ECP2056DataCommunicationsAndComputerNetworking2

ECP2056DataCommunicationsAndComputerNetworking2 - STUDENT...

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Unformatted text preview: STUDENT ID N O MULTIMEDIA UNIVERSITY SECOND TRIMESTER EXAMINATION , 2002 /2003 SESSION ( Main Paper) ECP2056 —— Data Communications and Computer Networking (All Groups) . ~ 18 November 2002 .- 2.30 AM — 4.30 AM (2 Hours ) INSTRUCTION TO STUDENT 1. This Question paper consists of 9 pages with 6 Questions only. 2. Attempt any FOUR out of SIX questions. All questions carry equal marks and the distribution of the marks for each question is given 3. Please print all your answers in the Answer Booklet provided. ECP 2056 Data Communications and Computer Networking 18 Nov 2002 (1) (a) Identify and explain three possible data communication modes in communication systems. Give one example for each 0f the communication modes.’ [ 6 marks ] (b) Determine the Huffman codes for the following group of symbols: ABBCACCDECDCCAAEAE Compare the Huffman code representation with the ASCII representation in terms of average bit per symbol. [ 8 marks ] (c) Illustrate how bit synchronisation can be achieved in asynchronous transmission. [ 6 marks ] (d) Error control in transmission systems can be divided into two categories: forward error control and feedback error control. Which of these categories do parity bit check, block sum check and cyclic redundancy check belong to? In a particular data bit transmission, the following bits were transmitted and received: 10010101 00011110 11101110 00111100 01011001 [transmitted] 10110001 00011110 11101110 00011000 01011001 [received] where block sum check is the error control techniques employed with underlined bits are parity bits and even parity check being adopted for both row and column parity check. Will there be any error detected at the receiver? [Smarks] Continued ..... .. SWL 2/9 ECP 2056 Data Communications and Computer Networking 18 Nov 2002 (2) (a) Two modems, A and B are available to connect a computer to the Internet service provider Via the telephone line. Modem A’uses the quadrature phase shift keying (QPSK) with 4 constellation points while modern B uses quadrature amplitude modulation (QAM) with 16 constellation points. The baud rate for the modems is 2400 baud/s. (i) Determine the bit rate of Modern A. [ 3 marks ] (ii) Determine the bit rate of Modern B. [ 3 marks ] (iii) A 256-level grey scale picture with 512x512 pixels is to be transmitted over the telephone line to the Internet using the modems. Assuming no other overheads are involved, determine how long it takes to transmit the picture. [4 marks ] (iv) Which modern has higher resilience to transmission channel noise? Explain. [ 4 marks ] (b) Explain why V.90 modem cannot provide an upstream bit rate of 56kbps as in the case of downstream data traffic. [4 marks] (0) What is the maximum achievable bit rate between two computers connected to the network Via V.9O modems and the telephone lines? Explain. [ 3 marks ] (d) Shanon’s theorem on information capacity of a transmission channel is given by C = B log2(1 + SNR) where C, B and SNR are the bit rate, bandwidth and signal-to- noise ratio respectively of the channel. Apply the theorem to determine the maximum achievable capacity of the analog telephone line with B = 2900Hz and SNR = 35dB. How does V.90 modem overcome the capacity limit to achieve downstream bit rate of 56kbps? [4marks] Continued ..... .. W 3/9 SWL ECP 2056 Data Communications and Computer Networking 18 Nov 2002 (3) (a) Identify and briefly explain layers in the Open System Interconnection (OSI) reference model developed by ISO. [ 7 marks ] (b) Explain the functions of the following fields in the TCP header (F 1g Q3.1): (i) Source port address [ 2 marks ] (ii) Sequence number [2 marks ] (iii) Window size [ 2 marks ] Header- ‘_~~ Source port addressiftfi bits] Destination port address.