But in last octet we always want value of 1 .. isnt it ? so i guess it would not be wrong to write 192.1.1.1 ..
do correct me if i miss something .. 192.1.1.1 0.0.14.0 will include 192.1.1.1 , 192.1.3.1 , 192.1.5.1 , 192.1.7.1 , 192.1.9.1 Gaurav Madan. On 1/27/08, Scott Morris <[EMAIL PROTECTED]> wrote: > > Because 192.1.1.0 would set your initial starting bits. Once you put > your "starting point" in place and then apply a mask (which tells you which > bits can change and which ones cannot) then the results you get will be > different! > > HTH, > > > > Scott Morris, *CCIE**4** (R&S/ISP-Dial/Security/Service Provider) #4713, > JNCIE-M #153**, JNCIS-ER, CISSP, et al. > **CCSI/JNCI-M/JNCI-ER > *VP - Technical Training - IPexpert, Inc. > IPexpert Sr. Technical Instructor > > A Cisco Learning Partner - We Accept Learning Credits! > > [EMAIL PROTECTED] > > > > Telephone: +1.810.326.1444 > Fax: +1.810.454.0130 > http://www.ipexpert.com > > > > ------------------------------ > *From:* [EMAIL PROTECTED] [mailto: > [EMAIL PROTECTED] *On Behalf Of *GAURAV MADAN > *Sent:* Saturday, January 26, 2008 12:39 PM > *To:* Ralph Olsen > *Cc:* [email protected] > *Subject:* Re: [OSL | CCIE_RS] Wildcard Q > > > 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 > > > > >
