IPv6 question.....

pjd007

So I understand IPv4 custom subnetting but today I had a question which I was unsure of.

The question was asking me to state the first and last subnet addresses that would be used on a network, the address was something like FD00:0AB1:AB11:0800::/54

So if I'm looking at this correctly two bits would need to be used from the last octet (ie. 16bits per octet) for the subnet ID but how do you calculate the subnet ID's ?

I wasn't expecting a IPv6 subnetting type of question in this exam.

Snow.bros

I am trying to understand your question here (been trying to figure out this the whole day),

FD00:0AB1:AB11:0800::
1111110100000000: 0000101010110001: 1010101100010001: 0000100000000000: 0000000000000000: 0000000000000000: 0000000000000000: 0000000000000000

or

FD00:0AB1:AB11:0800::/54

1= network
0= host

1111111111111111: 1111111111111111: 1111111111111111: 1111110000000000: 0000000000000000: 0000000000000000:0000000000000000
Which doesn't make any sense because the subnet musk are already built into the IP and this is an old ipv4 cider prefix.

6: You don't need a traditional subnet mask

In IPv4, every IP address comes with a corresponding subnet mask. IPv6 also uses subnets, but the subnet ID is built into the address.
In an IPv6 address, the first 48 bits are the network prefix. The next 16 bits are the subnet ID and are used for defining subnets. The last 64 bits are the interface identifier (which is also known as the Interface ID or the Device ID).
If necessary, the bits that are normally reserved for the Device ID can be used for additional subnet masking. However, this is normally not necessary, as using a 16-bit subnet and a 64-bit device ID provides for 65,535 subnets with quintillions of possible device IDs per subnet. Still, some organizations are already going beyond 16-bit subnet IDs.

So I guess in your case:

FD00:0AB1:AB11:0800:0000:0000:0000:0000 - the first three octet are the network prefixes.

and then, FD00:0AB1:AB11:0800:0000:0000:0000:0000 - is the subnets ID which defines the network.

The leading zero's are the interface identifier (device ID) FD00:0AB1:AB11:0800:0000:0000:0000:0000.

Hope I helped us get an idea.

Anyone else please intervene if this does not make sense to you.

pjd007

Snow.bros wrote: »

I am trying to understand your question here (been trying to figure out this the whole day),

FD00:0AB1:AB11:0800::
1111110100000000: 0000101010110001: 1010101100010001: 0000100000000000: 0000000000000000: 0000000000000000: 0000000000000000: 0000000000000000

or

FD00:0AB1:AB11:0800::/54

1= network
0= host

1111111111111111: 1111111111111111: 1111111111111111: 1111110000000000: 0000000000000000: 0000000000000000:0000000000000000
Which doesn't make any sense because the subnet musk are already built into the IP and this is an old ipv4 cider prefix.



So I guess in your case:

FD00:0AB1:AB11:0800:0000:0000:0000:0000 - the first three octet are the network prefixes.

and then, FD00:0AB1:AB11:0800:0000:0000:0000:0000 - is the subnets ID which defines the network.

The leading zero's are the interface identifier (device ID) FD00:0AB1:AB11:0800:0000:0000:0000:0000.

Hope I helped us get an idea.

Anyone else please intervene if this does not make sense to you.

Apologies, yeah I probably haven't described the question well and my brain was a bit scrambled at the time of posting.

They were asking for the first and last network ID's that you would use on your network, so I guess you would use them for different physical network segments of your environment.

So if FD00:0AB1:AB11:0800:: is the first subnet ID and /54 notation, how do you calculate the last possible network ID ie. what can be used in the fourth octet ?

I just want to know how to answer the question correctly, I think I answered it with the highest possible number so:

Answer A: FD00:0AB1:AB11:0AB2::
Answer B: FD00:0AB1:AB11:0EF7::

My answer would've been B.

Could you have FD00:0AB1:AB11:FFFF:: would that not be a network broadcast ?


I'm sure other people will see this question in the exam so hopefully someone can correct me or assist here as it's hurting my head thinking about it but I'm probably making more work of this than needed !

Hatch1921

Very good series... might be helpful?

[h=1]IPv6-01 Making sense out of an IPv6 Address Keith Barker[/h]https://www.youtube.com/watch?v=rljkNMySmuM

Snow.bros

pjd007 wrote: »

So if FD00:0AB1:AB11:0800:: is the first subnet ID and /54 notation, how do you calculate the last possible network ID ie. what can be used in the fourth octet ?

I don't know if it is just the complexity of ipv6 or I simply cannot understand the question.

I was it a multiple choice question, if so, what were your options? Maybe that might help answer the question.

pjd007

Snow.bros wrote: »

I don't know if it is just the complexity of ipv6 or I simply cannot understand the question.

I was it a multiple choice question, if so, what were your options? Maybe that might help answer the question.

