Hi fellas
Hope you're all fine. I am in the process unlocking the IPng mysteries and would be most thankful if someone can provide me with the tutorial of complete IPv6 subnetting techniques. I really need to master to increase my chances in the Job market. Scenarios in subnetting and VLSM and summarization would alos help as well.
Regards
Sms
The first question I'd ask you is whether or not you know how to subnet IPv4? And do you know VLSM for IPv4?
If the answers are yes, then awesome. You have nothing new to learn!
Same stuff, just bigger and uglier numbers! Oh yeah, it's in hex now.
But once you break it down to binary, everything works the same for splitting things up and carving them up!
If you have specific questions, I'd be happy to go over them, but seriously it's the same thinking as IPv4 subnetting.
Scott
Jeff Doyle's TCP/IP Volume I Second Edition now has a chapter on IPv6. I think one of the more difficult things is actually keeping all of the different types of addresses straight in my mind.
Hi Scott
hope you're good. Oh Yes Sir! I am a 10 second Subnetter and very comfortable with VLSM. but How would I do that in IPv6. Lets say we have 10.1.1.0/26. We can make 4 subnets namely 10.1.1.0, 10.1.1.64, 10.1.1.128 and 10.1.1.192 with 62 hosts per subnet. Straight forward, conventional, no issues. But lets say the ISP has assigned my company an addres of 2340:1111:AAAA::/48. I need to make 10 different subnets, every subnet has 200 hosts. How would I go about doing that? At this point it gets a little murky.
:-)
Thanks Paul
I'll try to get my hands on this book.
Hi Sms.
shujanaqvi escribió:
Hi Scott
hope you're good. Oh Yes Sir! I am a 10 second Subnetter and very comfortable with VLSM. but How would I do that in IPv6. Lets say we have 10.1.1.0/26. We can make 4 subnets namely 10.1.1.0, 10.1.1.64, 10.1.1.128 and 10.1.1.192 with 62 hosts per subnet. Straight forward, conventional, no issues. But lets say the ISP has assigned my company an addres of 2340:1111:AAAA::/48. I need to make 10 different subnets, every subnet has 200 hosts. How would I go about doing that? At this point it gets a little murky.
:-)
I will try to clear up this for you, given that the process it's the same as IPv4, first thing to take into account is that in this case we have 128bits - 48 bits from network prefix equals 80 available bits for subnets/hosts!!! so we have room for many many subnets and hosts.
To keep that into your example let's say that we want a subnets with 200 hosts, no VLSM, so like IPv4 we need 8 bits of room for that amount of hosts per subnet, that means that we still have 72 bits for subnets, you can use those bits not just only for 10 subnets.... many more can be created as you can see...
2340:1111:AAAA::/48
our network is ... 2340:1111:AAAA:0000:0000:0000:0000:0000/48
let's put apart the required subnet and hosts bits (72 bits for subnets)(8 For Hosts) that yields a /120 subnet prefix...
2340:1111:AAAA:(Network)0000:0000:0000:0000:00 (Subnet) 00 (hosts) /120 (Subnet prefix)
Basically the subnets are already done with hosts bits apart, now the trick could be Hex but if you can make subnneting there's no problem for you on Bin to Hex coversions sice they're simply nibbles...
Let's give results:
First subnet 2340:1111:AAAA::/120 Hosts range 2340:1111:AAAA::0001/120 - 2340:1111:AAAA::00FF/120
Second one 2340:1111:AAAA::0100/120 Hosts range 2340:1111:AAAA::0101/120 - 2340:1111:AAAA::01FF/120
Third one 2340:1111:AAAA::0200/120 Hosts range 2340:1111:AAAA::0201/120 - 2340:1111:AAAA::02FF/120
and so on, all it's bigger and hex, the logic remains the same. (Note.- Here the reserved anycast addresses are not considered)
Feel free to correct it as it's best-effort on the fly calculation...
Other approach with this "plentifulness" is to let 48 bits for network, 12 bits for subnets(4096), and 64 for hosts eui-64 like.
HTH,
Carlos.
I think you're thinking about 16 bits for subnet though.
The way things "are" in the global thinking right now is that ISPs will get /32's per allocation. They will assign /48's to enterprise customers. This leaves 16 bits for them to subnet.
An enterprise receiving a /48 will assign /64's to every LAN (for the EUI-64 "magic" to work). Again, leaving 16 bits for subnetting.
There's no REASON to do this other than organization/aggregation and the EUI-64 magic. You can make a subnet (as example above with /120's) of any size you feel like.
The big thing that gets people is creating summary addresses for routing. This is where you 'manually' aggregate the routes. And it uses the same binary concepts as subnetting just in the opposite direction!
So let's say you had:
2000:1234:5678:ABCD::/64
2000:1234:5678:ACDB::/64
How would you summarize them? Well, they are identical up through 2000:1234:5678:A... Cool, that's 52 bits. Then we'll look at binary.
B = 1011
C = 1100
So one more bit is ok, then changing. That means a /53 is what I need in order to summarize them. Now... Writing it out! 1000 is the binary I'll allow. That's "8" in hex!
2000:1234:5678:A800::/53
The process is no different than IPv4. But we have different ways of thinking about our addresses. (/32, /48, 64/EUI-64 are kinda like the "classes" in IPv6 in that they just exist so we need to remember them)
have fun!
