[AusNOG] /20 Available

[Paul Gear]

On 01/22/2013 08:37 AM, august forsakov wrote:
>
>     IPv6 addresses are assigned to organizations in much larger blocks
>     as compared to IPv4 address assignments—the recommended allocation
>     is a/48block which contains 2^80 addresses, being 2^48 or
>     about2.8×10^14 times larger than the entire IPv4 address space of
>     2^32 addresses and about7.2×10^16 times larger than the/8blocks of
>     IPv4 addresses, which are the largest allocations of IPv4
>     addresses. The total pool, however, is sufficient for the
>     foreseeable future, because there are 2^128 or about3.4×10^38
>     (340trillion <http://en.wikipedia.org/wiki/10%5E12>trillion
>     trillion) unique IPv6 addresses.
>
>     Each RIR can divide each of its multiple/23blocks into
>     512/32blocks, typically one for each ISP; an ISP can divide
>     its/32block into65536/48blocks, typically one for each
>     customer;^[16]
>     <http://en.wikipedia.org/wiki/IPv6_address#cite_note-16> customers
>     can create65536/64networks from their assigned/48block, each
>     having 2^64 addresses. In contrast, the entire IPv4 address space
>     has only 2^32 (about4.3×10^9 ) addresses.
>
>     ...
>
>
> However many trillion, quadrillion or quintillion ip addresses are 
> possible with IPv6 there will always be a shortage at some point...
> Any technological renewal wave will show that to be true, so we _will_ 
> run out of IPv6...
> Maybe we'll want to assign each gene an ip address as will as its 
> associated body cell... hahaha

I think quoting the number of addresses in IPv6 ranges rather than 
number of subnets is misleading at best and dangerous at worst.

No one is going to use 2^64 MAC addresses in a /64. They're going to use 
somewhere between 2^1 and maybe 2^24 at the outside in the very largest 
L2 data centre deployments.  (Assuming they find some way of dealing 
with broad-/multicast issues and are using something like MAC address 
rewriting [1].)

We should be talking in terms of numbers of /64 subnets.  Using the 
figures above, /23 == 512 ISPs, each of which has 65536 customers (/48), 
each of which has 65536 subnets (/64).  That corresponds exactly with 
the number of /24 subnets available to organisations using the 
10.0.0.0/8 block internally under IPv4 today (with the obvious advantage 
of not needing NAT).

When i see those sort of numbers i'm glad Jeff Doyle is suggesting that 
each building gets a /48 instead of each customer - i can see it not 
being enough even within a short space of time, given current industry 
talk about in-car networks (CANs?) and the "Internet of things".

Regards,
Paul

[1] 
http://www.cs.duke.edu/courses/fall10/cps296.2/lectures/16RoutingTopology.pdf

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