Fibre Channel (SAN)

Reply
Frequent Contributor
Posts: 90
Registered: ‎12-26-2010

2148 bytes equal 21480 bits when encoded.

"The average time between transmits which is calculated per second and then averaged over the time value. As a reference at 1G with full sized frames the minimum this value could be is 20.25us (a bit is transmitted every 941 picoseconds, full size frame with SOF, header, payload, CRC, & EOF is 2148 bytes, 21480 bits when encoded, and there is a requirement of at least 6 ordered sets between frames, each ordered set is 40 bits encoded, so a minimum of 21520 bits are transmitted)" Could someone help to clarify how 2148 bytes became 21480 bits?  This i saw in the bottleneck detection  best practices document.

Frequent Contributor
Posts: 160
Registered: ‎08-07-2009

Re: 2148 bytes equal 21480 bits when encoded.

I agree that it is a little unclear as to how these numbers were derived. I will look into this and get some clarification.

-Bill

Valued Contributor
Posts: 547
Registered: ‎03-20-2011

Re: 2148 bytes equal 21480 bits when encoded.

i believe that's because of 8/10 bits encoding. see 8b/10b encoding - Wikipedia, the free encyclopedia for more details...

Frequent Contributor
Posts: 90
Registered: ‎12-26-2010

Re: 2148 bytes equal 21480 bits when encoded.

looking forward to what Bill comes up with . surely, let me know if its different.

Frequent Contributor
Posts: 160
Registered: ‎08-07-2009

Re: 2148 bytes equal 21480 bits when encoded.

2148 bytes becomes 21480 bits due to the 8b/10b encoding. Essentially each eight-bit data byte is converted to a

10-bit transmission character. This is applicable to pre-16G. I hope this helps.


-Bill

Join the Community

Get quick and easy access to valuable resource designed to help you manage your Brocade Network.