GATE 2012 EXAM congestion control modified question

Question

Consider an instance of TCP’s Additive Increase Multiplicative Decrease(AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 12 MSS. Assume that a time out occurs during the seventh transmission. Find the congestion window size at the end of the 11th transmission. Please help i have confusion will it jump to 12 mss directly and then afterwards additive increase mechanism or it will additive increase after 8mss straight away to 9mss then 10mss.. please help

Comments

@Shubham: See this..

https://gateoverflow.in/2156/gate2012_45

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thanks bro...i got it...really helpful...

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@arjun sir

Shall we count the 10 packets as delivered or not when timeout occurs?
Like it in 5th transmission window size is 10 so 10 sent so will they be considered as received at receiver or not?

Asking for questions that requires sending 100 packets to receive kind of.
Pls explain.

Answer 1

2 4  8 9 10 11 12  TO (new thresold=12/2=6) 2 4 5 6(11 th transmission)   7   

7 answer...

we will not jump directly to even in slow phase...12(if thresold is not a power of 2 then increment linearly)

Comments

New threshold will be 15/2 = 7 (not 12/2 = 6).

See this...

https://gateoverflow.in/2156/gate2012_45

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 rfzahid 

i have visited given link but still i don't know how threshold is 15/2  ?

in question initial threshold is 12 mss......in slow start phase 2,4,8,9,10,11,then 12

but why u have taken....12 just after 4 ???

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2 --> 4 --> 8 --> 12 --> 13 --> 14 --> 15 (TO) --> 1 --> 2 --> 4 --> 7 --> 8 --> 9 --> 10 ................

TO is occurring during the 7th transmission, so new threshold = 15/2 = 7