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(elec315)[2011](s)midterm~548^_44291.pdf
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Midterm solution and marking schemes
1. (6pts)
(a)
RTT= 36000KM x 4 /3x108m =0.48 sec
(2pts for calculation of RTT)
100K bps = x bytes/0.48sec.\> x = 48 Kbits
(1 pts for final answer)


(b)
2 x RTT + 10K/100K = 1.06 sec (see the HTTP GET protocol in the handouts) (2pts for calculation process and 1 pts for final answer )



2. (6pts)
(a)
(a) nslookup Ctype=ns. (3 points) (1 pts for the .)

(b)
IP header: source address (1 pts),
protocol (1 pts)
TCP header: SYN bit and ACK bit. (1 points)




3. (6pts)
(a)
port 3.
(2 pts for correct answer)
(1 pts for explanations)


(b)
port 3
(2 pts for correct answer)
(1 pts for explanations)




4. (6pts)
(a)2x+300+ 100=1000.\> x = 300. (1 pts)
Thus A = 400K bps, B = 600K bps (1 pts)

(b)
2x+200+ 300= 1000.\> x = 250. (1 pts)
Thus A = 450K bps, B = 550K bps (1 pts)


(c)
3x + 200 +200 +300 = 1000.\> x =100. (1 pts)
Thus A = 300k bps, B = 300K bps, C = 400K bps (1 points)




5. (6pts)
(a)
Sending window is : 0,1,2,3,4 (1.5 pts)
Receiving window: 2,3,4 (1.5 pts)


(b)
Sendind window: 2,3,4,5,6 (1.5 pts)
Receiving window: 2,3,4. (1.5 pts)



6. (3 pts) use DNS load balancing. (1 pts for key words) Set up two type A records in the DNS server, one with ISP1 IP address, and the other with ISP2s
IP address. The DNS server will use a round robin scheme to select each of the two entries. (2 pts for explanation)

7. (6 pts)
(a)
(2 pts for caulculation of accessing delay)
delay of the access link (10K/1.5M)/ (1\0.99) = 0.67 sec.
(1 pts for total delay)
Thus the total delay = 1.67 sec


(b)
0.5(1 + 6.7 ms/0.5) = 0.5065 sec
(2 pts for formula and 1 pts for final answer)




8. (7pts)
(a) DST Distance Next Hop
B1 B
C2 B

D3 B
E3 G
F2 G
G1 G

(2 pts for correct answer)
(b)
After two exchanges, the table is already stable. Thus after t = 2, the routing table reaches the steady state.

(2 pts for correct answer)

(c)
Lets just consider the impact for destination C. The analysis for destination B can be analyzed similarly. As far as destination C is concerned, the routing tables of C,D,E,F will not be affected. Only the routing tables of A, B and G will be affected.


The distance to node C in the routing tables of A, B and G are given as below:
Aftert=101 A=4, B=2, G=3.
Aftert=102 A=3, B=5, G= 4(learnedfromF)
Aftert=103, A=5(learnedfromG),B=4, G=4
After t = 104, A = 5, B = 6, G = 4 Now the tables reach the final state.

The same thing canb e said about the routing tables for destination B (routing tables C,D, E will be affected)
(1 pts for correct answer and 2 pts for explanations)