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Data Communications & Computer Networks

Data Communications & Computer Networks. Assignment 1 Solutions Fall 2004. Reminder for the r ules for assignments (1). All assignments must be done individually You should NOT: Copy any part of another student's answers. Allow another student to copy your work.

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Data Communications & Computer Networks

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  1. Data Communications & Computer Networks Assignment 1 Solutions Fall 2004

  2. Reminder for the rules for assignments (1) • All assignments must be done individually You should NOT: • Copy any part of another student's answers. • Allow another student to copy your work. • Present the work of another as your own. If you use the idea of another in your work, you MUST provide appropriate attribution (that is, cite the work and the author).

  3. Reminder for the rules for assignments (2) • It is the student’s responsibility to make up for any missed work due to his/her absence(s). • Assignments should be handed-in on time • 10 November 2004 for assignment 1 • However, they can be submitted up to one week late, but will receive a 10% mark reduction penalty. • NO credit will be given for ANY assignment submitted later than one week from the due date, since we will go over the assignment in class.

  4. Answer to Problem 1 a) Give two reasons for using layered protocols? (4 marks) - Smaller more manageable pieces - Protocols can be changed without effecting higher or lower layers. b) List two main disadvantages to layered approach to protocols (4 marks) - More processing - More overhead due to the headers added by the layers c) A system has an n-layer protocol hierarchy. Applications generate messages of length M-bytes. At each of the layers, an h-byte header is added. What fraction of the network bandwidth is filled with headers? (6 marks) Each message is M-bytes long. There are n-layers each one adding an h-byte header, therefore, each message that is transmitted through the network is (nh+M) bytes long.The fraction of the network bandwidth that is filled with headers is given by nh/(nh+M) If it is assumed that the bottom layer (physical) does not generate any header, then there shall be (n-1)h headers generated, so the fraction of the network bandwidth that is filled with headers is given by (n-1)h/[(n-1)h+M]

  5. Answer to Problem 2 a) What is the difference between a confirmed service and an unconfirmed service? For each of the following, tell whether it might be a confirmed service, an unconfirmed service, both or neither: (7 marks)      (i)         Connection establishment      (ii)        Data transfer (iii)       Connection release In a confirmed service, there is a request, an indication, a response and a confirmation. In an unconfirmed service, there is just a request and an indication. Connection establishment: Confirmed service, since an explicit response is required Data transfer: Can be confirmed or uncorfirmed, depending whether or not the sender needs an acknowledgement Connection release: This is a disconnect service, therefore it is unconfirmed since there is no response

  6. Answer to Problem 2 b) What is the principal difference between connectionless communication and connection-oriented communication? Is it possible to have both of them in two adjacent layers? Discuss! (6 marks) Connection-oriented service is modelled after the telephone system. To use a connection-oriented network service, the service user first establishes a connection, uses the connection and then it releases the connection. Connectionless service is modelled after the postal system. Each message (letter) carries the destination address and each one is routed through the system independent of all the others. Yes, it is possible to have connection-oriented and connectionless service in adjacent layers. Example is TCP/IP. TCP is connection-oriented protocol that lies in transport layer (layer 4), IP is connectionless that lies in network layer (layer-3). c) What is the main difference between TCP and UDP protocols? Give an example of a service these protocols can support? (6 marks) TCP provides reliable connection-oriented protocol that delivers a byte stream from one node to another, guaranteeing delivery and provides flow control. TCP can be used in file transfer applications. UDP provides unreliable connectionless protocol for applications. UDP can be used in applications for carrying voice traffic over packet-switched networks.

  7. Answer to Problem 3 a) Contrast and criticize OSI and TCP reference models outlining their similarities and differences. (6 marks) Both models have network, transport and application layers with similar functionalities. Both are based on the concept of a stack of independent protocols. However, OSI supports both connectionless and connection-oriented communication in the network layer but only connection-oriented communication in the transport layer. TCP/IP has only connectionless mode in the network layer but supports both connectionless and connection-oriented communication in the transport layer. OSI model : • Has been devised before the corresponding protocols were invented • Has good definition of service, interface, and protocol • Fits well with object oriented programming concepts • Protocols are better hidden TCP model : • the protocols came first, the model was just a description of the protocols • the model isn't good for any other protocols part from TCP/IP. A critique of OSI model: • Bad Timing (TCP already in use by the time OSI came along) • Bad Technology (Layers don't really match. Dominated by phone company mentality) • Bad Implementation (Huge, unwieldy, slow). A critique of TCP/IP model: • Doesn't separate specification from implementation. • Model is only good for describing TCP. • Doesn't specify physical and data link layers.

  8. Answer to Problem 3 b) Which layer of the OSI reference model is responsible for the following: (6 marks) (i)Negotiating data transfer syntax Presentation (ii)Addressing devices and routing through an internetwork Network (iii)Framing Data Link (iv)Flow control, acknowledgement, windowing Transport (v)Coordinating communications between systems Session (vi)Synchronizing sending and receiving applications Application c) At which layer of the OSI reference model are the following components positioned? (4 marks)                   (i)        Router Layer 3 - Network                   (ii)        Repeater Layer 1 – Physical                   (iii)        Switch Layer 2 – Data Link                   (iv)        Hub Layer 1 - Physical

