[ntpwg] ntpwg Updated NTPv4 Protocol Specification/timestamp location

[David L. Mills]

Stuart,

In the famous equation you cite, add one second on the path from t1 to 
t2 and a coresponding second on the path from t3 to t4. The additional 
second cancels out. offset = {[(t2 + D) - t1] + [t3 - (t4 + D)]} / 2

Dave

STUART VENTERS wrote:

> Dave,
>
>>> That's the beauty of it. Herbie and Sue don't care. The delays are 
>>> still
>>> reciprocal and the fact Sue's delays are one packet longer doesn't 
>>> matter.
>>
>
> Sadly, this grasshopper appears to have a misconception of what you 
> mean when you say 'the delays are still reciprocal'. (Or perhaps is 
> having trouble drawing timelines or doing algebra.) Please help me 
> understand where I strayed from the path.
>
> If the 'the delays are still reciprocal' and the client and server 
> clocks are aligned, then shouldn't the offset equation equal zero?
>
>
> Regards,
>
> Stuart
>
>
> ps, From RFC-1305 section 3.4.4, setting i=4, offset = ((T2-T1) + 
> (T3-T4))/2.
> From the example, T2-T1 = T3-T4 = -(packet time)
> offset = -(packet time)
>
> pps, My working definition for matching reciprocal delays was that
> the propagation/queueing/transmission delay between T1 and T2
> is the same as between T3 and T4.
> Where same means same sign and magnitude.
> (In the example, they appear to be same magnitude, but different sign.)
>
> ppps, It's interesting that in this example, T2 happens before T1.
> Appendix H, section 8.4, says this can't happen in our universe.
> (Perhaps I should consider this clue further.)