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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 (111 points)
edited by | 82 views
0

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..
0
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.
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Why are we taking  at 7th RTT.Its 7th packet right?Means

In 1st RTT p1 is transmitted.

In 2nd RTT p2,3 are transmitted

In 3rd RTT p4,5,6,7 are transmitted

now timeout occurs.and so on right.

1 Answer

0 votes
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 (337 points)
0

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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