • N/S = 30 dB • C = 3000 log (1001) = 30 Kbps 2 CS 585

5. Encoding • Signals propagate over a physical medium • modulate electromagnetic waves • e.g., vary voltage • Schemes • NRZ • NRZI • Manchester • 4B/5B CS 585

6. 12. Data Link Layer Outline Framing Error Detection Sliding Window Algorithm Token rings (FDDI) Wireless CS 585

7. Data Link Layer • Data Link Layer • Framing • Error Detection • Sliding Window (reliable transmission, flow control) • Medium Access Control (MAC) Sub-Layer • Token ring (FDDI) • Wireless • Logical Link Control (LLC) : IEEE802.2 CS 585

8. Bits Node A Adaptor Adaptor Node B Frames Framing • Break sequence of bits into a frame • Typically implemented by network adaptor CS 585

9. 8 16 16 8 Beginning Ending Header Body CRC sequence sequence Approaches • Sentinel-based • delineate frame with special pattern: 01111110 • e.g., HDLC, SDLC, PPP • problem: special pattern appears in the payload • solution: bit stuffing • sender: insert 0 after five consecutive 1s • receiver: delete 0 that follows five consecutive 1s CS 585

10. Approaches (cont) • Counter-based • include payload length in header • e.g., DDCMP • problem: count field corrupted • solution: catch when CRC fails CS 585

11. Approaches (cont) • Clock-based • each frame is 125us long • e.g., SONET: Synchronous Optical Network • STS-n (STS-1 = 51.84 Mbps) CS 585

12. Error Detection • Parity • Two-dimensional parity • Internet Checksum • CRC CS 585

13. Cyclic Redundancy Check (CRC) • Add k bits of redundant data to an n-bit message • want k << n • e.g., k = 32 and n = 12,000 (1500 bytes) • 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 • Modulo 2 operation (XOR) CS 585

14. CRC (cont) • Transmit polynomial P(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 • Receiver polynomial P(x) + E(x) • E(x) = 0 implies no errors • Divide (P(x) + E(x)) by C(x); remainder zero if: • E(x) was zero (no error), or • E(x) is exactly divisible by C(x) CS 585

15. Selecting C(x) • All single-bit errors, as long as the xk and x0 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) CS 585

16. Sliding Window • Stop and wait • Selective repeat • Go back N CS 585

17. Acknowledgements & Timeouts CS 585

18. Stop-and-Wait Sender Receiver • Two sequence numbers (0 and 1) • Problem: keeping the pipe full • Example • 1.5Mbps link x 45ms RTT = 67.5Kb (8KB) • 1KB frames imples 1/8th link utilization CS 585

19. Sender Receiver … ime T … Sliding Window • Allow multiple outstanding (un-ACKed) frames • Upper bound on un-ACKed frames, called window CS 585

20. £ 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 CS 585

21. 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 £ RWS … … LFR LFA CS 585

22. 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 CS 585

23. Selective Repeat • At the receiver, it will send an ack for each frame received, even if the frame is out of order. • The sender sets a timeout value for each frame sent out, and will resend the frame upon timeout. The timeout flag is cleared when an ack for the frame is received. • Only those frames (or acks for them) are lost are retransmitted. • The receiver may need a large buffer to hold out of order frames. CS 585

24. Go Back N • At the receiver, it will send an ack for each frame received, only if the frame is out of order. Out of order frames are discarded. It also sends a NAK when it finds a frame is lost. • The timer mechanism is the same. • The acks are cumulative in the sense that an ack indicates that the receiver has got all the frames up to the frame acked. • The sender will resend all the frames starting from the frame for which a NAK is received. • The receiver does not need a large buffer to hold out of order frames. CS 585