Hi Perfect explanation . We want
192.1.1.1 192.1.3.1 192.1.5.1 192.1.7.1 192.1.9.1 This will make 192.1.1.1 0.0.14.0 . Which is why we write access-list 1 permit 192.1.1.1 0.0.14.0 Can any one comment on why they have written 192.1.1.0 0.0.14.0 instead of 192.1.1.1 0.0.14.0 Gaurav Madan. On 1/26/08, Ralph Olsen <[EMAIL PROTECTED]> wrote: > > Hi Khalid, > > You don't care about the 1st, 2nd and 4th octet, because that will be 0 > as it doesn't change. > You will have the 3rd octet being > 1 > 3 > 5 > 7 > 9 > > If you write that in binary you will get > > 1 00000001 > 3 00000011 > 5 00000101 > 7 00000111 > 9 00001001 > > So in the wildcard mask we know that a fixed bit will be marked as a 0 > and a wildcard bit will be marked as a 1. > > 1 00000001 > 3 00000011 > 5 00000101 > 7 00000111 > 9 00001001 > > 00001110 <-- Bit 5,6,7 (from left to right) is wildcard bits. And > converting 00001110 to decimal will give you 14. > > The wildcard mask being 0.0.14.0. > > Hint: Use Windows notepad and calculator and write all IP address' in > binary. That will give you a better visual overview. > > /Ralph Olsen > > Hi all, > > > > we have 2 routers R1.......> R2 and R1 has 10 Lo interfaces: > > Lo1: 192.1.1.1/24 <http://192.1.1.1/24> > > Lo2: 192.1.2.1/24 <http://192.1.2.1/24> > > Lo3L 192.1.3.1/24 <http://192.1.3.1/24> upto 192.1.10.1/24 > > <http://192.1.10.1/24> > > > > we need to configure R2 to ONLY receive the odd addresses from R1 > > > > the answer was: > > > > R2: > > access-list 1 permit 192.1.1.0 <http://192.1.1.0> 0.0.14.0 > > <http://0.0.14.0> > > router rip > > distance 255 150.50.17.1 <http://150.50.17.1> 255.255.255.255 > > <http://255.255.255.255> 1 > > Note: 150.50.17.1 <http://150.50.17.1> is R1 > > this Q is from the workbook. Can you give me a full explanation > > about 0.0.14.0 <http://0.0.14.0> > > > > thnx > >
