From: John Conzone (jkconzone@xxxxxxxx)
Date: Sat Sep 09 2000 - 10:36:30 GMT-3
Derek, from my early morning calculations the second list denies all
even routes not divisable(?) by 4.
> access-list 4 deny 192.168.2.0 0.0.252.255
> access-list 4 permit any
Third octet 0 0 0 0 0 0 1 0
Mask 1 1 1 1 1 1 0 0
So if the last 2 bits must always be 1 0, then no odd routes will match
the deny statement becasue of the last bit, and no routes that can be
evenly divided by four will match because of the second bit.
So this denies even routes in 192.168.X that can't be divided by four
evenly. Like 192.168.10.X, 192.168.14.X., 192.168.2.X.Check my math as I may
be off. I'm sure if I'm wrong someone will let me know!
----- Original Message -----
From: "Derek Buelna" <dameon@aracnet.com>
To: <ccielab@groupstudy.com>
Sent: Saturday, September 09, 2000 12:00 AM
Subject: access-list 3 deny 192.168.1.0 0.0.254.255
> Hi,
>
> I'm trying to understand this access-list:
>
> access-list 3 deny 192.168.1.0 0.0.254.255
> access-list 3 permit any
>
> Does it say that all odd routes within 192.168.0.0/16 will be denied
> because we match against the last bit in the third octet being one. I
guess
> this would cover all odd routes??
>
> Would then:
>
> access-list 4 deny 192.168.2.0 0.0.252.255
> access-list 4 permit any
>
> deny all even routes?
>
> Thanks,
>
> -Derek
>
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