Scott
shujanaqvi wrote:
Hi Scott
hope you're good. Oh Yes Sir! I am a 10 second Subnetter and very comfortable with VLSM. but How would I do that in IPv6. Lets say we have 10.1.1.0/26. We can make 4 subnets namely 10.1.1.0, 10.1.1.64, 10.1.1.128 and 10.1.1.192 with 62 hosts per subnet. Straight forward, conventional, no issues. But lets say the ISP has assigned my company an addres of 2340:1111:AAAA::/48. I need to make 10 different subnets, every subnet has 200 hosts. How would I go about doing that? At this point it gets a little murky.
:-)
Now, you DO realize that with a /48 you still have 80 bits for subnet/host addressing, right? That's WAY more addresses then you'll ever, ever want.
But we can play that game.
Same thing you do in IPv4. How many bits will I need on a subnet to get 200 hosts? (8).
If I have 128 bits, and use 8 (right-most) for hosts, what's my subnet bits? (128-8 = 120)
And just like Carlos laid out, just work the math from there. No difference!
Scott
I think you're thinking about 16 bits for subnet though.
Hello Scott,
The fact is that I was thinking about 12 bits, I was just sinking in maths , 128 bits are probably too much for my at his time. You're right and that 16 bits yields even more Subnets, wich supports, even more, the eui-64 election for the hosts as well, the example was meant as subnneting exercise, first for myself and maybe helpful for others but a little crazy....
Great explanation on summaries where the IPv6 binary action really occurs... really appreciated
Best,
Carlos.
Thanks Carlos, thanks Scott. This whole week I'll practice it and if there are any questions, I'll put it forward. :-)
this is the best conversation i can find and its the only thread where i felt I have understood this somewhat. Im still confused though
If I have 128 bits, and use 4 (right-most) for hosts, what's my subnet bits? (128-4 = 124)
which would mean a /124 ? and my hosts would equal 16 hosts per subnet?
my question is determining the amount of subnets
you know how the 16 bits after the first 48 bits are allocated to be the subnet section ? is this the bit where we specify how many subnets we want and then from there manipulate the prefix length to determine how many hosts to put in the subnets ?
example
2001:1111:1111:A::/124
does this mean im using 4 bits for host which is 16 hosts per subnet?
I then used A in the subnet field to specify 10 subnets ?
I somehow dont think this is correct.
I have a feeling theres way more hosts and networks than this.
please find my attach picture to help demonstrate what i am trying to say
If I have 128 bits, and use 4 (right-most) for hosts, what's my subnet bits? (128-4 = 124)
which would mean a /124 ? and my hosts would equal 16 hosts per subnet?
Yes, that is correct.
you know how the 16 bits after the first 48 bits are allocated to be the subnet section ? is this the bit where we specify how many subnets we want and then from there manipulate the prefix length to determine how many hosts to put in the subnets ?
Yes, but not how you explain it below. The subnet mask defines where the host bits start. With IPV6 the first 48 are specified for you so whatever bits remain in the middle give you the total number of subnets. IOW, if you used a /64 mask and the /48 was already given then you would have the middle 16 bits to define your subnets. Those 16 bits could give you up to 65,536 subnets.
2001:1111:1111:A::/124
does this mean im using 4 bits for host which is 16 hosts per subnet?
I then used A in the subnet field to specify 10 subnets ?
Yes you have given yourself 4 bits for the hosts. No, 'A' in the subnet field does not mean 10 subnets. You could say it was the 10th subnet I suppose. This might help: http://www.tcpipguide.com/free/t_IPv6GlobalUnicastAddressFormat-2.htm
thanks for taking the time to reply.
so you could say that the /48 is similar to the theory of ipv4 using a 172.16.10.0 / 24 meaning the default class is 255.255.0.0 but you are using 8 subnet bits which isnt default.
so the /48 is the default and ipv6 address is giving you 16 bits to have for subnet bits.
Oh so the A is just pretty much saying subnet 10 not the amount of subnets. I was playing around in a lab changing subnet bits and trying to ping each other and if i made the subnet section 0001 on host A and 0002 on host B they cant ping each other which i assume is because one is on subnet 1 and the other is on subnet 2 ?
ps - how would you specify that you wanted 10 subnets only. - not the way I had it where it just meant the 10th subnet?
example
2001:1000:1000:1::1/64 Host A
2001:1000:1000:2::1/64 Host B
(they both cant ping each other)
but if i change it to
2001:1000:1000:1::1/64 Host A
2001:1000:1000:1::2/64 Host A
(they CAN ping each other)
could you be kind enough to post up an example of another random ipv6 subnetting scheme with the ranges aswell. It helps to work it out and see how you came to the answers. something other than a /64 would be very helpful.
Thanks very much
so you could say that the /48 is similar to the theory of ipv4 using a 172.16.10.0 / 24 meaning the default class is 255.255.0.0 but you are using 8 subnet bits which isnt default.
so the /48 is the default and ipv6 address is giving you 16 bits to have for subnet bits.
Correct. For global unicast you will be given the first 48 bits. Really what you do with the rest is up to you, although the design is to use the first 48 for assigning via ISPs, the next 16 for subnets, and the last 64 for hosts. Could you use, say a 48/32/48 scheme? Sure. A 48/2/78? Why not? So you could limit your subnets to 8 or 16 if you want to, but why? The only thing you gain is more host bits, and you already have 64 bits for hosts using the standard format. You could use some of the subnet bits to increase your host range but why would you ever need to? If anything I am a bit suprised that they only left room for 65k subnets. It seems like some big companies may want to borrow a few host bits to expand their subnet availability but I suppose they can do that if they wish.
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