From: RIDNEY_M_LAUDIANO.TTSP@ts.tsuneishi.co.jp
Date: Fri Nov 25 2005 - 22:42:46 GMT-3
hi,
i'll give this a shot.
"Grabler, Ross \(IT\)" <Ross.Grabler@morganstanley.com> wrote on 11/26/2005
06:27:00 AM:
) Hi, I am trying to understand solution to the below problem
) Permit below in the minimal amount of lines in an acl, solution is
) below, I don't understand wildcard of 8, can someone explain? (this is
) from IE lab 17.
) 200.0.1.2
) 200.0.3.2
) 200.0.3.10
) 200.0.1.18
) 200.0.3.26
) 200.0.1.10
) 200.0.3.18
) 200.0.1.26
)
) solution
) access-list 1 deny 200.0.1.2 0.0.2.8
denies:
200.0.1.2
200.0.3.2
200.0.1.10
200.0.3.10
) access-list 1 deny 200.0.1.18 0.0.2.0
200.0.1.18
200.0.3.18
) access-list 1 deny 200.0.1.26 0.0.2.0
200.0.1.26
200.0.3.26
) access-list 1 permit any
explanation for access-list 1 deny 200.0.1.2 0.0.2.8
convert the 3rd and 4th octet of the addresses to binary and put into a table
decide what subnet and mask we will use for the access-list using the following
rules.
1. If the column is all 0’s, the subnet is 0, and the mask is 0
2. If the column is all 1’s, the subnet is 1, and the mask is 0
3. If the column is a mixture of 1’s and 0’s, the subnet is 0, and the mask is
1.
3rd octet (common numbers: 1 and 3)
1 00000001
3 00000011
s 00000001 (1)
m 00000010 (2)
4th octet (common numbers: 2 and 10)
2 00000010
10 00001010
s 00000010 (2)
m 00001000 (8)
where s: subnet, m: mask for corresponding octet
the result would be:
access-list 1 deny 200.0.1.2 0.0.2.8
you can also use one line to deny all the above addresses:
access-list 1 deny 200.0.1.2 0.0.2.24
hope this helps!
ridney
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