Brian,
The selection algorithm doesn't vote; all it does is establish the final
correctness interval. Servers within the interval are truechimers, while
those outside are falsetickers. "Voting" among the truechimers is done
by the clustering algorithm on the basis of stratum and root distance.
When I say "vote", what I mean is that a majority vote among three of
the four servers is possible in order to declare the fourth a traitor.
All this says is the three servers are truechimers and the fourt is a
falseticker. The "vote" is not for the winner, it's for the loser.
Dave
Brian Utterback wrote:
> Yes, to all of your points. I wasn't dealing with the dispersion
> intervals and jitter and all that. I was just trying to illustrate
> why three servers won't cut it. In reality the offsets shouldn't be
> so large and all that. But if the intervals don't overlap (and they
> often don't) then the example stands.
>
>
> Brian Inglis wrote:
>
>> On Thu, 17 Mar 2005 12:59:29 -0500 in comp.protocols.time.ntp, Brian
>> Utterback 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.
>
>
> |