From: DERY, FREDERIC (frederic.dery@xxxxxxxxxxx)
Date: Thu May 25 2000 - 17:31:04 GMT-3
Hi,
As your 3 networks are not contiguous you cannot create a supernet with
them.
But if you need to filter them in one entry in an access-list or other
kind of filtering, you can do something as these access-list are not
restrained by the contiguous requirement.
As we see, all third octet are odd.
If we check the value associated with each bit in an octet:
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 1 = 1
0 0 0 0 0 0 1 1 = 3
0 0 0 0 0 1 1 1 = 5
We see that the only way to play with odd/even number are related to the
last bit. If the last bit is set to 1 it will always gives us an odd
number and if it is set to zero, it will always give us an even number.
Also we see that 1, 3 and 5 are all values associated with the last 3
bits in the octet.
So we can contruct an access-list entry that include all theses network
:
access-list 1 permit 172.16.1.0 0.0.6.255
The 0 bits must match and the 1 are don't care bits, giving us for the
3rd octet :
128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 1 Network = 1
0 0 0 0 0 1 1 0 MASK
Which tell us that that the even/odd bit (the last bit) must match
exactly the network number stated, which is odd. So we could have used 3
or 5 as the 3rd octet in our access-list network with the same result.
If anything looks wrong or you need more explanation, just ask
Regards
Frederic Dery
Robert_Wang@toyota.com wrote:
>
> How do I create a supernet to cover 172.16.1.0/24 and 172.16.3.0/24 and
> 172.16.5.0/24??
>
> -Robert-
>
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