ipv6 subnetting (take 2)

mattau

After a lot of heavy study into hex conversions I am revisiting my try at ipv6 subnetting which seems like a nightmare

something did click to me today after much reading on various ipv6 books and cisco forums. alot of it was just finally getting it in my head that if you convert the hex into decimal you can see it alot clearer and now i sorta feel like i have some idea. (not all but some)

everyone says just treat it like ipv4 like this

If I have 128 bits, and use 4 (right-most) for hosts, what's my subnet bits? (128-4 = 124)

which would mean a /124 ? and my hosts would equal 16 hosts per subnet?

my question is determining the amount of subnets

you know how the 16 bits after the first 48 bits are allocated to be the subnet section ? is this the bit where we specify how many subnets we want and then from there manipulate the prefix length to determine how many hosts to put in the subnets ?

here is a picture to help explain better on what im trying to ask. its confusing but i wana get this so badly.

Priston

If you if you are given a /64 address 2001:1111:1111:000A::0 you have 64 bits to play with, if 4 bits are hosts, you have 60 network bits.
1152921504606846976 network addresses


If you have a /8 address
FD00::0
and want 987954853651556842951 subnets
70 bits gives you 1180591620717411303424 subnets
so you use a /78 mask
which leaves you 50 bits for host 1125899906842624

FD00::0 - FD00::3:FFFF:FFFF:FFFF
FD00::4:0:0:0 - FD00::7:FFFF:FFFF:FFFF
FD00::8:0:0:0 - FD00::B:FFFF:FFFF:FFFF