Kim, even if there is more than one octet you still can look at the number of bits that are different. Example:
192.168.0.0 192.168.0.1 192.168.1.0 192.168.1.1 The above addresses have 2 bits (bit 0 in the 3rd and 4th octets) that differ and we can combine them in one ACL. 3rd and 4th octets: 0000 0000 | 0000 0000 0000 0000 | 0000 0001 0000 0001 | 0000 0000 0000 0001 | 0000 0001 0000 0000 | 0000 0000 AND 0000 0001 | 0000 0001 XOR 192.168.0.0 0.0.1.1 would be the ACL entry. -hth Bryan Bartik CCIE #23707 (R&S), CCNP Sr. Support Engineer - IPexpert, Inc. URL: http://www.IPexpert.com On Mon, Jun 8, 2009 at 7:47 AM, Rodriguez, Jorge < [email protected]> wrote: > Jeremy this should help you in doing the calculating wildcard mask > > > > http://www.internetworkexpert.com/resources/01700370.htm > > > > > http://blog.internetworkexpert.com/2007/12/26/q-how-do-i-compute-complex-wildcard-masks-for-access-lists/ > > > > Rgds > > Jorge > > > > *From:* [email protected] [mailto: > [email protected]] *On Behalf Of *JEREMY FURR (RIT > Student) > *Sent:* Friday, June 05, 2009 10:12 AM > *To:* [email protected] > *Subject:* [OSL | CCIE_RS] ACL Wildcards > > > > Does anyone know of a website or book that explains well how ACL wildcards > work? I have been trying to filter out four blocks from a bunch of route > advertisments but just can't get the three I want through, this is what I > have R2 is originating 192.168.2.0/24 through 192.168.15.0/24 in RIP to > R1. I want to only accept blocks 192.168.5.0, 192.168.10.0, 192.168.13.0 and > 192.168.14.0 > > > > If I use acl with 192.168.10.0 0.0.4.0, I will get 10 and 14 but not > thirteen. For the 5 network I just use the 192.168.5.0 0.0.0.255. > > > > Any thoughts or help would be appreciated. > > > Jeremy Furr [email protected] > --
