IPv6 Subnetting questions

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IPv6 Subnetting questions

Post by Guest » Sat Sep 18, 2010 7:43 am

All, Im having a difficult time trying to find any information on subnetting a v6 address into smaller subnets. A coworker is studying for an MS exam and they had a question on there that frankly makes no sense. The subnet calculators that Im using doesn give the same answer either. Im going to put the question and answer in here, but Id like to know how to subnet this out manually if possible: Summarized question: You have 3 networks and need to add a fourth. The global address prefix 3ffa:ff2b:4d:b000::/41 is assigned to you. Assign 4 departments from this assignment. Correct Answer:3FFA:FF2B:4D:C800::/43 There is no explanation as to how the answer is derived. There is another answer thats not corret, but looks like 3FFA:FF2B:4D:C000:/43. I was under the impression that when you subnetted after your assignment, the numbers were arbitrary as long as they fit in your subnet, but apparently this isn the case. Please help  Thanks!John

Guest

Re:IPv6 Subnetting questions

Post by Guest » Sat Sep 18, 2010 9:13 am

* EDIT * Heres what Im coming up with, and please tell me if theres an easier way: 3FFA:FF2B:4D:B000::/41 Breaking down B000: 8421 8421 8421 84211011 0000 0000 0000 Their answer is 3FFA:FF2B:4D:C800::/43 43 - 41 = 2^2 (4 subnets which is what it asks for) Coming up with their answer: 8421 8421 8421 84211100 1000 0000 0000 That gives me :C800, but is flipping the bit the correct way of going about this? Thanks!

Guest

Re:IPv6 Subnetting questions

Post by Guest » Sat Sep 18, 2010 10:14 am

John, Actually, the prefix 3FFA:FF2B:004D:B000::/41 is not a network prefix at all, rather it is already a host IPv6 address: the /41 bits are actually inside the third field 4D. The real prefix is 3FFA:FF2B::/41, because the 3FFA:FF2B:00 part is fixed (40 bits), and moreover, from the 4 nibble (=0100 binary), the uppermost bit is fixed also. The remaining bits to the right are irrelevant for the /41 prefix. The task seems to be assigned badly from the start. Ive once tried to develop a method to compute IPv6 prefix numbers from memory - you may want to have a look although I am by no means saying that that method is usable or effective (or even comprehensible...) https://supportforums.cisco.com/message/3169385#3169385 Best regards,Peter

Guest

Re:IPv6 Subnetting questions

Post by Guest » Sat Sep 18, 2010 11:05 am

Peter, First of all, thanks for the response. Im still very confused even after following the link that you had sent. The part that confuses me is how many networks we can get out of subnetting: For instance, say I want 8 subnets out of a /64. For ease, lets say my global prefix is: 2001:233F:335D:355A::/64 How would I go about subnetting this out to have 8 subnets? Thanks,John

Guest

Re:IPv6 Subnetting questions

Post by Guest » Sat Sep 18, 2010 11:47 am

John, Okay, if you have the prefix 2001:233F:335D:355A::/64 and need to split it into 8 subnets, then: You need additional 3 bits for subnetting as 2^3=8, thereby making the prefix length /64+3=/67The additional 3 bits from the field immediately after the "355A" word is going to hold the subnet ID. This is going to be the interesting nibble.The interesting "mask" nibble would be 1110b = 14 (the additional 3 bits in the mask, padded to 4 bits to form a nibble)The networks will go in increments of 16 (constant) - 14 (the mask nibble) = 2 in the interesting nibbleThus we have:2001:233F:335D:355A:0000:/67, shorthand written as 2001:233F:335D:355A::/672001:233F:335D:355A:2000:/672001:233F:335D:355A:4000:/672001:233F:335D:355A:6000:/672001:233F:335D:355A:8000:/672001:233F:335D:355A:A000:/672001:233F:335D:355A:C000:/672001:233F:335D:355A:E000:/67 Confused alright? Good Me is neither fluent in this... IPv6 arithmetics  Best regards,Peter

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