FACULTY OF SCIENCE AND AGRICULTURE
SPRING SESSION EXAMINATION 2004
ITC242 INTRODUCTION TO DATA COMMUNICATIONS
DAY & DATE:
WRITING TIME:
TIME:
Three (3) Hours
READING TIME: Ten minutes
MATERIALS SUPPLIED BY UNIVERSITY:
1 x 24pp Answer Booklet
General Purpose Answer Sheet
MATERIALS PERMITTED IN EXAMINATION:
Pens and Pencils, Calculator
2B pencil, eraser
NUMBER OF QUESTIONS:
Part A: Thirty (30) multiple
choice questions
Part B: Ten (10) short answer
questions
Part C: One (1) essay question
INSTRUCTIONS TO CANDIDATES:
1.
2.
3.
4.
5.
6.
Mark your answers to the multiple choice questions, Part A, on the general purpose
answer sheet.
Write your answers to the written questions, Parts B and C, in the booklet provided.
Multiple choice questions are worth one mark each. (Total of 30 marks)
Short Answer questions are worth 5 marks each. (Total of 50 marks)
Essay question is worth 20 marks. (Total of 20 marks)
A total mark possible is 100.
INSTRUCTIONS TO INVIGILATORS:
1.
QUESTION PAPER MUST BE COLLECTED
STUDENT NAME:
STUDENT SIGNATURE:
STUDENT NO:
PART A: Multiple Choice Questions
(Each question is worth 1 mark each. Answer the questions on the general
purpose answer sheet provided).
1. At which layer of the TCP/IP protocol suite would you find an email program?
a.
b.
c.
d.
Transport
Presentation
Application
Session
Discrete
2. Using Shannon’s theorem, S(f) = f  log2 (1 + W/N), where S(f) is the maximum data
transfer rate, f is the frequency of the signal, W is the power of the signal in watts and N is the
power of the noise in watts, calculate the data transfer rate given the following information:
signal frequency = 10,000 Hz, signal power = 5000 watts, noise power = 1,666 watts.
a.
b.
c.
d.
10,000 bps
20,000 bps
30,000 bps
40,000 bps
3. Given Nyquist’s theorem, C = 2f  log2(L) where C is the channel capacity, f is the
frequency of the signal, and L is the number of signal levels, what is the channel capacity of
a signal that has 8 different levels and a frequency of 10,000 Hz?
a.
b.
c.
d.
18,062 bps
30,000 bps
60,000 bps
160,000 bps
4. Which term describes the process where data is packaged with necessary protocol
information as it passes through layers of the OSI reference model?
a.
b.
c.
d.
Wrapping
Encryption
Encapsulation
Protocolisation
5. The sharing of a medium and its link by two or more devices is called
a.
b.
c.
d.
Modulation
Encoding
Line discipline
Multiplexing
6. What component of a coaxial cable makes it less susceptible to noise than twisted-pair
cable?
a.
b.
c.
d.
Outer conductor shielding
A single inner conductor
Twisted core conductor pairs
Diameter of cable
7. What is the maximum baud rate of a digital signal that employs Differential Manchester
encoding and has a data rate of 1000bps?
a.
b.
c.
d.
500 baud
1000 baud
2000 baud
4000 baud
8. The bandwidth for a data transmission line that transmits within a frequency range of 100
Hz to 3,500 Hz is:
a.
b.
c.
d.
3,000 Hz.
3,400 Hz.
3,500 Hz.
3,600 Hz.
9. The ____________________ of a signal is the number of times a signal makes a complete
cycle within a given time frame.
a.
b.
c.
d.
Bandwidth
Frequency
Amplitude
Spectrum
10. In a sliding window Go-Back-N ARQ system, A sends packets 0,1,2,3,4,5 and 6. Packet 3
arrives at B corrupted. What do A and B send to each other next?
a.
b.
c.
d.
