[ntp:questions] Asymmetric Delay and NTP
Joe Gwinn
joegwinn at comcast.net
Mon Mar 24 13:38:51 UTC 2014
Magnus,
In article <532FA47B.7060204 at rubidium.dyndns.org>, Magnus Danielson
<magnus at rubidium.dyndns.org> wrote:
> Joe,
>
> On 23/03/14 23:20, Joe Gwinn wrote:
> > Magnus,
> >
> > In article <532E45DB.5000902 at rubidium.dyndns.org>, Magnus Danielson
> > <magnus at rubidium.dyndns.org> wrote:
> >
> >> Joe,
> >>
> >> On 21/03/14 17:04, Joe Gwinn wrote:
> > [snip]
> >>
> >>> It is interesting. I've now read it reasonably closely.
> >>>
> >>> The basic approach is to express each packet flight in a one-line
> >>> equation (a row) in a linear-system matrix equation, where the system
> >>> matrix (the A in the traditional y=Ax+b formulation, where b is zero in
> >>> the absence of noise), where A is 4 columns wide by a variable number
> >>> of rows long (one row to a packet flight), and show that one column of
> >>> A can always be computed from the two other columns that describe who
> >>> is added and subtracted from who. In other words, these three columns
> >>> are linearly dependent on one another. The forth column contains
> >>> measured data.
> >>>
> >>> This dependency means that A is always rank-deficient no matter how
> >>> many packets (including infinity) and no matter the order, so the
> >>> linear system cannot be solved.
> >>
> >> It is just another formulation of the same equations I provided.
> >> For each added link, one unknown and one measure is added.
> >> For each added node, one unknown is added.
> >
> > True, but there is more.
>
> Let's come back to that.
>
> >> As you do more measures, you will add information about variations the
> >> delays and time-differences between the nodes, but you will not disclose
> >> the basic offsets.
> >
> > Also true. The advantage of the matrix formulation is that one can
> > then appeal to the vast body of knowledge about matrixes and linear
> > systems. It's not that one cannot prove it without the matrixes, it's
> > that the proof is immediate with them - less work.
> >
> > And the issue was to prove that no such system could work.
>
> As much as I like matrix formulation, it ain't giving you much more in
> this case, rather than a handy notation. The trouble is that beyond the
> properties of the noise, there is no information leakage about the
> static time-errors and asymmetries. You end up having free variables.
Yes. You correctly noted the mathematical equivalence of the two
approaches, and I agree. My point was that the matrix approach is less
work to get to the desired proof because by formulation as a linear
solution with matrixes, one immediately inherits lots of properties and
proofs.
> The problem is that the unknown and the relationships builds up in an
> uneven rate, and the observations only relate to two unknowns. The only
> trustworthy fact we get is the sum of the delays, but no real hint about
> its distribution. If you do more observations along the same paths, you
> can do some statistics, but you won't get un-biased result without
> adding a prior knowledge one way or another. Formulate it as you wish,
> but as you add more observations, those will be reduced to by their
> linear properties to equations existing and noise. You need to add
> observations which does not fully reduce in order for your equation
> system to grow to such size that you can solve it.
Yes, this is a good statement of the consequences of the proof.
> Show me how you achieve it, and I listen.
I don't understand the challenge. There is no dispute.
> >>> The "no matter the order" part comes from the property of linear
> >>> systems that permuting the rows and/or columns has no effect, so long
> >>> as one is self-consistent.
> >>>
> >>> So far, I have not come up with a refutation of this approach. Nor
> >>> have the automatic control folk - this proof was first published in
> >>> 2004 into a community that knows their linear systems, and one would
> >>> think that someone would have published a critique by now.
> >>>
> >>> The key mathematical issue is if there are message exchange patterns
> >>> that cannot be described by a matrix of the assumed pattern. If not,
> >>> the proof is complete. If yes, more work is required. So far, I have
> >>> not come up with a counter-example. It takes only one to refute the
> >>> proof.
> >>
> >> It is only "by cheating" that you can overcome the limits of the system.
> >
> > Is GPS cheating? That's our usual answer, but GPS isn't always
> > available or possible.
>
> If you are trying to solve it within a network, it is. You can convert
> your additional GPS observation into an a prior knowledge, and once you
> done enough of those, then you can solve it completely. The estimated
> variables better stay static thought, or you have to start over again.
GPS is the usual answer, but isn't always available or useful.
Recall that the original question was random asymmetry due to
asymmetric background traffic in a PTP network. If the network is
controllable, a lab experiment is to simply turn the background traffic
off and see how much the clocks change with respect to one another.
But this tells one how much trouble one is in, but does not solve the
problem. The solution will be found in better choice of hardware, and
better top-level design.
The PTP field has not yet achieved maturity, and we will be the
pioneers.
Joe Gwinn
More information about the questions
mailing list