Gene Heskett on Fri 24/01 08:34 -0500: > > > But we have a 100mbit LAN, so there should be 10240 kbps, thats > > > right? > > > > $ bc -ql > > 100 * (2^10)^2 / 8 / 2^10 > > 12800.00000000000000000000 > > I don't believe this is quite correct. Thats a serial protocol, and > AFAIK it still uses start and stop bits, so the real /8/ divisor above > should probably be /10/, giving a 10 megabytes or 100 megabaud a > second rate, hence the naming convention.
Ethernet is synchronous, and framed; the overhead is nowhere near that. http://sd.wareonearth.com/~phil/net/overhead/ looks like it comes out to about 12051 KBps, or 11.8MBps.