Part I: Multiple Choice questions
1. In the downlink in CDMA, the codes that are used to define independent channels of
communication to different users are known as
a. Orthogonal codes
b. Scrambling codes
c. Channelization codes
d. Hamming codes
e. Channel codes
2. The view of the mobile network that describes the signaling protocols is referred to as
a) Signaling plane
b) Control plane
c) User plane
d) Signaling traffic plane
e) None of the above
3. The use of digital signal processing to compensate for distortions introduced by
the channels is referred to as
a) Power control
b) Doppler shift
c) Fading
d) Equalization
e) None of the above
4. The view of the mobile network that describes the signaling protocols is referred
to as
a. Signaling plane
b. Control plane
c. User plane
d. Signaling traffic plane
e. None of the above
5. The use of digital signal processing to compensate for distortions introduced by
the channels is referred to as
a. Power control
b. Doppler shift
c. Fading
d. Equalization
e. None of the above
1. In the downlink in CDMA, the codes that are used to define independent channels of
communication to different users are known as
a. Orthogonal codes
b. Scrambling codes
c. Channelization codes
d. Hamming codes
e. Channel codes
2. The term quality of service (QoS) refers to factors such as
a. Delay time
b. Variability of delay time
c. Packet loss in the delivery of packets.
d. All of the above
e. None of the above
Part II: Short essay/problem questions
4. Explain the concept of quality of service (QoS) and how UMTS classify the traffic
according to the QoS requirements (Fall 2011 – Final Exam)
Part III: Long essay/problems
3. The purpose of this question is to compare the cost of a regular international phone
call (Mobile telephony network) and the same call but made using VoIP software
installed on this phone using Cellular Data Connection.
Consider the case of a user subscribed to a Pay-As-You-Go cellular data service of a
Mobile operator where he is charged $0.5 per consumed Megabytes (whether upload
or download). This user wants to call a mobile number in UK. The rate of the
international call to UK set by the mobile operator is 0.3$/minute.
a. Taking into consideration that the data rate is 12.2 kbps in each direction of a
regular phone call, what is the cost per Megabyte of the regular international call? (3
marks)
Cost per MB = cost of one minute / size of a one minute call in MB
Size in MB = 2 * 12.2 * 1000 * 60 / (8 * 1024*1024)= 0.174 MB
Cost per MB = 0.3/0.174 = 1.7189 $/MB
b. The user decides to use VoIP software for international call. As you know, the total
cost of the phone call is the sum of the VoIP service provider (VSP) rate and the
fees that result from data usage (Mobile operator). The second part (data usage) is
directly related to the audio codec used in the VoIP software of this VSP. The bit
rate of the audio-codec used in the VoIP software is 40 Kbps (in each direction) and
the rate of a one minute call to UK is 0.2$.
i. Using the minute rate and codec bit rate, calculate the cost of a one-minute
call to UK using the VSP service. (4 marks)
The cost of a one minute call = minute rate + 0.5$/MB * (size of one minute
call suing VoIP software in MB)
size of one minute call using VoIP software in MB = 2 *40000 *
60/1024*1024*8= 0.5722 MB
cost of one minute = 0.5722 * 0.5 + 0.2 = 0.4861$
ii. What is the cost per Megabyte of the call for each VSP? (2 marks)
Cost per MB = total cost calculated in b / size of one minute call in MB
calculated in I = 0.4861 / 0.5722 = 0.85 $/MB
c. Compare the numbers obtained in questions a and b. Discuss the result of the
comparison (2 marks)
Although the cost per MB of a VoIP (0.85 $/MB) is cheaper compared to the cost per
MB (1.7189 $/MB) of a regular international call, but the regular international call in
the case mentioned is cheaper (0.3$/minute) than using the VoIP software (0.4861
$/minute)
4. A mobile user is sitting on a high-speed train. Given that the carrier frequency is
nominally 2142.5 MHz and that the Doppler shift is 793 Hz. Calculate the speed of
the train. We consider that the train direction is toward the base station. The speed
of light is 3*10^8 m/s (6 marks)
Doppler shift = fr – fc
Fr = fc (1 + v/c) => Doppler Shift = fr – fc = fc * v/c = 793 Hz
=> v = Doppler shift *c/fc = 793 * 3 * 10^8 / (2142.5 * 10^6 ) = 111 m/s = 400 Km/h
Question 3- (Spring 2011 – Final Exam)
The computing power of a certain machine is the number of instructions it can
execute per unit of time. The computing power of nowadays personal computers
(2011) is around 4500 MIPS (core i5 Intel processor or equivalent) where MIPS stand
for Millions Instructions per Second. In this exercise we will consider such computing
power in the calculations. Suppose that an algorithm to verify one 56-bit DES key
needs around 600 elementary instructions. Suppose that we have a couple of clear
text and encrypted text using DES and that we want to find the encryption key using
brute force attack; which means by testing all the keys once after another. We
suppose that all keys are equally probable.
Note: large numbers can be expressed in the form of powers of 2
a- What time it takes for the machine mentioned above to test a key?
b- What is the number of instructions necessary to find a DES key? Consider the
worst-case scenario where all keys should be verified.
c- What time it takes if the computing power of one billion (109) PCs worldwide are
grouped for this task? (we suppose that all the PCs have the same computing power
mentioned above)
d- Calculate the time it takes to find a 112 bits triple-DES key supposing that the
number of instructions to test one 3-DES key is the double of a DES key.
c. The computing power of a certain machine is the number of instructions it can
execute per unit of time. The computing power of nowadays personal computers
(2011) is around 4500 MIPS (core i5 Intel processor or equivalent) where MIPS stand
for Millions Instructions per Second.
In this exercise we will consider such computing power in the calculations. Suppose
that an algorithm to verify one 56-bit DES key needs around 600 elementary
instructions. Suppose that we have a couple of clear text and encrypted text using
DES and that we want to find the encryption key using brute force attack; which
means by testing all the keys once after another.
We suppose that all keys are equally probable.
Note: large numbers can be expressed in the form of powers of 2
i. What time it takes for the machine mentioned above to test a key? (3 marks)
Time = number of required instructions/ number of instructions per second =
600/4500 * 106 = 0.13 * 10-6s = 0.13 μs(2 marks)
ii. What is the number of instructions necessary to find a DES key? Consider the
worst case scenario where all keys should be verified. (3 marks)
Number of instructions = nb of keys * nb of instructions per key = 2^56 * 600
instructions
iii. What time it takes if the computing power of one billion (109) PCs worldwide are
grouped for this task? (we suppose that all the PCs have the same computing power
mentioned above) (2 marks)
Time = nb of instructions to break the key / total computing power = 2^56 * 600/ 10^9
* 4500 * 10^6 = 9.6 s
iv. Calculate the time it takes to find a 112 bits triple-DES key supposing that the
number of instructions to test one 3-DES key is the double of a DES key. (2 marks)
Time = nb of instructions to break the key / total computing power = 2^112 * 1200/
10^9 * 4500 * 10^6 = 9.6 * 2^57 s