UNIVERSITI MALAYSIA PERLIS
School of Computer and Communications Engineering
DPT224 Data Communications and Networking
Tutorial 1
_____________________________________________________________________________________________________
1. Identify the five components of a data communications system.
The five components of a data communication system are the sender, receiver,
transmission medium, message, and protocol.
2. Name the four basic network topologies, and cite an advantage of each type.
We give an advantage for each of four network topologies:
a. Mesh: secure b. Bus: easy installation c. Star: robust d. Ring: easy fault isolation
3. What are some of the factors that determine whether a communication system is a LAN
or WAN?
The general factors are size, distances (covered by the network), structure, and ownership.
4. For each of the following four networks, discuss the consequences if a connection fails.
a.
b.
c.
d.
Five devices arranged in a mesh topology
Five devices arranged in a star topology (not counting the hub)
Five devices arranged in a bus topology
Five devices arranged in a ring topology
a. Mesh topology: If one connection fails, the other connections will still be work- ing.
b. Star topology: The other devices will still be able to send data through the hub; there will
be no access to the device which has the failed connection to the hub.
c. Bus Topology: All transmission stops if the failure is in the bus. If the drop-line
fails, only the corresponding device cannot operate.
d. Ring Topology: The failed connection may disable the whole network unless it
is a dual ring or there is a by-pass mechanism.
4. What are headers and trailers, and how do they get added and removed?
Headers and trailers are control data added at the beginning and the end of each data unit at
each layer of the sender and removed at the corresponding layers of the receiver. They
provide source and destination addresses, synchronization points, information for error
detection, etc.
6. What is the difference between a port address, a logical address, and a physical
address?
The physical address is the local address of a node; it is used by the data link layer to deliver
data from one node to another within the same network. The logical address defines the
ol-02.fm Page 3 Saturday, January 21, 2006 10:27 AM
sender and receiver at the network layer and is used to deliver messages across multiple
networks. The port address (service-point) identifies the application process on the station.
7. In Figure below, computer A sends a message to computer D via LANl,3 router Rl, and
LAN2. Show the contents of the packets and frames at the network and data link layer
for d.
each
hop interface.
Responsibility
for carrying frames between adjacent nodes: data link layer
Page 3 Saturday, January 21, 2006 10:27 AM
3
19.
a. Format and code conversion services: presentation layer
d. Responsibility
for carrying
frames
adjacent
nodes:
data link
b. Establishing,
managing,
andbetween
terminating
sessions:
session
layerlayer
19.
c. Ensuring reliable transmission of data: data link and transport layers
a. Format
and code
services: presentation
d. Log-in
andconversion
log-out procedures:
session layerlayer
b. Establishing,
managing,
and terminating
sessions:
layer presentation layer
e. Providing
independence
from different
datasession
representation:
c. 20.
Ensuring
reliable
transmission
of
data:
data
link
and
transport
layers
See Figure 2.1.
d. Log-in and log-out procedures: session layer
e. Figure
Providing
from different
data representation: presentation layer
2.1 independence
Solution to Exercise
20
20. See Figure 2.1.
Figure 2.1 Solution to Exercise A/40
20
LAN1
LAN2
R1
A/40
Sender
LAN1
B/42
D/80
C/82
LAN2
R1
Sender
Sender
42 40B/42
A D Data T2C/82
80 82 AD/80
D Data T2
Sender
21. See Figure 2.2.
80 82 A D Data T2 is between a process running at
42 40 A D Data
8. In Figure 2.22, assume
thatT2the communication
computer
with port
address
Figure 2.2A Solution
to Exercise
21 i and a process running at computer D with port address j.
21. See
Figure
Show
the2.2.
contents of packets and frames at the network, data link, and transport layer
for each hop.
LAN1
LAN2
Figure 2.2 Solution to Exercise 21 A/40
R1
A/40
Sender
LAN1
B/42
C/82
LAN2
R1
Sender
42
40 A D i B/42
j Data
D/80
Sender
D/80
T2 C/8280 82 A D i j
Data
T2
Sender
22. If the corrupted destination address does not match any station address in the net42 40
80 82 A destination
A D is
i lost.
j Data
D i j Dataaddress
T2
work, the
packet
If theT2corrupted
matches one of the stations, the frame is delivered to the wrong station. In this case, however, the error
detection destination
mechanism,address
available
in not
most
data link
protocols,
will find
error and
22. If the corrupted
does
match
any station
address
in thethenetdiscard
the
frame.
