I posted yesterday about changes to RFC3177 for IPv6 allocation. I wonder how many have done the maths on this and realised just how many addresses this works out to be.
Lets hold hands and work through the maths:
- Lets say you are allocated a /48 by your favourite RIR.
- Lets also say you allocate this in /64 chunks (as is custom and recommendation for IPv6 at this time).
- that means you have 2^16 networks.
- that 65536 subnets.
Just for reference that’s as many subnets as there are individual IP Addresses are in B-Class IPv4 address.
If you are allocated a /48 from the RIR then this is quite a lot of subnets – even when allocating a /64 per point to point connection. I don’t understand why people are saying that this isn’t enough address space or complaining that using a /64 is wasteful.
I can accept that a carrier or a very corporate might need a bigger space, but if you ask they will allocate it to you (providing you can make a case for it).
Am I missing the point here ?
Other Posts in A Series On The Same Topic
- Why Allocating a /64 is Not Wasteful and Necessary (23rd January 2011)
- IPv6 - /48 allocation in /64 chunks - that's a lot of addresses (21st January 2011)
- IPocalypse: What's next for the 'End of the Internet' ? (20th January 2011)
- The IPocalyse is Nigh - Forced Allocation of IPv4 to RIRs next week ? (19th January 2011)
- IETF IPv6 address allocation policy being updated. (12th January 2011)
- Scheduling the IPocalypse (25th November 2010)