[ILUG] Linux Certification Exam Question

[Conor Daly]

On Tue, Jun 04, 2002 at 04:33:04PM +0100 or so it is rumoured hereabouts, 
John Lyons thought:
> 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 ;)
> 
> So the Subnet ID remains 192.168.0.0 with a broadcast of 192.168.0.7

Um, so what's the netmask for a subnet of 192.168.0.8 broadcast
192.168.0.15 ?

And why?

Conor
-- 
Conor Daly <conor.daly at oceanfree.net>

Domestic Sysadmin :-)
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