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VLSM question help...

Discussion in 'Routing & Switching' started by albertc30, Mar 4, 2009.

  1. albertc30

    albertc30 Kilobyte Poster

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    Hello everybody.
    I just done my CCNA module 3 and pass but have a questions that is doing my head in and for as much as i try I can't get round it.

    This is the question in my own words and diferently then in the exam so that I don't get into trouble;

    I am using the IP address 172.16.0.0 and want to create subnets using VLSM, which combination of addresses and masks will be valid?

    172.17.128.0 /18 and 172.17.128.192 /26
    172.17.160.0 /20 and 172.17.168.0 /20
    172.17.32.0 /19 and 172.17.64.128 /25
    172.17.192.0 /18 and 172.17.224.0 /19

    What is the correct answer as the exam don't allow us to know and why?
    Please help is much needed and appreciated as I did thought that I had this VLSM thing in my brain by now.

    Thanks to all.
     
    Certifications: CCNA
    WIP: 220-701 - A+
  2. BosonMichael
    Highly Decorated Member Award

    BosonMichael Yottabyte Poster

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    Didn't you pay an instructor your hard-earned money to help you learn this stuff? :blink

    Figure out which addresses fall within each address range and you'll see which ones can't be used with each other.
     
    Certifications: CISSP, MCSE+I, MCSE: Security, MCSE: Messaging, MCDST, MCDBA, MCTS, OCP, CCNP, CCDP, CCNA Security, CCNA Voice, CNE, SCSA, Security+, Linux+, Server+, Network+, A+
    WIP: Just about everything!
  3. MJN

    MJN Bit Poster

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    Hi Albert,

    Your question is not too clear, but I am guessing that by 'valid' you mean non-overlapping i.e. suitable as two separate subnets for separate networks? If so, as Michael says, the key to this is expanding each subnet shown and ensuring there is no overlap.

    Let me start you off...

    Start by expanding the first subnet out:

    /18 = an 18 bit mask, hence 2 bits taken from the third octet. The mask is therefore 11111111.11111111.11000000.00000000 or 255.255.192.0

    If the mask in the third octet is 192 then the subnet size is 256-192 = 64. Hence, the subnets are 172.17.0.0/18, 172.17.64.0/18, 172.17.128.0/18, 172.17.192.0/18

    Thus your subnet, 172.17.128.0/18 goes from 172.17.128.0 to 172.17.191.255 (i.e. 1 less than where the next one starts).

    We can immediately see the other subnet offered, 172.17.128.192/26, sits within the range above hence would not be 'valid' (assuming my interpretation of your question is as above).

    There are many ways to do this, and far quicker too, but it is always worth going back to first principles if you ever lose your way.

    Try and repeat with the three remaining options - I can confirm that only one of them is valid (just so you know - there's nothing worse than trying to chase a valid answer that doesn't exist!).

    VLSM is like that - just when you think you've sussed it it all falls over. It's a case of two steps forwards, one step back... you'll get there! Post your workings - and don't worry about getting it wrong... you'll learn far more that way!

    Mathew
     
    Certifications: CCNA
    WIP: Relaxing...

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