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(COMP361)[2007](s)final-sol.pdf
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COMP 361 Computer Networks
Spring Semester 2007
Final Examination
Date: May. 25, 2007 Time: 12:30 pm C 3:30 pm
Name:_____________________ Student ID: ________________Email:__________________
Instructions
1.
Please write your name, student ID and email address on this page.
2.
Please answer all questions within the space provided on the examination paper. You may use back of the pages for your rough work. Please be concise, and this is NOT an essay contest.
3.
This paper consists of 8 questions and 13 pages.
4.
You have total 180 minutes to complete exam, thus each 10 points question roughly requires 18 minutes, please manage your time wisely.
5.
Please read each question very carefully and answer the questions carefully and clearly to the point. Make sure your answers are neatly written, legible, and readable.
6.
Show all the steps used in deriving your answer, wherever appropriate.
Question
Points
Score
1
10
2
20
3
14
4
12
5
8
6
10
7
13
8
13
Total
100
1.
Answer the following true/false questions by circling either T or F [10 points]
a)
Maximum segment size (MSS) at the TCP transport layer should be related to maximum data link frame size in order to avoid IP fragmentation. T F
b)
ICMP messages can not be lost by the network. T F
c)
Two network interfaces can have the same IP address, while in principle each Ethernet adapter should have a unique MAC address. T F
d)
PPP uses a CSMA access-control protocol. T F
e)
BGP is a link-state algorithm. T F
f)
Both Router and Switch are plug-and-play devices. T F
g)
The IEEE 802.3 Ethernet protocol is an unslotted protocol. T F
h)
A two-dimensional parity check scheme can correct all two-bit errors in the original data. T F
i)
A router can run multiple routing protocols at the same time. T F
j)
Checksum of the IPV4 datagram change when it passes through router. T F
2.
Give quick answer to the following questions [16 points].
a)
Consider an Ethernet with 3 pairs of computers, 1a/1b, 2a/2b, 3a/3b, connected via a central device x, which each pair exchange data (bi-directional) at their maximum possible speed (i.e. 1a exchange data with 1b, 2a with 2b, 3a with 3b), and the collision is ignored (i.e. efficiency is 100%) [4 points]
What is the aggregated throughput if Device X is a
Switch [3]
Answer:
420Mbps =10M*2 + 100M*2 + 100M*2
Each pair of computer have full-duplex traffic, notice computer 3b is 100M, so the traffic between 3a and 3b is 100Mbps
210