[arin-ppml] IPv6 Allocation Planning

[michael.dillon at bt.com]

> 3.	Multiply x*y to get the number of /48s (ISP) needed. Convert this
> to a number (n)
> 	of bits (2^n=x*y).

This is unneccesarily complex.
First of all, there is no need to do this at all. An ISP can just appply
for a /32, and then reapply when they run out. For some ISPs this ma
be the best way to go because it does not require detailed forward planning
even if it ends up being inconvenient for them when they have to reapply.
But any really large ISP would likely want to use whatever they already
have in their database, i.e. count customers or customer sites, or
customer access circuits, and then use that as justification for an
appropriately large block bigger than /32.

It should not really matter to us how accurate such estimates are as long
as they are not outrageously pie-in-the-sky.

--Michael Dillon