Subnetting Troubles..

Question: What valid host range is the IP address 172.31.163.107/22 a part of?

Answer: 172.31.160.1 through to 172.31.163.254

The way I have been doing it...
We have a Class B classful address.
11111111.11111111.11111100.00000000
Subnets available = 64
Hosts available = 1022
Will be incrementing by 4

Now I am stuck. I am looking to find an easy way to do this for the exam. Feeling very frustrated about this, thanks for any help.

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Comments

Node Man

i can offer a partial suggestion:

when for ip ranges when the 'interesting octet' is the 3rd one, you know that the fourth octet range is going to be 1-254.

the '/22' means that you have to focus on the 3rd octet.

(unless i am wrong)

oli356

I just think if it goes up with the increment of 4 in the 3rd octet.

Then 4 goes into 80 so it must go into 160. The host isn't above 164 so it must be in the 172.31.160 subnet.

You're wasting time calculating the number of subnets and hosts unless you NEED to.

Jason R

Here is how I answer this problem. When asked what is the host range for the 172.31.163.102/22 host address is part of, I break it down step by step. Immediately you should that is is a private class b address. So you can start with 172.31.0.0. You are borrowing 22 bits, which gives you a 255.255.252.0 subnet address. 256 - 252 = 4. So whats the next network address ? 172.31.4.0. Keep in mind that each following network address will increment by 4. For example, 172.31.8.0 then 172.31.12.0 and so on. Now you need to find out what subnet does 163 follow under. 4x40 = 160 and 4x41= 164. This tell us that the first address will be 172.31.160.1 - 172.31.163.254 with a broadcast address of 172.31.163.255. Guess what the next network is? 172.31.164.0. I hope that helps.

Jason R

Node Man wrote: »

i can offer a partial suggestion:

when for ip ranges when the 'interesting octet' is the 3rd one, you know that the fourth octet range is going to be 1-254.

the '/22' means that you have to focus on the 3rd octet.

(unless i am wrong)

You are correct about focusing on the third octet. Keep in mind though, the /22 means that you are borrowing 22 bits which gives you a subnet mask of 255.255.252.0

goldenlight

Odom said there is way to solve a subnet problem like this in 15 seconds.

This is what I have been doing...
Do you write out the chart ???

128 64 32 16 8 4 2 1 <

forgot

128 192 224 240 248 252 254 255<

subnet bit

chart not aligning up. But write it out and you will see