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# Mnemonic for Subnetting

Posted on 2014-02-08
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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:

This chart is been calculated so :

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?

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Question by:*** Hopeleonie ***
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Expert Comment

ID: 39844514
2^(32-n)
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Author Comment

ID: 39844530
Hi ozo

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

Many thanks
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Accepted Solution

Darr247 earned 668 total points
ID: 39844583
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
``````

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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Assisted Solution

ozo earned 668 total points
ID: 39844705
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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Assisted Solution

Don Johnston earned 664 total points
ID: 39846163
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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LVL 19

Author Comment

ID: 39846487
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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Author Closing Comment

ID: 39863931
Thanks to all again!
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