[ntp:questions] ntpd -x
joelshellman at yahoo.com
Thu Jan 26 15:57:55 UTC 2006
"As the result of this behavior, once the clock has
been set, it very rarely strays more than 128 ms, even
under extreme cases of network path congestion and
jitter. Sometimes, in particular when ntpd is first
started, the error might exceed 128 ms. This may on
occasion cause the clock to be set backwards if the
local clock time is more than 128 s in the future
relative to the server. In some applications, this
behavior may be unacceptable. If the -x option is
included on the command line, the clock will never be
stepped and only slew corrections will be used."
But then at the bottom, in the discussion of command
line args, it says:
Normally, the time is slewed if the offset is less
than the step threshold, which is 128 ms by default,
and stepped if above the threshold. This option sets
the threshold to 600 s, which is well within the
accuracy window to set the clock manually. Note: Since
the slew rate of typical Unix kernels is limited to
0.5 ms/s, each second of adjustment requires an
amortization interval of 2000 s. Thus, an adjustment
as much as 600 s will take almost 14 days to complete.
This option can be used with the -g and -q options.
See the tinker command for other options. Note: The
kernel time discipline is disabled with this option."
This seems to imply that if the offset is greater than
600s a step would occur. Is this correct? I understand
that a difference of 1000s would result in ntpd
exiting anyway, but in that space between 600-1000s, a
step could occur even with -x specified? I'm asking
because it's a requirement that we never have the time
stepped (especially not backwards) except at startup.
I know we can run -q -g at startup, but I want to
ensure that it is not possible to step the clock after
startup. I would rather have ntpd exit.
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