Hi,
Comments inline.
--- Ishtiaque Mahbub <[EMAIL PROTECTED]> wrote:
> 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:
3rd octet in Binary:
1234 5678
---------
172.16.1.0/24 = 0000 0001
172.16.2.0/24 = 0000 0010
172.16.3.0/24 = 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
When summarizing, the similar bit is 1 and where the
bit-boundary is common you can summarize. Remember
1=network and 0=host.
In this example, 2 and 3 have a common bit-boundry on
the 7th bit (8+8+7=23), so 172.16.2.0/23 for networks
172.16.2.0/24 and 172.16.3.0/24.
The first network, 172.16.1.0 the 8th bit is 1 and
there are no common bits with the other networks
provided so can't summarize it with others.
Hope this helps... keep at it.
> 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.
- Erick B.
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