I Need help with a Subnetting Question

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Hey Forum. So, I've been studying for the CompTIA Network+ for a couple months now. But I still need help with subnetting. I downloaded this app called "Practice Subnetting" and been using it to learn to subnet. Sometimes I'm able to subnet correctly others not so much. Like this screenshot below. Why is the Subnet ID 172.25.149.184 and not 172.25.149.187. I'm able to figure out the CIDR and "Magic #" but the Subnet ID can sometimes throw me off.

Hey @Demetrius-Richardson great question and let's see if we can make sense of this as it can be very confusing.

If you look at the binary, you can narrow down the octet that has the bits we need to examine and work with to get our answer:

10101100.00011001.10010101.10111011 = IP address (172.25.149.187)
11111111.11111111.11111111.11111000 = subnet mask (255.255.255.248)

We find that it is the last octet in the IP address that we need to focus on:

10111011 (IP address 4th octet = .187)
11111000 (subnet mask 4th octet = .248)

Next, we determine how many bits we are working with:

10111 011 (.187)
11111 000 (.248) <<< these are the 3 bits (I separated the bits to make it easier)

If we apply the binary ANDing process to the last octet, we get the following:

10111 000 (.184)  <<< the subnet mask reveals the network ID/address

All binary zeros in the host portion of the IP address makes up the network ID if we add the rest of binary octets back, we can determine that value:

10101100.00011001.10010101.10111000 (172.25.149.184) 

All binary 1's in the host portion of the IP address gives us the value of the broadcast address for this network:

10111 111 (.191)

Let's restore the first three octets and convert the binary back into dotted decimal notation like we did for the network ID:

10101100.00011001.10010101.10111111 (172.25.149.191)

You will see that 172.25.149.187 is the third usable IP address in the 172.25.149.184 network:

172.25.149.184 <<<<< network address/ID
172.25.149.185
172.25.149.186
172.25.149.187 <<<<< IP address
172.25.149.188
172.25.149.189
172.25.149.190
172.25.149.191 <<<<< broadcast address

172.25.149.192 <<<<< next network address

This is time consuming, I know, however in the beginning try to work with the binary a little and this will be a great help in your training.

@wes-bryan Thanks so much. I think I got it so the step I was missing from before was the ANDing process as well as knowing you can get the broadcast ID in the binary just by converting the zeroes in the host portion to 1's. Thanks a lot you've been a huge help.

@Demetrius-Richardson you are most welcome! This is one of the most confusing topics and I am glad I could help!

Hey @wes-bryan So I did another practice question and please tell me if I'm on the right track with my train of thought in doing the binary calculations. So, I'm subnetting the IP address 10.151.220.178 with subnet mask 255.192.0.0

So, in binary the IP address will be
00001010.100010111.11011100.10110010
Subnet Mask in Binary
11111111.11000000.00000000.00000000

The Bolded numbers are the NetID and the string of zeroes are the HostID.

The Subnet ID after ANDing
00001010.10000000.00000000.00000000

So, in the decimal format the Subnet ID will be
10.128.0.0

The first host 10.128.0.1

Now for the broadcast ID I was confused about originally, but I think I got it with your help honestly.

In binary the Broadcast ID is
00001010.10111111.11111111.11111111
The bolded numbers are the NetId portion and stay the same to match the subnet mask if I'm correct. So in dotted decimal format the broadcast ID is
10.191.255.255

And the last host id is simple you just subtract 1 from the broadcast id
10.191.255.254

The next subnet would be 10.192.0.0

@Demetrius-Richardson yes you are right on! That will be a large network with a possibility of 4,194,302 hosts. And the next network with be 10.192.0.0. Great job!

@wes-bryan Awesome, Thanks again. I feel confident in my subnetting abilities now.