[Linuxptp-users] Doubts about PTP standard

[Namit Agrawal]

Hi All,
My question is not related to linuxptp, but its rather a
general question about PTP protocol. In the IEEE Std paper of 1588
protocol, from page 201 to 219, there is an appendix in which they have
given examples for use of TCs, BCs and OCs. One thing to notice is that
the mean path delay calculated by means of residence time correction,
asymmetry correction etc. comes out to be the same as the assumed mean
path delay. Then why exactly do we take into consideration all this
residence time, asymmetry etc. into our calculations??

Also, what exactly is the residence time because according to my
understanding difference of the time from master to slave and vice versa
will automatically give you the time (including residence time) it took
to propagate from one end to other.

For eg. Suppose the path is like this :
Master -----------> Router1 -----------> Slave
5s 1s 5s
(residence time)

So suppose the packet started at t1 = 9:00 AM and reached slave at t2 =
9:02 and 11 seconds. So master and slave clocks have an offset of 2
minutes. But t2 - t1 ultimately takes into consideration the residence
time also.
Similarly on the other way:
Slave -----------> Router1 -------> Router2 -----------> Slave
5s 1s 2s 5s
(residence time)

t4 - t3 will give us 13 seconds difference even if we do not calculate
residence times separately.

Why are we separately building TCs and BCs to calculate residence times
?? What is the use of that??

[Richard Cochran]

Why are we separately building TCs and BCs to calculate residence times ??

Because each and every message will have its own widely varying
residence time.

What is the use of that??

To correct for the variation by removing it.

HTH
Richard

[Namit Agrawal]

Namit Agrawal <***@...> writes:

But the widely varying residence time can be taken care of in a simple
(t2 - t1) and (t4 - t3). Because no matter what, the difference will
always tell you the time it took to propagate including the residence
time. I think the point that nobody explains to user is that when you
are calculating the delay and offset by just dividing them by 2, you are
assuming that propagation on both sides is the same which is not. So you
cannot take average of something which is not equal on both sides.
That's why you have to take into consideration the residence time. But
this leads me to another question which is regarding the calculation
done in the appendices of the IEEE Std. 1588 paper from page 201 to 219.
Q. How and why is assumed mean path delay equal to calculated mean path
delay??

Q. Why do we have to calculate separately mean path delay when a simple
assumed mean path delay is giving us the answer??