[arin-ppml] Draft Policy ARIN-2025-6: Fix formula in 6.5.2.1c

[William Herrin]

On Tue, Jul 1, 2025 at 11:34 AM ARIN <info at arin.net> wrote:
> Draft Policy ARIN-2025-6: Fix formula in 6.5.2.1c
>
> Problem Statement:
>
> Sections 6.5.2.1 explains the initial IPv6 ISP/LIR allocation in a way that is difficult to follow and the formula in section (c) does not match the remainder of the text.
>
> Policy Statement:
>
> In 6.5.2.1c, replace:
>
> "This calculation can be summarized as /N where N = P-(X+Y) and P is the organization’s Provider Allocation Unit X is a multiple of 4 greater than 4/3*serving sites and Y is a multiple of 4 greater than 4/3*end sites served by largest serving site."
>
> with:
>
> "This calculation can be summarized as /N where N = P-(X+Y) and P is the organization’s Provider Allocation Unit, X is a multiple of 4 greater than 4/3*log_2(serving sites) and Y is a multiple of 4 greater than 4/3*log_2(end sites served by largest serving site).


FYI, I'm the primary Advisory Council shepherd for this draft policy.
Here's some explanation:

Section 6.5.2.1c holds the criteria for the _maximum_ initial IPv6
allocation for ISPs. They qualify for the number of IPv6 addresses
described here and may request that much or a smaller block. The
section is frankly hard to read. Here's what that part of the NRPM
currently says:

"c. The maximum allowable allocation shall be the smallest
nibble-boundary aligned block that can provide an equally sized
nibble-boundary aligned block to each of the requesters serving sites
large enough to satisfy the needs of the requesters largest single
serving site using no more than 75% of the available addresses.
This calculation can be summarized as /N where N = P-(X+Y) and P is
the organization’s Provider Allocation Unit X is a multiple of 4
greater than 4/3*serving sites and Y is a multiple of 4 greater than
4/3*end sites served by largest serving site.

d. For purposes of the calculation in (c), an end site which can
justify more than a /48 under the end-user assignment criteria in
6.5.8 shall count as the appropriate number of /48s that would be
assigned under that policy.

e. For purposes of the calculation in (c), an LIR which has
subordinate LIRs shall make such reallocations according to the same
policies and criteria as ARIN. In such a case, the prefixes necessary
for such a reallocation should be treated as fully utilized in
determining the block sizing for the parent LIR. LIRs which do not
receive resources directly from ARIN will not be able to make such
reallocations to subordinate LIRs and subordinate LIRs which need more
than a /32 shall apply directly to ARIN."


Here's how ARIN staff explained the current implementation of NRPM 6.5.2.1c:

"ARIN staff implements 6.5.2.1.c based on the text. The summarized
formula is overly complex and inaccurate for your typical IPv6
requestor. The text alone is more easily understood by customers and
implemented by ARIN staff.

ARIN staff calculates Initial allocation sizes by verifying how many
serving sites the ISP has in the ARIN region, and how many customers
are served at the largest serving site. ARIN assumes each customer
will receive a /48 for simplicity and to promote IPv6 transition.

Once the sites and customers are provided by the requestor, ARIN staff
confirms what size is justified at the largest serving site based on
the 75% rule. That size is applied to all sites, then checked against
the 75% rule for the overall allocation justified by the ISP. The ISP
can opt to request a smaller size. They are not required to request
the largest justified size, though it is recommended to avoid future
renumbering.

For example:
An ISP has 7 sites and 30,000 customers at the largest site.

ARIN assumes each of the 30,000 customers receives a /48. There are
only 4,096 /48s in a /36, so a /36 is too small. The next
nibble-boundary aligned subnet is a /32 which has 65,536 /48s. 30,000
is less than 75% of 65,536, so the ISP’s largest serving site
justifies a /32.

Thus, each of the 7 sites receives a /32. The next nibble-boundary
after /32 is a /28. There are 16 /32s in a /28. 7 /32s of the total 16
/32s is less than 75%, so the organization justifies a total
allocation of a /28. 7 /32s for immediate allocation to each of their
7 sites and 9 additional /32s for future growth.

Example 2:

Building off the previous example, if the largest serving site had
60,000 customers, then a /32 would be too small. 60,000 is greater
than 75% of the available 65,536 /48s in a /32. The next
nibble-boundary aligned subnet is a /28, so the largest serving site
justifies a /28. Thus, each of the 7 sites receives a /28, so the
organization justifies a /24."


Regards,
Bill Herrin

--
William Herrin
bill at herrin.us
https://bill.herrin.us/