Once you do enough of them, you will find your own patterns and ways, but if you use simple subtraction and look for the difference to be a power of 2 that really helps! For instance in the first octet if you have say 192 and 200 ... 200 - 192 = 8 = 2^3 ...so you know you can match them both with 1 bit in the "8" place.
Regards, Joe Astorino CCIE #24347 (R&S) Sr. Support Engineer - IPexpert, Inc. URL: http://www.IPexpert.com -----Original Message----- From: [email protected] [mailto:[email protected]] On Behalf Of Kim Pedersen Sent: Monday, June 08, 2009 2:27 PM To: Tyson Scott Cc: [email protected] Subject: Re: [OSL | CCIE_RS] ACL Wildcards Thanks for all of your help... When you guys do it, do you start by writing it all out in binary, or make an educated guess on what groups together? and it is best to start with the first octet and going forward, or the last going backwards? Again, Thanks! Sincerely, Kim Pedersen Tyson Scott wrote: > Yes Correct Kim, > > 194 and 193 can defiantly be matched in one line if all the rest were > the same. In your example none of those could be combined into one > line without matching additional networks. > > Regards, > > Tyson Scott - CCIE #13513 R&S and Security Technical Instructor - > IPexpert, Inc. > > Telephone: +1.810.326.1444 > Cell: +1.248.504.7309 > Fax: +1.810.454.0130 > Mailto: [email protected] > > Join our free online support and peer group communities: > http://www.IPexpert.com/communities > > IPexpert - The Global Leader in Self-Study, Classroom-Based, Video On > Demand and Audio Certification Training Tools for the Cisco CCIE R&S > Lab, CCIE Security Lab, CCIE Service Provider Lab , CCIE Voice Lab and > CCIE Storage Lab Certifications. > > > -----Original Message----- > From: Kim Pedersen [mailto:[email protected]] > Sent: Monday, June 08, 2009 2:02 PM > To: Tyson Scott > Cc: 'Bryan Bartik'; [email protected] > Subject: Re: [OSL | CCIE_RS] ACL Wildcards > > Hi Tyson, > > In my example, those 4 bits are just in the first octet alone. So im > assuming we really need to treat the entire address, and not just by octet? > > So there's no "set-in-stone" rules to go by, you just sort of have > to group them, see if that matches and go from there? > > Finally, in my example, if i add the 193 prefix, I would have 6 bits > of difference, so the closest i could do in one line is by matching 64 > nets, and this would give an indication on whether i need to narrow it down? > > Sincerely, > Kim > > Tyson Scott wrote: > >> Kim >> >> When it has a large amount of differences you need to find >> similarities between them to put them together >> >> 194 is 11000010 >> 174 is 10101110 >> >> This is 4 bit differences so you would have to have 16 entries to >> match >> > them > >> as one line without matching additional subnets >> >> It is important to also note if they say to not match any additional >> networks or if they just say to combine them to as few lines without >> specifying that you can't match additional networks as well. >> >> Regards, >> >> Tyson Scott - CCIE #13513 R&S and Security Technical Instructor - >> IPexpert, Inc. >> >> Telephone: +1.810.326.1444 >> Cell: +1.248.504.7309 >> Fax: +1.810.454.0130 >> Mailto: [email protected] >> >> Join our free online support and peer group communities: >> http://www.IPexpert.com/communities >> >> IPexpert - The Global Leader in Self-Study, Classroom-Based, Video On >> > Demand > >> and Audio Certification Training Tools for the Cisco CCIE R&S Lab, >> CCIE Security Lab, CCIE Service Provider Lab , CCIE Voice Lab and >> CCIE Storage Lab Certifications. >> >> >> -----Original Message----- >> From: [email protected] >> [mailto:[email protected]] On Behalf Of Kim >> Pedersen >> Sent: Monday, June 08, 2009 11:28 AM >> To: Bryan Bartik >> Cc: [email protected] >> Subject: Re: [OSL | CCIE_RS] ACL Wildcards >> >> Hi Bryan, >> >> I guess I didnt point out the problem (sounds soo serious :) ), but >> what if the question states: "make these into as few entries as >> possible", and they are soo different that it might not end up in one >> entry (again, with difference in multiple octets). >> >> For example (no logic behind choosing these): >> 194.64.0.96/27 >> 174.34.87.64/26 >> 193.23.10.8/30 >> ... >> Next, imagine 32 addresses just like this :) >> >> How do you go about breaking all of this down? >> >> Sincerely, >> Kim Pedersen >> >> Bryan Bartik wrote: >> >> >>> 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] >>> <mailto:[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-compl > ex-wil > >> dcard-masks-for-access-lists/ >> >> >>> >>> >>> Rgds >>> >>> Jorge >>> >>> >>> >>> *From:* [email protected] >>> <mailto:[email protected]> >>> [mailto:[email protected] >>> <mailto:[email protected]>] *On Behalf Of >>> *JEREMY FURR (RIT Student) >>> *Sent:* Friday, June 05, 2009 10:12 AM >>> *To:* [email protected] >>> > <mailto:[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 >>> <http://192.168.2.0/24> through 192.168.15.0/24 >>> <http://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] <mailto:[email protected]> >>> >>> >>> >>> >>> -- >>> >>> >>> >> >> > > -- // Freedom Matters // Follow my progress on: http://kpjungle.wordpress.com No virus found in this incoming message. Checked by AVG - www.avg.com Version: 8.5.339 / Virus Database: 270.12.54/2158 - Release Date: 06/08/09 06:01:00
