COMPP303: Networks and Internet Systems
COMP3616: Networks and Internet Systems

Dr. Liam Murphy

Answers to Sample Exam, March 2002

1. (a) see lecture notes

   (b) efficiency = 0.2, or 20%

   (c) see lecture notes

   (d) see lecture notes

2. (a) Stop-and-wait throughput = 4,016.4 packets/second;
       Go-back-n throughput = 4,878.5 packets/second;

       So Go-back-n can be used, but Stop-and-wait can't.

   (b) see lecture notes

3. (a) if the length of the channel is increased, the
        throughput decreases;
       if the average frame length is increased, the
        throughput decreases.

   (b) see lecture notes

4. (a) initial routing tables:

       A:	Network ID	Cost	Next Hop
		1		1	--
		2		1	--

       B:	Network ID	Cost	Next Hop
		1		1	--
		3		1	--

       C:	Network ID	Cost	Next Hop
		2		1	--
		3		1	--

       When A receives B’s initial routing table, since
        B is 1 hop from A, A modifies B’s table as
        follows:

       Modified B:	Network ID	Cost	Next Hop
			1		2	B
			3		2	B

       The only entry in common is for Network 1, and A’s
        original entry is lower-cost, so A’s new table is

	A:	Network ID	Cost	Next Hop
		1		1	--
		2		1	--
		3		2	B

   (b) link-state packets:

       A:	Advertiser  Network  Cost  Neighbour
		A	    1	     1     B
		A	    2	     4     C

       B:	Advertiser  Network  Cost  Neighbour
		B	    1	     3     A
		B	    3	     1     C

       C:	Advertiser  Network  Cost  Neighbour
		C	    2	     1     A
		C	    3	     2     B

       A uses Dijkstra’s shortest-path algorithm to
        determine its shortest-path spanning tree (see
        lecture notes for details and similar example)

5. (a) Transmission no.  Sender’s Congestion Window (kB)
		0		4
		1		8
		2		16
		3		24
		4		24
		5		4
		6		8
		7		12
		8		16
		9		20
		10		20

   (b) see lecture notes