[ntp:questions] Re: Clock selection?
David L. Mills
mills at udel.edu
Mon Jan 19 03:57:56 UTC 2004
Not quite. While the list is ordered by increasing stratum/distance, it
could be and sometimes happens that the lowest one gets torpedoed as the
cluster is whittled down. Selection error as defined in the briefings is
the RMS offset differences between a each survivor and all the others.
David J Taylor wrote:
> "David L. Mills" <mills at udel.edu> wrote in message
> news:400ABDE2.F767B4D9 at udel.edu...
> > David,
> > It doesn't work like you suspect. Rather:
> > 1. Order the servers by increasing synchronization distance (first
> > stratum, then one-half delay plus dispersion). This is the initial set
> > of survivors.
> > 2. Compute the mean incidental offset error r[i] for the ith survivor.
> > This is shown in the ntpq billboard as jitter. Compute the mean
> > selection error s[i] of the ith survivor relative to all the others.
> > 3. Toss out the survivor with the largest s[i]. If the minimum s[j]
> > remaining is less than the maximum r[k], or if the number of survivors
> > is three or less, terminate the procedure; otherwise, continue in step
> > 2.
> > The first survivor in the remaining list becomes the synchronization
> > source and the others participate only in the combining algorithm. The
> > details are of course in the briefings.
> > Ordinarily, delay is not a factor in the algorithm, as satellites can
> > make very good sources. The usual problem is not the satellite itself,
> > but asymmetric paths, one way by satellite the other by landline.
> > Dave
> Thanks very much for that - I'm not sure what you mean by "mean selection
> error" in step 2, but apart from that I follow what you describe.
> What you be the effect of choosing the server with the lowest
> synchronisation distance rather than "the first survivor" in the final
> step? From what you say, the list may indeed be synchronisation
> distance-ordered, so the lowest synchronisation distance server is
> automatically chosen....
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