Just want to make sure I understand /48 subnetting. If I have been assigned a routed /48 by HE that means I can do whatever I want with the fourth octet?
My assigned ip is 2001:470:e853::/48 so for example I could use one subnet as an inside address of:
2001:470:e853:a::/64
and another...
2001:470:e853:b::/64
and another
2001:470:e853:c::/64
Do I have this right? I have sixteen bits to do whatever I want with my /48?
Exactly.
You "own" those 16 bits. Well, actually, the 80 bits, but because IPv6 pretty much mandates 64 bit LANs you can't really do much w/ the lower 64.
65,536 subnets with ~18.4 billion billion hosts each. And it will be one of the more common, mundane allocations for IPv6. :)
So my above addressing was correct and I could assign one of those addresses to my inside interface?
Quote from: fonestar on April 17, 2010, 01:05:51 PM
So my above addressing was correct and I could assign one of those addresses to my inside interface?
Yes.
It works exactly like IPv4 CIDR addressing/subnetting except:
- Way more bits.
- Uses hexadecimal instead of decimal (easier than decimal IMHO).
I typically start addressing from 0 though, not a, b, c, so my subnets would be like 2001:db8:1234::/64, 2001:db8:1234:1::/64, 2001:db8:1234:2::/64 ... 2001:db8:1234:ffff::/64.