IPv6 Summarization Question

[foosterh]

Hi,

reading the book on ROUTE exam 642-902 I hit upon the following. Page 774 and Page 780

There are 3 Loopbacks on a router with following addresses:

3:1::/64

3:2::/64

3:3::/64

The author (page 780) decides to summarise these into 3::/16

I would have summarised as follows:

3:1::/64  = 0003:0001::/64   0003:00000000 0000 0001::/64

3:2::/64  = 0003:0002::/64   0003:00000000 0000 0010::/64

3:3::/64  = 0003:0003::/64   0003:00000000 0000 0011::/64

These subnets are thus bit-equal up to and including bit nr 16+8 +4+2 = 30

I would have summarised as 3::/30

Am I missing something?

Thanks

Fred

[Michael Vincent]

Yes, you are correct and aren't missing anything.  The author was a bit "sloppy" summarizing subnets that aren't actually in the aggregate.

The author's aggregate summarizes:

3::

...

3:ffff:ffff:ffff:ffff:ffff:ffff:ffff

A bit more than just 3:1::, 3:2:: and 3:3::.

Likewise, your example is a bit overreaching - technically.

Your aggregate summarizes:

3:0::

3:1::

3:2::

3:3::

Perhaps:

3:1::/64

3:2::/31

is more "technically" correct.

I've been using /128's for loopback addresses all carved from a single /64.  Just another way to do it.

cheers.