[ntp:questions] Re: Clock selection?
David J Taylor
david-taylor at blueyonder.co.uk.not-this-bit
Sun Jan 18 18:05:34 UTC 2004
"David L. Mills" <mills at udel.edu> wrote in message
news:400ABDE2.F767B4D9 at udel.edu...
> 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
> 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.
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
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