First of all, thanks to all those who responded.
The last octet mask is three bit but .148 which is 1001 0100.
My curiosity was aroused when during our class discussion of subnetting, our
instructor (I just don't know if I heard it correctly) mentioned that the
bits (for the subnet) can be anywhere in the octet. We were discussing the 3
bit mask which is .224.
So it the subnet bits can be located anywhere in the octet (although our
instructor mentioned not to waste our time on this as Cisco does not support
this kind of subnet masks), I made up a different mask for the 3 bit which
in this case is 148 (1001 0100).
So, what now? For a three bit mask, there would be 6 usable subnets (I am
excluding subnet zero). My problem now is, what is my 1st subnet and what
are the hosts for this subnet. And what are the other subnets (and their
corresponding hosts)?
It made my head spin and kept me awake. So, this is the reason why I asked
the list.
Until this time, it keeps me thinking.
Arthur
>From: "Bob Vance" <[EMAIL PROTECTED]>
>Reply-To: "Bob Vance" <[EMAIL PROTECTED]>
>To: "CISCO_GroupStudy List \(E-mail\)" <[EMAIL PROTECTED]>
>Subject: RE: Discontiguous networks
>Date: Thu, 15 Mar 2001 14:20:17 -0500
>
> >The binary mask for last octet would be 10100000.
>
>Actually, it's worse than that!!
>
> 10100000 = 128 + 32 = 160, not 148 :|
>
> 148 = 128 + 20 = 128 + 16 +4 = 1001 0100
>
>I'll let you fill the rest in :)
>
>
>-------------------------------------------------
>Tks | <mailto:[EMAIL PROTECTED]>
>BV | <mailto:[EMAIL PROTECTED]>
>Sr. Technical Consultant, SBM, A Gates/Arrow Co.
>Vox 770-623-3430 11455 Lakefield Dr.
>Fax 770-623-3429 Duluth, GA 30097-1511
>=================================================
>
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