Thanks for your answer but this triggered a few more questions(!)
My summary of thoughts are like this:
1. If there had not been the choice no. 2, my arithmatic is correct?
2. The question that I had just asked, there was a similar question on the
book, which proposed almost the same network summarasation:
172.16.12.0/24, 172.16.13.0/24, 172.16.14.0/24:
and the summarasation of 172.16.12.0/22 was the correct answer......(choice
was a) 172.16.0.0/24 b)172.16.14.0/24 c)172.16.12.0/22 d)172.16.14.0/22)
3. if the purpose of the route summarasation is to reduce routing table
entry so why not a single entry rather than two?
hope you dont mind my asking, just trying nail the idea into my thick skull!
Regards,
Ishtiaque
>From: "Kenny Sallee" <[EMAIL PROTECTED]>
>Reply-To: "Kenny Sallee" <[EMAIL PROTECTED]>
>To: "Ishtiaque Mahbub" <[EMAIL PROTECTED]>, <[EMAIL PROTECTED]>
>Subject: Re: Route Summarisation, once again!
>Date: Tue, 27 Jun 2000 23:49:32 -0700
>
>If you chose the answer C you would have to change the network statement to
>172.16.0.0 255.255.252.0. This would summarize networks 172.16.0.0,
>172.16.1.0, 172.16.2.0, and 172.16.3.0. So you would summ an extra
>etwork( the 172.16.0.0 /24 ). The answer B then is more correct. The
>second statement ( 172.16.2.0/23 ) would cover only networks .2 and .3 and
>the first would cover .1 of course.
>
>If it were my network and it was private addressing - I would choose C.
>But
>for a test question it would be B.
>
>Kenny
>
>----- Original Message -----
>From: "Ishtiaque Mahbub" <[EMAIL PROTECTED]>
>To: <[EMAIL PROTECTED]>
>Sent: Wednesday, June 28, 2000 4:56 AM
>Subject: Route Summarisation, once again!
>
>
> > Hello Group!
> >
> > Could someone be kind enough to explain a dilemma that I have been
>facing
> > with route summarisation? In Todd's book I found the following question:
> >
> > How the following networks should be summarised?
> > 172.16.1.0/24
> > 172.16.2.0/24
> > 172.16.3.0/24
> >
> > a) They cant be summarised
> > b) 172.16.1.0/24 and 172.16.2.0/23
> > c) 172.16.1.0/22
> > d) 172.16.0.0
> >
> > Well I chose C.
> >
> > Here is my explanation:
> > Considering third octet Binary format of 1: 0000 0001
> > Considering third octet Binary format of 2: 0000 0010
> > Considering third octet Binary format of 3: 0000 0011
> > So the highest number of similar bits for this octet is 6
> > The total number of bits similar for the networks: 8+8+6=22
> > That summarises the network to 172.16.1.0/22
> >
> > But the answer says B with the explanation: Networks must share the same
> > high-order bits. Look at the binary values to understand more clearly.
> >
> > What am I missing?
> >
> > Regards,
> >
> > Ishtiaque
> >
> > ________________________________________________________________________
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