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R.T.T=10 microsec.

M.S.S=100 Bytes

Slow  start protocol used to send data.Sender needs to send data till it completes sending window size  8MSS.Find:-

a.  Total Time taken

b. Throughput for 1st RTT

c. Total Throughput

d. Average Throughput
asked in Computer Networks by Boss (9.7k points)   | 67 views
BW not given?
No.Why do we need bandwidth here?I think parameters given are sufficient .
40 micro s

80 Mbps

1200 Mbps

300 Mbps ???
@joshi:- Actually i dont have answer. First two options i got the same answer but last two are not matching.

How did u get 1200 for c?

I solve it like 15 MSSS are transferred in 40 us then total throughput => 300 Mbps.

Please share your approach for c option

1 Answer

+2 votes
Best answer
For slow start :

1st RTT Send 1 mss
2nd RTT Send 2 mss
3rd RTT Send 4 mss
4th RTT Send 8 mss

so total 4 RTT Required..

a.  So total time take  = 4*RTT = 4*10 = 40 usec.

b.  Throughput for 1st rtt= 100/(10*10^6) = 10MBPS

c.  Total Throughput = (Throughput for 1st rtt+Throughput for 2nd rtt + Throughput for 3rd rtt + Throughput for 4th rtt)
                       = [100/(10*10^6)  + 2*100/(10*10^6)  + 4*100/(10*10^6) + 8*100/(10*10^6)]
                       = [10 +20+40+80] MBPS = 150 MBPS

d.  Average Throughput = (Throughput for 1st rtt+Throughput for 2nd rtt + Throughput for 3rd rtt + Throughput for 4th rtt)/4
                       = [100/(10*10^6)  + 2*100/(10*10^6)  + 4*100/(10*10^6) + 8*100/(10*10^6)]/4
                       = [10 +20+40+80]/4 MBPS = 150/4 = 37.5 MBPS
answered ago by Junior (967 points)  
selected ago by
for throughput you are considering

(no of MSS* Size of each mss)/ Size of each RTT.

But why you are considering T/P for every RTT for calculating TOTAL T/P?
(a) and (b) are correct but (c) and (d) are incorrect , i think !


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