# [802.3ae] RE: Question on the Maximum number of packets per second

The tolerance is +/-100 ppm, so the conformant maximum number of frames
would be 0.01% higher.
If though you are looking at the receive function in a DTE, you basically
double the number because a conformant receiver could be at minus tolerance
while the transmitter is at plus tolerance (the receive rate can be .020002%
faster than the local clock) . The latter analysis is used to budget clock
strokes and size fifos if you want to assure wire-rate, error-free operation
with any conformant tolerances.
--Bob Grow
-----Original Message-----
From: Joshua J. Brickel [mailto:joshua.b@xxxxxxxxxx]
Sent: Thursday, May 31, 2001 1:06 AM
To: stds-802-3-hssg@xxxxxxxx
Subject: Question on the Maximum number of packets per second
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