[time-nuts] Divide by five
magnus at rubidium.dyndns.org
Mon Nov 10 03:00:51 EST 2014
On 11/09/2014 09:44 PM, Poul-Henning Kamp wrote:
> In message <CABbxVHtommjwSWq1i=oH-u1S=G6P=xu8E0YzeKaDg-VchgkGwQ at mail.gmail.com>
> , Chris Albertson writes:
>> NTP does not pick the best clock. NTP finds
>> the subset of clocks that track each other.
> NTP does indeed find the best clock from the subset of clocks
> which pass its sanity check, and then it uses only that one.
> There are several problems with that, and as we speak I'm developing
> a new algorithm which at least so far, gives much superior performance.
> You can read my musings about this here:
> My goal is to release the new NTP client before X-mas.
Just as you point out, there is no easy way around the fact that delays
may be asymmetric. When one analyzes the problem, the actual time-error
and delays in both directions cannot be solved with the two measurements
one do, three unknowns and two equations. The sum of the observations,
forming the Round-Trip-Time, is however the only value we can trust, and
worst case asymmetry will be that of the RTT, regardless of statistical
distribution. Using RTT as a worst-case estimator is thus enough if you
don't know better, and using RTT as a simple stability estimator isn't
all that bad for a simple system.
The NTP scatter-plot wedge is another way to present it, and finding the
"tip" of that wedge is really about finding the min-value of delay in
each direction prior to doing the two-way time-transfer equations.
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