IPv6

sina2011

Hi Guys,

im trying to get my head around ipv6 subnetting but there is no sites out there that have any tutorials or anything like that.

does anyone recommend anything resources?

thanks.

Zartanasaurus

It's done exactly the same way as IPv4. If you know how to subnet IPv4 using binary to come up with the mask you can do it in IPv6.

WilyOne

That's true, the process is the same but there are differences between when subnetting is done in IPv6 vs IPv4.

In IPv6, unless you're a service provider you're really not going to subnet anything. Unless you work for a huge corporation, your ISP will issue you one or more /64 networks, which you do not want to subnet further for reasons that will become clear as you progress in your IPv6 studies. Where subnetting occurs is upstream, at the ISP and RIR levels, or only for the biggest of companies that can justify getting a /56, /48, or /32 block.

sina2011

Thanks for you reply guys.

i know how to subnet ipv4 but i just dont know how to calculate for e.g how many hosts per subnet e.g its just the hexdecimal positioning is confusing me.


would you guys be able to for e.g give me a sample question and show me how you calculate for e.g the hosts per subnet in ipv6.

I just dont know what format to write the subnets and hosts if that makes sense.


thanks.

Zartanasaurus

Do you know how to convert hex to binary and back? Convert it to binary and do it just like you would in IPv4.

WilyOne

I doubt that subnetting a /32, for example, is something they would ask at CCNA level. There are different types of subnets (unicast, multicast, link local) that you need to know; these are the kinds of questions I would expect.

The concept of "number of hosts" does not apply to IPv6. Each /64 subnet can contain 18,446,744,073,709,551,616 hosts, and /64 is smallest subnet you should ever have to deal with. So for all intents and purposes the number of hosts is irrelevant. /64 splits the 128-bit address exactly in half, so the first four sexdectets define the network prefix and the last four sexdectets define the host portion.

Example: 2001:db8:abcd:3f00::1 would be the gateway address for the 2001:db8:abcd:3f00::/64 network. (if you use the first available IP as your gateway)

2001:db8:abcd:3f00::1 =

[FONT=courier new][B][COLOR=#0000ff]
2001:0db8:abcd:3f00[/COLOR][/B]          |  [B][COLOR=#ff0000]0000:0000:0000:0001[/COLOR][/B] [COLOR=#0000ff]
^^^^ ^^^^ ^^^^ ^^^^[/COLOR] [COLOR=#0000ff]network[/COLOR]  |  [/FONT][COLOR=#ff0000][FONT=courier new]^^^^ ^^^^ ^^^^ ^^^^ host[/FONT][/COLOR]

fredrikjj

WilyOne wrote: »

That's true, the process is the same but there are differences between when subnetting is done in IPv6 vs IPv4.
In IPv6, unless you're a service provider you're really not going to subnet anything. Unless you work for a huge corporation, your ISP will issue you one or more /64 networks, which you do not want to subnet further for reasons that will become clear as you progress in your IPv6 studies. Where subnetting occurs is upstream, at the ISP and RIR levels, or only for the biggest of companies that can justify getting a /56, /48, or /32 block.

I have a feeling that an ISP prefers pointing an entire /48 to you, rather than subnetting those nets manually and then assigning you a certain number of /64s. You'll then be free to subnet the /48 into /64s according to whatever scheme you prefer.

sina2011

Thanks for you reply guys as zartanasaurus suggested i put my ipv4 subnetting skills into practice for ipv6

I have a sample address below please tell me if im on the right track in understand this.

3FA3:6666:2342:5666:: /64

IP: 00111111101000110110011001100110001000110100001001010110011001100000000000000000 Network Address in Binary
SM: 11111111111111111111111111111111111111111111111111111111111111110000000000000000 Network Prefix in Binary /64
AND:00111111101000110110011001100110001000110100001001010110011001100000000000000000 AND Process result which is the same as network address

So there is 64 bits left over from the network prefix so in this case if i do 2^64=18446744073709551616 hosts

if thats correct the next step in regards with my network ranges in what format do i write them out for e.g in ipv4 when you had for e.g

172.168.1.0
172.168.1.4
172.168.1.8

going all the way to 254 then jumping into 172.168.2.0
172.168.2.4
etc.

Now i know the subnets are probably wrong i just gave an example of the format on how its going to be written, now how would i do that in ipv6.

I hope it makes sense in what I'm trying to say.

Thanks Guys,

EdTheLad

If you still need to write out addresses and masks in binary, then i would say keep working on ipv4 until you can look at the address and instantly know the subnet. Once you are that comfortable with v4, v6 requires no additional work. As per what WillyOne is saying about not subnetting ipv6, he's wrong, i work with /127's on ptp links alot, my network has various levels of v6 subnetting, i think he's getting confused between supernetting and supnetting.

As per you example above if i wanted to supernet you ipv6 address i could do
3FA3::/16
Or if i wanted to subnet it i could have 2 subnets:

3FA3:6666:2342:5666:1:: /80 (dont forget about leading zero's)
3FA3:6666:2342:5666:2:: /80

and if i wanted to aggregate these i could have

3FA3:6666:2342:5666:2:: /78
which would include
3FA3:6666:2342:5666:0000::/80
3FA3:6666:2342:5666:0001::/80
3FA3:6666:2342:5666:0002::/80
3FA3:6666:2342:5666:0003::/80

WilyOne

Something like a p2p link is one of the few cases you might want to ever use anything smaller than /64, since with p2p you don't care about autoconfig. That being said, the concept of "conserving IP space" is old thinking in IPv6.

As for ISP assigning /48, it depends on the size of the company.

Routerronin

Zartanasaurus wrote: »

Do you know how to convert hex to binary and back? Convert it to binary and do it just like you would in IPv4.

