27 August 2008 1

EEE442COMPUTER NETWORKS

Test results&

analysis

27 August 2008 2

Overall Performance

Average = 62.82% (Level 3+) Good application of techniques with some ability in analysis)

Highest mark: 91Lowest mark: 2790+ = 1 ; 80+ = 12, 70+ = 29, 60+ = 51; 50+=27; 40+=12, <40=6

0

10

20

30

40

50

60

<40 40+ 50+ 60+ 70+ 80+ 90+

27 August 2008 3

Section A: Test for Knowledge

• Average = 8.4/10

• Strong knowledge of the subjects

• Most students score 20/20

• Majority scores > 5/10

• Some observations:– Many students have difficulties writing the

definition of the given terms in own words.

27 August 2008 4

Section A: Answers

• Section A1: Multiple choices: a, c, a, d, d• Section A2.1:

– Circuit switching-a data transfer technique that moves the data between networked computers via dedicated paths along intermediate switches

– Packet switching – a data transfer technique that breaks data into smaller packets before moving them between network computers through either predetermined routes (virtual circuit) or no predetermined route (datagram)

27 August 2008 5

Section A: Answers

27 August 2008 6

Section B: Test for Understanding

• Average = 6.8/10

• Good understanding of the operation

• Only 6 students score perfect 20/20

• Most students score >15/20

• Quite a number score <10/20 = need more understanding

27 August 2008 7

Section B: Test for Understanding

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7F0

F1

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

___

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

___

F3

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7___

F4

F5

F6

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 70 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

___

F7

Computer A Computer B

RR2

F2

RR4

F0

27 August 2008 8

Section C: Test for Application

• Average = 8.45/10

• Very good ability to apply techniques

• Majority score perfect 20/20

• Quite a number got it wrong = need more exercise

27 August 2008 9

Section C: Test for Application

C0C1C2 +Message

+C3 +

27 August 2008 10

Section C: Test for Application

C0C1C2 +Message

+C3 +

C3 C2 C1 C0 C3’

=C2

C2’

=C3+C1

C1’

=C3+C0

C0’

=C3+M

M

Initial 0 0 0 0 0 0 0 1 1

1 0 0 0 1 0 0 1 1 1

2 0 0 1 1 0 1 1 1 1

3 0 1 1 1 1 1 1 0 0

4 1 1 1 0 1 0 1 1 0

5 1 0 1 1 0 0 0 1 0

6 0 0 0 1 0 0 1 0 0

0 0 1 0

27 August 2008 11

Section D: Problem 1&2 test for Analysis

• Average = 5.5/10

• Some ability to do analysis, giving reason, point out the root cause of the problem, recognising different parts of the complete system, anticipate similar problem elsewhere in the system.

• 4 students score perfect 10/10

27 August 2008 12

Section D: Problem 1&2 test for Analysis

• Model answer:Problem 1• Possible reasons (any sensible reasons)

– Node C is connected to 5 other nodes compared to the others (A=3, B=3, D=4, E=3, F=2).

– One of its connections (between C&E) offers the highest bandwidth (1Mbps)

– There is also a bypass link to and from A at C.– This bypass link may provide a least cost path for packets to go through– At peak hour, packets are queued in the buffer and arriving packets that

could not find available buffer space are discarded. These packets will be retransmitted after time out and further aggravates the situation

• The source of the problem– Due to the least cost algorithm that attaches a very low cost to the

1Mbps link and because the routing is fixed, packets cannot be routed away from congestion.

• Other nodes = A and E (high occurrence in the matrix)

27 August 2008 13

Section D: Problem 1&2 test for Analysis

• Model answer: Problem 2• Possible reasons (any sensible reasons but must relate to the

bandwidths)– Link C-E usage is very high because this is one of the links with the

highest bandwidth (1Mbps). This is the fastest link for packets to travel through

– Additionally, packets going to node F will not choose C-F route (slower at 128 kbps), they will choose C-E-F instead (1Mbps and 512kbps)

– A-C, A-D, E-F high because • A-C is a bypass link that can offer lower cost• A-D is one of the fastest link (1 Mbps)• E-F is faster (512 kbps) compared to C-F(128kbps)

– B-D, C-D, C-F, D-E unused because• C-D and D-E the slowest link (64kbps)• B-D (384kbps) is slower compared to B-A-D (512kbps and 1 Mbps)• C-F (128kbps) is slower compared to C-E-F (1Mbps and 512kbps)

27 August 2008 14

Section D: Problem 3 test for synthesis

• Average = 4.2/10

• 4 students score perfect 20/20

• 39 students (~28.3% ) score more than 10/20=are able to handle synthesis

• A good number of students are able to synthesize what they have learnt and use them to propose creative solutions to problems).

27 August 2008 15

Section D: Problem 3 test for synthesis

• Solutions: Possible ideas (any sensible ideas as long as retaining the links (with the given bandwidth of course and fixed routing).

• Examples :1. Attach cost according to link usage with high usage =

high cost and recalculate the link cost considering both bandwidth and usage

2. Combine minimum hop and bandwidth-based least cost path routing. For neighbouring nodes- use direct link. Non direct nodes choose least cost path

27 August 2008 16

Section D: Problem 3 test for synthesis

• Solutions: Possible ideas (any sensible ideas as long as retaining the links (with the given bandwidth of course and fixed routing).

• Examples :1. Attach cost according to link usage with high usage = high cost and

recalculate the link cost considering both bandwidth and usage 2. Combine minimum hop and bandwidth-based least cost path routing.

For neighbouring nodes- use direct link. Non direct nodes choose least cost path

3. Add bias information so that packets can be directed to the destination and avoid high usage link

Cannot change the bandwidth or exchange the links: the question requires you to restrategise the routing

Simple one-by-one substitution is acceptable but with less marks because it is not possible to ask router to decide one by one

27 August 2008 17

Section D: Problem 3 test for synthesis

• Solutions: detail proposal– A good proposal must show follow the

principal ideas set in the previous answer. Contradicting elements will costs marks.

– Must show how the nodes are decided using examples to guide the reader to understand your proposal

– If you attach new costs to the link, the new costs must be shown

27 August 2008 18

Section D: Problem 3 test for synthesis

• Solutions: detail proposal– A good proposal must show that the routing follow

the principal ideas set in the previous answer. Contradicting elements will cost marks.

– Must show how the nodes are decided using examples to guide the reader to understand your proposal

– If you attach new costs to the link, the new costs must be shown

– At this point you do not need to evaluate how good is your proposal in redistributing the traffic. Enough to show how it does the work and the proposed matrix.

27 August 2008 19

Section D: Problem 3 test for evaluation

• Average score = 2/10.• Solutions: EvaluationSteps: • Estimate the current usage and compare to previous usage. Must show

perhaps using table of comparison that the proposed method has redistributed the traffic, no more unused links, no more overused links. (Simple explanation without proof will get fewer marks)

• A stronger evaluation (level 6) will also touch about consequences to individual nodes (eg node C). Although the question requires redistribution of traffic but someone with high level evaluation ability will tend to assess the impact of his proposal to other possible but unspecified requirements. (You may observe that some of your proposed solutions will increase the number of C’s in the matrix)

• Must comment for possible good or bad consequences but not necessarily give more solution (just evaluate). This is to be able to recognise that there are rooms for improvements in the future.