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[fonestar]

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?

[jimb]

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. 

[fonestar]

So my above addressing was correct and I could assign one of those addresses to my inside interface?

[jimb]

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.