IPV6 subnetting

[dveney001]

I'm having trouble with IPV6 subnets how does a ::/56 prefix produce 256 subnets? Can someone clarify this for me? I was taught that the ::/64 prefix produces 65,536 subnets so wouldn't the ::/56 prefix produce more than 256 subnets? I'm trying to grasp the bit positions.

[Gregory Leeson]

The /64 (CIDR notation) says how many bits make up the network.  You start from the left and count 64 bits.  So the following address :

2001:1111:CAFE:9876:ABCD:1234:DEAD:BEAD /64

Your network would be 2001:1111:CAFE:9876

Your PC's address would be the ABCD:1234:DEAD:BEAD part of the address.

So your printer down the hall might be 2001:1111:CAFE:9876:5555:6666:7777:8888 /64

So for a /56, you're giving up 8 bits.  So your network would be :

2001:1111:CAFE:9800:0000:0000:0000:0000 /56

Each character in IPv6 is 4 bits.

So if you received the IP range 2001:1111:CAFE::/48 from your ISP, you'd have 16 bits to play with for subnets, while still leaving the last 64 bits for your hosts.

Soooo:

2001:1111:CAFE:[your 16 bits, or 65536 subnets]:0000:0000:0000:0000 /48

So you get 16 bits in between those brackets to break up into your different subnets.

But if you only got a /56 from your ISP, you'd only have 8 bits to play with in there for your subnets, so that's where you'd get 256 subnets:

2001:1111:CAFE:98[your 8 bits, or 256 subnets]:0000:0000:0000:0000

I hope this helps.

Greg

[dveney001]

Thanks for your help. So does this mean you get 32 subnets per bit? when I divide 256 by 8 I get 32. If my prefix was ::/60 then that would make 384 subnets correct?

[Gregory Leeson]

Nope,

 You can't divide by 8 to figure it out.  It's 2 to the 8th power, not 2 times 8.

Also, you're going the wrong direction.  

/64 has zero subnets.  

/63 has 2.  

/62 has 4

/61 has 8

/60 has 16

/59 : 32

/58 : 64

/57 : 128

/56 : 256

[dveney001]

Thank you so much you have solved a very big problem for me.

[Gregory Leeson]

Cool. Glad I could help.