IPv6 Summarization Example

Summarizing IPv6 prefixes is similar to IPv4 summarization, the big difference is that IPv6 uses 128 bit addresses compared to 32 bits for IPv4 and IPv6 uses hexadecimal addresses.

In this lesson, I’ll explain how to create IPv6 summaries and we’ll walk through some examples together.

Example 1

Let’s start with a simple example:

• 2001:DB8:1234:ABA2::/64
• 2001:DB8:1234:ABC3::/64

Let’s say we have to create a summary that includes the two prefixes above. Each hextet represents 16 bits. The first three hextets are the same (2001:DB8:1234) so we have 16 + 16 + 16 = 48 bits that are the same so far. To find the other bits that are the same we only have to focus on the last hextet:

• ABA2
• ABC3

We’ll have to convert these from hexadecimal to binary to see how many bits are the same:

ABA2	1010101110100010
ABC3	1010101111000011

I highlighted the bits in red that are the same, the first 9 bits. The remaining blue bits are different. To get our summary address, we have to zero out the blue bits:

AB80	1010101110000000

When we calculate this from binary back to hexadecimal we get AB80. The first three hextets are the same and in the 4th octet we have 9 bits that are the same. 48 + 9 = 57 bits. Our summary address will be:

2001:DB8:1234:AB80::/57

That’s how you can create a summary address for IPv6.

Example 2

This time we have the following 3 prefixes:

• 2001:DB8:0:1::/64
• 2001:DB8:0:2::/64
• 2001:DB8:0:3::/64

And our goal is to create the most optimal summary address. The first three hextets are the same so that’s 16 + 16 + 16 = 48 bits that these prefixes have in common. For the remaining bits, we’ll have to look at the 4th hextet in binary:

0001	0000000000000001
0002	0000000000000010
0003	0000000000000011

Keep in mind that each hextet represents 16 bits. The first 14 bits are the same, to get the summary address we have to zero out the blue bits:

0000	0000000000000000

When we calculate this from binary back to hexadecimal we get 0000. The first three hextets are the same and in the 4th octet we have 14 bits that are the same. 48 + 14 = 62 bits. Our summary address will be:

2001:DB8::/62

Example 3

Let’s try one more:

• 2001:DB8:0:7::/64
• 2001:DB8:0:12::/64

Let’s see what the most optimal summary address is that has these two prefixes. The first three hextets are the same so that’s 16 + 16 + 16 = 48 bits in common. Let’s look at the 4th hextet for the remaining bits:

0007	0000000000000111
0012	0000000000010010

Be careful that you don’t accidently convert number 12 from decimal to binary. We are working with hexadecimal values here! We have 11 bits that are the same, let’s zero out the remaining 5 bits:

0000	0000000000000000

We have 48 + 11 bits that are the same so our summary address will be:

2001:DB8::/59

I hope these examples have been useful. If you have any questions, feel free to leave a comment.