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But why 2 ^ t integral can be as gain?
I can understand f (t) = (2 ^ t) of the independent variable t represents the RTT, so what is the physical meaning of the integral?
And why the minimum?
Looking forward to your reply
Hi,The explanation is very detailedThanks.
But will future versions replace 2.89 with 2.77? Just because of strict doubling, right?
Thank you very much!
That's a great explanation
But can you tell me the reason for the change in gain from 2/ln2 to 4ln2 and the original derivation of 2/ln2.
> On 26 Jun, 2018, at 3:02 am, 赵亚 <maryw...@126.com> wrote:
>
> The reason for "the window of silence" seems to be the coefficient high_gain.
The gain is greater than unity in order to permit the bandwidth estimate to converge to the true path bandwidth as soon as possible. In a TCP that was entirely dependent on managing cwnd, you could more reasonably set the pacing rate to transmit the window over exactly one RTT.
The real reason for the "window of silence" is that the cwnd is not similarly scaled in this initial phase, and this initial cwnd is still respected by BBR. You can actually see smaller silences, with the same cause, after the packet pairs resulting from the first few acks. Later on, the cwnd ceases to be a constraint under normal conditions, as long as the bandwidth estimate matches reality on the wire.
- Jonathan Morton
(lowest gain that will allow the pacing rate to double each round trip)
but doesn't 3.465 [5ln(2)] instead of 2.77 [4ln(2)] give the exact 2x pacing_rate each round trip?
On Thursday, August 26, 2021 at 6:50:38 PM UTC+8 muhamm...@umt.edu.pk wrote:kindly read this.On Friday, June 8, 2018 at 8:42:30 AM UTC+5 maryw...@126.com wrote:Hi ALL!
I encountered a problem while reading the source code. I hope I can get replies. Thanks!
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In the following statement, there is a magic number, as noted, which is 2/ln(2) :
static const int bbr_high_gain = BBR_UNIT * 2885/1000 + 1;
But I don't know how it came about.
I have read the paper of BBR and have the following explanation:
To handle the Internet link bandwidths span 12 orders of magnitude,Startup implements a binary search for BtlBw by using a gain of 2 / ln2 to double the sending rate while delivery rate is increasing. This discovers BtlBw in log2BDP RTTs but creates up to 2 BDP excessThe queue in the process.
But I still don't understand the details.Why not 3/ln(3), why not 2.5, why not e?I think there might be a smooth curve that satisfies the startup constraint, such as a two-fold increase in speed per round.And then you take the derivative of this curve, and you get some value...There must be a mathematical explanation behind this, but I don't know what it is.Regards.
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