Re: The RTO=4*RTT thing in TFRC

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Sorry,

> 1) s became 8 s
> 
> 2) b became 1
> 
> 3) RTO became 4 (not 4 RTT!)
> 
> I can understand 1) above, as this means that a non-DelACK
> receiver is assumed. I can also understand 2), because you need
> the throughput per 8 seconds (not 1 second) for the protocol -
> but I can't understand the reasoning behind replacing RTO with 4.

This is obviously a mistake - I swapped 1) and 2) here.
(but still I don't understand 3)  :-)  )

Cheers,
Michael




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