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IPv6 Subnetting - The easy way
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- เผยแพร่เมื่อ 15 ส.ค. 2024
- In this video, I demonstrate how to do basic subnetting on the nibble (4 bits) boundary using a quick hack, no maths involved ;)
Nibble bit boundary subnetting gives clean increments, i.e. 0, 1, 2, 3, 4.........e, f and are easy to read. Reverse DNS in IPv6 also works on the nibble bit boundary.
Of course, there's a formula to get the subnet increments [2^16 - (L’%16)] 😩 where L' is the sub-prefix length, but trust me, there's no maths in this one :)
There are lots of handy IPv6 subnet calculators online, so don't try subnetting by hand past the Nibble boundary......oh, and don't use a spreadsheet for your IPv6 address plan, the address space is huge 😉
The rules of representing ipv6 address says we should omit reading zeros not trailing zeros
That's correct! 👍
Hi Bashir, thanks I enjoyed the video. in IPv6 we can only omit the leading zeros.
I also have a question for you: We divided the hexademal digits by 4 equal parts because we needed 4 subnets, in case we need 5 or 7 or 9 subnets how do we do it?
Hi, I'm glad you enjoyed the video, and yes we only omit the leading zeroes :)
The method works perfectly for 2, 4, and 8 subnets because 16 is exactly divisible i.e. without a remainder with these numbers :)
2, 4 and 8 can give you a nice reference point for subnets in between. If you need 5 subnets then the next available option is 8 subnets. If you need 9 subnets, then the next available option is 16
Explained mathematically, 2 bits wouldn’t be sufficient for 5 subnets 2^2 = 2 hence you’ll need 3 bits which gives us the possibility of 8 subnets. You can use 5 subnets and have 3 extra subnets.
If you were to use the method in the video to split up the hexadecimal digit range, then a requirement of 5 subnets means that you’ll split the range into 8 equal parts, and have 3 extra subnets since 4 subnets i.e. 2 bits (2^2) won’t be enough for your required 5 subnets :)
Same thing applies to 9 subnets, splitting the range into 8 equal parts (i.e. 2^3) isn’t enough and you’ll need 4 bits (i.e. 2^4) which gives you 16 subnets. You’ll have your 9 subnets plus 7 extra subnets :)
The “shortcut” :) in the video aims to help you understand how we get subnets in the nibble bit boundary quickly. There are many IPv6 subnet calculators that do the heavy lifting, and as usual in IPv6, always use a calculator :)
Hope this helps!
Would you please try this one to explain FDBE:A661:24D4::/50 should be devided in 6 subnets ?????
Hi, to break down the /50 into 6 subnets you will need 3 bits i.e 2^3 = 8 subnets (2 extra subnets). 2^2 = 4 and would not be enough.
Since we'll use 3 bits, then our subnets will be /53s i.e. our original /50 + 3 bits
Next step is to identify the location of these 3 bits, once you do this, getting the actual subnets becomes very easy.
I'll expand your example slightly for easy readability, from FDBE:A661:24D4::/50 to FDBE:A661:24D4: *0* 000::/50. Our /50 lies in the nibble (4 bit) boundary highlighted in bold i.e. We are using 2 bits out of the available 4 bits to represent the /50 ( *00* 00). This just represents the *0* in *bold* from the /50 (FDBE:A661:24D4: *0* 000::/50)
Remember, we need 3 bits to cover your required 6 subnets right? :) so from the first 4 bits we only have 2 bits available i.e. 00 *00* since as mentioned in the previous line, the first 2 bits are used for the network prefix (/50)
This means that we'll need 1 extra bit from the second nibble/4 bit boundary i.e. the second *0* . So we'll need 2 bits from the first 4 bits and 1 additional bit from the second 4 bits FDBE:A661:24D4: *00* 00::/50
Here are the highlighted bits needed in binary 00 *00 0* 000. Take note of the bits in bold :)
So let's write down the subnets. I'll show you the bits first, then the subnet next to the bits. I'm assuming you can comfortably convert from Binary to Hex :)
1. Bits 00 *00 0* 000 - Subnet FDBE:A661:24D4: *00* 00::/53
2. Bits 00 *00 1* 000 - Subnet FDBE:A661:24D4: *08* 00::/53
3. Bits 00 *01 0* 000 - Subnet FDBE:A661:24D4: *10* 00::/53
4. Bits 00 *01 1* 000 - Subnet FDBE:A661:24D4: *18* 00::/53
5. Bits 00 *10 0* 000 - Subnet FDBE:A661:24D4: *20* 00::/53
6. Bits 00 *10 1* 000 - Subnet FDBE:A661:24D4: *28* 00::/53
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Additional subnets :)
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7. Bits 00 *11 0* 000 - Subnet FDBE:A661:24D4: *30* 00::/53
8. Bits 00 *11 1* 000 - Subnet FDBE:A661:24D4: *38* 00::/53
I hope this helps!
