A /48 is not "just slightly larger" than the current internet. It is 16 bits larger - >65,000 times larger. I don't know about you, but for me, that is not "slightly". That's enough to allocate the entire internet address space separately to every country - no - every province of every country - and still have about 60,000 times more address space leftover than the total we have today. Your concept of scale is completely off.
I don't care to quibble over language, but I've shown that the factor of 65536 is not necessarily large enough to work in a realistic case. No "concept of scale" was required in this reasoning. I used mathematics and what I think are realistic estimates of various effects. If you have a problem with any of those estimates, say so. If you think I made a mistake in the math, show where. Otherwise, why are we still arguing?
These things you say about countries would be just fine, if you use the IANA reserved bits for unicast addressing (I'm not sure this will or should necessarily happen) and assume perfect allocation at the level of countries and provinces (laughable). But even then, you wouldn't have enough addresses in the end. One internet for each province sounds like a lot until you look at population growth and the growth of the number of people on the internet. Only a tiny percentage of the world's people are online now. That's why I factored the assumption that everyone will have internet access eventually into my math. Once you factor in that, and the continual growth in the number of internet-enabled devices a typical person uses, you'll see exactly what I showed: things get tight, especially if something unexpected comes up. Did you look at my numbers?
Furthermore, you are comparing network addressing in v6 to host addressing in v4. Since the smallest non-subnetted network in v4 is a class C, a /48 is actually 24 bits larger, making the available space a few million times larger than what we had even with what you argue is 'inefficient' allocation.
Nope. Currently, most residences have ONE address. If they have a subnet, they use NAT, not a class C. The IPv6 policy is to assign a /48 to each of these people, so each ONE IPv4 address is being replaced by a /48. The situation may be different for businesses, but I'm not sure, and I don't know how many businesses are really just small entities with one or two computers.
/48s for every multi-LAN user are fine.
So are /56s for (practically) every multi-LAN user. So why allocate /48s? No reason at all? Milk is cheap, so why not just pour a gallon down the drain? So far, every single person on this thread seems to be in full agreement that /48s offer no advantage whatsoever over /56s for all but the very largest businesses. So why allocate them, and why argue for them, if they have no benefit whatsoever? Isn't it irrational to argue stubbornly for a policy choice that you believe will have no effect at all? I don't understand this.
xezlec - after reading this entire thread, my opinion is that you are trying to "realize" or "quanitfy" the IPv6 address space in a way that you can understand. ive seen many people try to do this.
the address space of IPv4 is easy for the human brain to understand - 4.2x billion addresses. theyre all allocated now. But the IPv6 address space is simply not quantifiable by the human brain. no one truely has any idea just how big 340 undecillion addresses is.
Uh... it is very easy to quantify the IPv6 address space. It is, in total, 2^128 addresses. They are then broken down and allocated in a specific way. And yes, everyone with a basic understanding of arithmetic knows precisely how big 2^128 is. I know how to add numbers together and I know they don't add up to a small enough value to make this a good idea. I showed this. Why does everyone here keep insisting on talking about things using hand-waving and intuition? Why not just do the math and see for yourself?
Also, you misunderstand a few things if you think you can really put 2^128 computers on IPv6. Each subnet gets a /64. That means, no matter how big a number 2^128 is, there can only be 2^64 subnets. The only way you can have 2^128 computers is if every subnet has 2^64 computers on it. The maximum possible number of computers on a subnet is much, much smaller than this, and the majority of subnets will have only 1 computer. Similar reasoning applies for /48s. Once you've assigned a /48, all the numbers under it are allocated. They're gone. It doesn't matter how many computers that home user has, all of those addresses are gone now and can never be used again (barring renumbering, which won't likely happen). Since most of these users only have 1 computer, that means you only get 2^48 computers, plus a few.