From: Conor Daly (conor.daly at domain oceanfree.net)
Date: Thu 06 Jun 2002 - 02:31:44 IST
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 domain oceanfree.net> Domestic Sysadmin :-) --------------------- Faenor.cod.ie 9:48pm up 13 days, 7:06, 0 users, load average: 0.00, 0.00, 0.00 Hobbiton.cod.ie 9:51pm up 13 days, 7:12, 2 users, load average: 0.12, 0.05, 0.01
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