VLSM - Advise needed...

Discussion in 'Routing & Switching' started by albertc30, Jan 10, 2009.

  1. albertc30

    albertc30 Kilobyte Poster

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    Hello everybody.
    Just done my first exam on CCNA Discovery module 3 chapter 2 - INTRODUCTION TO CLASSLESS ROUTING and I thought it was going really well only to find out at the end that I passed with a bad 83.3%, I can do better then this but that is why I am asking you guys for your advice.

    I had the following networks;

    192.1.1.64/26
    192.1.1.128/26
    192.1.1.192/28
    192.1.1.208/28
    192.1.1.244/30
    192.1.1.248/30

    Now the question, using VLSM what would be the result network address?

    I my calculations here we have;

    192.1.1.64 = last octet=0100 0000
    192.1.1.128= last octet=1000 0000
    192.1.1.192= last octet=1100 0000
    192.1.1.208= last octet=1101 0000
    192.1.1.244= last octet=1111 0100
    192.1.1.248= last octec=1111 1000

    As none of the the bits in the last octet are equal whe then have 192.1.1.0/26, am I wrong?

    Please help.
    Thanks guys.
     
    Certifications: CCNA
    WIP: 220-701 - A+
  2. pedwar1

    pedwar1 New Member

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    If none of the bits match in the last octet, wouldn't the summarised route would be 192.1.1.0 /24 ?
     
    Certifications: CCENT
    WIP: CCNA
  3. albertc30

    albertc30 Kilobyte Poster

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    You'd be correct. In any case I might have put this issue down maybe not in the best way possible anyways I shall ask my tutor to let me see the question again and see where did I went wrong.
    I'll post it after the class which is wednsday night.
    Cheers,
     
    Certifications: CCNA
    WIP: 220-701 - A+
  4. albertc30

    albertc30 Kilobyte Poster

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    Hello everybody.
    I am sorry but I have misled you all a bit with my question.
    Now I have the real on the paper scenario.



    The routers in the diagram use the subnet assignments shown. What is the most efficient route summary that can be configured on Router3 to advertise the internal networks to the cloud?

    The options were:
    192.1.1.0/26
    192.1.1.128/25
    192.1.1.0/23 and 192.1.1.64/23
    192.1.1.0/24
    192.1.1.0/25
    192.1.1.0/24 and 192.1.1.64/24

    I took the last octet in the already assigned networks on the routers 1 and 2 and converted them to binary and ended up with this:

    96 = 0110 0000
    104 = 0110 1000
    112 = 0111 0000
    120 = 1000 0010
    64 = 0100 0000
    80 = 0101 0000

    Now checking the bits from left to right I have none that do match, therefore my answer would be:
    192.1.1.0 with a default subnet mask for class C of 255.255.255.0.

    I actually got it correct by choosing the 192.1.1.0/25 but in my calculations I have a /24. I remember being in doubt at the time of the exam between the /24 and /25.
    Where am I going wrong here please anyone?
    Cheers,
     

    Attached Files:

    Certifications: CCNA
    WIP: 220-701 - A+
  5. BosonMichael
    Honorary Member Highly Decorated Member Award 500 Likes Award

    BosonMichael Yottabyte Poster

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    You might want to check your binary conversions on these again... ;)
     
    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!
  6. albertc30

    albertc30 Kilobyte Poster

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    BosonMichale you are the man.
    Thanks,

    Got the 120 wrong so 0111 0100 which gives me the first bit to be equal therefore /25

    I am still battling with the access list I now have the DHCP requests from the pcs on the FINANCE network going nowhere as I had applied a standard access list on port fa1/0 for inbound traffic.

    I am still trying, it has been 6 months without classes and no real certainty that we'd finish our course at the college so I am a bit lost...

    Cheers once again.
     
    Certifications: CCNA
    WIP: 220-701 - A+

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