more IPv6 route summarization

[John_S209]

so as a 'youthful' netadmin looking to expand what I know, I'm reading CCNP routing content & going through some review questions.   admittedly I don't have deep IPv6 experience but as I said, trying to learn, expand my mind, etc.

that said, I have a review task:  summarize these 3 routes:

2001:DB8:0:7::/64 [110/20} via Eth0/0

2001:DB8:0:8::/64 [110/100] via Eth0/0

2001:DB8:0:8::/64 [110/20] via Eth0/0

so, following conventional instruction, we easily summarize the 1st 3 hextets 2001:DB8:0 .  and the rest becomes

0000 0007 hex - 0000 0111 binary

0000 0008        - 0000 1000

0000 0009        - 0000 1001 - correct (so far)?

so the 1st 8 bits of the 4th hextet are also summarized from immediately above - yes?  would make the summary network 2001:DB8:0::/ 56 (16+16+16+ 8=56)

did I do this right?  if not, can you see where I got off track?

thanks in advance.

JohnS

[luis_cordova · Accepted solution]

Hi @John_S209 ,

Just one detail to correct:

2001:DB8:0: 0000 0000 0000  0111::/64 = 2001:DB8:0:7::/64

2001:DB8:0: 0000 0000 0000  1000::/64 = 2001:DB8:0:8::/64

2001:DB8:0: 0000 0000 0000  1001::/64 = 2001:DB8:0:9::/64

Then summarization:

2001:DB8:0: 0000 0000 0000 | 0000::/60 = 2001:DB8:0:0::/60

Regards

[John_S209]

excellent!  thanks very much!