1. 2. Getting connected (part 1)
Rocky K. C. Chang
Department of Computing
The Hong Kong Polytechnic University
7 February 2017

2. Many links on a network path

3. 1. Overview of this chapter
The lowest two layers:
Provide a virtual link for (unreliable) packets to above
Datalink layer
Datalink layer
Provide a virtual (unreliable) bit pipe to above
Physical layer
Physical layer

4. 1. Overview of this lecture
Direct link networks: point-to-point links or shared media
Five problems:
Bit synchronization
Frame synchronization
Error detection
Reliable link service
Multiple access control problem (for shared media only)
LAN: Ethernet and Wireless

5. 2. Some networking hardware
Network nodes
Switches and routers (highly specialized hardware)
Hosts (general-purpose computers):
CPU
Network
To network
Cache
adaptor
I/O bus
Memory

6. 2.1 Network hosts Network adaptors and device drivers
All the networking functionality described in this chapter, except ARQ schemes, is implemented in the network adaptor.
The device driver interfaces between the network adaptor and the OS.
Unparalleled performance improvement of memory latency and processor speed
A network host runs at memory speeds, not processor speeds.
Intel and Micron Produce Breakthrough Memory Technology

7. 2.2 Network links
A network link is a physical medium carrying signals in the form of electromagnetic waves.
Cabling:
Local links (office, lab, campus): Twisted pairs, coaxial cables, and optical fibers
Leased links from carriers: T1, T3; and OC-N, where N = 1, 3, 9, 12, 18, 24, ….
Links to residential home: POTS, ISDN, xDSL, CATV, ATM, etc.
Wireless: radio, microwave, infrared, light; small area, cellular network, satellite network.

8. 2.2 Network links
Common services available to connect your home

9. 2.2 Network links
Generally, a packet may go through different types of network link.
The slowest link determines the end-to-end throughput.
Bandwidth
Narrowband vs. broadband (ISDN vs. B-ISDN)
Increase bandwidth either by operating in a higher frequency band or increase the signal-to-noise ratio
Upper bound on the link’s data rate (Shannon-Hartley Theorem): Blog2(1+S/N) in bits per second

10. 2.2 Network links
Bluetooth is based on a global radio-frequency (RF) standard, operating on the 2.4 GHz ISM band.
3G mobile phone: 2,500MHz-2,690MHz band

11. 2.2 Network links
Sending binary bits onto a physical link (bit-serial transmission):
Encoding the bits before transmission (for the purpose of bit synchronization to be discussed later).
Modulation: a process by which a property or a parameter (frequency, amplitude, phase) of a signal is varied in proportion to a second signal.
For encoding binary data onto electromagnetic signals
Signals are carried in the form of electromagnetic waves.

12. 3. The five problems Assume modulation schemes given. Five problems:
Bit synchronization (need additional encoding data, such as from Manchester encoding, to delineate bits)
Frame synchronization (need additional protocols to delineate frames)
Error detection (need additional algorithms to detect errors, if occurred)
Reliable link service (need additional schemes to recover from errors)
Multiple access control problem (for shared media only; need additional protocols to share the medium)

13. 3.1 An Ethernet network adaptor
LAN controller
Manchester codec
Transceiver
HOST
BUS
NETWORK
TX FIFO
Data out TX
System bus interface
Medium attachment unit
802.3 MAC
Physical signalling
Data in RX
RX FIFO
Control in

14. 4. Problem 1: Bit synchronization (BS)
Problem: How does a receiver synchronize with a sender, so that bits can be decoded correctly from the signals?
Signalling component
Signal
Node
Adaptor
Adaptor
Node
Bits

15. 4.1 BS: four encoding methods
Assume that two discrete signals: high and low are used to encode 0s and 1s.
Solutions: NRZ, NRZI, Manchester, and 4B/5B:
Bits 1 1 1 1 1 1 1
NRZ
Clock
Manchester
NRZI

16. 4.2 BS: NRZ and NRZI Non-return to zero (NRZ)
0: a low signal; 1: a high signal
Problem: A consecutive of 0s or 1s causes baseline wander, as well as not providing enough signal transitions for bit synchronization.
Non-return to zero inverted (NRZI)
1: making a transition from the current signal; 0: staying at the current signal.
Problem: The problem of having consecutive 0s remains.

