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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 2 KB and the threshold at the start of first transmission is 24 KB. Assume that 3 duplicate ACK are received during the 5th transmission what is the congestion window size at the end of 10th transmission?

The answer given is 16KB which is correct but getting doubt on how they solve the question. 

I think the steps should be:- 

1MSS = 2KB.

(in MSS) = 1, 2, 4, 8, 12, 1, 2, 4, 6,7

So after 10th transmission it will be 8 which is 16KB.

And they solved (in KB) = 2,4,8,16, 24(TO then Threshold = 24/2 = 12) 12,13,14,15,16.

This is how they did.

Doubt :- Here it is mentioned that "3 duplicate ack". Now, if we go along TCP Tahoe then we follow what is given in above. But what steps to follow when TCP Reno is asked?


An instruction pipeline consists of following 5 stages:
                       IF = Instruction Fetch, ID = Instruction Decode, EX = Execute,
                       MA = Memory Access and WB = Register Write Back
Now consider the following code:

Assume that each stage takes 1 clock cycle for all the instructions. The number of clock cycles are required to execute the code, without operand forwarding over a bypass network ________.

I am getting 12, by doing overlapping between WB and ID stage i.e. in one half of cycle doing WB and in another half doing ID.

Given answer is 14.

asked in Programming by Boss (15.2k points) | 97 views
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 Shubhanshu  TCP reno,avoids slow start phase and makes
ssthresh = ssthresh /2
cwnd = ssthresh instead of setting cwnd to 1MSS

 2,4,8,16, 24(TO then Threshold = 24/2 = 12) 12,13,14,15,16.

but MSS = 2KB
 

2,4,8,16, 24(TO then Threshold = 24/2 = 12) 12,14,16,18,20 

i think this is more correct. 

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But as it is given that mss is 2KB shouldn't the increment after 12 be like this 12 14 16 ...
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  • Tahoe: If three duplicate ACKs are received (i.e. four ACKs acknowledging the same packet, which are not piggybacked on data and do not change the receiver's advertised window), Tahoe performs a fast retransmit, sets the slow start threshold to half of the current congestion window, reduces the congestion window to 1 MSS, and resets to slow start state.[12]
  • Reno: If three duplicate ACKs are received, Reno will perform a fast retransmit and skip the slow start phase by instead halving the congestion window (instead of setting it to 1 MSS like Tahoe), setting the slow start threshold equal to the new congestion window, and enter a phase called fast recovery.

Taken from Wikipedia:-

@Mk Utkarsh 

Tahoe Steps:- (In KB) = 2,4,8,16,24,2,4,8,12,14,16. 

Answer : - 16KB // Note that they are asking about AFTER 10th transmission what will be the congestion window.

Reno Steps:- (In KB) = 2,4,8,16,24,12,14,16,18, 20, 22.

Answer : - 22KB // Note that they are asking about AFTER 10th transmission what will be the congestion window.

And data will always transfer in terms of MSS, which is 2KB, so there should not be any window size which is an odd number like 13, 15 KB....

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They solved it wrong

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