Subnet Speed Practice #1 – Answers

certskills
By certskills May 20, 2011 08:19

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
Subnetting Speed Practice #1
Subnetting Speed Practice at CCENTSkills
certskills
By certskills May 20, 2011 08:19
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4 Comments

  1. KevLev May 27, 15:12

    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.

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  2. el3ctron November 26, 09:56

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

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  3. Chris March 14, 20:21

    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?

    Reply to this comment
    • CCENTSkills March 15, 09:48

      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

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