Data Communications

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 students 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.

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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 (nhM) bytes long. The fraction of the network
  bandwidth that is filled with headers is given by
  nh/(nhM)
• 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)hM

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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

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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.
•  

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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.

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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

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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 SNRdB30 dB then we must use
  Shannons equation, that is
• Maximum bit rate BW x log2(1SNR).
• For a SNRdB of 30dB, the SNR ratio is 1031000
  since 10 log10(SNR)30 dB
• So, the maximum bit rate is 10kHz x
  log2(11000)99.67kbps

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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 n8 states.
•  

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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/p)sin(20000t)(1/3)sin(60000t)(1/5)si
  n(100000t)
•  
• This equation is in analogy with the general
  Fourier series equation for k1, 3, 5 i.e.
• s(t)(4/p)sin(2pft)(1/3)sin(2p(3f)t)(1/5)sin
  (2p(5f)t)
•  
• Bandwidth is the highest frequency component
  minus the lowest frequency component of the
  signal. The highest frequency component is given
  by sin(2p(5f)t) and the lowest by sin(2pft). So,
• 2pft20000t gt 2pflowestt 20000t gt
  flowest20000/2p 3.183kHz
• 2p(5f)t100000t gt 2pfhighestt 100000t gt
  fhighest 100000/2p 15.915kHz
• Bandwidth fhighest - flowest 15.915 3.183
  12.732kHz
•  
• For a SNRdB of 13dB, SNRdB10 log10(SNR)13 dB,
  so SNR ratio is 101.319.95
  
• Maximum data rate as given by Shannons equation
  is
• BW x log2(1SNR) 12732 x log2(119.95)
  55.88kbps

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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)
•  
• SNRdB10 log10 (SNR) 10 log10(Pout/Nout)
  10log10(7x10-3/3.5x10-6) 33dB
•  

Metric Fiber Copper Bandwidth
High Low Interference Low High Flow of
Information Uni-directional Bi-directional Cost
High Low
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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
  (4pd1)2/?12 (4p f1 d1)2/c2
• If frequency is halved, i.e. f2f1/2, and
  distance is halved, i.e. d2d1/2 then
• (Pt/Pr)2 (4pd2)2/?22 (4p f2 d2)2/c2 (4p
  (f1/2) (d1/2))2/c2
• (1/16)(4p f1 d1)2/c2 (1/16)(Pt/Pr)1
• Therefore, signal reduces by 10log10 (Pt/Pr)2 /
  (Pt/Pr)1 10log10 (1/16) -12dB

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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 (4pd)2/?2
  (4p f d)2/c2
• or, in decibels LdB 10 log10 (Pt/Pr)
  -20log(?) 20log(d) 21.98 dB
•  
•  where Pt signal power at tx antenna, ?
  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 f11.75GHz, then ?c/f 3x108/11.75x109
  0.02553 meters
• Hence, LdB -20log(?) 20log(d) 21.98 dB gt
  -20log(0.02553)
  20log(40000x103) 21.98
• 31.87 152.04 21.98
  205.9dB

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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
• PowerdB10 log10Power-87.9 dB gt
• 
  Power10(-87.9/10)1.62x10-9 1.62nW