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Ah ha subnetting question??

Discussion in 'General Cisco Certifications' started by 16v6n, May 12, 2007.

  1. 16v6n

    16v6n New Member

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    I've just stumbled accross this question :



    Question: What is the last valid host on the subnetwork 172.23.170.0/23?

    Answer: 172.23.171.254



    Just cant seem to work my head around this answer..

    I know the maximum subnets I can have is 2 (172.23.128.0)and (172.23.0.0) If you use subnet 0 hosts are 32766


    What I dont understand is why is the last valid host 171.254 and not 170.254?

    any ideas?

    Thanks
     
  2. hbroomhall

    hbroomhall Petabyte Poster Gold Member

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    First - welcome to CF!

    The clue here is that it is a /23 and *not* on a byte boundary!

    I don't understand what you are saying there - the two networks you say here are not *subnets* of the original. They seem to be /16 or /17 - which is 'larger' than a /23.


    This is because it is a /23 and not a /24. Write out the IP address in binary and match it with the mask of 23 '1' bits and you will see why the given answer is right.

    Harry.
     
    Certifications: ECDL A+ Network+ i-Net+
    WIP: Server+
  3. 16v6n

    16v6n New Member

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    Thanks for the reply , that helps explains things much more clearly
     
  4. Headache

    Headache Gigabyte Poster

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    The answer is simple, really.

    The subnet mask for a 23 bit mask is 255.255.254.0

    This signifies that the subnets are rising in increments of 2

    The current subnet is 170.0

    So, because the subnets are rising in increments of 2, the next subnet would be 172.0

    Therefore to find out the broadcast address of the current subnet, you simply subtract 1 from the next subnet i.e

    172.0 - 1

    This will give you 171.255

    And to find the last valid host of the current subnet, simply subtract 1 from the broadcast address i.e

    171.255 - 1

    This will give you 171.254
     
    Certifications: CCNA
    WIP: CCNP

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