IPV6 Subnetting Practice Sites?

Beany

Hi All,

I'm looking for a website that will allow me to practice ipv6 subnetting questions. I'm referring to questions like eg:

1. Identify the first four /36 address blocks out of
2406:6400::/32
1._____________________
2._____________________
3._____________________
4._____________________

If i can find a website with these questions and the correct answers it will be helpful. Anyone know of any sites?

Thanks

EdTheLad

I don't know any sites but subnetting v6 is the same as v4 with a longer address space and some hex.

2406:6400::/32
same as
2406:6400:0:0:0:0:0:0/32
or
2406:6400:0000:0000:0000:0000:0000:0000/32

Total length is 128 bytes, 8 sections of 16 bits, each character is a nibble i.e. 4 bits represented in hex

So 36 bit mask, 36/4 = 9

2406:6400:0

That last 0 is a nibble, i.e. in binary 0000; first 4 blocks 0000,0001,0010,0011 i.e. 0,1,2,3
(Remember you can drop leading zero's but not trailing zero's)
2406:6400:0/36
2406:6400:1000/36
2406:6400:2000/36
2406:6400:3000/36

cwshellhamer

I thought it was all CIDR notation

Beany

EdTheLad wrote: »

I don't know any sites but subnetting v6 is the same as v4 with a longer address space and some hex.

2406:6400::/32
same as
2406:6400:0:0:0:0:0:0/32
or
2406:6400:0000:0000:0000:0000:0000:0000/32

Total length is 128 bytes, 8 sections of 16 bits, each character is a nibble i.e. 4 bits represented in hex

So 36 bit mask, 36/4 = 9

2406:6400:0

That last 0 is a nibble, i.e. in binary 0000; first 4 blocks 0000,0001,0010,0011 i.e. 0,1,2,3
(Remember you cant drop leading zero's but not trailing zero's)
2406:6400:0/36
2406:6400:1000/36
2406:6400:2000/36
2406:6400:3000/36

I don't get the last bit, how did you know it has to go up by 1, like 0,1,2,3 ?

EdTheLad

So 36 bit mask, 36/4 = 9

2406:6400:0/36

The last 0 is a nibble i.e 4 bits which is represented in an ipv6 address as hex(16 possible values from 0-F)

0 = 0000
1 = 0001
2 = 0010
"
"
15 = FFFF

The /32 original network contains 16 /36 networks.

Your question asked for the first 4 i.e.
0000 = hex 0
0001 = hex 1
0002 = hex 2
0003 = hex 3

King_84

EdTheLad wrote: »

So 36 bit mask, 36/4 = 9

2406:6400:0/36

The last 0 is a nibble i.e 4 bits which is represented in an ipv6 address as hex(16 possible values from 0-F)

0 = 0000
1 = 0001
2 = 0010
"
"
15 = FFFF

The /32 original network contains 16 /36 networks.

Your question asked for the first 4 i.e.
0000 = hex 0
0001 = hex 1
0002 = hex 2
0003 = hex 3

I can't get my head around IPV6 Subnetting. I understand most of the concepts of IPV6 but cannot subnet like this. Can you please explain more examples like this? I might eventually get the idea.

Thanks

mikeybinec

EdTheLad wrote: »

(Remember you cant drop leading zero's but not trailing zero's)

You can drop leading zeros?

thanks

EdTheLad

Can, edited!

Beany

EdTheLad wrote: »

So 36 bit mask, 36/4 = 9

2406:6400:0/36

The last 0 is a nibble i.e 4 bits which is represented in an ipv6 address as hex(16 possible values from 0-F)

0 = 0000
1 = 0001
2 = 0010
"
"
15 = FFFF

The /32 original network contains 16 /36 networks.

Your question asked for the first 4 i.e.
0000 = hex 0
0001 = hex 1
0002 = hex 2
0003 = hex 3

what would the last two subnets be? I take it there will be 16 subnets in total with this?

EdTheLad

2406:6400:E000/36
2406:6400:F000/36

btw, 15 = 1111, not FFFF as per my previous post.

1110 = E (hex)
1111 = F (hex)

Beany

EdTheLad wrote: »

2406:6400:E000/36
2406:6400:F000/36

btw, 15 = 1111, not FFFF as per my previous post.

1110 = E (hex)
1111 = F (hex)

Oh jeez, i missed that on your last post.

Thanks for your replies. Appreciate it.

