I have thoroughly enjoyed reading portions of your book "Building Storage Networks." It will take some more time for me to digest it all.
My question pertains to Fibre Channel. When transmitting at full speed (100 MByte/s), the encoded bits are transmitted on the line at 1.0625 Gbits/s. I have heard that out of 1.0625 Gbits, 800 Mbits comprise of data (payload), --> --> --> MBits comprise of 8b/10b overhead penalty of 25%. Where does the remaining 62.5Mbits come from? Is it framing overhead and if so, how is it calculated.
The signaling for FC is as you say, 1.0625 Gb/sec. 8b/10b encoding consumes 20% of the total bandwidth, or ~200Mb/sec. Therefore, FC has 80% bandwidth available for payload transmission rates, or 850 Mb/sec. The missing Gb/sec - are not worth worrying about - it's simple rounding.
Editor's note: Do you agree with this expert's response? If you have more to share, post it in our Storage Networking discussion forum at http://searchstorage.discussions.techtarget.com/WebX?replyToMessage@200.MullaECzaUO^1@.ee83ce4!viewtype=convdate or e-mail us directly at email@example.com.
Dig Deeper on SAN technology and arrays
Related Q&A from Marc Farley
Have a question for an expert?
Please add a title for your question
Get answers from a TechTarget expert on whatever's puzzling you.