On Fri, 28 Nov 2003 14:52:11 +0100, "Anthony G. Atkielski" <anthony@xxxxxxxxxxxxx> said: > A 128-bit field contains 2^128 addresses. If you divide that into two > 64-bit fields, you may get as few as 2^64*2 addresses; that's 18 > million trillion times smaller than the 128-bit field. Exactly. And the *reason* why IPv6 has 128 bit addresses is because the designers realized that such losses happen, and ruled out 64-bit addresses because of that effect. I suggest you figure out just how much bigger 2^65 is than the current 2^32, and factor in the current burn rate on the unused IPv4 addresses, and from there figure out if there's really a problem there at all....
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