| 4th Octet Prefix length | 3rd octet prefix length | 2nd octet prefix length | 1st oct prefix length | Octet Value | Group Size |
|---|---|---|---|---|---|
| /32 | /24 | /16 | /8 | 255 | 1 |
| /31 | /23 | /15 | /7 | 254 | 2 |
| /30 | /22 | /14 | /6 | 252 | 4 |
| /29 | /21 | /13 | /5 | 248 | 8 |
| /28 | /20 | /12 | /4 | 240 | 16 |
| /27 | /19 | /11 | /3 | 224 | 32 |
| /26 | /18 | /10 | /2 | 192 | 64 |
| /25 | /17 | /9 | /1 | 128 | 128 |
How to Write The Subnetting Cheat Sheet
If you are taking a certification exam, you can’t bring a note. So it is very important that you know how to write the cheat sheet so that you can quickly do that once your exam starts.
Here are steps to write the subnetting cheat sheet.
Group Size: Write “1” on the first row and double it to get a value for the second row. Then double the value of second row to get the value for the third row. Keep doing this till you get to 128 (8th row).
Octet value: Write 255 on the first row and subtract the corresponding group size to get the value for the next row. E.g., 1 was subtracted from 255 to 254, which is the value for the next row. 2 was subtracted from 254 to get the value for the next row, and so on.
Prefix length: For the 4th octet column, Start with /31 and keep decreasing by 1 till you exhaust the whole rows, then move to the 3rd octet column and then to others.
Here is a video on how to write the subnetting cheat sheet;
How to use the cheat sheet to solve subnetting problem
Example
Find the network address, broadcast address, first host, last host, next network address, number of hosts, and the subnet mask (dotted decimal notation) for 10.1.1.55/28.
Step 1: Find the subnet mask
From the subnet cheat sheet table, /28 lies in the fourth octet, and the octet value is 240. Hence, the subnet mask is 255.255.255.240.
Step 2: Locate the group size
For /28 prefix length, the group size is 16.
Step 3: Add the group size to the network address
The first network address is 10.1.1.0, and the target IP address is 10.1.1.55, so we will keep adding the group size till we pass our target IP (10.1.1.55).
10.1.1.0 + 16 = 10.1.1.16
10.1.1.16 + 16 = 10.1.1.32
10.1.1.32 + 16 = 10.1.1.48
10.1.1.48 + 16 = 10.1.1.64
Hence;
Network address = 10.1.1.48 (highest network address before the target IP address)
Next network = 10.1.1.64 (Network address after our target IP address)
Broadcast IP address = 10.1.1.63 (IP address before our next network)
First host IP address = 10.1.1.49 (Network address + 1)
Last host IP address = 10.1.1.62 (broadcast IP – 1)
Number of IP addresses = 16 (Same as group size)
Number of usable IP addresses = 14 (Group size – 2)
Here is a video on how to solve this example;
That is all for now, if you want to learn more tricks to subnetting, then I recommend checking out this playlist.
I am a passionate Networking Associate specializing in Telecommunications.
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