17. 4.3 BS: Manchester encoding
0: low-to-high transition; 1: high-to-low transition (the result of exclusive-OR of the NRZ-encoded data with the clock)
Advantage: provide enough signal transitions for bit synchronization.
Disadvantage: double the rate of signal transitions.
Baud rate is the rate at which the signal changes. In this case, baud rate is double of data rate.

18. 4.4 BS: 4B/5B encoding 4B/5B encoding
Insert a bit to every four bits of data, such that consecutive 0s or 1s will be broken up.
(breaking up consecutive 0s) Each code must have no more than one leading 0 and no more than two trailing 0s.
E.g., and are not used.
As a result, the encoded data would not contain more than 3 consecutive 0s.
See Table 2.5 in the text for the encoding.
(breaking up consecutive 1s) The resulting 5-bits are transmitted using the NRZI encoding.

19. 4.4 BS: 4B/5B encoding Data 5-bit codes 0000 11110 0001 01001

20. 5. Problem 2: Frame synchronization (FS)
Problem: Given that a receiver can synchronize bits sent by a sender, how does the receiver recognize bits belonging to the same frame?
Signalling component
Signal
Node
Adaptor
Adaptor
Node
Bits
Frames

21. 5.1 FS: IP over Ethernet
For example, when an IP datagram is handed down to an Ethernet network adaptor, the IP datagram is sent out in an Ethernet frame.
Dest address
Src address
Type
Data
CRC
Type 0800
IP datagram

22. 5.2 Several solutions Byte-oriented protocols (e.g. PPP)
Data unit in terms of bytes (ASCII, EBCDIC)
Sentinel approach vs. byte counting approach
Bit-oriented protocols (e.g. HDLC, Ethernet)
Sentinel approach
Clock-based framing (SONET)
It addresses both the framing and encoding problems.

23. 5.3 Byte-oriented protocols
Sentinel approach to framing
A special character (0x7e) serves as a start-of-text character (flag field)
Character stuffing: escape this special character in the data by prepending it with an escape character 0x7d.
0x7e escaped to 0x7d, 0x5e
0x7d escaped to 0x7d, 0x5d

24. 5.3 Byte-oriented protocols
PPP’s frame structure:
Flag 7E
Addr FF
Control 03
Flag 7E
Protocol
Information
CRC
Protocol 0021
IP datagram

25. 5.3 Byte-oriented protocols
Byte-counting approach to framing, e.g. DECNET’s DDCMP
SYN character serves as a synchronization character.
The COUNT field specifies how many bytes are contained in the frame.

26. 5.4 Bit-oriented protocols
For example, High-Level Data Link Control (HDLC) protocol.
A special bit sequence is transmitted when the link is idle.
When this sequence appears in the data, it is escaped using bit stuffing. 8 16 16 8
Beginning
Ending
Header
Body
CRC
sequence
sequence

27. 5.4 Bit-oriented protocols: bit stuffing
On sending side:
Except for transmitting the special sequence, 0 is inserted after 5 consecutive 1s.
On receiving side:
If five consecutive 1s arrive,
if the next bit is 0, it must be a stuffed bit. It is therefore removed.
If the next bit is 1,
If the next bit is 0, then the sequence indicates the end of the frame.
If the next bit is 1, there must be an error, and the whole frame is discarded.

28. 5.4 Bit-oriented protocols: bit stuffing
For example,
Original bit stream:
After bit stuffing by source:
After bit de-stuffing by receiver:
What happens if transmission errors
occurred to the stuffed bits?
occurred to the nonstuffed bits?

