[ntpwg] ntpwg Updated NTPv4 Protocol Specification/timestamp location

[Kurt Roeckx]

On Thu, Feb 07, 2008 at 05:34:08PM +0000, David L. Mills wrote:
> Stuart,
> 
> No, compare your equation with mine. The view has to be from A, so the 
> delay cancels.


I'm still not convinced.  So I'll try to make an example.  Let's assume
that it takes 2 second for the packet to travel from A to B and the
other way around.  That is, if A sends a physical signal from low to
high, B will see that change 2 seconds later.  Let's also assume that it
takes A 1 second to transmit the packet.  Then let's assume that A's and
B's clock are in sync, and that A takes the timestamp at the start of
the packet, and B and the end.

We start with A sending a packet at time 0.0.  A takes the time
when the packets starts to be transmitted, so sets T1 to 0.0.

The packet will start to arrive at B at 2.0, and stop at
3.0.  B takes the timestamp at the end of the packet
and will set T2 to 3.0.

It takes B 1 second to process the packet.  It will start to transmit
the packet at 4.0.  But since it timestamps the packets after
it's completly send, it will set T3 1 second later to 5.0.  B will inform
A about T3 with an other packet it will send later.

The packet starts to arrive at A at 6.0, which is when A takes
the timestamp, getting us T4 = 6.0.  The packet completly arrives at
7.0, but that doesn't matter.

This gives:
offset = 1/2 * [(T2-T1) + (T3-T4)] = 1/2 * [(3-0)+(5-6)] = 1.

While we expected the result to be 0.

Note that for the delay we don't have the problem.
delay = (T4-T1) - (T3-T2) = (6-0) - (5-3) = 4.


Kurt


> STUART VENTERS wrote:
> 
> > Dave,
> >
> >>> Let t1, t4 belong to A and t2, t3 belong to B. From A's point of view,
> >>> t2 is late by D and t3 is late by D. Prestatus equationatus.
> >>
> >
> > Yeeeesssss, finally, your words match the example.
> > I had given up all hope.
> > Now it appears there is a glimmer of light in the long tunnel.
> >
> > A consequence of your words appears to be that
> > the resulting offset eqn calculated at A is:
> > offset = {[(t2 + D) - t1] + [(t3 + D) - t4]} / 2
> >
> > for the example, offset = D = -(packet time)
> >
> >
> > If you agree with this baby step, then let's pop the question stack 
> > for the example.
> >
> > If the 'the delays are still reciprocal' and the client and server 
> > clocks are aligned, then shouldn't the offset equation equal zero?
> >
> >
> > 73's,
> >
> > Stuart
> >
> > ps, My morse is rusty, I meant to say QRT for I had truly given up.
> >
> >
> 
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