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1 votes
1 votes

Initially (as slow start )

full window 12MSS

cwnd = 1MSS

ssthresh = 6MSS

1MSS sent 

after 1st RTT -> cwnd = 2 ,

and so on subsequent RTT -> 4 , 5 ,6, 7, 8,9,10,11,12

so 11RTT

so answer should be 110ms

but none of the option matches

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5 Answers

Best answer
3 votes
3 votes
Sender window = 24KB
Thresold = 12KB

2KB --> 4KB --> 8KB --> 12KB --> 14KB --> 16KB --> 18KB --> 20KB -- > 22KB --> 24KB
Total 10 RTT..

Time = 10*10 = 100ms
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1 votes
1 votes
With slow start, the first RTT sends out 1 segment (or 2KB), the 2nd RTT sends out 2 segments (or 4KB), the 3rd 4 segments (or 8KB), the 4th 8 segments (or 16KB). The 5th RTT would have sent out 16 segments (or 32KB), however, it'll exceed the receiver's window. Therefore, the amont of time it takes BEFORE the 5th RTT (or full window, that is, 24KB) is 4*10=40msec.
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0 votes
Ans(C)

MSS=2kB and sender window size is 24kB and RTT=10ms

using SLOW START ALGO ,we calculated threshold as 24kB/2=12Kb=6MSS

NOW,

(1) 1MSS(2kB)

(2) 2MSS(4kB)

(3) 4MSS(8kB):we can not send 8MSS as threshold is 6MSS therefore additive increase or congestion avoidance algo. applied

(4) 5MSS(10kB):(2kB+4kB+8kB+10kB=24kB) complete window size is transmitted

it requires 4 RTT's to send complete packet 4*10ms=40ms
–1 votes
–1 votes
Th=12

2->4->8->10->12->14->16->18->20->22->24

So total time=10*time=10*10ms=100

So above option is wrong.

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