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Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD)
algorithm where the window size at the start of slow start phase is 1 MSS and the
threshold at the start is 1st transmission is 16 MSS. Assume TCP use over a lossy
link i.e., timeout occur after transmission of 7th packet . What is the congestion
window size at the end of 14 RTT (in MSS)?

A) 9   B) 11 C) 12 D) 14
asked in Computer Networks by (159 points) | 37 views
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1  || 2 || 4 || 8 || 16  || 17 || 18(TO threshold=18/2=9mss)  ||  1 ||  2 || 4 || 8  || 9  || 10   || 11 || 12 .

=12 answer.

0

but here its saying there is a timeout after 7th packet so:

1  || 2 || 4 || 8 || 16  || 17 || 18(TO threshold=18/2=9mss)  ||  1 ||  2 || 4 || 8  || 9  || 10   || 11 TO threshold=11/2=6mss)   || 1  || 2 || ........

so ans should be 1??

+1
they haven't said that it happens after every 7th transmission.

so why to take it again.. we will have TO only once..
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Why u are taking threshold as 8 MSS pls explain?
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At 7th packet time out occured ,and at that time mss is 16 ,so we divide by 2 i.e 8 to become the new threshold.

1 Answer

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I got this question the answer is 11

at loss of 7 packet the cnwd size is 18 MSS time out occurs therefore threshold reduce to 9 and cnwd starts from  and now exponentially grow till 9  then till 11 u will get 14 th packet therefore CNWD will be 11MSS
answered by (339 points)
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I think answer should be 12.

1  || 2 || 4 || 8 || 16  || 17 || 18(TO threshold=18/2=9mss)  ||  1 ||  2 || 4 || 8  || 9  || 10   || 11 || 12 .

=12 answer.

What is the ans given in key ?

0
Yes u r correct answer will be 12 at end of 14MSS

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