Stanislav Ganin, F44122
1.Suppose Ethernet was the only existing LAN technology, so every host in the Internet was part of a
local Ethernet and therefore had a globally unique Ethernet address. Would you now recommend to
get rid of IP addresses by simply using Ethernet addresses instead of IP addresses? What about the
other way around, why do we not simply assign network adapters IP addresses instead of Ethernet
addresses so we don’t have to deal with both, IP and Ethernet addresses?
Answer: No, that would be a bad idea for a number of reasons. First the IP address is smaller in size
than the MAC address, thus less overhead network traffic. A more important reason is that IP
addresses are assigned in a hierarchical fashion, which allows for hierarchical routing. MAC
addresses are assigned by the manufacturers of the devices and thus contain company specific
information only(usually encodes the manufacturer's registered identification number).
2.Ethernet is sometimes said to be inappropriate for real-time computing because the worst case
retransmission interval is not bounded. Briefly describe a scenario in which a machine connected on
an Ethernet (a local network using CSMA/CD for conflict recovery) ends up waiting a very long amount
of time before it is allowed to send its next packet.
Answer: The problem is that the backoff period is random and if that happens to be the same number
for both transmiting machines, another collision will ocure and the back-off times are exponentially
increased thus leading to a high amount of delay, which cannot be pre-calculated.
3.Compare the frame structures for 10BaseT, 100BaseT and Gigabit Ethernet. How do they differ?
Answer: The main difference is the transimission speeds, which are 10, 100 and 1000 Mbits per
seconds for the three Ethernet varieties. Use standard Ethernet frame format
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802.3i 1990 10BASE-T 10 Mbit/s (1.25 MB/s) over twisted pair
802.3u 1995 100BASE-TX, 100BASE-T4, 100BASE-FX Fast Ethernet at 100 Mbit/s (12.5
MB/s) w/autonegotiation
802.3ab 1999 1000BASE-T Gbit/s Ethernet over twisted pair at 1 Gbit/s (125 MB/s)
4. Suppose your machine is connected to a token ring network. There are n machines connected to
the ring network. The circulating token takes time t to circulate between one node and the next one in
the ring. Whenever a node receives the token, if it has nothing to send at that time, it passes the token
on immediately to the next node in the ring. If, on the other hand, the receiving node has a packet to
send, then it sends that packet instead of the token. The packet sent circulates around the ring taking
time p. In the meantime, the node to which the packet is addressed has the opportunity to read it.
When the packet returns to the original sender, the sender removes the packet from the ring and
reinserts the token. The token will then flow to the next node in the ring and the above algorithm will be
repeated. Describe the worst case scenario in the case of the above token ring. What is the maximum
time that a node will have to wait before it is allowed to send its next packet? Why is the worst case
behavior of a token ring considered to be superior to that of an Ethernet network?
Answer: The worst case scenarion would be if the machine right before the current token holder
wants to send a packet, because it would have to wait for all of the previous machines to get the
token, send any packets they have and eventually pass the token to the “last” machine. The maximum
time would be achieved if every node has a packet to transmit and that time can be calculated by the
formula: n*t + n*p. The Token Ring topology has the deterministic property of being able to calculate
the maximum time before each machine can transmit data which makes it better than Ethernet for
cases in which delay needs to be accounted for.