On Jun 8, 2011, at 6:09 AM, Cameron Byrne wrote:
On Wed, Jun 8, 2011 at 6:04 AM, Owen DeLong <owen@delong.com> wrote:
On Jun 8, 2011, at 5:47 AM, Cameron Byrne wrote:
On Wed, Jun 8, 2011 at 12:09 AM, Owen DeLong <owen@delong.com> wrote:
On Jun 7, 2011, at 9:59 PM, Martin Millnert wrote:
Owen,
On Tue, Jun 7, 2011 at 11:47 PM, Owen DeLong <owen@delong.com> wrote:
LSN is required when access providers come across the following two combined constraints:
1. No more IPv4 addresses to give to customers. 2. No ability to deploy those customers on IPv6.
2 has little bearing on need of LSN to access v4. Insufficient amount of IPv4 addresses => LSN required.
Regards, Martin
No, if you have the option of deploying the customers on IPv6, you don't need LSN.
The problem is that until the vast majority of content is dual-stack, you can't deploy customers on IPv6 without IPv4.
cough cough NAT64/DNS64 ...
Doesn't solve the problem unless your users are all on cell-phone browsers that don't do a lot of the things most users do with real internet connections.
Most of my users are on cell phone browsers :)
Furthermore, i can choose which ones get ipv4-only NAT44 and which get ipv6-only + NAT64
Now, only if there was major cell phone OEM support ....
Also, i would like to extend the idea that as IPv6 becomes dominant in the next few years (pending access networks), the need for IPv4 access will wane and LSN for the IPv4 will become more acceptable as IPv4 is just the long tail.
Agreed... However, where I differ is that I believe it is content and services which will drive the ability for IPv4 to be considered long tail. If all of the content and services were IPv6-capable today, the need for LSN would be very near zero (limited to the consumer devices that need to be upgraded/replaced to understand IPv6.) However, as it stands currently, a consumer would not consider an IPv6 connection with NAT64 or other LSN to be equivalent to what they expect today (unless they're on a cell-phone where they already expect the internet experience to be completely degraded). Owen