CS 457, Fall 2004
Computer Networks
Homework 2
Due Thursday, Sept 23rd, in class
Please work on this homework by yourself. Homework is subject to the honor code.
Please sign: I pledge that I have neither given nor received help on this assignment.
______________________________________
Please mark the single best answer for each of the following multiple choice questions:


1. The bandwidth of an OC-48 link is approximately 2.4Gbps. Assume packet
size is 1500 bytes (12,000 bits). A 3-packet message is sent from Bar Harbor,
Maine to San Diego, CA on a datagram packet switched network. The path
contains four hops, each 600 miles long. Assume the speed of light is 125,000
miles/s. Neglecting processing delays, what is the time elapsed between sending
the first bit of the message and receiving the last?
(2 points)
Answer: (Datagram packet switched network!)
o Number of hops M=4;
o Number of packets N=3;
o Per-hop processing delay P=0s;
o Link propagation delay L = distance/speed of light = 600/125,000;
o Packet size = 1500 bytes = 12,000 bits;
o Packet transmission delay T = packet size/transmission speed =
12,000bits/2.4Gbps.
Delay = ML + NT + (M-1) T + (M-1) P =0.1923s=19.23ms.
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2. ADSL (please read the ADSL tutorial on the class web page as a reading
assignment before answering this question):
A subscriber lives 5 km away from the ADSL distribution center. According to
the ADSL tutorial on the class web page:
(a) What is the maximum possible data rate of this subscriber's ADSL connection
given the current ADSL technology?
Answer: 2 Mbps.
According to the following materials:
Capabilities: An ADSL circuit connects an ADSL modem on each end of a twistedpair telephone line, creating three information channels -- a high speed downstream
channel, a medium speed duplex channel, depending on the implementation of the
ADSL architecture, and a POTS (Plain Old Telephone Service) or an ISDN channel.
The POTS/ISDN channel is split off from the digital modem by filters, thus
guaranteeing uninterrupted POTS/ISDN, even if ADSL fails. The high speed channel
ranges from 1.5 to 6.1 Mbps, while duplex rates range from 16 to 832 kbps. Each
channel can be submultiplexed to form multiple, lower rate channels, depending on
the system.
ADSL modems provide data rates consistent with North American and European digital
hierarchies and can be purchased with various speed ranges and capabilities. The
minimum configuration provides 1.5 or 2.0 Mbps downstream and a 16 kbps duplex
channel; others provide rates of 6.1 Mbps and 64 kbps duplex. Products with downstream
rates up to 8 Mbps and duplex rates up to 640 kbps are available today. ADSL modems
will accommodate ATM transport with variable rates and compensation for ATM
overhead, as well as IP protocols.
Downstream data rates depend on a number of factors, including the length of the copper
line, its wire gauge, presence of bridged taps, and cross-coupled interference. Line
attenuation increases with line length and frequency, and decreases as wire diameter
increases. Ignoring bridged taps, ADSL will perform as follows:
Distance
Data Rate
Wire Gauge
Distance
Wire Size
1.5 or 2 Mbps
24 AWG
18,000 ft
0.5 mm
5.5 km
1.5 or 2 Mbps
26 AWG
15,000 ft
0.4 mm
4.6 km
6.1 Mbps
24 AWG
12,000 ft
0.5 mm
3.7 km
6.1 Mbps
26 AWG
9,000 ft
0.4 mm
2.7 km
(b) What is the likely thickness of the wire connecting this subscriber to the
distribution center?
Answer: 24 AWG or 0.5 mm
(c) What is the likely downstream connection bandwidth assuming a noiseless
channel?
Answer: Since the maximum possible data rate of this subscriber's ADSL connection
given the current ADSL technology is 2 Mbps, thus, the likely downstream
connection bandwidth assuming a noiseless channel will be half of it, which means 1
MHz.
(d) The user's model assigns one band for upstream data and another band for
downstream data. Does the modem use echo cancellation or FDM?
(2 points)
Answer: FDM.
To create multiple channels, ADSL modems divide the available bandwidth of a
telephone line in one of two ways -- Frequency Division Multiplexing (FDM) or
Echo Cancellation. FDM assigns one band for upstream data and another band for
