[sp332]

For perspective: IANA instructs RIRs to allocate no less than a /32 to each customer. That leaves 32 bits for those customers to resell smaller subnets, usually /64. 32 bits is as much as the entire IPv4 Internet, so they can sell as many /64 subnets as there are individual computers on the IPv4 Internet. So even if Linode only bought 1 of these /64 subnets, they would have 64 bits of address space to allocate to their customers. I think they can spare more than one :)

[jfr]

Note: in your case, customer means ISP.

From what I remember, the instructions were that ISPs receive no less than /32, which should provide end-users either /48 or /56. End users use the remaining 8-16 bits to partition their networks in /64 subnets. /64 is the minimum routeable prefix, and is the prefix used for automatic configuration.

Linode offering /128 for free and /64 at a cost is ridiculous. Users will have problems to forward since that would require breaking the /64 into smaller networks (to have at least one distinct subnet at each end of the tunnel), which is not possible for radvd.

[joeyh]

And it's not just academic, consider if you want to run lots of https and support browsers that don't support https on one address. Not to mention all the applications for many addresses that should develop if/when ipv6 is widespread.

As a further datapoint, I have two systems with native ipv6, and each have a /64.

[marshray]

consider if you want to run lots of https and support browsers that don't support https on one address

Then Linode expects you to pay more money for the privilege of doing that. They got paid per-https-site over IPv4, they want their cut on IPv6 too.

[bdonlan]

Linode has a /30: http://whois.arin.net/rest/net/NET6-2600-3C00-1

[sp332]

That's equivalent to 4 /32 subnets, or 2^34 = 17,179,869,184 /64 subnets. I think they could afford to give each customer a /64.

IPv6 subnets don't really get smaller than /64. The way the routing is configured, the first 64 bits are "network address" and the last 64 bits are the "interface address". One /64 subnet has 2^64 addresses, about 18.4 quintillion.

[bdonlan]

Sure, but remember each /64 takes up space in their router's routing table. Assigning every linode a /64 would take up quite a lot of memory, then.

[danellis]

What you're saying implies that they have a single routing entry for all their /128s currently, otherwise that would be the same number of routing entries. So, why can't they just have a single routing entry that aggregates all the /64s?