IPv6 subnetting

[Lim Seow Hua]

Hi,

I have came across some note about the IPv6.

"

How about /47s in 2001:1::/32?

Hex 2001 = Binary 0010 0000 0000 0001 = /16
Hex 0001 = Binary 0000 0000 0000 0001 = /32
Hex 0000 = Binary 0000 0000 0000 00X0 = /47
So bit “X” keep changing:

Binary 0000 0000 0000 0000 = Hex 0000
The first /47 is 2001:0001:0000::/47
-----------------------------------------------------------------

Binary 0000 0000 0000 0010 = Hex 0002
So the second /47 is 2001:0001:0002::/47
-----------------------------------------------------------------

Binary 0000 0000 0000 0100 = Hex 0004
So the third /47 is 2001:0001:0004::/47
----------------------------------------------------------------

Binary 0000 0000 0000 0110 = Hex 0006
So the forth /47 is 2001:0001:0006::/47
----------------------------------------------------------------

Binary 0000 0000 0000 1000 = Hex 0008
So the fifth /47 is 2001:0001:0008 ::/47

"

I do not understand that what is meant the statement "So bit "X" keep changing". Can anyone kindly explain to me on this?

Thanks

[Mahesh Gohil]

Hello Lim,

Let me try subnetting of ipv6

2001:1:/32 and you want subnets of /47 so the difference between 47-32=15 so you need to work upon the 15 bits

now which 15 bits from 128 bits that we decide as below

write full ipv6 address.

2001:0001:0000:0000:0000:0000:0000:0000  << address in hex
16     16    16   16      16     16     16    16    <<  no of bits each octate contain

We have to concentrate upon bits from 32 to 47 so as above it lies in between octate 2nd so let's write it in binary

0000 0000 0000 0000  so we have first 15 bits for subnetting

hence the subnets will be like

0000 0000 0000 0010  >>  2001:1:2::/47
0000 0000 0000 0100  >>  2001:1:4::/47
0000 0000 0000 1000  >>  2001:1:8::/47

.

.

etc

Hope this helps

Regards

Mahesh

[Lim Seow Hua]

Hello Mahesh,

It is the subnet will increase in the multiple of 2 like below:-

0000 0000 0000 0010  >>  2001:1:2::/47
0000 0000 0000 0100  >>  2001:1:4::/47
0000 0000 0000 1000  >>  2001:1:8::/47

0000 0000 0000 1010  >>  2001:1:10::/47
0000 0000 0000 1100  >>  2001:1:12::/47
0000 0000 0000 1110  >>  2001:1:14::/47

.

.

.

etc

But until which subnet will this subnetting stop?

[cadet alain]

Hi,

Can't we  only subnet in bits from 48 to 64  in IPv6?

Regards.

Alain.

[cadet alain]

Hi,

subnetting in IPv6 is same as in IPv4  so if you have a /48 and you want to subnet to /50 then you gonna use bits 49 and 50 to subnets

so you'll have 50-48=X ,  2 to the power of X  subnets= 4 subnets 00, 01, 10, 11 for bits 49-50.

Regards.

Alain.

[Mahesh Gohil]

Hello Lim,

Yes your assumption is ok

let me explain it differently...first let's take small example

say you want to know subnets of /38 from given /32 ip. Please look at below two steps

1- First decide how many subnets will be there...substract 32 from 38 =6 so 2^6=64 , it will yield 64 subnets

2- Now let's decide that subnets will be in multiple of what. To do this consider how many 1's bits will be there in /38

1111 1111 1111  1100 ...yes the last 6 bits..Now if you write last 8 bits in decimal it will give you 16+12=28

To decide the subnets will be in multiple of what 28 from 32 (16+16) which gives you 4. so your subnets will be in multiple of 4 something like

2001:1::/38
2001:1:400::/38
2001:1:800::/38
2001:1:c00::/38

If you convert above in binary it will give you how you going to shift 1's in binary format to achieve subnets

0000   0000 0000 0000 0000 >>>last two bits

0400   0000 0100 0000 0000
0800   0000 1000 0000 0000
0c00   0000 1100 0000 0000  >> we are done with last two bits and now we do not have any more combination with 2 bits so now consider 3 bits
1000   0001 0000 0000 0000  >> last 3 bits
1400   0001 0100 0000 0000
1800   0001 1000 0000 0000
1c00   0001 1100 0000 0000  >> we are done with three bits so now onwards consider last 4 bits

and so on

I hope above is meaningful explanation

Regards

Mahesh