IPv6 Subnetting questions

[John Blakley]

All,

I'm 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 I'm using doesn't give the same answer either. I'm going to put the question and answer in here, but I'd 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 that's 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't the case.

Please help

Thanks!

John

[John Blakley]

* EDIT *

Here's what I'm coming up with, and please tell me if there's an easier way:

3FFA:FF2B:4D:B000::/41

Breaking down B000:

8421 8421 8421 8421

1011 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 8421

1100 1000 0000 0000

That gives me :C800, but is flipping the bit the correct way of going about this?

Thanks!

[Peter Paluch]

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.

I've 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

[John Blakley]

Peter,

First of all, thanks for the response. I'm 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, let's say my global prefix is:

2001:233F:335D:355A::/64

How would I go about subnetting this out to have 8 subnets?

Thanks,

John

[Peter Paluch · Accepted solution]

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=/67
• The 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 nibble
• Thus we have:
  • 2001:233F:335D:355A:0000:/67, shorthand written as 2001:233F:335D:355A::/67
  • 2001:233F:335D:355A:2000:/67
  • 2001:233F:335D:355A:4000:/67
  • 2001:233F:335D:355A:6000:/67
  • 2001:233F:335D:355A:8000:/67
  • 2001:233F:335D:355A:A000:/67
  • 2001:233F:335D:355A:C000:/67
  • 2001:233F:335D:355A:E000:/67

Confused alright? Good Me is neither fluent in this... IPv6 arithmetics

Best regards,

Peter

[John Blakley]

This is making my head spin for some reason

What happens when you are trying to figure out the mask and the network portion is 'F' and the mask portion is 'F'? Does it and itself out, or do you use that as the 16 - 15? Also, I was trying to work through a couple of random networks and using a subnet calculator. What I figured to be 1, it was incrementing to be 2....is it not possible to have a subnet increment by 1?

I see what you were talking about on the OP. That answer for the question isn't correct; it's not even close to being correct. The closest I could make it get was by using a /54 (I think).

Thanks,

John

[lgijssel]

The maths are fine however, I read somewhere that:

The prefix length for an IPv6 subnet will always be /64; no more, no less. It allows you to place as many IPv6 devices as the underlying network medium allows.

This seems very wasteful but still makes some sense because it simplifies routing algorithms. (No VLSM anymore)

It also allows using the EUI64 host adresses.

http://www.tcpipguide.com/free/t_IPv6InterfaceIdentifiersandPhysicalAddressMapping-2.htm

So instead of subnetting, one should rather request a number of /64's.

Current practice is to assign /32 or /48 to ISP's who can give subsets to their customers.

When a customer is assigned a /60 for example, this would allow him to assign 2^4 networks of size /64.

regards,

Leo