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.