On Tue, Jun 04, 2002 at 04:33:04PM +0100, John Lyons wrote:
> On Tuesday 04 June 2002 10:02, John Bolger wrote:
> > Subtract the last number from 256 (in this case 256-248). This gives an
> > answer of 16, which will always be the size of the subnet (including
> > network address and broadcast address).
> > Next get a list in your head of products of this number (in this case they
> > will be 0,16,32,48,64, etc.). These are your possible network addresses.
> > At that point is obivious which network your host belongs to (0, as
> > 0<1<15). So your answer is 192.168.0.0
>> This is all wrong of course as 256 -248 is 8 and not 16 ;)
Argh - you would think after all these years of working this stuff out
Id know how to subtract :). Anyway 8 is the correct answer and the
"algorithm" will still work ....
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