ipv6 help

jamesleecoleman

So I got the library set and I'm really excited to study for the CCNA (again). I'm reviewing the material for the stuff that I already know but the thing is that the IPv6 addresses is going to be another pain. I'm having trouble trying to figure out why I'm not understanding the subnetting.

I understand the/64 prefix but I'm not understanding anything that comes out to be like /80. The way that I'm thinking is that every quartet is 16. So for a /64 prefix, 16 times 4 quartets would get me to the prefix of /64. But for a prefix of 80, I'm getting confused.

So for example, in the ICND1 book that I have (I took the older ICND1 exam), the address/length is 34BA:B:B:0:5555:0:6060:707/80.
The way I look at is that each spot in the quartet is 4. So the first quartet four spots with each spot being equal to four. So for the first quartet, I have 16. But if I try to do the same method for the other quartets, I don't get /80.

In the book it has 34BA:B:B:0:5555::/80 for the prefix. I'm totally not understanding how that's working out. I add up the different placements so the way I'm looking at it is, 16+4+4+0+16 but I'm not getting /80 out of it.

EdTheLad

You're forgetting the leading zero rule, i.e.
34BA:B:B:0:5555::/80
is the same as
34BA:000B:000B:0000:5555::/80

jamesleecoleman

I thought they didn't count because they were zero's.

Thanks EdTheLad!!!

TechGuru80

EdTheLad wrote: »

You're forgetting the leading zero rule, i.e.
34BA:B:B:0:5555::/80
is the same as
34BA:000B:000B:0000:5555::/80

Blah beat me to it. Remember :: for groups if zeros too.

theodoxa

jamesleecoleman wrote: »

I thought they didn't count because they were zero's.

Thanks EdTheLad!!!

Each group (between the colons) is 16 bits regardless of how many characters are actually shown. Additionally, as stated below, double colons can be used in place of groupings that are all zeroes [but only once.]

jamesleecoleman

Alright, I'm trying to figure out these two.

34BA:B:B:0:5555:0:6060:707/36 and the answer that they have is: 34BA:B::/36.
I thought it would be 34BA:B:B because of 16+16+4

The second one is 210F:A:B:C:CCCC:B0B0:9999:9009/40
I'm thinking that it's 210F:A:B::/ but the answer in the book has 210F:A:/40

So confused and I'm still trying to find tutorials.

theodoxa

34BA:B:B:0:5555:0:6060:707/36 (34BA:000B::/36)
210F:A:B:C:CCCC:B0B0:9999:9009/40 (210F:000A::/40)

If there are less than 4 digits in a group, one or more leading zeroes have been removed. Each digit = /4, so /36 would mean the first 9 digits or first 2 groups + first digit of the 3rd group. Since the 3rd group is a single digit, that means 3 leading zeroes have been removed (the first 3 digits are zeroes)

jamesleecoleman

Thanks theodoxa!