[ntp:questions] Re: Sufficient # servers to sync to

Brian Inglis Brian.Inglis at SystematicSW.Invalid
Fri Mar 18 20:15:30 UTC 2005


On Thu, 17 Mar 2005 12:59:29 -0500 in comp.protocols.time.ntp, Brian
Utterback <brian.utterback at sun.removeme.com> wrote:

>Richard B. Gilbert wrote:
>
>> Take an extreme case:
>> 
>> Server A says it's 11:53
>> Server B says it's 11:55
>> Server C says it's 23:52
>
>To add to what Brad said, what you really have is the following
>candidates for the time:
>
>1. The interval 11:53-11:55
>2. The interval 11:55-23:52
>3. The interval 11:53-23:52

If the uncertainty for all servers was 2 minutes, the candidate
intervals would be:

peer  low   mid   high
A    11:51 11:53 11:55
B    11:53 11:55 11:57
C    23:50 23:52 23:54

>Let's see... number 1 has two servers on it, number 2 has
>two servers, and number 3 has all three servers. So, with
>three servers voting for number 3, I guess it is the winner.
>
>This means that all three servers are allowed to proceed to
>the next stage in the selection, so server C may still get the nod.

The overlap interval would be 11:53-11:55 with servers A and B in the
clique and C flagged as a falseticker. 

>Just to clarify, the vote goes to the shortest interval that has
>at least n/2 servers on it.

Okay. 

>So, if we add one more server right
>around 11:5x, we see that this would then create a new interval
>with three servers, not allowing server C.

Only if the new server interval overlapped the existing interval. 

>The original long
>interval will have all four, but is trumped by the shorter one
>with three, since 3 is still more than n/2 = 4/2 = 2 servers on
>it.

Depending on the overlap, it could join the existing clique, or form a
new clique with another server. 

-- 
Thanks. Take care, Brian Inglis 	Calgary, Alberta, Canada

Brian.Inglis at CSi.com 	(Brian[dot]Inglis{at}SystematicSW[dot]ab[dot]ca)
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