IPv6 route summarization

[Steph1963]

Hi to all,

I am presently studying IPv6 and I try to understand how we can summarize IPv6 route. Is there any concept of subnetting in IPv6 like in IPV4 where we could summarize the following route with a /28 route that would include anything ending from 0 to 15 or is this a totally different concept. Is more like the first 48 bits are all the same? As for example, how could we summarize the following IPv6 route.

2001:0DB8:0:0:7::/64

2001:0DB8:0:0:8::/64

2001:0DB8:0:0:9::/64

Thanks for your help

Stéphane

[lgijssel]

As you have probably guessed, these three routes cannot be easily summarized.

However, the ISP who has been delegated this block can for summarize the whole address range as follows:2001:0DB8::/32

Customers generally receive a /48. Just like with ipv4, it is your own responsibility to allocate adresses in a clever way so that you can summarize wherever possible or useful. Instead of allocating prefixes ending with 7,8,9 you should use 4,5,6 instead. But I guess you already understood this(?)

regards,

Leo

[Steph1963]

Hi Leo,

I spend some times looking at why it would be easier to summarize 4,5 & 6. I am unfamiliar with IPv6 and I I am trying to make some sense out of this

2001:0DB8:0:0:4::/64

2001:0DB8:0:0:5::/64

2001:0DB8:0:0:5::/64

Can I replace the previous IPv6 address by the following:

2001:0DB8:0:0:(000)[0100]::/64

2001:0DB8:0:0:(000)[0101]::/64

2001:0DB8:0:0:(000)[0110]::/64

Number in parantheseis are hexadecimal, bits are between bracket

and from there determine that the first 79 (16+16+16+16+12+1) bits are similar and concluse that we can used

2001:0DB8:0:0:(000)0xxx or 02001:0DB8:0:0:0000/79 as summary

Thanks for your help

Stéphane

[lgijssel · Accepted solution]

Summarizing 4, 5, 6 is easier because the range starts with a multiple of 2.

In binary this would be 0100, 0101, 0110. This can be summarized on the first two bits as these are equal for the whole range.

You know that the standard host range is /64? It can of course be summarized but this will only have local significance.

According to my calculation there should be 78 bits in the network part of your (theoretical) summary: 5 x 16 minus 2.

This summary includes only one additional subnet: 2001:0DB8:0:0:7::/80. Summary is 2001:0DB8:0:0:4::/78.

When summarizing 7, 8, 9, your summary automatically includes subnets 0 till 15. Your mask would become /76.

In your example, the subnets actually are /80 in size so the notation using /64 is not correct.

This would actually represent different host addresses in the same network because the first 64 bits are identical.

In contradiction with IPv4, IPv6 generally takes the subnets from the network part. This implies the smallest mask is /64.

regards,

Leo

[Steph1963]

Hi Leo,

This is perfectly clear.

Thanks for your help

Stephane

[johng231]

I have a similar question about what is the best possible shortest match for the following prefixes below: 

2001:0:0:7::/64

2001:0:0:8::/64

2001:0:0:9::/64

2001:0:0:10::/64

I thought it would be 2001::/60 but I am confused on how it would be /59. Can you draw out the bits for me to line it up where /59 comes into play? 

This what I came up with for the binary bit boundary for the 4th hextet.

8 4 2 1  8 4 2 1  8 4 2 1  8 4 2 1
0 0 0 0  0 0 0 0  0 0 0 0  0 1 1 1 = 7
0 0 0 0  0 0 0 0  0 0 0 0  1 0 0 0 = 8 
0 0 0 0  0 0 0 0  0 0 0 0  1 0 0 1 = 9
0 0 0 0  0 0 0 0  0 0 0 0  1 0 1 0 = 10
1 1 1 1  1 1 1 1  1 1 1 1  0 0 0 0 = /60 ?

Thanks! 

[Harold Ritter]

2001:0:0:10/64 is not the same as 2001:0:0:A::/64 (hexadecimal 10)

therefore

0 0 0 0  0 0 0 0  0 0 0 0  1 0 1 0 is not equal to 10 (equals A)

but rather 0 0 0 0  0 0 0 0  0 0 0 1  0 0 0 0 = 10

Hence you indeed need a /59 to summarize the prefixes you mentionned.

Regards,