[ntp:questions] strange behaviour of ntp peerstats entries.
david at ex.djwhome.demon.co.uk.invalid
Sat Feb 2 16:40:25 UTC 2008
> Worse than that . Only if the latest sample is the one with the min delay
> is it chosen Otherwise it is not. You can go for 16 or more samples never
> using any of thembefor one fits the criteion. (actually the samples are
> aged as well-- ie the delay is increased as they get olderbut that seems to
> have little effect.)
The increase is a pessimistic estimate of the error due to not having
quite the right frequency, so has a larger effect with larger polling
intervals. The delay is a pessimistic estimate of the affects of link
asymmetry. Incidentally, are you sure it is delay; I would have
expected the root dispersion to be included.
If you use that logic, whilst the last good sample is still in the
filter, only the head sample can actually override it.
The real questions seem to be whether one should use such worst case
estimates of error and whether one should rescan the whole filter when
the current value drops off the end.
If it really is failing to rescan, that is either an implementation
error or a change from NTP V3's formal specification. NTPv3 requires a
sort by distance, which includes dispersion terms. ntpd 4.2.0, does
seem to implement the sort that chooses the lowest distance, and does
seem to use distance, rather than delay, so it looks to me that you've
either misread the code, or it has changed between 4.2.0 and 4.2.4,
apparently for the worse.
I wonder if you are getting confused by the bit that says a peer sample
must be younger than the last one used? It seems to me that that is
only there as a tie breaker, if two samples have equal aged differences.
More information about the questions