Subnet Speed Practice #2: Answers

By certskills May 31, 2011 12:47

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 8 15 9 510 16 9 7 126 24 5 3 6 8 18 6 62 16 11 5 30
  Subnet ID 1st Addr. last Addr. B’cast
Subnetting Speed Practice 2
Serial Link Interview Question
By certskills May 31, 2011 12:47
Write a comment


  1. Karma March 16, 18:02

    Is 2min per question too slow?

    Reply to this comment
  2. JAMES December 21, 11:58

    Took me 12 min – 30 sec; practice makes perfect!
    Will keep at it!

    Reply to this comment
    • CCENTSkills January 18, 10:28

      James – that is the right attitude!
      Helps to do a little every day! Do it!

      Have fun, hope you’re sticking with it!

      Reply to this comment
  3. Bryant January 22, 20:36

    It took me 8 mins. I am going to keep practicing

    Reply to this comment
  4. parvesh June 21, 20:33
    is broadcast address right
    isn’t it should be

    Reply to this comment
    • Chris June 26, 07:58

      Hello Parvesh

      /29 –
      256 – 248 = 8 – Network address – 1st host – last host – Broadcast

      Reply to this comment
  5. Jorge November 4, 17:42

    Huff, took me 14:18:81

    Reply to this comment
  6. Albert November 17, 07:47

    I think the answer for the first address for sub is wrong. It must be but you put

    Are you agree with me?

    Reply to this comment
    • Wendell Odom November 17, 19:51

      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,, .3, .4, etc to, then,,, .3, etc to (last usable), the as the subnet broadcast address.

      Hope this helps.

      Reply to this comment
  7. Albert November 19, 14:56

    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.

    Reply to this comment
View comments

Write a comment

Comment; Identify w/ Social Media or Email