  9. Answer to Problem 4 a) A noiseless channel has a bandwidth of 10kHz. If digital quaternary signaling is used (i.e. four voltage levels per symbol)  (i)   What is the maximum bit rate (capacity) of the channel? (4 marks) The maximum bit rate for a noiseless channel of 10kHz bandwidth with quaternary signalling is 2 x BW x log2n = 2 x 10 kHz x 2 = 40kbps, using Nyquist equation. (ii) How would (i) change if the channel signal to noise ratio is 30dB? (4 marks) If the channel SNRdB=30 dB then we must use Shannon’s equation, that is Maximum bit rate = BWx log2(1+SNR). For a SNRdB of 30dB, the SNR ratio is 103=1000 since 10 log10(SNR)=30 dB So, the maximum bit rate is 10kHz x log2(1+1000)=99.67kbps

  10. . . . . . . . . Answer to Problem 4 (iii) One way to increase the maximum bit rate is to change the encoding and use phase modulation together with amplitude modulation. This will increase channel capacity. For the constellation pattern shown below, find the maximum bit rate of the channel, assuming that the channel is noise free. (4 marks) Assuming a noiseless channel, the maximum bit rate is 2 x BW x log2(n) = 2 x 10kHz x 3 = 60kbps, where n=8 states.

  11. Answer to Problem 4 (b) A signal described by the following equation is inserted through a noisy channel of 13dB signal-to-noise ratio. What is the maximum achievable data rate? (6 marks) where k= 1, 3, 5 The above equation can be expanded as follows s(t)=(4/π){sin(20000t)+(1/3)[sin(60000t)]+(1/5)[sin(100000t)]} This equation is in analogy with the general Fourier series equation for k=1, 3, 5 i.e. s(t)=(4/π){sin(2πft)+(1/3)[sin(2π(3f)t)]+(1/5)[sin(2π(5f)t)]} Bandwidth is the highest frequency component minus the lowest frequency component of the signal. The highest frequency component is given by sin(2π(5f)t) and the lowest by sin(2πft). So, 2πft=20000t => 2πflowestt = 20000t => flowest=20000/2π = 3.183kHz 2π(5f)t=100000t => 2πfhighestt = 100000t => fhighest = 100000/2π = 15.915kHz Bandwidth = fhighest -flowest = 15.915 – 3.183 = 12.732kHz For a SNRdB of 13dB, SNRdB=10 log10(SNR)=13 dB, so SNR ratio is 101.3=19.95 Maximum data rate as given by Shannon’s equation is BW x log2(1+SNR) = 12732 x log2(1+19.95) = 55.88kbps

  12. Answer to Problem 5 a) Briefly differentiate between copper and fiber optic cables for establishing a communication channel in terms of bandwidth, interference, flow of information and cost. (8 marks) b)A leased line is known to have a loss of 40dB. The output signal power is measured as 7mW and the output noise level is measured as 3.5μW. Using this information calculate the output signal-to-noise ratio in dB. (6 marks) SNRdB=10 log10 (SNR)= 10 log10(Pout/Nout)= 10log10(7x10-3/3.5x10-6)= 33dB MetricFiberCopper Bandwidth High Low Interference Low High Flow of Information Uni-directional Bi-directional Cost High Low

  13. Answer to Problem 6 (a) By halving the transmission frequency as well as halving the distance between transmitting and receiving antennas by how many decibels is the received signal power improved or reduced? (8 marks) Assume that tx and rx antennas are d1 meters apart at a frequency f1. Then (Pt/Pr)1 = (4πd1)2/λ12 = (4π f1 d1)2/c2 If frequency is halved, i.e. f2=f1/2, and distance is halved, i.e. d2=d1/2 then (Pt/Pr)2 = (4πd2)2/λ22 = (4π f2 d2)2/c2 = (4π (f1/2) (d1/2))2/c2 = (1/16)(4π f1 d1)2/c2 = (1/16)(Pt/Pr)1 Therefore, signal reduces by 10log10 [(Pt/Pr)2 / (Pt/Pr)1 ] = 10log10 (1/16)= -12dB

  14. Answer to Problem 6 b)You are receiving television signals from a synchronous geostationary satellite 40,000 km away at 11.75 GHz using a parabolic antenna.    (i) What is the free space loss in decibels? (5 marks) The free space loss in ratio is Pt/Pr = (4πd)2/λ2 = (4π f d)2/c2 or, in decibels: LdB = 10 log10 (Pt/Pr) = -20log(λ) + 20log(d) + 21.98 dB  where Pt = signal power at tx antenna, λ = carrier wavelength in m Pr = signal power at rx antenna, f = carrier frequency, c = speed of light (~3x108 m/s), d= distance between antennas in m Since f=11.75GHz, then λ=c/f = 3x108/11.75x109 = 0.02553meters Hence, LdB = -20log(λ) + 20log(d) + 21.98 dB => -20log(0.02553) + 20log(40000x103) + 21.98 = 31.87 + 152.04 + 21.98 = 205.9dB

  15. Answer to Problem 6 (ii) Assuming that the antenna gain of both the satellite and ground-based earth station are 45dB and 50dB, respectively, and that the earth station transmits at an output power of 200W, what is the power received at the satellite antenna? (6 marks) Assuming that the antenna gain of both the satellite and ground-based earth station are 45dB and 50dB the free space loss is LdB = 205.9 – 45 – 50 = 110.9dB Since the Earth station transmits at an output power of 200W, a power of 200W translates into 10log(200) = 23dBW, so the power received at the receiving satellite antenna is 23-110.9 = -87.9 dB PowerdB=10 log10Power=-87.9 dB => Power=10(-87.9/10)=1.62x10-9= 1.62nW

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