downstream data. The downstream path is then divided by time division multiplexing
into one or more high speed channels and one or more low speed channels. The
upstream path is also multiplexed into corresponding low speed channels. Echo
Cancellation assigns the upstream band to over-lap the downstream, and separates the
two by means of local echo cancellation, a technique well know in V.32 and V.34
modems. With either technique, ADSL splits off a 4 kHz region for POTS at the DC
end of the band.
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3. Bluetooth (please research Bluetooth technology on the web before answering
this question):
What is the maximum number of devices that can be connected to a piconet?
(1 point).
Answer: The maximum number of devices that can be connected to a piconet is 8.
Bluetooth can provide interconnection for local devices
Bluetooth is able to simultaneously interconnect up to 8 transceivers in a “piconet” over
a short range. Each Bluetooth transceiver costs the same (~$10) and there is no
requirement for any infrastructure (see diagram). The simultaneous connectivity limit of
8 devices may seem to be a serious constraint, however several piconets can operate
in close proximity and Bluetooth devices can rapidly move from one piconet to another.
In fact Bluetooth devices need only remain a member of a piconet for the period of
time required to complete a communication transaction. So devices can join and leave
a local piconet frequently, effectively overcoming the 8-device constraint.
What is the typical communication range?
(1 point).
Answer: 10m.
Bluetooth is a de-facto open standard for short range digital radio. It is designed to
operate in the unlicensed ISM (Industrial, Scientific, Medical applications) band which is
generally available in most parts of the world (see table overleaf). The specification
includes air interface protocols to allow several Bluetooth applications to
intercommunicate simultaneously, and to overcome external sources of interference such
as domestic and commercial microwave ovens. The short range referred to above is
defined as up to 10m in normal operation although greater range/penetration can be
achieved through higher output powers under some circumstances.
(Please provide reference.)
http://www.cs.virginia.edu/~zaher/classes/CS457/lectures/bluetooth.pdf
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4. High fidelity music has a bandwidth of 16kHz. According to Nyquist Theorem,
in order to be reconstructed, what is the minimum rate at which such music
should be sampled?(1 point)
Answer: Such music should be sampled at a minimum rate of 2*16kHz = 32kHz.
Sampling Theorem: If a signal f (t) is sampled at regular intervals of time and at a rate
higher than twice the highest significant signal frequency, then the samples contain all the
information of the original signal. The function f (t) may be reconstructed from these
samples by the use of a low pass filter.

5. Amplitude modulation is:
(a) a way of encoding digital data
(b) a way of encoding analog data into an analog signal
(c) a digital representation of analog data
(1 point)
Answer: (b) a way of encoding analog data into an analog signal.
According to the following materials (Data Encoding at the Physical Layer); There
are three types of modulation (Analog Data / Analog Signal): Amplitude Modulation
(AM), Frequency Modulation (FM) and Phase Modulation (PM).

6. A binary channel with a signal to noise ratio of 15, has bandwidth H. What is
its capacity (in bps)?
(1 point)
Answer: Capacity of a noisy channel:
C = H log base2 (1+S/N) bps = H log base2 (1+15) bps = 4H bps
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7. In the channel in problem 6, if data is sampled at twice the channel's
bandwidth, what is the maximum number of discrete levels that each sample may
be represented by?
(1 point)
Answer: The channel in problem 6 has a capacity of 4Hbps (the result of the problem
6), and if data is sampled at twice the channel’s bandwidth (2H samples/sec), then we
need 2 bits for per sample (4H/2H=2 bits), which means that each sample should be
represented by 4 discrete levels(2=log base2 4).