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
J.
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