B sends REJ-3 , A then sends packets 3,4,5,6,7,0 and 1
B sends REJ-2, A then sends packets 3,4,5,6,7,0 and 1
B sends REJ-3, A then sends just packet 3
B sends REJ-2, A then sends just packet 3
11. In a sliding window scheme, acknowledgements contain a value _____ the number of the
next expected packet.
a.
b.
c.
d.
greater than
equal to
less than
equal to or greater than
12. The _____ is the data transfer rate agreed on by both customer and frame relay carrier.
a.
b.
c.
d.
constant bit rate
customer transfer rate
committed information rate
approved data rate
13. A logarithmic loss as a function of distance and the resistance within the wire is:
a.
b.
c.
d.
attenuation
noise
amplitude
spectrum
14. _____ uses frequency hopping and direct sequence techniques.
a.
b.
c.
d.
Frequency modulation
Differential Manchester
Pulse code modulation
Spread spectrum
15. Which of the following is FALSE regarding transmission impairments?
a. Thermal noise is distributed across the frequency spectrum and is therefore difficult
to eliminate.
b. Impulse noise is often referred to as white noise.
c. Attenuation causes an electromagnetic signal to become gradually weaker over
distance.
d. Guided transmission media such as physical cable can be designed to reduce the
affects of Cross Talk noise.
16. A network style where the application and presentation logic resides on the client and the
data storage logic resides on a server is called:
a.
b.
c.
d.
Client/Server
Client Based
Host Base
Peer-to-Peer
17. The physical layout of a Local Area Network is called
a.
b.
c.
d.
Contention
Nodes
Topology
None of the above
18. Ethernet is said to be non-deterministic because of which of the following?
a. It is not possible to determine how long it will take to get a frame from one device to
another.
b. It is not possible to determine whether an error has occurred during the transmission
of a frame.
c. It is not possible to determine if another device wishes to transmit.
d. It is not possible to determine the maximum time a device will have to wait to
transmit.
19. When a bridge receives a packet that has a destination MAC address located on a different
segment from which it came, the packet is:
a.
b.
c.
d.
Forwarded
Blocked
Repeated on all segments
All of the above
20. The purpose of the token in a token ring is:
a.
b.
c.
d.
To control who gets to transmit next
To listen to medium, if no one is transmitting, then allow transmission
To listen for collisions
To eliminate broadcasts
21. ISDN basic rate interface (BRI) multiplexes _______________ separate channels.
a.
b.
c.
d.
One
Two
Three
Four
22. The start and stop bit in Asynchronous transmission is used to ensure:
a.
b.
c.
d.
that the total number of bits is a factor of 10
a continuous stream of data
that single bit errors do not occur
the receiver maintains synchronization with the sender
23. ____________________ is a very powerful error detection technique and should be
considered for data transmission systems.
a.
b.
c.
d.
Vertical redundancy check
Cyclic redundancy checksum
Simple parity
Horizontal parity
24. The CSMA/CD network is described by which frame specification?
a.
b.
c.
d.
IEEE 802.2
IEEE 802.3
IEEE 802.4
IEEE 802.5
25. In a ____________________ subnet, no unique dedicated physical path is established to
transmit the data packets across the subnet.
a.
b.
c.
d.
Circuit-switched
Packet-switched
Large
Heavily loaded
26. The length of an IP address is:
a.
b.
c.
d.
8 bits
16 bits
32 bits
48 bits
27. What is the protocol that allows dynamic assignment of IP Addresses to workstations?
a.
b.
c.
d.
ICMP
DHCP
SMTP
SNMP
28. The equation to calculate availability is?
a.
b.
c.
d.
(Total Available Time + Downtime) * Total Available Time
(Total Available Time - Downtime)
(Total Available Time + Downtime)
(Total Available Time - Downtime) / Total Available Time
29. With Public Key Encryption:
a. messages that cannot be encoded are returned to the sender
b. data compression is provided
c. knowledge of the encryption algorithm and the encryption key determines the
decryption key;
d. a pair of keys are needed for encryption and decryption of messages
e. none of the above
30. A _____ cipher replaces a character or group of characters with a different character or
group of characters.
a.
b.
c.
d.
polyalphabetic substitution-based
monoalphabetic substitution-based
transposition-based
network-based
PART B: Short Answer Questions
(Each question is worth 5 marks each. Answer the questions in the Answer
booklet provided. Use diagrams where appropriate to enhance your answers.)
