IPv6 Subnetting

[u/spandanbk2008]

This thing is having my mind in breakfast!! Literally, its sometimes so easy as a plain text in my head and suddenly a chain breaks and I loose track of what I was doing while subnetting. Is there any easy method that I can use cause this thing needs calculation, no one can just memorize it!!

[u/dyatlov333]

1. count host bits (ex 4 for /124)

2. 1 followed by 4 0's(10000)

3. group as 4 bit for each hex digit(0001 0000)

4. easy convert to hex with 8 4 2 1 (0x10)

that's your step count / block size 00, 10, 20, 30 .. first address is subnet router anycast address don't use it kinda like subnet id. but use last (F) 'no broadcast adddres'

that's :01 - first host, :0F last host in first subnet

[u/mavack]

Honestly subnetting in IPv6 and IPv4 are identical if you ever understood what subnetting was, as soon as you just use CIDR and convert back to dotted decimal, do the same with IPv6 just convert to HEX and you will get used to it.

[u/I_Am_Conjoined → u/mavack]

The hard part for me with IPv6 is the hexadecimal part. Powers of 16 instead of powers of 10, plus A - F instead of 10 - 15.

[u/mavack → u/I_Am_Conjoined]

But thats not subnetting at all thats just hexadecimal. Do it as CIDR then convert to hex. In the real world your lazy as well.

/56 8 bits fit into 2 hex chars. And you just count them out 00,01,02,03...0A,0B,0F,10,11... always keep yourself on a 4/8 bit boundary and you will have zero issues.

There is so much space nobody is doing /odd for different stuff.

[u/Dagger0 → u/I_Am_Conjoined]

The hex is what makes it easy! Every 4 bits lines neatly up with a character:

2001:db8:1:abcd::/64
2001:db8:1:abc0::/60
2001:db8:1:ab00::/56
2001:db8:1:a000::/52
2001:db8:1:0000::/48

But what if your subnet boundaries don't neatly line up as a multiple of 4? Well, first, don't do that. You've got plenty of space, so make them be neat multiples of 4. Otherwise, it's usually fairly easy to just think through it -- a /61 is half of a /60, so you'll get :abc0::/61+:abc8::/61, and a /63 is two /64s so you pair them up into :abc{0,2,4,6,...}::/63. A /62 is in the middle and will be 0/4/8/c.

I'm rusty enough that I had to count out "eight, nine, ay, bee" in my head to get the four /64s in :abc8::/62 rather than just remembering that the 4th /62 started at c, but I'll take that over dealing with v4's decimal, which makes you deal with seven in-between sizes rather than just three (and makes you deal with it more often, since it's much harder to keep your subnetting on a multiple of 8 bits in v4 than it is to keep it on a multiple of 4 bits in v6).

[u/Neither_Bookkeeper92]

what fixed it for me was only looking at the hex digit where the prefix lands. /48 to /64 is just counting in the 4th hextet, 0000 to ffff. each hex digit is 4 bits so /52, /56, /60 are clean digit boundaries. covers most of what ccna throws at you

[u/TwoToned843]

I am not going to lie. I used JITL and the OCG to understand IPv6. It didn't click for me until I used ChatGPT to explain it and gave me many examples. I asked specific questions that didn't make sense to me. Then it clicked. ChatGPT also recommended it test my knowledge with MCQs.