I have no personal knowledge of this situation, so this is wild speculation. http://news.cnet.com/verizon-completes-alltel-purchase/ Verizon Wireless is going to be soon selling operations in 105 markets. It may well be that the IP addresses for those markets will be transfered to the new company as well, and you'll see some of these blocks leave their name soon. It could also be that AllTel had a much lower percentage of subscribers using data, and Verizon is fixing to change that soon. With the merger complete Verizon Wireless will have 83.7 million subscribers (per the article). I see 27,371,520 IP's in all their advertised blocks now, add in the 8,388,608 they just got, for a total of 35,760,128. If we assume across all blocks they can get 80% USAGE efficiency (which would surprise me) that's enough IP's to feed data to 28,608,102 subs. That would mean they can serve about 34% of their customers with data. Lastly, you've assumed that only a "smart phone" (not that the term is well defined) needs an IP address. I believe this is wrong. There are plenty of simpler phones (e.g. not a PDA, touch screen, read your e-mail thing) that can use cellular data to WEP browse, or to fetch things like ring tones. They use an IP on the network. By the same math they have 55.1 million (83.7 million subs - 28.6 they can serve now) they can't serve data to yet, and using the same 80% effiency that will take another 68.9 million addresses to do that. A /6 has 67.1 million addresses, so I suspect you'll see over time another /6, or two /7's, or four /8's, or eight /9's....... Which leaves us with two take aways: 1) The comment is weird. 2) If one company is likely to need four more /8's, and there are now 32 in the free pool man is IPv4 in trouble. At this point it would only take eight companies the size of verizon wireless to exhaust the free pool WORLDWIDE. No matter how much effort we put into reclaiming IPv4 space there's just no way to keep up with new demand. Is your network IPv6 enabled yet? -- Leo Bicknell - bicknell@ufp.org - CCIE 3440 PGP keys at http://www.ufp.org/~bicknell/