1. A 30 minutes telephone conversation is recorded and stored digitally using Pulse
Code Modulation at 8000 samples per second. If each sample is encoded as 8-bits,
how many floppy disks are required to store the conversation if each floppy disk can
take 1.44 MB? (Show all working and assumptions)
Storage Required = (30*60) * 8000 * 8
=115,200,000 bits
Floppy Size = 1,440,000 * 8
= 11,520,000 bits
* Assuming 8 bits in a byte, and 1.44Mb is 1,440,000 bytes.
Number of floppies required = 115,200,000 / 11,520,000
= 10.
5 marks for correct answer, working and assumptions.
3 marks for just correct answer without working.
Between 1 and 4 marks for working but without correct answer.
2. On a certain network 1,000,000 bytes of data needs to be transmitted using
Asynchronous transmission, using 7 data bits, 1 start bit, 2 stop bits and 1 parity bit. If
the transmission is at 56kbps, how long would it take to complete the transmission
and what would be the overhead in time? Assume there are 8 bits in a byte. (Show all
working and assumptions)
Data to transmit = 1,000,000 * 8
= 8,000,000 bits
Overhead = int ((data to transmit / no. chars )) * bits of overhead per char
= (8,000,000 / 7) * 4
= 1,142,858 * 4
= 4,571,432 bits
Time to transmit = (data + overhead) / rate
= (8,000,000 + 4,571,432) / 56,000
= 224.5 seconds
= 3 minutes 45 seconds
Time to transmit overhead = overhead / rate
= 4,571,432 / 56,000
= 81.6 seconds
= 1 minute 22 seconds
2.5 marks for each correct answer with working and assumptions.
1.5 marks for each correct answer without working.
Between .5 and 2 marks for each question for working but without correct answers.
3. On a certain network 1,000,000 bytes of data needs to be transmitted using
Synchronous transmission where each frame is 8000 bits in total of which 48 bits are
overhead. If the transmission is at 56kbps, how long would it take to complete the
transmission and what would be the overhead in time? (Show all working and
assumptions)
Data to transmit = 1,000,000 * 8
= 8,000,000 bits
Overhead = int ((data to transmit / no. frames )) * bits of overhead per char
= (8,000,000 / (8000 – 48)) * 48
= 1007 * 48
= 48,336 bits
Time to transmit = (data + overhead) / rate
= (8,000,000 + 48,336) / 56,000
= 143.7 seconds
= 2 minutes 24 seconds
Time to transmit overhead = overhead / rate
= 48,336 / 56,000
= .86 second
= 1 second
2.5 marks for each correct answer with working and assumptions.
1.5 marks for each correct answer without working.
Between .5 and 2 marks for each question for working but without correct answers.
4. What do we mean when we say that a signal is “analogue”?
Analogue: (maximum of 5 marks)
 “represented as a continuous waveform that can be at an infinite number of
points between some given minimum and maximum” (White 2002) 3 marks
 Difficult to separate noise from the original waveform. 1 mark
 Has amplitude, frequency and phase. 1 mark
 Examples include audio such as voice, music. 1 mark.
5. Describe the function of a router. At what layer of the OSI model does it operate?
Router: (maximum of 5 marks)
 Main function of a router is path determination, or path selection. 1.5 marks
 Operates at layer 3 – Network Layer. 1.5 marks
 Also can perform packet filtering. 1 mark
 Uses Layer 3 addresses, or IP addresses. 1 mark
 Can connect a LAN to a WAN. 1 mark
 Other correct points. 1 mark
6. List the types of transmission impairment that unshielded twisted pair cable can
suffer from. Provide a clear definition for each impairment.
Types of transmission impairment:





White noise, or thermal noise or Gaussian noise. A constant noise, always
present, depends on temperature. 1 mark
Impulse noise. A spike of non-continuous noise, one of the most difficult to
detect, occurs at random, caused by external influences. An analogue burst of
energy. 1 mark
Crosstalk. Unwanted coupling between two different signal paths. Relatively
constant, can be reduced by proper precautions such as correct terminations.
Types of crosstalk include near end cross talk, far end crosstalk. 1 mark
Jitter. Small timing irregularities during the transmission of digital signals
that become magnified as the signals are passed from one device to another. 1
mark
Attenuation. Continuous loss of signal strength as it travels through a medium.
