Subnet Speed Practice #2: Answers
By certskills
This post makes no sense without reading this post first. The earlier post lists 5 subnetting problems, and tells you to time yourself. The answers are below the fold in this post. Don’t look til you try it for yourself! Post questions if you have them.
| Problem | Network Bits | Subnet Bits | Host Bits | # Hosts |
| 10.1.1.1/23 | 8 | 15 | 9 | 510 |
| 172.16.203.203/25 | 16 | 9 | 7 | 126 |
| 192.168.1.161/29 | 24 | 5 | 3 | 6 |
| 10.1.99.101/26 | 8 | 18 | 6 | 62 |
| 172.16.77.177/27 | 16 | 11 | 5 | 30 |
| Subnet ID | 1st Addr. | last Addr. | B’cast | |
| 1 | 10.1.0.0 | 10.1.0.1 | 10.1.1.254 | 10.1.1.255 |
| 2 | 172.16.203.128 | 172.16.203.129 | 172.16.203.254 | 172.16.203.255 |
| 3 | 192.168.1.160 | 192.168.1.161 | 192.168.1.166 | 192.168.1.167 |
| 4 | 10.1.99.64 | 10.1.99.65 | 10.1.99.126 | 10.1.99.127 |
| 5 | 172.16.77.160 | 172.16.77.161 | 172.16.77.190 | 172.16.77.191 |
By certskills
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Is 2min per question too slow?
Took me 12 min – 30 sec; practice makes perfect!
Will keep at it!
James – that is the right attitude!
Helps to do a little every day! Do it!
Have fun, hope you’re sticking with it!
Wendell
It took me 8 mins. I am going to keep practicing
192.168.1.161/29
is broadcast address right
isn’t it should be 192.168.1.163
Hello Parvesh
/29 – 255.255.255.248
256 – 248 = 8
192.168.1.160 – Network address
192.168.1.161 – 1st host
192.168.1.166 – last host
192.168.1.167 – Broadcast
Huff, took me 14:18:81
Hey – you will improve! Like getting back into shape! You can do it!
I think the answer for the first address for sub 10.1.0.0/23 is wrong. It must be 10.1.1.0 but you put 10.1.0.1
Are you agree with me?
Hi Albert,
Thanks for the post! I disagree, but that’s what these exercises are great for – give it a try, struggle a bit, and learn.
Thinking it decimal, whatever the subnet ID is, add 1 to the 4th octet to get the numerically lowest address.
Conceptually, I think you’re thinking about the /23 mask as creating 1 host bit in the 3rd octet, but ignoring the 8 host bits in the 4th octet.
The usable addresses in the subnet are 10.1.0.1, 10.1.0.2, .3, .4, etc to 10.1.0.255, then 10.1.1.0, 10.1.1.1, 10.1.1.2, .3, etc to 10.1.1.254 (last usable), the 10.1.1.255 as the subnet broadcast address.
Hope this helps.
Wendell
Dear Wendell,
I just noticed my logic was totally upside down, so yes your explanation helps.
Now it’s burned in my memory : ” Add 1 to the 4th octet whatever there is before to get the first address”.
Thank you for help.
You’re quite welcome, Albert.