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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