Re: Question on the Maximum number of packets per second
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.
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 =
>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