On Jun 21, 2011, at 11:32 PM, Marko Milivojevic wrote: > Bits are actually: 128 64 32 16 8 4 2 1 (powers of 2, starting from > 2^0 on the right).
Right we use these for normal subnets, but for wildcard masks the values are minus 1 correct? > > The reason this doesn't work is that in this case, wildcard mask needs > to be contiguous, i.e. the real inverse netmask. Netmask > 253.255.255.255 is *not*: > > 00000010.11111111.11111111.11111111 ===> 111111101.00000000.00000000.00000000 > So you're saying AutoRP doesn't support non contiguous wildcard masks? That seems odd, why does it support the wildcard mask in the first place then ? It does explain what Joe was talking about though, thanks! Makes sense. > AutoRP therefore has no idea what you have in mind and (I assume) > finds the closest match, which in this case is 228.0.0.0/7. If you > wanted to advertise both 228.0.0.0/7 and 230.0.0.0/7, you'd either > need to lines in your ACL: > > access-list 1 permit 228.0.0.0 1.255.255.255.255 > access-list 1 permit 230.0.0.0 1.255.255.255.255 What about just 228.0.0.0 0.255.255.255 and 230.0.0.0 0.255.255.255, this seems redundant ?(no offense) > > or you'd need to advertise 228.0.0.0/6 using a single line ACL, like this: > > access-list 1 permit 228.0.0.0 3.255.255.255.255 > > Inverted, 3.255.255.255 is 252.0.0.0, which is a correct mask. This i still don't get sorry. Again is 00000010 not equal to .2 in wildcard speak ? or /6 in prefix notation? how does that become .3 ? is there simply no 2 value at all, just 3? but i just want 228 and 230, i only flipped the second last bit on, now i get 4 networks ? i want 2.255.255.255, not 3.255.255.255 if i do 11100100.00000000.00000000.00000000 228.0.0.0 00000010.11111111.11111111.11111111 2.255.255.255 ---------------------------------------------------------- 11100110.11111111.11111111.11111111 230.255.255.255 or 228 OR 230 as the second last bit is a don't care bit Am i mistaken? maybe i should just forget about these wildcard mask values, i understood it before (or so i think) :-) > > -- > Marko Milivojevic - CCIE #18427 > Senior Technical Instructor - IPexpert > > FREE CCIE training: http://bit.ly/vLecture > > Mailto: [email protected] > Telephone: +1.810.326.1444 > Web: http://www.ipexpert.com/ > > On Tue, Jun 21, 2011 at 18:12, Alef <[email protected]> wrote: >> Ok maybe i haven't been proper with my question >> This was in relation to lab 24.5, Vol1. Joe says in the video that normally >> a mask of 2.255.255.255 would work here, to encompass both networks, >> 228.0.0.0 and 230.0.0.0. >> and that makes sense >> 11100100.00000000.00000000.00000000 228.0.0.0 >> 00000010.00000000.00000000.00000000 2.255.255.255 >> >> so i would cover either 228 or 230 when i flip the second bit on. would this >> not be a /6 in a wildcard ? >> >> It's just that now i went over these wildcard values i got slightly confused >> as if i followed these it would not be a 2 for that bit value, but a 1 (as >> per 127 63 31 15 7 3 1 0) >> >> On Jun 21, 2011, at 9:21 PM, Marko Milivojevic wrote: >> >>> ------------------------------------------------ >>> IP Entered = ..................: 228.0.0.0 >>> CIDR = ........................: /6 >>> Netmask = .....................: 252.0.0.0 >>> Netmask (hex) = ...............: 0xfc000000 >>> Wildcard Bits = ...............: 3.255.255.255 >>> ------------------------------------------------ >>> Network Address = .............: 228.0.0.0 >>> Broadcast Address = ...........: 231.255.255.255 >>> Usable IP Addresses = .........: 67,108,862 >>> First Usable IP Address = .....: 228.0.0.1 >>> Last Usable IP Address = ......: 231.255.255.254 >>> >>> -- >>> Marko Milivojevic - CCIE #18427 >>> Senior Technical Instructor - IPexpert >>> >>> FREE CCIE training: http://bit.ly/vLecture >>> >>> Mailto: [email protected] >>> Telephone: +1.810.326.1444 >>> Web: http://www.ipexpert.com/ >>> >>> On Tue, Jun 21, 2011 at 14:56, Alef <[email protected]> wrote: >>>> Hi Guys, >>>> I've got a bit of a specific question regarding wildcard mask values: >>>> >>>> We write the multicast range 224-239 as a /4, using the 0 bits 127 63 31 >>>> 15 and write it out as .15 or 15.255.255.255 >>>> 226 and 227 we would as a /7, using the first 7 0 bits and write it out as >>>> 1.255.255.255 >>>> >>>> If that's the case, how do we write out a /6 ? It seems it's .2 (only >>>> talking first octet here,just think the other 3 octets are 255.255.255 for >>>> now) >>>> >>>> but if i follow the inverse values 127 63 31 15 7 3 1 0 should it not be >>>> written as .1 if i have 00000010 in the first octet? >>>> >>>> slightly confused.. >>>> _______________________________________________ >>>> 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 >>>> >> >> _______________________________________________ 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
