Home Work Assignment 3
Total Points : 87
1. Briefly explain the different types of impairments faced by signals traveling over
a medium.
(8 points)
2. a medium.
(10 marks)
•Attenuation –loss of signal strength over distance
•Amplitude Distortion –different losses at different frequencies
•Delay Distortion –occurs when signals travel over different paths and arrive at various
times at the receiving end, thereby distorting the signal
•Noise –distortions of signal caused by interference
some types of noise are
•Thermal (aka “white noise”)
–Uniformly distributed, cannot be eliminated
•Induced Noise
–Caused by high power element around - motors
•Crosstalk
–Overlap of signals
•Impulse noise
–Irregular spikes, less predictable
A comprehensive explanation of most of the impairment should be sufficient.
3. What are the advantages and disadvantages in using
a. Twisted pairs
b. Co-axial cables
c. Fibre optic
d. Wireless media
Twisted pair advantages
•Inexpensive and readily available
•Flexible and light weight
•Easy to work with and install
Twisted pair disadvantages
•Susceptibility to interference and noise
•Attenuation problem
–For analog, repeaters needed every 5-6km
–For digital, repeaters needed every 2-3km
•Relatively low bandwidth
Coaxial cables advantages
•Higher bandwidth
–400 to 600Mhz
–up to 10,800 voice conversations
•Can be tapped easily (pros and cons)
•Much less susceptible to interference than twisted pair
Coaxial cables disadvantages
(12 points)
•High attenuation rate makes it expensive over long distance
•Bulky
fibre advantages
•greater capacity (bandwidth of up to 2 Gbps)
•smaller size and lighter weight
•lower attenuation
•immunity to environmental interference
•highly secure due to tap difficulty and lack of signal radiation
fibre disadvantages
•expensive over short distance
•requires highly skilled installers
•adding additional nodes is difficult
wireless advantages
•no cabling needed between sites
•wide bandwidth
•Multi-channel transmissions
wireless disadvantages
•line of sight requirement
•expensive towers and repeaters
•subject to interference such as passing airplanes and rain
4. Explain the LRC scheme step by step using an example of the following bit
stream 00111110 – 11110011 – 11111000 - 00001111
(7 points)
•Checks parity by the vertical method and the data “horizontally” as well
•data is organized into columns and rows, each column is checked and given a parity bit,
LRC is attached to the end of the data
•Data units, each with their own VRC (rows), are blocked together
•Sender determines LRC: parity of each position in the data unit•LRC bits are appended
to the data unit and transmitted
•Receiver processes all parity bits: if VRC bits and LRC bits are not the same, the data is
rejected
example
5. Use a schematic and explain the operation of the ‘Stop and Wait’ protocol. In the
schematic explain what happens when a damaged frame is received?
(8 points)
2 points for the diagram, 2 ½ points explanation – 1st part, 2 ½ points
explanation 2nd part.
T
R
Data 0
WAIT
TIME
ACK 1
Data 1
WAIT
TIME
ACK 0
Data 0
WAIT
TIME
NAK
Data 0
WAIT
TIME
ACK 1
...
In the stop and wait protocol, the sender sends a frame and waits till the receiver
acknowledges the sent frame before sending the next frame. The frames are
numbered 0 and 1, to identify the acknowledgments to the sent frames. This scheme
does not overload the receiver, but is very slow as a long time is wasted, waiting for
the acknowledgements to be returned.
In the case a frame is received damaged. The receiver sends a NAK, this is an
intimation to the sender to resend the last sent frame.
6. Use a schematic and explain the operation of the “Sliding window” protocol,
which uses a window size of 7 and a mod 8 counter for numbering the frames. In
the schematic explain what happens when a frame is lost en-route and the systems
implement “Go-back-n” ARQ. If the window size is maintained at 1, can you
identify the type of flow-control that you would get?
(10 points)
st
nd
3 points – picture, 4 points 1 explanation, 3 points 2 explanation, 1 point – last
answer
T
R
Data 0
Data 1
Data 2
Data 3
Data 4
Data 5
NAK, 3
Data 6
Data 3
ACK 3
In the sliding window protocol, the systems are allowed to send frames up to the
number identified in the window size. That is if the window size is 7 then the sender
can send up to a maximum of seven frames, before it waits for an acknowledgement
to send any more frames. The frames are numbered using mod 8 then the frame
numbers can go from 0 to 7 and start again from 0. As the sender sends frame, the
window size goes down, as acknowledgements are received the window size keeps
increasing.
