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The initial congestion window size over a TCP is 1. If slow start algorithm is used and the size of congestion window incremented by 1 whenever an ACK is received i.e. after first rounda trip time congestion window size is 2 segments. Assume that connection never leaves slow start. The number of RTT’s to send 3999 segments are _______.

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Verify this logic:

For every packet sent the window size increases by 2. So when

wind=2...... RTT =1

wind=4..... RTT= 1+2

wind=8...... RTT=1+2+3

.....

wind=2048..... RTT=1+2+3+4+....11=66

Now after this for every ack 2 packets so to send remaining 1951/2

when window size reaches 2048 66 round trips. Now for remaining packets 3999-2048=1951. For every ack one packet will be sent so required RTT = 1951/2=975.5. so total would be  1041

is it right approach?
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6 votes
6 votes
Ans. is 12.

1st RTT- 1 segment sent

2nd RTT- 2 new segments sent

3rd RTT-4 new segment sent

4th RTT-8 new segments sent

5th RTT-16 new segments sent

6th RTT-32 new segments sent

7th RTT-64 new segments sent

8th RTT-128 new segments sent

9th RTT-256 new segments sent

10th RTT-512 new segments sent

11th RTT-1024 new segments sent

12th RTT-2048 new segments sent

Total segments sent=

1+2+4+8+16+32+64+128+256+512+1024+2048=4095>3999

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