[16 bits] Sequence number [32 bits] : Acknowledgement. number [32 bits} HLEN Reserved U A- P R ’5 F . ;Z' R c S '5 Y i W -‘ 1 1:. this] I N N WWI “"51 1.; [4 bits} Checksum {16 bits} . Urgent pointer [16 bits} 5' Optional [if any] Fig. Q3.1 (c) For the timing diagram of the Idle RQ scheme given in Fig. Q32 show that the link 1 l+2a utilisation, U can be approximated as U : where a = Tp/Tix. State the assumptions that you make in deriving U. Determine the link utilisation for the following data link types assuming an average data frame size of 1024 bits, data transmission rate of 2.4kbps and link signal propagation velocity of 2.5 x 108 ms". (i) Twisted pair cable of 1km in length. [ 3 marks] Continued ..... .. ECP 2056 Data Communications and Computer Networking 18 Nov 2002 (ii) A leased line of IOOOkm in length. [ 3 marks ] (iii) What can you conclude from the results in (c)(i) and (c)(ii). [ 2 marks ] Timer stopped Timer started Secondary, S T» 9 = Frame propagation delaytP-és] Ti,r 1".: = Frametransmission time {P631 Tic Tip : Frame processing time in S 7": T9 = ACK propagation delay {3-913} Tax Ta = ACK transmission time [89?] Tap Tap : ACK gracessing time is P Fig. Q32 Continued . .. SWL 5/9 ECP 2056 Data Communications and Computer Networking 18 Nov 2002 (4) (a) Identify and describe briefly four different local area network (LAN) topologies. [8marks] (b) Carrier sense multiple access / collision detection (CSMA/CD) is a multiple access technique used in LAN. Describe briefly how it works. Explain how collision can occur under the scheme. [6marks] (0) Consider a bus topology as shown in Fig. Q4. 1. Assume the bit rate of the network system is lOMbps, the signal propagation speed on the medium is 2 x 108 ms'1 and CSMA/CD access scheme is adopted. (i) What is the minimum MAC frame length in terms of L? [ 3 marks] (ii) Determine the efficiency of the network if the frame length determined in (c)(i) is used. [2 marks ] (iii) What are the effects of the frame length and L on the efficiency of the network? If a throughput of 6Mbps is to be achieved, frame length. “ propose possible values of L and [6marks] Fig. Q4.l Hint : The efficiency of the network is U = 1 15 where a is the ratio of propagation + a delay to frame transmission time. Continued ..... .. M 6/9 ECP 2056 Data Communications and Computer Networking 18 Nov 2002 overhead in circuit switching. Because there is no overhead in circuit switching, line utilisation must be more eflicient than packet switching. ” Comment on the above statement. [ 6 marks ] (c) One key design decision for ATM was whether to use fixed- Let us consider this decision from the point of View of efficie transmission efficiency as or variable-length cells. ncy. We can define N _ Number of information octets Number of information octets + Number of overhead octets the header octets. Define L = Data field size of the cell in octets H=Header size of the cell in octets X: Number of information octets to be transmitted as a single message Derive an expression for transmission efficiency, need to use the operator where rYl equal to Y. N. Hint : The expression will = the smallest integer greater than or Determine when the optimum transmission efficiency, Nap, occurs. [4marks] (ii) If cells have variable length, then overhead is determined by the header, plus the flags to delimit the cells or an additional length field in the header. Let Hv = additional overhead octets required to enable the use of variable-[en Derive an expression for the transmission efficiency, NW Hv. Assume that the entire X octets can fit into a single gth cells. in terms of X, H, and variable-length cell. [4marks] (iii) For L=48, H=5 and Hv = 2, plots of N, Nap, and NW , versus message size are given in Fig. Q5. 1. Comment on the results. [ 5 marks] Continued ..... .. SWL ECP 2056 Data Communications and Computer Networking 18 Nov 2002 8/9 Continued ECP 2056 Data Communications and Computer Networking 18 Nov 2002 (b) State and explain 3 major operational issues in intemetworking. [9 marks] (0) Based on the Fig. Q6.l, determine the IP address classes of the following IP addresses: 57.24.32.33 129.220.13.78 191720.118 223.128.128.128 225.0.119.22 243.77.99.23 [ 6 marks ] Bilorder 1 8 16 24 32 “:34 7 i i 2. i i 12! 14 i re 1 3 I 21 I 8 i 4 I 28 i "a multicast address = Multicast i 4 I 28 I = Reserved netid = Network identifier hostid = Host identifier Fig. Q6.l (d) Describe the principle of subnetting. For a network with Class B IP address of 144. 220.X.X subnets within the network. Propose a suitable subnet m network. , it is desired to have 68 ask to be deployed in the [4marks] End of Page ...
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ECP2056DataCommunicationsAndComputerNetworking2 - STUDENT...

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