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Sunday, December 1, 2024

community – TCP congestion management – quick restoration algorithm examples


I am studying TCP congestion management and I am wanting over explicit 2 slides which I am undecided I am getting all the way down to the main points of them.

So right here is the primary:

Slide1

and right here is the second:

Slide2

I do not perceive why within the first it says that just one,7 and eight are pending, and thus with cwnd of 4 after 3 duplicate acks the sender can ship the 1 and thus get ack(9) and proceed. What I do perceive that after that quick restoration it could proceed with cwnd=4 the place all the pieces is okay.

If I have a look at the arrows rigorously and what despatched and acknowledged at what level, I additionally perceive that 0-6 are initially despatched, then 0 is acquired (the primary black arrow with ACK(1) signifies this to the sender), then cwnd is elevated to eight, and that is why 7 and eight despatched, making the entire cwnd [1,8]. Then after the ship of 8, the sender cannot ship no extra (cnwd is reached to max pending packets). We get 3 duplicate ACKS which corresponds to segments 2,3,4, and in line with quick restoration protocol (this instance as a lot as I perceive no selective ack is but examined) it ought to re-transmit the 1 and minimize the cwnd by 2 from 8 to 4 (right me if I am incorrect so far).

How at this level 1,7,8 are pending and never additionally 5,6 ? I see on the graph that evidently acks for five and 6 additionally acquired by the sender however within the graph it appears to occur 1 retransmitted, however how it’s re-transmitted if the window appears to be full ? Maybe the order of the arrows is not correct ? In any other case what do I miss at that time ?

It appears to be perhaps if acks from 5 and 6 gotten earlier than the pink line of retransmission of 1, then perhaps certainly just one,7,8 had been pending and 1 might be retransmitted, is not it ?

Perhaps if I get the primary one, I can perceive how the second situation differ from it and why the sender is caught and solely a timeout may help.

I believe I get the larger image there, the place there are an excessive amount of pending segments and the cnwd is full, so it could’t retransmit section 2 after it received ack(2), however I do not get the main points of the method and what occurs precisely at which level, which I discover a profound challenge I need to perceive.

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