On Friday 24 January 2003 05:35, Scott Mcdermott wrote:
>dalton on Fri 24/01 09:59 +0100:
>> But we have a 100mbit LAN, so there should be 10240 kbps, thats
>> right?
>
>just being picky,
>
>$ bc -ql
>100 * (2^10)^2 / 8 / 2^10
>12800.00000000000000000000
>
>:) of course if you actually hit that wire speed I'll be amazed.
>
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.

-- 
Cheers, Gene
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