Questions on how to partition subnetspace
Shadow Hawkins on Friday, 18 January 2008 16:35:17
Hi,
I've a tunnel from 2001:7b8:2ff:196::1 to 2001:7b8:2ff:196::2. This are the one IPv6 addresses I can use in the subnet without a subnet, can't I?
So I have ordered a subnet now: 2001:7b8:3cd::/48
Is it a good idea to use the tunnels IP just for routing my subnet ans nothing else (i.e. don't provide services on 2001:7b8:2ff:196::2)?
And now another question for my subnet 2001:7b8:3cd::/48: This means that the *network* has 48 bits and I can use 128-48=80 bits for my own network, can't I?
Is there any convention on HOW to divide up the 2^80 IP ranges? I've heard that 64 Bits are always the host part...does this mean that I have "only" 80-64=16 bits for my own subnets?
Or can I use (as with CIDR in IPv4) e.g. 72 (from my 80) bits for partitioning my network and use just the last 8 bits for the host part?
Is there any recommendation on this?
Thank you very much!
Questions on how to partition subnetspace
Shadow Hawkins on Sunday, 27 January 2008 03:36:25
You should use 16 bits for your own subnets, and 64 bits for your hosts. That's what is expected by most IPv6 software I've found. If 64K subnets isn't sufficient for you :-) then you can always get another allocation.
Questions on how to partition subnetspace
Shadow Hawkins on Monday, 28 January 2008 00:13:23
I would imagine that since you have a subnet routed to you, you can utilize the address space however you wish. You were allocated a /48 leaving you 80 bits to do whatever you want. You could make a network like 2001:7b8:3cd:0:0:0:1::/112 and only have the last 16 bits set aside for hosts. 2^16 = plenty of room. This approach may not be recommended as I am not an "expert". However, I see no reason why it wouldn't work. Design your network however you want within your /48 of course. 8-)
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