Good point, plus one can learn to do this method in ten minutes despite the fact that most cisco press books dont describe how to do it.

sina2011

Thanks for your reply "EDTheLad"

I dont use binary to figure out my subnets etc i just showed an example to see if it was the same process as this is the first time im doing ipv6.

In relations with the bottom part of your reply where you have the following subnets:

3FA3:6666:2342:5666:0000::/80
3FA3:6666:2342:5666:0001::/80
3FA3:6666:2342:5666:0002::/80
3FA3:6666:2342:5666:0003::/80

what happens when for example you reach:

3FA3:6666:2342:5666:0009::/80

would your next subnet be 3FA3:6666:2342:5666:001A::/80
3FA3:6666:2342:5666:001B::/80
etc.. then reaching 3FA3:6666:2342:5666:001F::/80 then going on to
3FA3:6666:2342:5666:002A::/80

Am i writing it in the right formation?


Thanks.

sina2011

Thanks for your reply "EDTheLad"

I don't use binary to figure out my subnets etc i just showed an example to see if it was the same process as this is the first time im doing ipv6.

In relations with the bottom part of your reply where you have the following subnets:

3FA3:6666:2342:5666:0000::/80
3FA3:6666:2342:5666:0001::/80
3FA3:6666:2342:5666:0002::/80
3FA3:6666:2342:5666:0003::/80

what happens when for example you reach:

3FA3:6666:2342:5666:0000:0000:0000:0009 /80

would your next subnet be 3FA3:6666:2342:5666:0000:0000:0000:001A /80
3FA3:6666:2342:5666:0000:0000:0000:001B

etc.. then reaching 3FA3:6666:2342:5666:0000:0000:0000:001F /80 then going on to

3FA3:6666:2342:5666:0000:0000:0000:002A /80

Am i writing it in the right formation?

Really hope im on the right track.

Thanks Guys.

sina2011

ok guys I think im starting to get it after a few hours of intensive researching.

I just have 3 Questions:

Given this network address 3FA3:6666:2342:5666:2324:5444:A224:AD00 /120

1)in which octet do i start the Subnet address from

2)what formation is the subnet address written in for e.g in ipv4 we had

192.168.3.0
192.168.3.4
192.168.3.8
etc

3)where is my increment.


Regards,

goldenlight

subnetting in ipv6 is way easier then subnetting in ipv4. google danscourse. Then go to youtube and search for ipv6 cbtnuggets and the cisco god will explain everything to you.

Its all about counting in hex. remember each hex is 4 bits . Oh by the way most ipv6 start with hex 2-3


3FA3:6666:2342:5666:2324:5444:A224:AD00 /120
3FA3:6666:2342:5666:2324:5444:A224:AE00 /120
3FA3:6666:2342:5666:2324:5444:A224:AF00 /120
3FA3:6666:2342:5666:2324:5444:A224:B100 /120


Too time intensive to do it this way. Plus Have a greater chance of making a mistake

Real world you will be given a /48 subnet one of the 4 bits and use The IPv6 Interface ID and EUI-64 Format


so for example range of host

3FA3:6666:2342:5666:2324:5444:A224:AD01

---

>3FA3:6666:2342:5666:2324:5444:A224:ADFF

EdTheLad

sina2011 wrote: »

what happens when for example you reach:

3FA3:6666:2342:5666:0000:0000:0000:0009 /80

would your next subnet be 3FA3:6666:2342:5666:0000:0000:0000:001A /80
3FA3:6666:2342:5666:0000:0000:0000:001B

etc.. then reaching 3FA3:6666:2342:5666:0000:0000:0000:001F /80 then going on to

3FA3:6666:2342:5666:0000:0000:0000:002A /80

Am i writing it in the right formation?

Really hope im on the right track.

Thanks Guys.

my /80 subnets would be:

3FA3:6666:2342:5666:0000::/80
3FA3:6666:2342:5666:0001::/80
3FA3:6666:2342:5666:0002::/80
" "
3FA3:6666:2342:5666:0009::/80
3FA3:6666:2342:5666:000a::/80
" "
3FA3:6666:2342:5666:000f::/80
3FA3:6666:2342:5666:0010::/80
3FA3:6666:2342:5666:0011::/80
" "
3FA3:6666:2342:5666:001f::/80
3FA3:6666:2342:5666:0020::/80

etc etc

blatini

EdTheLad wrote: »

my /80 subnets would be:

3FA3:6666:2342:5666:0000::/80
3FA3:6666:2342:5666:0001::/80
3FA3:6666:2342:5666:0002::/80
" "
3FA3:6666:2342:5666:0009::/80
3FA3:6666:2342:5666:000a::/80
" "
3FA3:6666:2342:5666:000f::/80
3FA3:6666:2342:5666:0010::/80
3FA3:6666:2342:5666:0011::/80
" "
3FA3:6666:2342:5666:001f::/80
3FA3:6666:2342:5666:0020::/80

etc etc

It's a good thing the actual CCNA didn't really spend too much time subnetting in hexadecimal... I thought you would increase subnets such that:

0001
....
0009
000A
000B

sina2011

hey guys,


thanks for all the input


I can feel it that im nearly getting there.


Here's a sample question:


2001:db8:c001:ba00::/56 5 Subnets Required


In Order to get 5 Subnets I did 2^3=8 now the Network Prefix will change to /59


The Increment now will be:2


The reason is because the 1 is a value of 2


1000000000000000000000000000000000000000000000000000000000000000000000


please tell me I'm on the right track im dieing here

EdTheLad

2001:db8:c001:ba00::/56

Yes you are correct.

2001:db8:c001:ba00:0000::/59
2001:db8:c001:ba00:2000::/59
2001:db8:c001:ba00:4000::/59
" "
2001:db8:c001:ba00:E000::/59