172.16.1.0/22 is not correct. 172.16.0.0/22 would be the correct syntax,
but since the 172.16.0.0 is not included, I would aggregate these 3 networks
as:
172.16.1.0
172.16.2.0/23
Geoff
----- Original Message -----
From: "thangs" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Thursday, October 19, 2000 10:22 PM
Subject: Re: Summarization
> I feel the summarized supernet route should be 172.16.1.0/22
>
> from the below fig ..its clear that the no of common bits in the MSB part
is
> 22
>
> |
> If you expand 172.16.1.0 in bin ----->10110110.00010000.000000
|01.00000000
> 172.16.2.0 in bin------>10110110.00010000.000000
> |10.00000000
> 172.16.3.0 in bin------>10110110.00010000.000000
> |10.00000000
>
> |
>
> Thanks
> Thangavel
>
> ----- Original Message -----
> |
> From: <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Cc: <[EMAIL PROTECTED]>
> Sent: Thursday, October 19, 2000 3:57 AM
> Subject: Re: Summarization
>
>
> > In a message dated 10/19/00 4:33:48 AM Eastern Daylight Time,
> > [EMAIL PROTECTED] writes:
> >
> >
> > > Hi All,
> > >
> > > There are 3 contiguous networks:
> > > 172.16.1.0/24
> > > 172.16.2.0/24
> > > 172.16.3.0/24
> > > What is the supernet ?
> > > Is it 172.16.1.0/22 ? Would you pls explain to me ?
> > >
> > > TIA,
> > >
> >
> > That is not the supernet because 172.16.2.0 does not have a common bit
> place
> > in 172.16.1.0/22. I believe that this would be superneted to
> 172.16.0.0/16.
> > This is where I see them all matching up that is. It would be real hard
> for
> > me to explain this so I'll let someone else do that who is more
> experienced
> > with this. Hope I helped a little.
> >
> > Mark Zabludovsky ~ CCNA, CCDA, 1/4-NP
> > <A HREF="mailto: [EMAIL PROTECTED]">[EMAIL PROTECTED]</A>
> >
> > "If you need luck, apparently you're not prepared...Go study!"
> >
> > ~Mark Zabludovsky~
> >
> > _________________________________
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