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RE: Proposal for accomodating 10.0000 and 9.58464 line rates



Title: RE: Proposal for accomodating 10.0000 and 9.58464 line rates

Ed,

You may wish to re-work your numbers.  If you assume that there is no stripping of the preamble, only the IPG is stripped, then the maximum size packet is 1530 bytes.  This would require a minimum 67 byte IPG.  Based on the OC-192 payload transfer rate of 9.58464 Gb/s, the equation for calculating the required IPG is:

0.043336 * x = y

where:
x = the packet size (in bytes)
y = the IPG size (in bytes)

To help offset the loss of throughput due to selecting on IPG size (i.e. 67 byte IPG with 64 byte frame), the "standard" could state the equation used for the WAN (OC-192) PHY but leave it up to the implementor as to the granularity of the implementation.  For example, a step function based on the standardized minimum IPG (12 bytes) could be implemented, or another implementor could use a smaller granularity step function.

Cheers,
Brad

Brad Booth
bbooth@xxxxxxxxxx
Level One Communications, Austin Design Center
(512) 407-2135 office
(512) 589-4438 cellular