Thanks for ur reply Patrik let me go through it, I'll get back to u. -RS
On Jan 22, 2012, at 20:43, <[email protected]> wrote: > Rostam, > > > When trying to do the same, see below the binary numbers for 192, 205, 207 > and 208. > > 192 11000000 > 205 11001101 > 207 11001111 > 208 11010000 > > First to decide what type of statement we want to have: > Inclusive, all networks in 1 statement but might be including networks not in > the list > Exclusive, all networks in possibly more statements, but never including > networks not in the list. > > In the Inclusive case > 205.49.166.0 11001101.00110001.10100110.00000000 > 207.49.166.0 11001111.00110001.10100110.00000000 > 208.49.166.0 11010000.00110001.10100110.00000000 > 205.49.167.0 11001101.00110001.10100111.00000000 > 207.49.167.0 11001111.00110001.10100111.00000000 > 192.49.166.0 11000000.00110001.10100110.00000000 > Result of XAND 11000000.00110001.10100110.00000000 192.49.166.0 > Result of XOR 00011111.00000000.00000001.00000000 31.0.1.0 or > 31.0.1.255 (depending on the application) > > In the Exclusive case > 205.49.166.0 11001101.00110001.10100110.00000000 > 207.49.166.0 11001111.00110001.10100110.00000000 > 208.49.166.0 11010000.00110001.10100110.00000000 > 205.49.167.0 11001101.00110001.10100111.00000000 > 207.49.167.0 11001111.00110001.10100111.00000000 > 192.49.166.0 11000000.00110001.10100110.00000000 > > First octet has 2 sets that only differ 1 bit (making sure that only these > will be included in the statement) > 205 11001101 > 207 11001111 > 205 11001101 > 207 11001111 > Result of XAND 11001101 205 > Result of XOR 00000010 2 > > 192 11000000 > 208 11010000 > Result of XAND 11000000 192 > Result of XOR 00010000 16 > > Second octet has no differences anywhere. > > Third octet could make things more interesting, but it does not. > 205.49.166.0 10100110 > 207.49.166.0 10100110 > 205.49.167.0 10100111 > 207.49.167.0 10100111 > Result of XAND 10100110 166 > Result of XOR 00000001 1 > > > 192.49.166.0 10100110 > 208.49.166.0 10100110 > Result of XAND 10100110 166 > Result of XOR 00000000 0 > > Now what could have made it more interesting? > In the exclusive case, no other networks than the one mentioned are supposed > to be in the statement. If the 192 OR 208 had .167 as its third octet, 2 > statements would be needed to make the statement exclusive. Same would count > for the 205 and 207 ranges if one of the 167 or 166 statements was missing. > > Now the last octet are all 0's, yet in the example they gave was a .255 at > the end of the mask. It will depend on the application whether the mask > should show .0 or .255. > If the statement is used to describe the networks, then a 0 at the last octet > will make sure the statement only covers the major network number and not any > subnets, .255 will also include all possible subnets. > If the statement is used in an ACL to allow or deny hosts in that network to > send traffic, .255 will be used to cover all hosts in that network. > > 192.49.166.0 8.0.0.255 or 8.0.0.0 > 205.49.166.0 2.0.1.255 or 2.0.1.0 > > Just let me know if you have any questions. > > Kind regards, > > Patrick Keja > > -----Original Message----- > From: [email protected] > [mailto:[email protected]] On Behalf Of Rostam Sohrab > Sent: 22 January 2012 10:04 > To: [email protected] > Subject: [OSL | CCIE_RS] Network & Wildcard masks for multiple networks > > Sending this mail again as an attached word document to protect the format. > > -RS > > I was working on to get a network and wildcard for the below mentioned > networks and I got this answer --> 192.49.166.0 4.255.1.255 which is untrue! > The correct answer is 205.49.166.0 2.0.1.255 > > > > http://blog.ipexpert.com/2009/06/10/wild-card-masks/ > > > 205.49.166.0/24 > 207.49.166.0/24 > 208.49.166.0/24 > 205.49.167.0/24 > 207.49.167.0/24 > 192.49.166.0/24 > > I'm writing the process I am following to do the ANDing and XORing operations > of these networks, I'll take the first octet of all the networks. > > First the AND operations > > 208 -> 1 1 0 0 1 1 0 1 208 -> 1 1 0 0 1 1 0 1 207 > -> 1 1 0 0 0 1 1 1 > 207 -> 1 1 0 0 0 1 1 1205 -> 1 1 0 0 0 1 0 1 192 -> 1 1 0 0 0 0 0 0 > ---------------------------------------------------------------- > -------------------------------------- > AND->1 1 0 0 0 1 0 1 -> 208AND->1 1 0 0 0 1 0 1 ->208AND-> 1 1 0 0 0 0 0 0 -> > 192 > > Now I am ANDing the results of first two and the result of this I'll AND with > the result of third AND operation done above which is 192. > > 208 -> 1 1 0 0 0 1 0 1196-> 1 1 0 0 0 1 0 0 > 208-> 1 1 0 0 0 1 0 0192-> 1 1 0 0 0 0 0 0 > ----------------------------------------------------------------- > AND-> 1 1 0 0 0 1 0 0 -> 196AND->1 1 0 0 0 0 0 0 -> 192-> And hence the first > octet of the network would be 192! > > Now the XOR operations. > > 208 -> 1 1 0 0 1 1 0 1 208 -> 1 1 0 0 1 1 0 1 207 > -> 1 1 0 0 0 1 1 1 > 207 -> 1 1 0 0 0 1 1 1205 -> 1 1 0 0 0 1 0 1 192 -> 1 1 0 0 0 0 0 0 > ---------------------------------------------------------------- > -------------------------------------- > XOR-> 0 0 0 0 1 0 1 0 ->10 AND-> 0 0 0 0 1 0 0 1 ->9AND-> 0 0 0 0 0 1 1 1 -> > 7 > > Now I am XORing the results of first two and the result of this I'll XOR with > the result of third XOR operation done above which is 7. > > 10 -> 0 0 0 0 1 0 1 0 3-> 0 0 0 0 0 0 1 1 > 9-> 0 0 0 0 1 0 0 1 7-> 0 0 0 0 0 1 1 1 > ----------------------------------------------------------------- > XOR->0 0 0 0 0 0 1 1 -> 3 AND->0 0 0 0 0 1 0 0 -> 4 -> And hence the > first octet of the wildcard mask would be 4! > > So the first Octet of Network statement I get is 192 and the first octet of > wildcard mask I get is 4! > >>>> The other question I have is about the wild card mask, in the answer it is >>>> 2.0.1.255, here the second octet which has all zeros is written as a Zero >>>> "0" rightly, but the last octet which is also all zeros is written as 255? >>>> I found this a rather strange phenomenon!! > > While I am sure I am doing the AND and XOR operations correctly, I might be > inadvertently deviating somewhere during the calculation! > > Requesting comments from all the experts in this forum. > > -RS _______________________________________________ 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? 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