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CSC358 Tutorial 10 Notes

Below are some notes and sample solutions to the tutorial questions. These
notes are only meant to provide the necessary amount of information for you to
verify your own work and to help you recall the discussions in the tutorial.
Simply reading these solutions does NOT convey the same learning experience as
attending a tutorial (not even close). The tutorial materials are a mandatory
and important component of this course, so don't miss any of them!
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Question 1

(a) MAC: 2^48 = 2.8e14 (280 trillion), IPv4: 2^32 = 4 billion, IPv6: 2^128 = huge

(b) hub vs switch: hub has all links on the same collision domain therefore the
bandwidth is just the bandwidth of a single link, whereas switch has each link
on its own collision domain, the bandwidth is multiplied.

switch vs router: see Week 11 slides, Page 15

(c) CD lets collisions happen then detects them; CA tries to avoid collisions
before they happen, using control messages (RTS/CTS).

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Question 2

Note: The "aggregate throughput" means, while the hosts within the local
network are sending data between each other, the total amount of traffic (i.e.,
sum of rates of all on-going transmissions) that can take place simultaneously.

You may assume that all switches are "full-duplex", i.e., each switch has
separate channels for sending and receiving, so a link is 100 Mbps, A-to-B and
B-to-A can transmit at the same time, both with rate 100 Mbps.

(a) If all the 15 nodes send out data at the maximum possible rate of 100 Mbps,
a total aggregate throughput of 15*100 = 1500 Mbps is possible.

(b) Each of the three "branches" of the hosts is a single collision domain that
can have a maximum throughput of 100 Mbps. The links connecting the web server
and the mail server has a maximum throughput of 100 Mbps. Hence, if the three
collision domains and the web server and mail server send out data at their
maximum possible rates of 100 Mbps each, a maximum total aggregate throughput
of 500 Mbps can be achieved.

(c) All of the 15 end systems will lie in the same collision domain. In this
case, the maximum total aggregate throughput of 100 Mbps.

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Quesiton 3

TTL is ignored in each entry.

(a) A sends a frame to G
S1: (A, 1)
S2: (A, 0)
S3: (A, 0)
S4: (A, 0)

(b) G replis a frame to A.
S1: (A, 1), (G, 0)
S2: (A, 0)
S3: (A, 0), (G, 1)
S4: (A, 0), (G, 2)

(c) D sends a frame to F
S1: (A, 1), (G, 0), (D, 0)
S2: (A, 0), (D, 1)
S3: (A, 0), (G, 1), (D, 0)
S4: (A, 0), (G, 2), (D, 1)

(d) F replies a frame to D
S1: (A, 1), (G, 0), (D, 0)
S2: (A, 0), (D, 1), (F, 3)
S3: (A, 0), (G, 1), (D, 0)
S4: (A, 0), (G, 2), (D, 1)

(e) A sends a frame to D
(no change)
S1: (A, 1), (G, 0), (D, 0)
S2: (A, 0), (D, 1), (F, 3)
S3: (A, 0), (G, 1), (D, 0)
S4: (A, 0), (G, 2), (D, 1)

(f) D replies a frame to A
(no change)
S1: (A, 1), (G, 0), (D, 0)
S2: (A, 0), (D, 1), (F, 3)
S3: (A, 0), (G, 1), (D, 0)
S4: (A, 0), (G, 2), (D, 1)

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Quesiton 4

(a) A host will associate with one of the SSIDs (that is, one of the APs).
After association, there is a virtual link between the host and the AP. Label
the APs AP1 and AP2. Suppose the host associates with AP1. When the host sends
a frame, it will be addressed to AP1's MAC address. Although AP2 will also
receive the frame, it will not process the frame because the frame is not
addressed to it.  Thus, the two APs can work in parallel over the same channel.
However, the two APs will be sharing the same wireless bandwidth. If hosts in
different APs transmit at the same time, there will be a collision. For
802.11b, the maximum aggregate transmission rate for the two APs is 11 Mbps.

(b) Now if the two APs use different channels, when two hosts in different APs
transmit at the same time, there will not be a collision. Thus, the maximum
aggregate transmission rate for the two APs will be doubled (22 Mbps for
802.11b).