I will try to explain my way of working out summarisation. I'm not among the
Cisco genius group, but never have problems with it, because of the method I
use.
You need to start thinking of things as a power of 2.
You've taken two bits from the third octet to summarise your networks.
2 to the power of 2 = 4. so your summarisation will be in 'chunks' of 4.
(172.16.0.0, 172.16.4.0, 172.16.8.0, 172.16.12.0) all with /22
If you take 3 bits from an octet for summarisation - 2 to the power of 3 =
8. so your summarisation will be in 'chunks' of 8. (172.16.0.0, 172.16.8.0,
172.16.16.0, 172.16.24.0) all with /21
Similarly with a standard mask of say 255.255.252.0, take the third octet
(or whichever octet is not 0 or 255) away from 256.
This number (4) is the size of the 'chunks' to be used. (10.1.0.0, 10.1.4.0,
10.1.8.0, ..........10.1.100.0, 10.1.104.0 etc)
Just remember to start from 0 every time and see which 'chunk' your networks
fit into.
It could probably be explained a little clearer, but if you get your head
round the explanation, it never fails and is so simple.
Regards,
Gareth
""arthurx4"" <[EMAIL PROTECTED]> wrote in message
931rv9$cqv$[EMAIL PROTECTED]">news:931rv9$cqv$[EMAIL PROTECTED]...
> When you summarize these addresses to a /22 it also includes
>
> 172.16.0.0/24
>
> Joe
>
> "suaveguru" <[EMAIL PROTECTED]> wrote in message
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> | In route aggregation you normally match the number of
> | bits from left to right until a mismatch occurs and
> | that's how you get 172.16.0.0 /22 since for the third
> | octet
> |
> | 00000001
> | 00000010
> | 00000011
> |
> | the common bits are the first 6 bits
> |
> |
> | hope this helps
> |
> | suaveguru
> | --- Hunt <[EMAIL PROTECTED]> wrote:
> | > Can anyone please explain to me how to aggregate the
> | > following 3 routes?
> | >
> | > 172.16.1.0 /24
> | > 172.16.2.0 /24
> | > 172.16.3.0 /24
> | >
> | > into...
> | >
> | > 172.16.0.0 /22
> | >
> | >
> | >
> | >
> | > _________________________________
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