1 mark
7. What is the difference between a deterministic and a non-deterministic protocol?
A deterministic protocol is on where it is possible to calculate the maximum time a
device may have to wait before being able to transmit. Example – token ring. 2½
marks
In a non-deterministic protocol you can not determine how long a device might have
to wait before being able to transmit. Example – CSMA/CD or Ethernet. 2 ½ marks
8. What are three relative advantages and three relative disadvantages of circuit
switching and packet switching?
Advantages
Disadvantages
Circuit switching
 Once circuit established data
travels quickly – low latency
 Simple – data does not need
addressing
 Guaranteed bandwidth when
circuit established
.5 each maximum 1.5 marks

Takes time to establish and tear
down circuit
 Inefficient, each circuit dedicated
to one connection
 Inefficient, circuit may not be
fully utilized
 Fixed path – does not deal with
congestion or link failure as well
.5 each maximum 1.5 marks
Packet switching
 Efficient – packets share link
bandwidth
 Variable paths may be used – greater
flexibility in path selection
 Dynamic path selection
 Does not require circuit establishment
at start and tear down at end
.5 each maximum 1.5 marks
 Longer latency
 Processing required at each node –
more complex
 Data requires addressing
.5 each maximum 1.5 marks
Maximum total 5 marks.
9. Name four (4) types of multiplexing giving a brief explanation of each.




TDM: Sharing of a signal by dividing available transmission time on a
medium among users. 1 mark
Can have Synchronous TDM and Statistical TDM. 1 mark
FDM: Frequency Division Multiplexing. The assignment of non-overlapping
frequency ranges to each user of a medium. 1 mark
CDM: Code Division Multiplexing allows multiple users to share a common
set of frequencies by assigning unique digital codes to each user. 1 mark
DWDM: Dense Wavelength Division Multiplexing. Multiplexes multiple data
streams onto a single fiber optic line using different wavelength lasers
(lambdas) to transmit multiple signals. 1 mark
10. Compare and contrast conventional encryption and public-key encryption.
Conventional Encryption 2.5 marks
 Plaintext is converted to ciphertext using an encryption algorithm.
 Early cryptography algorithms used the same key for both encryption and
decryption.
 Types of encryption include monoalphabetic substitution, polyalphabetic
substitution and transposition based ciphers.
Public Key Encryption 2.5 marks
 Encoding and decryption keys are different but mathematically related.
 It is very difficult to deduce one key from the other.
 One key is kept private the other key is made public.
PART C: Essay Question
(This question is worth 20 marks. Answer the question in the Answer booklet
provided Use diagrams where appropriate to enhance your answer.)
1. Name and describe in as much detail as you can each of the seven layers of the OSI
reference model?
7. Application Layer. Layer closest to user. Provides network services to users
applications. 2 marks
6. Presentation Layer. Translates between multiple data formats also responsible for
compression and encryption. Common data format. 2 marks
5. Session Layer. Establishes manages and terminates sessions between end points. 2
marks
4. Transport Layer. Concerned with data transport issues - error detection and
recovery, flow control. Can provide a reliable service. Examples of protocols at this
layer include TCP and UDP. The Protocol Data Unit of this layer is a Segment. 4
marks
3. Network Layer. Conectivity and path selection between nodes. Path selection,
routing, and logical addressing. Protocol used at this layer is IP. PDU of this layer is
a Packet. 4 marks
2. Data Link Layer. Provides transit of data across physical links. Concerned with
physical addressing, error detection, network media access. Protocols at this layer
include CSMA/CD, Ethernet, token ring. PDU of this layer is a Frame. 4 marks
1. Physical Layer. Defines the electrical, mechanical, procedural and functional
specifications for the physical link. Defines characteristics such as voltages,
frequencies, transmission distances, physical connectors, etc. PDU of this layer is
bits. 2 marks
Part A – Multiple Choice Concise Answers
1–c
11 – b
2–b
12 – c
3–c
13 – a
4–c
14 – d
5–d
15 – b
6–a
16 – a
7–c
17 – c
8–b
18 – d
9–b
19 – a
10 – a
20 – a
21 – c
22 – d
23 – b
24 – b
25 – b
26 – c
27 – b
28 – d
29 – d
30 – a, b or c