[time-nuts] Divide by five ->Ensemble

Bob Camp kb8tq at n1k.org
Sun Nov 9 17:22:31 EST 2014


What is going on is that people are confusing the estimation process that is used by the selection process (which does look at a lot of stuff) and how that is described. They are then making the leap to the locking process, which is something else altogether.  It’s easy to see that this is what’s going on by looking at the performance of an NTP implementation. If you give it multiple good clocks, the output is no better than one of the clocks in the group. In certain cases, with equally good  / very good clocks, the selection process falls apart and the output actually degrades compared to the best clock in the group due to “best clock” selection changes. That’s not any sort of fault in NTP, you are giving it a situation that (by it’s rules) it really does not matter which one it picks. Any pick will result in an output that it considers to be “good enough”. 


> On Nov 9, 2014, at 4:54 PM, Chris Albertson <albertson.chris at gmail.com> wrote:
> On Sun, Nov 9, 2014 at 12:45 PM, Bob Camp <kb8tq at n1k.org> wrote:
>> Hi
>> The main point is that NTP picks *one* source from among it’s batch of
>> inputs and uses that. The ADEV of the output can be no better than the ADEV
>> of the output.
> The statement above is not correct.  NTP does not select just one clock.
> The statement below is correct and is  what NTP actually does.   If you
> look at the output from "ntpstat" you might think NTP selects one clock but
> internally it's not going that.  The display is misleading.
> In the case of an ensemble of clocks combined with a better approach the
>> ADEV of the output can be better than the ADEV of the best clock in the
>> group.
> It's best to read this http://www.eecis.udel.edu/~mills/ntp/html/warp.html
> In simple terms it searches for consensus range of time where all the error
> bars of the various clock overlap and then eliminate clocks who d't agree
> with the consensus.  Of those still "in" it figures out and kind of
> weighted average.
> I think you could do the same thing.  First find the set of 10MHz
> oscillators who are in phase with each other to within some statistical
> limit and then you compute the weighted average phase
> Maybe start with "Clock Select Algorithm".
> -- 
> Chris Albertson
> Redondo Beach, California
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