Beany

What about this problem:

Get 16 subnets from: 2001:db8:3eff::/48

the answer was: 2001:db8:3eff::/52

1 2001:db8:3eff:0000::/52
2 2001:db8:3eff:1000::/52
3 2001:db8:3eff:2000::/52
4 2001:db8:3eff:3000::/52
5 2001:db8:3eff:4000::/52

i was under the impression that the 52 bit suppose to increment not the 49 bit like above? im confused.

Can someone please shed some light on this?

EdTheLad

You want 16 subnets, which means 4 bits are used for subnetting 2^4 = 16

If your allocated address is a /48, that means you need to add 4 to 48 = 52

1 2001:db8:3eff:0000::/52
2 2001:db8:3eff:1000::/52
3 2001:db8:3eff:2000::/52
4 2001:db8:3eff:3000::/52
5 2001:db8:3eff:4000::/52

That 0,1,2,3,4 that is incrementing is not a bit, its a nibble = 4 bits, 0 =0000 1=0001; 2 =0010 etc
Remember every character in the ipv6 address is 4 bits.

If you wanted /49 subnets, it would mean the 49th bit can be high or low.

0000 = 0
1000 = 8

2001:db8:3eff:0000::/49 or 2001:db8:3eff::/49
2001:db8:3eff:8000::/49

King_84

EdTheLad wrote: »

You want 16 subnets, which means 4 bits are used for subnetting 2^4 = 16

If your allocated address is a /48, that means you need to add 4 to 48 = 52

1 2001:db8:3eff:0000::/52
2 2001:db8:3eff:1000::/52
3 2001:db8:3eff:2000::/52
4 2001:db8:3eff:3000::/52
5 2001:db8:3eff:4000::/52

That 0,1,2,3,4 that is incrementing is not a bit, its a nibble = 4 bits, 0 =0000 1=0001; 2 =0010 etc
Remember every character in the ipv6 address is 4 bits.

If you wanted /49 subnets, it would mean the 49th bit can be high or low.

0000 = 0
1000 = 8

2001:db8:3eff:0000::/49 or 2001:db8:3eff::/49
2001:db8:3eff:8000::/49

Wow you're good at ipv6. Hey you've been excellent and very patient so thank you so far.

The more you explain the closer I get to understanding this concept. I understand the nibble part now but if you can explain to me how you determine what number it goes by like the one above , 0, 8 ? Or sometimes when it goes up from 0, 800, 1000, 1800 ( another example I saw). I know u tried explaining this a few times but I still can't get around it.

You must be sick of me now

EdTheLad

An ipv6 address consists of 128bits, the base address provided is 48 bits so we have 80 bits left to allocate as we choose.
If i want 2 subnets, i will use 1 bit, the 49th bit.

2001:db8:3eff::/48
same as
2001:db8:3eff:0:0:0:0:0/48

Each section between the ":" is 16 bits i.e. 4 nibbles, there are 8 sections, therefore 8 x 16 =128 bits

2001 = 16 bits
db8 = 16 bits; its also written as 0db8 ; leading zero's can be dropped
3eff = 16 bits
0 = 16 bits; it can also be written as 0000

16+16+16 = 48; so we know we cannot touch 2001:db8:3eff; the 49th bit is the first 0 of the next 16 bits.
The next 16 bits = 0000 with each of those 0's being 0000, therefore it's actually 0000 0000 0000 0000

If we want to use the 49th bit for subnetting, the 16 bit section between ":" would look like:
0000 0000 0000 0000
1000 0000 0000 0000

The first bit above which can be 0 or 1 is 49th bit. Now we need to convert this binary string back into hex, as the ipv6 address is represented in hex.

0000 = 0 hex
1000 = 8 hex

The 2 subnets are therefore:

2001:db8:3eff:0::/49
2001:db8:3eff:8000:/49

Now lets imagine we wanted 4 subnets rather than 2.

Base address is 2001:db8:3eff::/48, 4 subnets means 2 bits required. 48 +2 = /50 mask

2001:db8:3eff = 48 bits
We need to use the next 2 bits from the next 16 bit section after the ":"

0000 0000 0000 0000
0100 0000 0000 0000
1000 0000 0000 0000
1100 0000 0000 0000

0000 = 0 hex
0100 = 4 hex
1000 = 8 hex
1100 = C hex

0000 0000 0000 0000 = 0000 hex
0100 0000 0000 0000 = 4000 hex
1000 0000 0000 0000 = 8000 hex
1100 0000 0000 0000 = c000 hex


So our subnets are:

2001:db8:3eff:0::/50
2001:db8:3eff:4000:/50
2001:db8:3eff:8000:/50
2001:db8:3eff:c000:/50

King_84

EdTheLad wrote: »

An ipv6 address consists of 128bits, the base address provided is 48 bits so we have 80 bits left to allocate as we choose.
If i want 2 subnets, i will use 1 bit, the 49th bit.

