"Even /20 would work. The increment goes in ranges of 8:" Incorrect, a /20 would be increments of 16.
To start you have to know how many networks/hosts are in a given bit range. 1 bit (1000 0000) is .128 or /17 and produces 2 ranges of 128, so 0-127 or 128-255 2 bits (1100 0000) is 192 or /18 and produces 4 ranges of 64, so 0-63, 64-127, 128-191, 192-255 3 bits (1110 0000) is 224 or /19 and produces 8 ranges of 32, so 0-31, 32-63, 64-91, etc 4 bits (1111 0000) is 240 or /20 and produces 16 ranges of 16, so 0-15, 16-31, 32-47, etc 5 bits (1111 1000) is 248 or /21 and produces 32 ranges of 8, so 0-7, 8-15, 16-23, 24-31, *32-39, etc* see there? I think you can figure out the rest of the pattern. On Thu, Jul 25, 2013 at 7:30 PM, Wilberth E. Lemaître < [email protected]> wrote: > Even /20 would work. The increment goes in ranges of 8: > > 10.0.8.0 > 10.0.16.0 > 10.0.24.0 > 20.0.32.0* > 20.0.40.0 > > *In that range the networks will fit perfectly and the summarization would > be more efficiently. > I think I need to practice ;) > > From: [email protected] > To: [email protected] > Subject: FW: OSPF Summarization > Date: Thu, 25 Jul 2013 17:55:15 -0600 > > > > > I was able to see the logic, if I use 3 bits the ranges would go like > these: > > 10.0.0.0 - 10.0.31.255 > 10.0.32.0 - 10.0.63.255 > 10.0.64.0 - 10.0.95.255 > > As we can see, they will all catch in the second range. Correct me if I am > wrong? > > > From: [email protected] > To: [email protected] > Subject: OSPF Summarization > Date: Thu, 25 Jul 2013 17:27:49 -0600 > > > > > Hello community, > > I have a question in regards summarization. > Let's say I have area 2 configured on a router, the advertised networks > are the following: > > router ospf xxx > network 10.0.32.1 0.0.0.0 area 2 > network 10.0.33.1 0.0.0.0 area 2 > network 10.0.34.1 0.0.0.0 area 2 > network 10.0.36.1 0.0.0.0 area 2 > network 192.168.0.2 0.0.0.0 area 0 > > I want to summarize area 2, the mechanism I use for summarization is the > following and correct me if I am wrong. I look for the octet where the > decimal number changes, or where the bits are not equal. As we can see this > happens on the 3rd octet and I start using increments with each octet value: > > 128 64 32 16 8 4 2 1 > > I wonder myself which increment or which number would catch all of those > into one advertisement counting from zero, in other words, how many binary > digits I have to convert to 1 in order to do the summarization. > Based on this scenario, 32 would catch less than what we need, 64 would be > the perfect one, and I will need to convert only 2 bits (the first 2 bits) > and my subnet mask would look like 11111111.11111111.11000000.00000000 and > the network range would go like this: > > 10.0.0.0 - 10.0.63.255 > 10.0.64.0- 10.0.127.255 > > As we can see, if I use a subnet mask of 18, I will be including ranges > that are NOT necessary. > > What would be the efficient summarization in order to accomplish this > scenario? Please explain ;) > > Best regards, > Wilberth > > > > > > > _______________________________________________ > For more information regarding industry leading CCIE Lab training, please > visit www.ipexpert.com > > Are you a CCNP or CCIE and looking for a job? Check out > www.PlatinumPlacement.com > > http://onlinestudylist.com/mailman/listinfo/ccie_rs > -- Marc Abel CCIE #35470 (Routing and Switching) _______________________________________________ For more information regarding industry leading CCIE Lab training, please visit www.ipexpert.com Are you a CCNP or CCIE and looking for a job? Check out www.PlatinumPlacement.com http://onlinestudylist.com/mailman/listinfo/ccie_rs
