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Pharos University in Alexandria
Faculty of Engineering
Communications and Computer Networks
Sheet 3
Pharos University in Alexandria
Faculty of Engineering
Computer Engineering Department
Communications and Computer
Networks Course Code: CE 363
Lecturer: Dr. Amr Elsaadany
T.A: Eng. Nehal Fathi
Sheet (3)
State which of the following statements are true and which are false, stating your
reasons for false statements.
1. Hubs help in enhancing the performance of star networks under collisions.
2. If the number and load of computers in a shared Ethernet and a switched Ethernet are the
same, then the average speed of the two networks will be the same.
3. Cut-through switches provide better filtering of erroneous packets than hubs.
4. A bridge can carry out traffic filtering based on Layer-2 addressing.
5. In a switched star Ethernet, one might expect the performance of the network to improve
later than at the instant the network is started up.
6. Bridging can be used to increase the number of hosts connected to a shared Ethernet without
significant reduction in average network speed.
7. Bridges or switches may be used to interconnect 2 Ethernet segments if one segment is a 10
Mbps segment and the second segment is a 100 Mbps segment.
8. The preamble portion of an Ethernet frame is responsible for synchronizing frame transfer
between different hosts.
9. A destination's logical address may be determined from the destination address field of an
Ethernet frame.
10. If the network addresses of some computers on a network are changed, there is no need to
change these computers' entries in the switch's forwarding tables.
Short Response Questions
1. Describe the components of an IEEE 802.3 frame, and then answer the following related
questions:
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
What layer is responsible for defining this frame format?
What is the purpose of the preamble field (PRE)?
What is the purpose of the start of frame delimiter (SFD)?
What are the minimum and maximum lengths of the data field?
What is the reason behind placing the destination address field before the source address
field?
What is the purpose of the CRC field? How is it used at the receiving side?
What data would you expect to find inside the data field of the frame?
Why is there a maximum limit for the length of the data field?
Why is there a minimum limit for the length of the data field?
What is the ratio of useful data to the entire packet for the smallest Ethernet frame? What is the
ratio for the largest frame?
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Pharos University in Alexandria
Faculty of Engineering
Communications and Computer Networks
Sheet 3
2. An LLC frame passed down to the Mac layer consists of 20 bytes. Byte 20 happened to consist
only of zeros:
How many padding bytes will be added by the Mac layer?
3. If a switch is required to forward a packet of data to a host that was just connected into the
network, how will the switch determine the appropriate port to which the packet should be
forwarded?
4. Suppose the length of a l0Base5 cable is 2500 m. If the speed of propagation in a thick coaxial
cable is 200,000,000 mps, how long does it take for a bit to travel from the beginning to the
end of the network? Assume there are 10 μs delay in the equipment.
5. What is the maximum length of traditional 10-Mbps Ethernet given that the time required for
a station to send 512 bits 51.2 microseconds? What is the typical length used for such
network.
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Pharos University in Alexandria
Faculty of Engineering
Communications and Computer Networks
Sheet 3
Network Analysis
1. The figure to the right shows a typical shared
Ethernet with a star topology. The Ethernet is
a 10Base-T network. Assume that the group
speed of the signal in enhanced CATS twisted
pair wires is two thirds the speed of light (i.e.,
the propagation speed is 200 Mega mps). The
CSMA/CD protocol declares the maximum
allowable radius of such network to be
approximately 2500 meters. By considering
the Mac layer frame without the PRE and SFD fields, show that the maximum radius of this
network for this network to work properly using the CSMA/CD protocol is exactly 2560
meters, then explain why the back-off time takes multiple values of 512 bit durations.
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