It was a drag and drop style question so there was a list of IPV6 addresses and you had to select what would be the first and last network addresses that could be used.

I can't remember the details in terms of the addresses used except /54 part of it.

I'm going to watch some more vids and try to get my head round this.

pjd007

Hatch1921 wrote: »

Very good series... might be helpful?

IPv6-01 Making sense out of an IPv6 Address Keith Barker

https://www.youtube.com/watch?v=rljkNMySmuM

I've watched the first vid and will watch some of the others, first one was good and to the point.

Snow.bros

Awesome, come back to us once you get your head around it.

subnettingpractice

Hmm, not sure I understand the question completely. The usable range for that network is:

fd00:ab1:ab11:800:0:0:0:0 through fd00:ab1:ab11:bff:ffff:ffff:ffff:ffff

Snow.bros

subnettingpractice wrote: »

Hmm, not sure I understand the question completely. The usable range for that network is:

fd00:ab1:ab11:800:0:0:0:0 through fd00:ab1:ab11:bff:ffff:ffff:ffff:ffff

Please share how you calculated this, I don't get it

pjd007

subnettingpractice wrote: »

Hmm, not sure I understand the question completely. The usable range for that network is:

fd00:ab1:ab11:800:0:0:0:0 through fd00:ab1:ab11:bff:ffff:ffff:ffff:ffff

Seems like you may understand the answer even if you don't get the question !

Why can't the fourth octet be afff, cfff, dfff or efff ?

Jon_Cisco

This is probably not a good explanation but I'll try quickly to explain the starting point.
Without seeing the real question I think I might just confuse the question more if I guessed.


So IPv6 is 128 bit address. First 64 bits are the network and the last 64 bits are the hosts.
IPv6 does not use broadcasts they have been replaced by a multicast that anyone can listen to I believe.

FF00:0AB1:AB11:0
This is the first 52 bits

FF00:0AB1:AB11:08
The 8 is the hex number that we need to examine. We need it's first two bits.

1000 = 8

the 10 is fixed and the 00 changes

10 00 = 8
10 01 = 9
10 10 = 10(A)
10 11 = 11(b)


FF00:0AB1:AB11:0800:0000:0000:0000:0000 - FF00:0AB1:AB11:08ff:ffff:ffff:ffff:ffff
FF00:0AB1:AB11:0900:0000:0000:0000:0000 - FF00:0AB1:AB11:09ff:ffff:ffff:ffff:ffff
FF00:0AB1:AB11:0A00:0000:0000:0000:0000 - FF00:0AB1:AB11:0Aff:ffff:ffff:ffff:ffff
FF00:0AB1:AB11:0B00:0000:0000:0000:0000 - FF00:0AB1:AB11:0Bff:ffff:ffff:ffff:ffff

So your overall range is:
FF00:0AB1:AB11:0800:0000:0000:0000:0000 - FF00:0AB1:AB11:0Bff:ffff:ffff:ffff:ffff

Network range is the first 64 bits:
FF00:0AB1:AB11:0800:: - FF00:0AB1:AB11:0Bff::



Sorry if this is confusing I was trying to think of a way to make if visual.
Good Luck

pjd007

Jon_Cisco wrote: »

This is probably not a good explanation but I'll try quickly to explain the starting point.
Without seeing the real question I think I might just confuse the question more if I guessed.


So IPv6 is 128 bit address. First 64 bits are the network and the last 64 bits are the hosts.
IPv6 does not use broadcasts they have been replaced by a multicast that anyone can listen to I believe.

FF00:0AB1:AB11:0
This is the first 52 bits

FF00:0AB1:AB11:08
The 8 is the hex number that we need to examine. We need it's first two bits.

1000 = 8

the 10 is fixed and the 00 changes

10 00 = 8
10 01 = 9
10 10 = 10(A)
10 11 = 11(b)


FF00:0AB1:AB11:0800:0000:0000:0000:0000 - FF00:0AB1:AB11:08ff:ffff:ffff:ffff:ffff
FF00:0AB1:AB11:0900:0000:0000:0000:0000 - FF00:0AB1:AB11:09ff:ffff:ffff:ffff:ffff
FF00:0AB1:AB11:0A00:0000:0000:0000:0000 - FF00:0AB1:AB11:0Aff:ffff:ffff:ffff:ffff
FF00:0AB1:AB11:0B00:0000:0000:0000:0000 - FF00:0AB1:AB11:0Bff:ffff:ffff:ffff:ffff

So your overall range is:
FF00:0AB1:AB11:0800:0000:0000:0000:0000 - FF00:0AB1:AB11:0Bff:ffff:ffff:ffff:ffff

Network range is the first 64 bits:
FF00:0AB1:AB11:0800:: - FF00:0AB1:AB11:0Bff::



Sorry if this is confusing I was trying to think of a way to make if visual.
Good Luck

Thanks that is the answer I was looking for and makes sense so I'll know how to answer this if I see it again, as will others