When a frame is lost and the systems implement go-back-n ARQ, then in the picture
if frame 3 was lost and frames 4, 5 were received the receiver recognizes the loss of
frame 3 and sends a NAK 3 frame, on receiving this nak frame the sender has to
resend from frame 3 onwards.
When the window size if 1, the sliding window becomes a ‘stop and wait’ scheme.
1 point
7. Explain the function of the different bits in the control field of a supervisory
frame
(8 points)
The control field in the supervisory frame has the field S, which identifies, the type of
the supervisory frame. The first two bits are 10, which identifies it as a supervisory
frame. The supervisory frames can be used for various types of operations, like
acknowledgement, negative acknowledgment, polling, select, selective reject etc.
2 points
The N(R) field holds the number of the data frame, for which this supervisory frame
is an acknowledgement or NAK. 2 points
Normally the value is one higher than the sequence number of the correctly received
frame, which informs the sender that the previous frame to the number in this
supervisory frame was received correctly. 2 points
The poll final bit is set either for a command or a response supervisory frame. In the
command frame the bit is called the poll bit and in the response frame it is called the
final bit, which identifies that response frame as being in response to some command
frame. 2 points
8. Explain how the supervisory frame can be used to accommodate a ‘polling’ and a
‘select’ operation between primary and secondary stations. Provide the
appropriate response frames from the secondary. State how the address field get
used in this situation.
(8 points)
In the supervisory frame, an RR frame can be used for polling of the secondary
stations by the primary station. The poll bit is set to 1. The address field will be set to
the stations address which is being polled. The polled stations if it has no data to send
will reply with a similar RR supervisory frame, with the final bit set and its address in
the address field. If the polled station has data to send then it will use an I frame, with
the final bit set.
For ‘selecting’ the primary station uses an RNR frame with the poll bit set and the
address field holding the address of the station to which it wants to send the data. If
the secondary station can receive the data it replies with an RR frame final bit set. If it
can not receive any data then it uses the RNR frame with the final bit set.
4 points each.
9. What are the window sizes at the check points specified below in Fig. 1 9 points
Station 1
Sending
window = 8
Station 2
time
Data frame, 0
Receiving
window = 8
Data frame, 1
Data frame, 2
Check point B
Sending Window =?
Check point A
Receiving Window =?
Ack, 1
Check point C
Receiving Window =?
Check point D
Sending Window =?
Data frame, 3
Data frame, 4
Check point F
Sending Window =?
Ack, 4
Check point E
Receiving Window =?
check point A - 5
check point B - 5
check point C - 6
Check point D - 6
Check point E - 7
Check point F - 7
10. What is a switched network? What is the difference between circuit switching and
packet switching? You may use a schematic to explain.
(9 points)
Switched networks make use of switches to reduce the number of direct connection
required in a mesh network. On a requirement basis the switches make connection to
the end points , by providing resources - 2 ½ points
In the case of circuit switching, a physical path exists between the sender and
receiver. This physical path may be in the form of a time slots to frequency, but it is
reserved resources that user uses from one end to the other for sending his data. The
connection or path between sender and receiver is there always for the duration of the
call. 4 points
In the case of packet switching, the switches provide buffering, the packets which
arrive at the input ports are stored and forwarded as and when resources are available
in the outgoing links. 2 ½ points
11. Explain the difference between virtual circuits and datagram approach. What are
the highlights of the virtual circuit approach
(8 points)
In the case of virtual circuit, a connection or virtual circuit is said to be maintained
between the sender and receiver. As in the case of the switched circuits, resources are
booked along the way from sender to receiver. This requires a call set up. Once the
virtual circuit is set up all packet for that session between the sender and the reciver
follow the same path. 4
In the datagram case each packet is handled individually and the decision for it
routing is done at each node based on the destination address. Hence the packet in one
session can follow different routes and arrive at the destination out of order. 3
Virtual circuits provide in-sequence delivery and identifying lost packet and recovery
becomes easier. 1