@@bmudeheri How did you get 0000, 0800, 1000, 1800, 2000, 2800 ánd two additional subnets 3000 3800 ? I didnt get this
@ameladukic6232 By converting the binary values to hex for each of the 4 bit positions
Here's a quick recap of binary to hex conversion, take note of what's in *bold* :)
*0000 = 0*
*0001 = 1*
*0010 = 2*
*0011 = 3*
0100 = 4
0101 = 5
0110 = 6
0111 = 7
*1000 = 8*
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1111 = F
So from my example in binary, here's what we'll get:
00 *00* and *0* 000 = 0 & 0 hence *00* 00
00 *00* and *1* 000 = 0 & 8 hence *08* 00
00 *01* and *0* 000 = 1 & 0 hence *10* 00
00 *01* and *1* 000 = 1 & 8 hence *18* 00
00 *10* and *0* 000 = 2 & 0 hence *20* 00
00 *10* and *1* 000 = 2 & 8 hence *28* 00
00 *11* and *0* 000 = 2+1(3) & 0 hence *30* 00
00 *11* and *1* 000 = 2+1(3) & 8 hence *38* 00
Now if we combine all the information, here are our subnets:
1. Binary 00 *00 0* 000 = *00* | Subnet = FDBE:A661:24D4: *00* 00::/53
2. Binary 00 *00 1* 000 = *08* | Subnet = FDBE:A661:24D4: *08* 00::/53
3. Binary 00 *01 0* 000 = *10* | Subnet = FDBE:A661:24D4: *10* 00::/53
4. Binary 00 *01 1* 000 = *18* | Subnet = FDBE:A661:24D4: *18* 00::/53
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8. Binary 00 *11 1* 000 = *38* | Subnet = FDBE:A661:24D4: *18* 00::/53
Once you understand binary to hex conversion then the rest fits in perfectly.
Your example involves subnetting past the nibble/4 bit boundary and is usually confusing. It's always recommended to stick to nibble/4 bit boundaries whenever possible :) the IPv6 address space is huge!
Ensure that you understand how to subnet using 4 bits first and then expand to the non-nibble bit boundary subnetting.
Always use the online calculators or an IP Address Management System, when you understand the logic behind IPv6 subnetting :)
I hope this helps!
Because the 8 comes first, we can't leave the zero's
Correct! 👍
I think this method only works with /16 /32 /48 /64 /80 /96 and so on...
Yes! The nibble bit boundary to be specific, but you'll be subnetting on the nibble bit boundary in most cases anyways :)
@@bmudeheri Yea sorry. Should have read the description. Going for your non-nibble video now. I need that for school and I don't like to convert to binary. Thanks for the vids. :)
@@L057_50UL No worries and you are welcome :) All the best!
We don't omit trailing zeros
That's correct! I'm glad you saw my question at the end of the video :)
Also with zero suppression, 2001:db8:8:: = 2001:db8:0008:: ... that's not a /33! ;)
no we cant write :8
That's correct!