# A Quick Mathematical Formula for Subnetting

March 24, 2016 | Views: 23110

Let’s begin.

Steps:

1. Write powers of 2 left to right
2. Write CIDR subnet masks from 1 to 32. This will take 4 lines, one for each octet in the IP addresses
3. Write the subnet mask: add powers of 2 from left to current column

You can invert 2 and 3 if you think it’s more logical (I’m used to this way).

You an add the number of usable IP by doing (power of 2) -2

A more mathematical way to express the the subnet mask calculation is: current value = value on the left (previous value) + power of 2 of this column.

It should take about 1 minute to do this when the exam starts. It will be extremely useful all through the exam to double check any IP and subnet. The only missing part would be a list of possible subnet values (i.e 0, 16, 32, … for a /28 CIDR), but I found it’s best to do this when you need it rather than pre-calculate it. It’s not needed that often during the exam, as most of the exam is being “nice” with you and gives you easy values such as 0, 128, 64, …

You should be able to generate this table:

 2x 128 64 32 16 8 4 2 1 1st octet 1 2 3 4 5 6 7 8 2nd 9 10 11 12 13 14 15 16 3rd 17 18 19 20 21 22 23 24 4th 25 26 27 28 29 30 31 32 mask 128 192 224 240 248 252 254 255

A few examples of how to use this table and concept:

– Find the last usable IP in the subnet 172.16.1.0/30

/30 gives a 4 total IP’s (0 to 3), meaning 2 usable IPs (1 and 2), 0 is the subnet and 3 is the broadcast. Answer is 172.16.1.2

– Can 192.168.15.23 use 192.168.16.35 as default gateway if subnet is 255.255.254.0?

254 as a mask gives us 2 IP in the third octet, so:

• 192.168.0.x and 192.168.1.x are in the same subnet
• 2.x and 3.x
• 4.x and 5.x
• 6.x and 7.x

The answer is “no,” as both IP’s are not in the same subnet.

– What is the CIDR subnet mask for /12?

– What mask do you need to have 64 hosts in a subnet?

64 total IP’s won’t be enough (don’t forget broadcast and subnet IP’s remove usable IP’s from the total). You need to choose a subnet that can accommodate 126 hosts (128-2) hence a mask of 255.255.255.128, or /25

Thanks and I hope this is helpful to you.

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1. Of all the charts I have seen, this one is the most practical. You essentially have all 4 octets which is perfect. Thanks!

2. Of all the charts I have seen, this one is the most practical. Thanks!

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