In
both
cases,
the
source
will
somehow
be
informed
using
packet
lost. If
theeach
corrupted
destination
address
matches one of the sta- one
9. work,
Whatthe
isthe
thedata
bitislink
rate
for
of the
following
signals?
of
control
mechanisms
discussed
in
Chapter
11.
tions,
frame
is in
delivered
to1 the
wrong
station.s In this case, however,
the error
a.23. the
A
signal
which
bit
lasts
0.001
Before
using
the
destination
address
in
an
intermediate
or
the
destination
node, the
detection mechanism, available in most data link protocols, will find the error and
b. packet
A signal
inthrough
whicherror
1 bitchecking
lasts 2 ms
goes
that
may
help
the
node
find
the
corruption
discard the frame. In both cases, the source will somehow be informed using one
A signal
inprobability)
which
10and
bits
last 20
(with
a high
discard
theinJ-ls
packet.
Normally
the upper layer protocol
ofc.the data
link
control
mechanisms
discussed
Chapter
11.
will
inform
the
source
to
resend
the
packet.
23. Before using the destination address in an intermediate or the destination node, the
packet goes through error checking that may help the node find the corruption
(with a high probability) and discard the packet. Normally the upper layer protocol
will inform the source to resend the packet.
a. bitrate=1/(bitduration)=1/(0.001s)=1000bps=1Kbps
b. bitrate=1/(bitduration)=1/(2ms)=500bps
c. bitrate=1/(bitduration)=1/(20μs/10)=1/(2μs)=500Kbps
10.
A signal travels from point A to point B. At point A, the signal power is 100 W. At
point B, the power is 90 W. What is the attenuation in decibels?
dB=10log10 (90/100)=–0.46dB
11.
If the bandwidth of the channel is 5 Kbps, how long does it take to send a frame of
100,000 bits out of this device?
100,000 bits / 5 Kbps = 20 s
12.
A line has a signal-to-noise ratio of 1000 and a bandwidth of 4000 KHz. What is the
maximum data rate supported by this line?
4,000 log2 (1 + 1,000) ≈ 40 Kbps
13.
Define FHSS and explain how it achieves bandwidth spreading.
The frequency hopping spread spectrum (FHSS) technique uses M different carrier frequencies that are modulated by the source signal. At one moment, the signal
modulates one carrier frequency; at the next moment, the signal modulates another carrier
frequency.
14.
Define DSSS and explain how it achieves bandwidth spreading.
The direct sequence spread spectrum (DSSS) technique expands the bandwidth of the
original signal. It replaces each data bit with n bits using a spreading code.
15. Assume that a voice channel occupies a bandwidth of 4 kHz. We need to multiplex 10
voice channels with guard bands of 500 Hz using FDM. Calculate the required
bandwidth.
To multiplex 10 voice channels, we need nine guard bands. The required band- width is then
B = (4 KHz) × 10 + (500 Hz) × 9 = 44.5 KHz
16. We need to use synchronous TDM and combine 20 digital sources, each of 100 Kbps. Each
output slot carries 1 bit from each digital source, but one extra bit is added to each frame for
synchronization. Answer the following questions:
a. What is the size of an output frame in bits?
b. What is the output frame rate?
c. What is the duration of an output frame?
d. What is the output data rate?
e. What is the efficiency of the system (ratio of useful bits to the total bits).
a. Each output frame carries 1 bit from each source plus one extra bit for synchro- nization.
Frame size = 20 × 1 + 1 = 21 bits.
b. Each frame carries 1 bit from each source. Frame rate = 100,000 frames/s.
c. Frame duration = 1 /(frame rate) = 1 /100,000 = 10 μs.
d. Data rate = (100,000 frames/s) × (21 bits/frame) = 2.1 Mbps
e. In each frame 20 bits out of 21 are useful. Efficiency = 20/21= 95%
17. Ten sources, six with a bit rate of 200 kbps and four with a bit rate of 400 kbps are to be
combined using multilevel TDM with no synchronizing bits. Answer the fol- lowing questions
about the final stage of the multiplexing:
a.
What is the size of a frame in bits?
b. What is the frame rate?
c.
What is the duration of a frame?
d. What is the data rate?
We combine six 200-kbps sources into three 400-kbps. Now we have seven 400- kbps
channel.
a. Each output frame carries 1 bit from each of the seven 400-kbps line. Frame size = 7 × 1 =
7 bits.
b. Each frame carries 1 bit from each 400-kbps source. Frame rate = 400,000 frames/s.
c. Frame duration = 1 /(frame rate) = 1 /400,000 = 2.5 μs.
d. Output data rate = (400,000 frames/s) × (7 bits/frame) = 2.8 Mbps. We can also
calculate the output data rate as the sum of input data rate because there is no synchronizing
bits. Output data rate = 6 × 200 + 4 × 400 = 2.8 Mbps.