[802.3_NGAUTO] Question on PHY/MAC and pause_quanta and interleaving latency

[William Lo <will@xxxxxxxxxx>]

All,

 

I spec some latency numbers in yesterday’s ad hoc and depending on the interleaving L=1, 2, 4  that is selected

the pause_quanta changes which is what we want since we are trading longer latency for better error correction.

The pause_quanta is used by the MAC for the pause function.

 

This raises 2 questions for me in light of the topic raised yesterday that the MAC and PHY only has some predefined signaling

to tell each other what is going on and the PHY cannot willy-nilly affect parameters that impacts the MAC without a way to inform

the MAC.  If someone knowledgeable with the MAC can chime in on the following questions it would be great.

 

Question 1:

Let there be PHY A and PHY B and corresponding MAC A and MAC B.  PHY A tells PHY B what interleaving L = 1, 2, 4 that it wants during training.

So how does PHY B tell MAC B what the pause_quanta is since this value is dependent on what PHY A wants and is not known a-priori.

 

Question 2:

The max PHY latency (which is converted into units of pause_quanta for MAC use) is defined as the sum of PHY A transmit delay + PHY A receive delay

      which with unequal interleaving becomes  PHY A tx delay (interleaved per B’s request) + PHY A rx delay (interleaved per A’s request)

Since PHY A and PHY B do not have to request identical interleaving values how is the pause quanta computed in the mixed case since

the real latency in one path is PHY A tx delay (interleaved per B’s request) + PHY B rx delay (interleaved per B’s request) and the

other path is PHY B tx delay (interleaved per A’s request) + PHY A rx delay (interleaved per A’s request)

 

Question 2B:

A related question is when we compute PTP delays, based on my limited understanding, the round trip delay is computed and we

divide by 2 to estimate the delay on the link.  If we have unbalanced latency because of different interleaving, will this affect PTP?

The delay difference between L = 1 vs L = 4 (i.e. 1us) is much much larger than the propagation delay across 15m cable.

 

One way to solve this is that if the interleaving values requested are not identical then the larger L is the

one that has to be used by both PHYs. 

 

Thanks,

William

 

 

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