2001:db8:3eff::/48
same as
2001:db8:3eff:0:0:0:0:0/48

Each section between the ":" is 16 bits i.e. 4 nibbles, there are 8 sections, therefore 8 x 16 =128 bits

2001 = 16 bits
db8 = 16 bits; its also written as 0db8 ; leading zero's can be dropped
3eff = 16 bits
0 = 16 bits; it can also be written as 0000

16+16+16 = 48; so we know we cannot touch 2001:db8:3eff; the 49th bit is the first 0 of the next 16 bits.
The next 16 bits = 0000 with each of those 0's being 0000, therefore it's actually 0000 0000 0000 0000

If we want to use the 49th bit for subnetting, the 16 bit section between ":" would look like:
0000 0000 0000 0000
1000 0000 0000 0000

The first bit above which can be 0 or 1 is 49th bit. Now we need to convert this binary string back into hex, as the ipv6 address is represented in hex.

0000 = 0 hex
1000 = 8 hex

The 2 subnets are therefore:

2001:db8:3eff:0::/49
2001:db8:3eff:8000:/49

Now lets imagine we wanted 4 subnets rather than 2.

Base address is 2001:db8:3eff::/48, 4 subnets means 2 bits required. 48 +2 = /50 mask

2001:db8:3eff = 48 bits
We need to use the next 2 bits from the next 16 bit section after the ":"

0000 0000 0000 0000
0100 0000 0000 0000
1000 0000 0000 0000
1100 0000 0000 0000

0000 = 0 hex
0100 = 4 hex
1000 = 8 hex
1100 = C hex

0000 0000 0000 0000 = 0000 hex
0100 0000 0000 0000 = 4000 hex
1000 0000 0000 0000 = 8000 hex
1100 0000 0000 0000 = c000 hex


So our subnets are:

2001:db8:3eff:0::/50
2001:db8:3eff:4000:/50
2001:db8:3eff:8000:/50
2001:db8:3eff:c000:/50

Superb post, now I get how to work out subnets. You're a star. BTW there must be a quicker way of working out the subnet numbers, binary first the hex is painfully long?!!!

Jon_Cisco

Great Post EdTheLad. I found reading this to be a nice refresher from when I studied it last year.

@king_84
I bet IPv4 addresses were a little confusing at first. Most of us don't use binary every day so we would need to write it out on scrap at first. Eventually I could convert anything up to 256 into binary in my head just by looking at it. Then I learned a few common numbers that made subnetting even faster.

I believe if you were interested you could do this with hex also. It's a bit more complicated but if you look at individual characters rather then each octet as you did in IPv4 then it can almost be seen as easier. Since you are only dealing with 4 bits per character rather then 8 bits per octet.

I know that when I was studying last year and focusing on IPv6 I was able to do this in my head for a while. Now I am much slower but I remember the ideas. Your study will likely be the same. Learn the concepts and get good enough for testing but don't go overboard with IPv6 unless you are currently using it for work. Someday we will all be using IPv6 and you will be ahead of the curve since you already learned the basics.

King_84

Jon_Cisco wrote: »

Great Post EdTheLad. I found reading this to be a nice refresher from when I studied it last year.

@king_84
I bet IPv4 addresses were a little confusing at first. Most of us don't use binary every day so we would need to write it out on scrap at first. Eventually I could convert anything up to 256 into binary in my head just by looking at it. Then I learned a few common numbers that made subnetting even faster.

I believe if you were interested you could do this with hex also. It's a bit more complicated but if you look at individual characters rather then each octet as you did in IPv4 then it can almost be seen as easier. Since you are only dealing with 4 bits per character rather then 8 bits per octet.

I know that when I was studying last year and focusing on IPv6 I was able to do this in my head for a while. Now I am much slower but I remember the ideas. Your study will likely be the same. Learn the concepts and get good enough for testing but don't go overboard with IPv6 unless you are currently using it for work. Someday we will all be using IPv6 and you will be ahead of the curve since you already learned the basics.

Thank you for your reply.Your'e completely correct, the more you practice the easier it becomes and you begin to see a pattern which gives you opportunities to subnet faster. With me sometimes understanding a topic just falls down to me finding the right explanation like EdTheLad provided.