ccm4010 - revision introduction, computer networks, standards, osi, tcp/ip, lans, mans, wans...
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CCM4010 - Revision
• Introduction, computer networks, standards, OSI, TCP/IP, LANs, MANs, WANs
• Wireless LANs• Data Compression• Error Detection/Correction• Routing• Network Management• Network Security
OSI Reference Model
• Seven layered structure– Application– Presentation– Session– Transport– Network– Data Link – Physical
OSIRM
• Functions of each layer• Relationships between various networking
standards and the OSIRM– LAN standards
• IEEE802.x
• FDDI
• DQDB
– ATM
– etc.
LANs/MANs/WANs
• IEEE802.11 – wireless LAN• HiperLAN - wireless LAN (Europe)• IEEE802.x – LANs• 10Base2/10Base5/10Base T/100Base T etc.• ATM• DQDB• FDDI – SAS/DAS• SDH/SONET• Selection criteria
Wireless LANs and Mobile Systems
• Infrastructure
• Ad-hoc
• WAP
• i-mode
• UMTS
• Location technology:– Global Positioning System (GPS)
Transmission Media and supporting technologies
• ISDN
• ADSL
• PSTN
• Satellite link
• Data link using mobile telephony
• Microwave
• The Internet
Delay calculations
• Propagation time = distance/speed
• Transmission time = Number of bits in a frame/bit rate
• Total time = outward propagation time + outward transmission time + inward propagation time + inward transmission time
Data Compression
• Packed decimal
• Relative encoding
• Character suppression
• Huffman coding– Static – dynamic
• Limpel-Zev coding
Huffman coding
• Statistical encoding
• Analyse the characters to be transmitted
• Determine character types and their relative frequency
• Create an unbalanced tree (Huffman code tree/ binary tree)
• Obtain codeword for each character
• Calculate the number of bits needed to transmit these characters
• Calculate the average number of bits per codeword needed to transmit these characters
Huffman coding
• Example:– “AGAIN AND AGAIN AND AGAIN”
• Determine character types and their relative frequencyA 8; N 5; Sp 4; G 3; I 3; D 2. Total 25
• Create an unbalanced tree (Huffman code tree/ binary tree)• Obtain codeword for each character
A 00; N 11; Sp 010; G 011; I 100; D 101
• Calculate the number of bits needed to transmit these charactersTotal bits 62
• Calculate the average number of bits per codeword needed to transmit these characters62/25 = 2.48 b/character
Limpel-Zev Coding
• Example:– “AGAIN AND AGAIN AND AGAIN”– Ad a space to make the number of characters even (26)– “AGAIN AND AGAIN AND AGAIN ”
• Take pair of characters together and determine their frequencyAG 3; AI 3; NSp 3; AN 2; DSp 2; Total 13
• Create binary tree• Obtain codeword for each character
AG 11; AI 00; NSp 01; AN 100; DSp 101
• Calculate the number of bits needed to transmit these charactersTotal bits 30
Compression
• A very simple stratagem applied to the string could reduce the number of bits needed for transmission of the string still further. Suggest and validate, a suitable stratagem
• AnswerPut the extra space at the beginning of the string. There
are now only four pairs and the total number of bits is 26.
SpA; GA; IN; ND;
Huffman - decoding
• A code table must be available
• Read data as received, compare to known codes, – if it is a match, identify code, – if not keep reading and concatenating bits
received
Huffman - decoding
• Example001101000000010110011010011010001
Character coding:
E = 1; T = 01; G = 001; C = 0000; A = 0001
001 1 01 0000 0001 01 1 001 1 01 001 1 01 0001
G E T C A T E G E T G E T A
Error control
• ASCII• Parity bit: additional bit added to each
character– Even parity
• Total number of bits (inclusive) in a digital word is even; e.g. 10110111
– Odd parity• Total number of bits (inclusive) in a digital word is
odd; e.g. 10110110
Error control
• Hamming distance: The Hamming distance of the code is the minimum number of bit positions in which two valid codewords differ.00000000
00000011
00000101
00000110
Hamming distance?
Odd/even parity?
Hamming Codes
• The parity bits are inserted into the positions which numerically are the powers of 2 (i.e. 1,2,4,8,16 etc.)
• Parity bit 1 covers bits 1,3,5,7,9,…• Parity bit 2 covers bits 2,3,6,7,10,11,14,15,..• Parity bit 4 covers bits
4,5,6,7,12,13,14,15,20,21,22,23,…• Parity bit 8 covers bits
8,9,10,11,12,13,14,15,24,25,26,27,28,29,30,31,…
Network Management
• Functions– Fault management– Performance management– Security– Accounting– etc.
• Characteristics of network managers• Other network management topics
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