29. 6. Problem 3: Error detection (ED)
Transmission errors do occur, with different probabilities in different media.
Two general approaches:
Error correction code (forward error correction)
Error detection code + an error correction mechanism when errors are detected.
Insert redundancy for error correction or detection.
Common error detection methods:
Cyclic redundancy check (CRC)
Checksum

30. 6.1 ED: error detection codes
Error detection codes are usually inserted in more than one layer, e.g.
HTTP
TCP (16-bit checksum for the TCP header and data)
IPv4 (16-bit checksum for the IP header)
PPP/Ethernet (CRC-16, CRC-32 for the whole frame)
Why don’t we just have CRCs on the datalink layer?

31. 6.2 ED: two-dimensional parity
The 2-dim parity catches all 1-, 2-, and 3-bit errors, and most 4-bit errors.
Parity
bits 1 1
Data 1 1
Parity
byte

32. 6.3 ED: Checksum
Add up all the words that are transmitted and then transmit the result of that sum.
If any transmitted data, including the checksum itself, is corrupted, then the results will not match.
Internet checksum:
A sender add 16-bit words using ones complement arithmetic, and then take ones complement of the result. The result is stored in the checksum field.
A receiver performs 16-bit checksum. If no error, the checksum should give all 1s.

33. 6.3 ED: Checksum Ones complement arithmetic:
A negative integer -x is represented as the complement of x.
A carryout from the most significant bit needs to be added to the result.
For example: 4-bit word checksum
Data: , and its checksum: 1000.
Data sent:
Receiver adds all the data sent in 4-bit word in ones complement, which gives 1111.

34. 6.3 ED: Checksum Advantages: Disadvantages:
Use a relatively small number of bits.
Easy to implement in software.
Disadvantages:
Not a strong error detection algorithm, as compared with CRC

35. 6.4 ED: cyclic redundancy check (CRC)
Think of an (n+1)-bit message, M(x), as a polynomial of degree n, e.g. as
1x7 + 0x6 + 0x5 + 1x4 + 1x3 + 0x2 + 1x1 + 0x0.
Given a divisor polynomial of order k, C(x), find a code word of k bits, such that M(x) concatenated with the code word is divisible by C(x).
P(x) is a concatenation of M(x) and the code word.

36. 6.4 ED: cyclic redundancy check
For example, C(x) = 1x3 + 1x2 + 0x1 + 1x0.
The computed code word is 101.
P(x) is given by concatenated with 101, or 1x10 + 0x9 + 0x8 + 1x7 + 1x6 + 0x5 + 1x4 + 0x3 + 1x2 + 0x1 + 1x0.
One may verify that P(x) is divisible by C(x), i.e. 0 remainder.
Errors can be detected when the remainder is not zero when a receiver divides P(x) by C(x).

37. 6.4 ED: cyclic redundancy check
To obtain the code word,
Divide 1x10 + 0x9 + 0x8 + 1x7 + 1x6 + 0x5 + 1x4 + 0x3 + 0x2 + 0x1 + 0x0 by 1x3 + 1x2 + 0x1 + 1x0 .
Subtraction is performed by an exclusive-OR operation (XOR).
The result is a 3-bit remainder, one bit less than the divisor polynomial.
It can be implemented efficiently in hardware.

38. 6.4 ED: cyclic redundancy check
Generator
1101
Message
1101
1001
1101
1000
1101
1011
1101
1100
1101
1000
1101
101
Remainder

39. 6.4 ED: cyclic redundancy check
Choice of the divisor polynomial:
Express a corrupted message received by a receiver as P(x) + E(x), where E(x) is an error polynomial.
Errors go undetected if E(x) is divisible by C(x).
For example, the CRC can catch
all single-bit errors if the xk and x0 terms in C(x) have nonzero coefficients, and
any odd number of errors if C(x) contains the factor (x+1).

40. 6.4 ED: cyclic redundancy check
Six generator polynomials that have become international standards are:
CRC-8 = x8+x2+x+1
CRC-10 = x10+x9+x5+x4+x+1
CRC-12 = x12+x11+x3+x2+x+1
CRC-16 = x16+x15+x2+1
CRC-CCITT = x16+x12+x5+1
CRC-32 = x32+x26+x23+x22+x16+x12+x11+x10+x8+x7+x5+x4+x2+x+1

41. This will be continued in part 2.