# Subnet Speed Practice #1 – Answers

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 10.1.1.1/24 8 16 8 254 172.16.203.203/24 16 8 8 254 192.168.1.161/26 24 2 6 62 10.1.99.101/22 8 14 10 1022 172.16.77.177/28 16 12 4 14
 Prob Subnet ID 1st Addr. last Addr. B’cast 1 10.1.1.0 10.1.1.1 10.1.1.254 10.1.1.255 2 172.16.203.0 172.16.203.1 172.16.203.254 172.16.203.255 3 192.168.1.128 192.168.1.129 192.168.1.190 192.168.1.191 4 10.1.96.0 10.1.96.1 10.1.99.254 10.1.99.255 5 172.16.77.176 172.16.77.177 172.16.77.190 172.16.77.191
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KevLev
May 27, 2011 3:12 pm

Doing it my way, I’m getting faster and faster. Now even though I’m getting the correct answers, SOME of the problems takes me longer than it should as in the case of
172.16.77.177/28
It takes me a long time because once I figure out my “block” size, which is (16), to count blocks up to 177 to figure out what subnet it comes from can take a while. But then again(I remembered) – instead of adding blocks of 16 numerous times until I get up to 177 to figure out what subnet it is in, I remembered a much quicker way is to just “and” the new mask, and the IP, and that will give you the subnet. Much, much quicker than adding blocks of 16 or whatever. Especially for the test.

Sara
April 6, 2020 7:06 pm

Hi Kevlev ,,
I suggest another fast method for the same problem of taking much time counting the blocks of size 16 until 177 , and that is : try subtracting : 256 – 16 in each time , and you will reach the 177 faster .
Good luck

el3ctron
November 26, 2014 9:56 am

Just that you know the multiplication table, and it is even faster than “and”

Chris
March 14, 2017 8:21 pm

For 192.168.1.161 /26, I got the subnet ID as 192.168.1.160 rather than 192.168.1.128

I did get the same broadcast address of 192.168.1.191 though. This means that the next subnet ID would be 192.168.1.192

A /26 mask provides a magic number of 32. If you subtract 32 from the subnet ID 192.168.1.192, you should get 192.168.1.160 as the correct subnet ID for this problem.

Am I overlooking something?

lyjo
March 15, 2017 9:48 am

Hi Chris,
/26 leaves 6 host bits. 2^6 is 64, not 32, for a magic number of 64. That’s the issue. Sounds like you have the process down, so re-worked with magic number 64, you’d get 192.168.1.128 as the subnet ID.
Wendell

christian_elias
December 18, 2020 4:15 pm

13 minutes 2 seconds

Sincap
November 17, 2022 7:02 am

Hi,

There is a mistake for the problem n°4 :

The 1st address for subnet 10.1.96.0 is 10.1.97.0 rather than 10.1.96.1 because the two last bits in the 3rd byte is reserved for hosts (Prefix = 22).

November 17, 2022 7:48 pm

Hi Sincap,
Thanks for the post! However, I disagree. I think you’re focusing on the host bits in octet 3, but ignoring those in octet 4. With a /22 mask, the last 10 bits are host bits. The numerically-lowest usable address is indeed 10.1.96.1, then .2, .3, etc, all the way to 10.1.96.255, then 10.1.97.0, 10.1.97.1, .2, to .255, and so on, all the way to 10.1.99.254. Hope this helps.
Wendell

GJM
December 29, 2022 12:30 am

Man I haven’t done subnetting like this in a long time. It took me a really long time to complete by memory and had to bust out the calculator a few times but I finished them all and got them all correct on the first try.

It helped once I remembered an old technique to get all the subnet ID’s quickly once. Take the subnetted portion of the mask and subtract that from 256 to get the network ID’s.

ie: 255.255.240.0; 256-240 = 16; the 3rd octet would be x.x.0.0, x.x.16.0, x.x.32.0, and so forth all the way to 256