Hacker News Reader

Comment by acqq | Hacker News Reader

Anybody knows how the 128 bits in the addresses are really going to be allocated? If everybody gets "lower 64-bit for a subnet," and every device simply considers that it should allow that the subnet is whatever is a slave of it, we end up using only the upper part, and then if you again give big chunks of upper 64 bits to different entities like it happened with IPv4, there's still real chance that even if the space is big it remains inefficient, especially if upper 64 bits are separated to encode some special assumptions about routing or whatever. Is there any text about this whole subject, which just doesn't brag that 128 bits is "a lot?"

View on HN

This Wikipedia section seems relevant: http://en.wikipedia.org/wiki/IPv6_address#General_allocation. A choice quote:

IPv6 addresses are assigned to organizations in much larger blocks as compared to IPv4 address assignments: the recommended allocation is a /48 block which contains a number of addresses that is 248 or 7.9×1027 times larger than the entire IPv4 address space.

How much the /48 is larger is again what I don't care as I can calculate that easily but it doesn't mean anything in how it's to be used. The relevant piece is:

"5 RIRs, /23 to /12, smallest gets 512 /32 blocks, one per ISP, ISP divides into 64K /48 blocks, typically one for each customer"

It's obvious that the "customer" here is not "every man on Earth" or "every phone" but probably should be something like the smaller ISP or the big company. The lower 64 bits are it seems effectively "unique device id." So the idea is to use upper 64 bits to fix the location of the device and lower 64 to address the device. And yes, I think I've understood that the "device id" (lower 64-bits) doesn't have to be fixed and encoded in hardware.

You can get a free tunneled /48 right now from Hurricane Electric.

http://tunnelbroker.net/

The smallest RIP is probably Africa, and if it has /23 and if magamiako is right:

http://news.ycombinator.com/item?id=1743268

and every /48 goes to every home user, that means that you can't ever have more than 33 million home users in Africa! (512 * 64 * 1024) The 33 million limit is why for me seemed "obvious" that /48 doesn't go to "every man on Earth." Does anybody know something more exact than what's written in Wikipedia article? Some real example how the space is going to be distributed? That was why I started this thread, and sadly still most of the answers are hand-waving "it's big, don't care," even if that doesn't carry any real information.

Those are some pretty big assumptions. But, even if only 64 bits were used, that's still enough space for every human on earth to have 2.75 billion IPv6 addresses.

Only if there are no distribution rules in upper 64 bits, which is not going to happen. That's why I'd like to know the planned allocation process. Do you know anything about it? I posted my question to get exactly that info.

As the wikipedia article states, and as the IANA assignments are coming through--you, as your personal self, gets a /48 block. Or rather, your "premises" gets a /48 block. If you're a small business owner with 10 computers, you get a /48 block. If you're a home user, you get a /48 block. If you're a large enterprise business, you get a /48 block.

It's a rather significant waste of addressing and I really wish they had moved it over a few blocks, but that's the current plan. That said, you have to figure.

Right now, for internet alone, we're using only 1/8th the space of the IPv6 address space. The plans are 2000::/3. Which essentially means 2000:: through 3FFF::. that's a rather ridiculous amount of hosts. It comes out to 2^125 hosts. 4.25352959 × 10^37 hosts on the internet. It's unfathomably large. So it doesn't really matter if huge chunks are wasted. You won't ever really use all of it anyway.

I agree that the IPv6 space is ridiculously large, but that's kind of the point. They realized that the internet infrastructure will only continue to grow--and they want to, at all costs, avoid any problem with hitting the limit of addressing at any point in the future. The larger it grows, the more of a problem it will be to change it later on.

https://www.arin.net/policy/nrpm.html#six

Right now they are only allocating out of one /8, so if they are being too broad in their allocation policies now, then they can start cleaning up the mess when the need the next /8.

Even with this generous distribution it's still about /32 per ISP. Imagine IPv4 Internet where each existing ISP has just a single v4 address.

Maybe when we'll have a colony at Alpha Centauri that won't be enough anymore, but this should be perfectly sufficient for quite a while.

Even then it will be sufficient for the Centaurians to have their own Internet and reuse the whole address space. Until someone invents ansible, any communication between us and them will be too slow for any real-time purposes.

So do you actually know that every ISP exactly gets /32? Can you point to some material about that? I'm still asking for some exact information.

For IANA -> RIR allocations, http://lacnic.net/documentos/lacnicvii/POLITICA-IPV6-IANA-RI... and for an example of RIR allocation to ISP/end user: https://www.arin.net/policy/nrpm.html

Each RIR has it's own policy though, so check with the one you're concerned with (ARIN vs APNIC, etc).

Reply on news.ycombinator.com