RE: Question on the Maximum number of packets per second
As noted earlier, the nominal rate is the 14.88095 Mp/s (i.e. 10x1.488095
from 1Gb/s ethernet). The +/- ppm is just that, +/-, not 2X. 2X is the
potential, in-spec, delta between what you could be transmitting, and what
you could be receiving, not the delta of what you could be receiving WRT the
nominal.
I'll stay out of the 20ppm vs. 100ppm thrash here... :)
Best Regards,
Mike Bly
Senior ASIC Engineer
World Wide Packets
115 N. Sullivan Rd.
PO Box 950
Veradale, WA 99037
-----Original Message-----
From: Stefan M. Wurster [mailto:stefan@xxxxxxxx]
Sent: Thursday, May 31, 2001 2:28 PM
To: Sanjeev Mahalawat; Joshua J. Brickel; stds-802-3-hssg@xxxxxxxx
Subject: RE: Question on the Maximum number of packets per second
with the only problem, that the +/-20ppm capable timing recovery of the
other end will not lock to my +/-100ppm offset data. This is the whole point
of Roy's discussion. We need to resolve this.
Stefan M. Wurster
Director, Engineering
TDK-Semiconductor
465 Fairchild Drive, Suite 130
Mountain View, CA 94043
Voice: 650-988-4812
FAX: 650-988-0582
-----Original Message-----
From: owner-stds-802-3-hssg@xxxxxxxxxxxxxxxxxx
[mailto:owner-stds-802-3-hssg@xxxxxxxxxxxxxxxxxx]On Behalf Of Sanjeev
Mahalawat
Sent: Thursday, May 31, 2001 13:28
To: Joshua J. Brickel; stds-802-3-hssg@xxxxxxxx
Subject: Re: Question on the Maximum number of packets per second
Joshua,
In LAN PHY the remote (transmitter) clock can be 100ppm faster and the local
(receiver) clock can be 100ppm slower than the nominal and be compliant.
This
results in 0.1024 bits more (for 64byte packets) that would reduce
the ipg by this number resulting in 14.88322 (round to 14.8833) mpps.
In WAN PHY I guess it has been changed to +/-20 ppm clock difference allowed
but if one designed for +/-100ppm clock difference packet throughput it will
cover
+/-20ppm clock difference.
Thanks,
Sanjeev
At 10:06 AM 05/31/2001 +0200, Joshua J. Brickel wrote:
>
>I was wondering if to find for 10GbE the maximum number of packets that
will
>enter on an interface per unit time can be calculated as follows...
>
>64bytes for minimum packet + 8 Bytes Preamble/SFD + 12 Bytes IPG/EFD = 84
>bytes = 672 bits
>
>Then the maximum number of these minimum sized packets would be 10E+10/672
=
>14.881 MP/s
>
>This would seem correct, except that I have not included how much off the
>transmitters clock can be. Can this significantly alter the numbers I have
>above? If anyone has any insight as to how fast a clock could be off of
the
>nominal 10 gigabit rate and still be considered compliant I would
appreciate
>it posted.
>
>Thanks,
>
>Joshua
>