1 / 37

An Introduction to Computer Networks

An Introduction to Computer Networks. Lecture 4: Point-to-Point link. University of Tehran Dept. of EE and Computer Engineering By: Dr. Nasser Yazdani. Outline. Concepts: Data signal Links Encoding Framing Error Detection Error correction or reliable transmission Stop and wait

ruby
Télécharger la présentation

An Introduction to Computer Networks

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. An Introductionto Computer Networks Lecture 4: Point-to-Point link University of Tehran Dept. of EE and Computer Engineering By: Dr. Nasser Yazdani Introduction to Computer Network

  2. Outline • Concepts: • Data • signal • Links • Encoding • Framing • Error Detection • Error correction or reliable transmission • Stop and wait • Sliding Window Algorithm Introduction to Computer Network

  3. Data • Discrete data: an instance is binary. Computer works with discrete data. Discrete is encoded in 0s and 1s. • Continuous data: change with time or space. • It is converted to discrete data by sampling • The sampling rate must be at least 2 times the maximum frequency. • Data is delivered by signals in the links Introduction to Computer Network

  4. Signals Electromagnetic waves propagating in the light speed. • Frequency • Wavelength • A (periodic) signal can be viewed as a sum of sine waves of different frequencies and strengths. • Every signal has an equivalent representation in the frequency domain. » What frequencies are present and what is their strength (energy) Introduction to Computer Network

  5. Signals (cont) Example- The following signal is the sum of sine waves. Introduction to Computer Network

  6. Modulation • Sender changes the nature of the signal in a way that the receiver can recognize. » Similar to radio: AM or FM • Digital transmission: encodes the values 0 or 1 in the signal. » It is also possible to encode multi-valued symbols • Amplitude modulation: change the strength of the signal, typically between on and off. » Sender and receiver agree on a “rate” » On means 1, Off means 0 • Similar: frequency or phase modulation. • Can also combine method modulation types. Introduction to Computer Network

  7. Channel capacity • Every transmission medium supports transmission in a certain frequency range. This is called channel capacity. » The channel bandwidth is determined by the transmission medium and the nature of the transmitter and receivers • A noiseless channel of width H can at most transmit a binary signal at a rate 2 x H. » E.g. a 3000 Hz channel can transmit data at a rate of at most 6000 bits/second Assumes binary amplitude encoding • Shannon extended this result by accounting for the effects of noise. • More aggressive encoding can increase the channel bandwidth. » Example: modems Introduction to Computer Network

  8. Limits in sending signals • Noise: “random” energy is added to the signal. • Attenuation: some of the energy in the signal leaks away. We need repeaters. • Dispersion: attenuation and propagation speed are frequency dependent. » Changes the shape of the signal Introduction to Computer Network

  9. Copper Wire • Unshielded twisted pair » Two copper wires twisted - avoid antenna effect » Grouped into cables: multiple pairs with common sheath » Category 3 (voice grade) versus category 5 » 100 Mbps up to 100 m, 1 Mbps up to a few km » Cost: ~ 10cents/foot • Coax cables. » One connector is placed inside the other connector » Holds the signal in place and keeps out noise » Gigabit up to a km Introduction to Computer Network

  10. Microwaves • High frequency electromagnetic waves (>1GHz) q Line of sight terrestrial transmissions and for communications via satellites. q Some atmospheric interference occurs but reliable transmission can be obtained over distances up to 50 Km. q Microwaves is absorbed by rain and does not penetrate obstacles. Introduction to Computer Network

  11. Fiber Optic q Thin thread of glass or plastic q Lightweight. q Fibers act as wave-guides for light which is usually produced by lasers. q Visible light has frequency around 5*10 15 Hz, which ensures an extremely high bandwidth. q The raw materials are cheap. q Immune to electrical interference. q Difficult to join and tap. q Security advantages Introduction to Computer Network

  12. Encoding • Encode binary data onto signals • Bits flow is between network adaptors. • 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. Introduction to Computer Network

  13. Bits 0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 0 NRZ Encoding • How to send clock information? • Extract from changes in signals. • The simplest way is 0 as low signal and 1 as high signal. This is known as Non-Return to zero (NRZ) Introduction to Computer Network

  14. 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 Introduction to Computer Network

  15. Alternative Encodings • Try to solve the clock recovery problem. • 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. Introduction to Computer Network

  16. Bits 0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 0 NRZ Clock Manchester NRZI Encodings (cont) Introduction to Computer Network

  17. Encodings (cont) • 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 • 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. Introduction to Computer Network

  18. Bits Node A Adaptor Adaptor Node B Frames Framing • How to distinguish between data and garbage. • Break sequence of bits into a frame • Typically implemented by network adaptor Introduction to Computer Network

  19. 8 16 16 8 Beginning Ending Header Body CRC sequence sequence 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. Introduction to Computer Network

  20. 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 Introduction to Computer Network

  21. 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. 16 8 16 8 Introduction to Computer Network

  22. 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) Introduction to Computer Network

  23. 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, adding one extra bit, parity bit, to code and also find the parity for each bit position for total data. Introduction to Computer Network

  24. 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); } Introduction to Computer Network

  25. 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) T R M m MSB i.e. T = M.2r + R Modulo-2 addition (XOR) Introduction to Computer Network

  26. 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 Introduction to Computer Network

  27. 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. Introduction to Computer Network

  28. 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 Introduction to Computer Network

  29. 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) Introduction to Computer Network

  30. Reliable Transmission • Error detection code. • Very costly • Acknowledgment and timeout • Send Acks in opposite direction. • Retransmit if sender did not receive Ack. Introduction to Computer Network

  31. Acknowledgements & Timeouts Introduction to Computer Network

  32. Stop-and-Wait Sender Receiver • 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. Introduction to Computer Network

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

  34. £ 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 Introduction to Computer Network

  35. £ 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 Introduction to Computer Network

  36. 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 Introduction to Computer Network

  37. 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 Introduction to Computer Network

More Related