Click here for Part 1 of this series.

**WARNING**: You must master subnetting using our course or some other trusted materials before you start using these shortcut approaches. It is a common issue for Cisco candidates to move directly to subnetting shortcuts for the exams without fully understanding exactly how subnetting functions.

*ICND1 (CCENT)*

**Question 2: **You have run the ipconfig command and discovered your IP address and mask are 192.168.20.102 and 255.255.255.224. How many hosts are permitted on your subnet?

**Step 1:** I reference the Powers of Two chart I created on my scratch paper when I encountered the first question. Adding 128 + 64 + 32 = 224. There are 3 bits used for subnetting and that leaves 5 bits for hosts.

**2^7=128 | 2^6=64 | 2^5=32 | 2^4=16 | 2^3=8 | 2^2-=4 | 2 ^1=2 | 2^0=1**

**Step 2:** The equation for the number of hosts per subnet is 2^h – 2 where h is the number of host bits. From the chart I see that 2^5 = 32. 32-2 = 30 hosts per subnet! Too easy!

As always, let us know in the comments if you have a quicker approach.

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### 6 Responses to “Speed Subnetting, Part 2”

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Take a last mask octet(224) and subtract it from 256 = 32. Minus 2 = 30 permited host in subnet.

I do what Olga does, another neat trick if you need to convert subnet mask to an ACL is to do 255 – the octect in the mask. So say 255.255.255.224 turns into 0.0.0.31.

Also say that you have a mask like 255.255.252.0 (/22). How many subnets does that give us? 256-252 = 4 subnets, which will be 0-3 if the IP is 192.168.0.1.

[...] For Part 2 of this series – click here. [...]

I just take 256 minus the subnet. for example this one would be 256-224 which is 32 and then drop the broadcast and network and you come up with 30. Subnet 255.255.255.240 would be 256-240=16-2=14. Subnet 255.255.255.128 would be 256-128=128-2=126.

This was such a helpful way to conduct sub-netting to find the number of host.

Thanks Olga and Daniel

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