Number of RTTs

Question

Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of slow start phase is 2 KB. The receiver advertise its window at the starting of communication to be 24 KB. Assume the size of advertise window is never changed by the receiver throughout the
communication. The time taken to send the full window is _________ msec, considering RTT of 10 msec.

I think it should be 9rtts but the solution says 10rtts since after 9 RTTs we would be able to send full 24kb window  

Can somebody provide some refrence links for this

Answer 1

I assume that communication is being set up. SO, their is no value of ssthreshold set up. So, we need to increase exponentially(or linearly in avoidance phase) till we get timeout.

We start with 1 MSS

1 MSS

2 MSS

4 MSS

8 MSS - 16KB

16MSS - 32KB

Now, this will create timeout So, ssthreshold = 24/2 = 12KB = 6MSS

Note: Here, they have given receiver window and it has nothing to do with congestion window. Just assumed that 24KB is also congestion window.

1MSS

2MSS

4MSS

5MSS

6MSS

7MSS

8MSS

9MSS

10MSS

11MSS

12MSS

So, 15 RTTs

Comments

threshold should be 12

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why 12?

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24/2

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@Pankaj. We are wrong. They havent given whether its starting of communication or  it has already started. If So, I assume that  communication is just starting.So, window size will reach 32KB and timeout will occur.

CHeck my answer.

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The question is wrong actually. They should have said congestion window and not receiver window.

Also, they havent given ssthreshold value.

I suggest just ignore the question.

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Ok that I can understand but my doubt was something different can you point me to some other question which has a verified solution so I can clear my doubt on rtts

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Could you browse the network Qs? Even I dont have.

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Why 5 MSS after 4 MSS ? I did not get this Sushant. Please clarify. I am also getting same answer as Pankaj. I think it should be 9 RTT. How did you get 15 RTT?

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Thats what I told . They havent given ssthreshold. So, should we assume that this is setting up of the communication (which is I have assumed) where you increase the window size until timeout

OR

assume that 24KB is the window size where timeout occured?

So, Insufficient info for me.

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@sushant,full widow size is 24KB.and thresold is min(rec window,congestion wid) /2

so ,here it should be 12KB

why are u taking timeout at 32KB??

i also thnk 9rtts are required

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@Akriti. I really dont know whats the actual meaning of threshold. I suppose its just an estimate from where you can expect congestion to occur.

Now, they are its start of communication.

Do you have any reference for folowing formula?

min(rec window,congestion wid) /2

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@Akriti.

See this

See the "operation" part of it. 

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@sushant,in above link,in "slow start" part,it is given that "The transmission rate will be increased with slow-start algorithm until either a loss is detected, or the receiver's advertised window (rwnd) is the limiting factor, or the slow start threshold (ssthresh) is reached. "

and generally in such questions,we initially take threshold = cpngestion window/2

but how come 32KB come into picture???

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@Akriti.

So, I took exponential increase until timeout occured/loss detected.

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and how do u know that timeout occured atr 32KB?

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because receiver advertized window size = 24KB. SO, remaining spilled out by receiver