Mnemonic for Subnetting

Hi

I’m teaching our trainees how to find out Subnets by mental arithmetic as they need it for School.
This is the way I learned how to.

First we where told to learn and write this chart before the Exams:

chart1
This chart is been calculated so :

chart2
I just learned for MS Exams Subnet Mask /20 - /32. But the trainees must learn /8 - /32. I think this chart can be easily continued till /8. I also told them to lean the power of 2 (till 12). Is this the easiest way? Alternatively, do you have any other better mnemonic’s?

Many thanks in advance
LVL 19
*** Hopeleonie ***IT ManagerAsked:
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Darr247Connect With a Mentor Commented:
If you work the first chart backwards, you might notice the mask subtracted from 256 tells you how many addresses are in a subnet (then you would subtract 1 for the network ID and 1 for the broadcast IP).
So, for a /29 the mask is
255.255.255.248
256-248=8 -1 (for network ID), -1 (for broadcast) = 6 usable IPs.

Not a mnemonic, but you may also find this to be a useful visualization:
00000000.00000000.00000000.00000000 = /0  = 0.0.0.0
10000000.00000000.00000000.00000000 = /1  = 128.0.0.0
11000000.00000000.00000000.00000000 = /2  = 192.0.0.0
11100000.00000000.00000000.00000000 = /3  = 224.0.0.0
11110000.00000000.00000000.00000000 = /4  = 240.0.0.0
11111000.00000000.00000000.00000000 = /5  = 248.0.0.0
11111100.00000000.00000000.00000000 = /6  = 252.0.0.0
11111110.00000000.00000000.00000000 = /7  = 254.0.0.0
11111111.00000000.00000000.00000000 = /8  = 255.0.0.0

11111111.10000000.00000000.00000000 = /9  = 255.128.0.0
11111111.11000000.00000000.00000000 = /10 = 255.192.0.0
11111111.11100000.00000000.00000000 = /11 = 255.224.0.0
11111111.11110000.00000000.00000000 = /12 = 255.240.0.0
11111111.11111000.00000000.00000000 = /13 = 255.248.0.0
11111111.11111100.00000000.00000000 = /14 = 255.252.0.0
11111111.11111110.00000000.00000000 = /15 = 255.254.0.0
11111111.11111111.00000000.00000000 = /16 = 255.255.0.0

11111111.11111111.10000000.00000000 = /17 = 255.255.128.0
11111111.11111111.11000000.00000000 = /18 = 255.255.192.0
11111111.11111111.11100000.00000000 = /19 = 255.255.224.0
11111111.11111111.11110000.00000000 = /20 = 255.255.240.0
11111111.11111111.11111000.00000000 = /21 = 255.255.248.0
11111111.11111111.11111100.00000000 = /22 = 255.255.252.0
11111111.11111111.11111110.00000000 = /23 = 255.255.254.0
11111111.11111111.11111111.00000000 = /24 = 255.255.255.0

11111111.11111111.11111111.10000000 = /25 = 255.255.255.128
11111111.11111111.11111111.11000000 = /26 = 255.255.255.192
11111111.11111111.11111111.11100000 = /27 = 255.255.255.224
11111111.11111111.11111111.11110000 = /28 = 255.255.255.240
11111111.11111111.11111111.11111000 = /29 = 255.255.255.248
11111111.11111111.11111111.11111100 = /30 = 255.255.255.252
11111111.11111111.11111111.11111110 = /31 = 255.255.255.254
11111111.11111111.11111111.11111111 = /32 = 255.255.255.255

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With IPv6 becoming more prevalent, I see less and less usefulness for subnet masks in the future. It's like teaching kids how to do long square roots manually instead of using the SqRt key on their calculators. Not quite as useful as learning long division. You can't memorize EVERYthing, so just remember where to look up some things.
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ozoCommented:
2^(32-n)
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*** Hopeleonie ***IT ManagerAuthor Commented:
Hi ozo

Can you please explain 2^(32-n)  in an example?
I was never good in maths :-)

Many thanks
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ozoConnect With a Mentor Commented:
2^0 = 1
2^1 = 2
2^2 = 2*2 = 4
2^3 = 2*2*2 = 8
2^4 = 2*2*2*2 = 16
2^5 = 2*2*2*2*2 = 32
...
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Don JohnstonConnect With a Mentor InstructorCommented:
With IPv6 becoming more prevalent, I see less and less usefulness for subnet masks in the future.
Next time you take a Microsoft, Cisco, HP, Juniper exam, tell them you're not going to answer any IPv4 questions.  Let us know how that works out for you.  :-)
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*** Hopeleonie ***IT ManagerAuthor Commented:
Next time you take a Microsoft, Cisco, HP, Juniper exam, tell them you're not going to answer any IPv4 questions.  Let us know how that works out for you.  :-)

LOL :-)
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*** Hopeleonie ***IT ManagerAuthor Commented:
Thanks to all again!
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