> MAC address allocations are paid for by the Ethernet chipset/card vendor.
>
> They're not paid for by an ISP, or by any other Ethernet end-user, except
> as a pass-through, and therefore it's considered a fixed cost. There are
> no RIR fees, and there is no justification. You buy a gizmo with this
> RJ45 and you get a unique MAC. This is simple and straightforward. If
> you buy one device, you get one MAC. If you buy a hundred devices, you
> get one hundred MAC's. Not 101, not 99. This wouldn't seem to map well
> at all onto the IPv6 situation we're discussing.How many ISP customers pay RIR fees? Near enough to none, if not none.
Who said anything about ISP customers paying RIR fees? Although they
certainly do, indirectly.
I never have when I've been an ISP customer.
(Must be one of those legacy ISP's?)
Why are you pretending they do?
I don't recall bringing them into the discussion, BUT...
I think your taking an end-user perspective when discussing
ethernet but an RIR fee paying ISP position when discussing IPv6 subnet
allocations. That's not a valid argument, because you've changed your
viewpoint on the situation to suit your position.
Oddly enough, I'm one of those rare people who've worked with both ISP's
and OEM's that have been assigned MAC's. You can think as you wish, and
you're wrong.
Anyway, the point I was purely making was that if you can afford to
spend the bits, because you have them (as you do in Ethernet by design,
as you do in IPv6 by design, but as you *don't* in IPv4 by design), you
can spend them on operational convenience for both the RIR paying
entity *and* the end-user/customer. Unnecessary complexity is
*unnecessary*, and your customers won't like paying for it if they
discover you've chosen to create it either on purpose or through
naivety.
Okay, here, let me make it reaaaaaaallllllyyyyy simple.
If I am going out and buying an Ethernet card today, the mfr will pay $.NN
for my MAC address, a cost that is built into the retail cost of the card.
It will never be more or less than $.NN, because the number of MAC
addresses assigned to my card is 1. Always 1. Always $.NN.
If I am going out and buying IPv6 service today, the ISP will pay a
variable amount for my address space. The exact amount is a function of
their own delegation size (you can meander on over to ARIN yourself) and
the size they've delegated to you; and so, FOR PURPOSES OF ILLUSTRATION,
consider this.
If your ISP has been delegated a /48 (admittedly unlikely, but possible)
for $1,250/year, and they assign you a /56, their cost to provide that
space is ~$5. They can have 256 such customers.
If your ISP has been delegated a /48 (admittedly unlikely, but possible)
for $1,250/year, and they assign you a /48, their cost to provide that
space is ~$1,250. They can have 1 such customer.
If your ISP has been delegated a /41, for $1,250/year, and they assign
you a /48, their cost to provide that space is ~$10. They can have 128
such customers.
There is a significant variation in pricing as the parameters are changed.
You do not just magically have free bits in IPv6 by design; the ISP is
paying for those bits. There will be factors MUCH more real than whether
or not customers "like paying for it if they discover you've chosen to
create [complexity]," because quite frankly, residential end users do not
typically have a clue, and so even if you do tick off 1% who have a clue,
you're still fine.
Now, seriously, just who do you think is paying for the space? And if
$competitor down the road is charging rock bottom prices for Internet
access, how much money does the ISP really want to throw at extra address
space? (Do you want me to discuss naivety now?)
And just /how/ is this in any way similar to Ethernet MAC addresses,
again? Maybe I'm just too slow and can't see how "fixed cost" ==
"variable cost." I won't accept any further hand-waving as an answer,
so to continue, please provide solid examples, as I've done.
Perhaps more on-topic, how many IP addresses can dance on the head of
a /64?
... JG