> Marc Vanbrabant wrote: > I would say the follow (though consider me newbie): > 193.109.122.62 = ::ffff:193.109.122.62/96;
No. First, you did not mention a subnet mask for the IPv4 address, but the only thing that would match /96 would be /0 for v4, no go. What you should have written is: 193.109.122.62/32 = ::ffff:193.109.122.62/128 > If you want a subnet like: > 193.109.122.62/29 This has never been a subnet. The network address for this IP and subnet mask is 193.109.122.56. > that would be > ::ffff:193.109.122.62/125; Wrong. That would be ::ffff:193.109.122.56/125 > /125 from /96 + /29 This part is correct. Michel. _ ____ __ __ ____ _ _ _ _ Michel Py | | | _ \ \ \ / / / ___| | \ / | | | | | Sr. Network Engineer | | | |_| | \ \ / / | |__ | \/ | | |__| | CNE, MCSE, CNP | | | __/ \ \/ / | _ \ | \ / | | __ | CCIE #6673 | | | | \ / | |_| | | |\/| | | | | | [EMAIL PROTECTED] |_| |_| \/ \___/ |_| |_| |_| |_| IPv6 Multihoming Solutions http://arneill-py.sacramento.ca.us/ipv6mh http://arneill-py.sacramento.ca.us/public/ipv6mh/ --------------------------------------------------------------------- The IPv6 Users Mailing List Unsubscribe by sending "unsubscribe users" to [EMAIL PROTECTED]