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Question

My solution: As reciever's window is 36 KB (also 18 MSS as MSS is 2KB), so threshold is 1/2(18 MSS)=9MSS.

Now the algorithm will execute like this:

1MSS->2MSS->4MSS->8MSS->9MSS->10MSS->11MSS->12MSS->13MSS->14MSS->15MSS->16MSS->17MSS->18MSS

So. total time is 13RTT's ie 65ms.

Where am I doing mistake??

Comments

Reciever's window=36KB and threshhold=1/2*(receivers window)

                            =1/2*36=18KB.

=2KB--4KB---8KB--16KB---32KB--36KB

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Threshold is not asked, but full size is asked.

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k,no timeout

so to transmit entire window using AIMD

Answer 1

In the question given that there is no congestion. So that initially sender starts with 2mss

(2->4->8->16->32)->36. so there are 4 gaps(RTT) to reach 36. 4*5ms=20ms

Comments

How come threshold is 32KB??