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Which one is correct?

Discussion in 'Routing & Switching' started by albertc30, Aug 16, 2009.

  1. albertc30

    albertc30 Kilobyte Poster

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    Hello everybody.

    I have this question to which I got the wrong answer to, so I kindly ask anybody for an explanation please.

    The question is as follow;

    Which of the following class C IP address is reserved for broadcasts if 3 bits have been borrowed from the network portion?
    219.129.32.5
    219.129.32.63
    219.129.32.97
    219.129.32.167

    I had two answers for this, the second and last option, but last option was wrong?
    Any explanation to this please?

    If we borrow 3 bits from the network portion we get only 21 bits for the network address so, we get networks by increments of 8.

    Network 0, 8, 16, 24, 32, 40 and so on where 64 and 168 are network addresses so 63 and 167 are broadcast addresses.

    I am confused.

    Any comments are more than welcome.

    Cheers,
    Albert
     
    Certifications: CCNA
    WIP: 220-701 - A+
  2. Bonneville

    Bonneville Bit Poster

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    If you are borrowing 3 bits from the host portion (last octet), the new networks will be 0, 32, 64, 96 etc not 0, 8 etc.
     
    Certifications: MCSE, CCNA
    WIP: CCNA Security, CCNP
  3. soundian

    soundian Gigabyte Poster

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    Cross purposes?
     
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  4. soundian

    soundian Gigabyte Poster

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    upload problem.
     
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  5. soundian

    soundian Gigabyte Poster

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    Let's get binary on this.

    borrowing 3 bits from network address. must mean subnet mask is 255.255.255.224
    11111111.11111111.11111111.11100000
    Valid addresses for broadcast are:
    X.X.X.xxx11111
    Last octet, we're looking for a number with either:
    xxx(binary)+11111(binary=decimal 31)=answer
    001+31=63
    010+31=95
    100+31=159
    011+31=127
    101+31= 191
    110+31=223
    Only one there is .63
    and the question says "Which ....is..." not "which ...are..." (sometimes little clues are helpful)

    Sneaky: they've used variations of 192 so you think 168 comes next. I bet you're really used to thinking 192.168 so 219.129 might be confufuddled with 192, which might make you think 167 is a valid number for a classless network address since it's one less than a number you associate with a classful class C address.
    I think that's why I made the same mistake on first reading the question.
    edit: the subnet mask can't be 255.255.224.0
    11111111.11111111.11111000.00000000
    there's no way the 3rd octet can be 32 if the 3 host bits in it are equal to 31.
     
    Certifications: A+, N+,MCDST,MCTS(680), MCP(270, 271, 272), ITILv3F, CCENT
    WIP: Knuckling down at my new job
  6. Bonneville

    Bonneville Bit Poster

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    I was trying to point out the the subnetting was incorrect and that by using the correct subnetworks only option 2 would have been correct (by a process of elimination). You always borrow host bits and not network bits when subnetting a classful network.
     
    Certifications: MCSE, CCNA
    WIP: CCNA Security, CCNP
  7. soundian

    soundian Gigabyte Poster

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    I see what you mean now. I blame my confusion on the long conversation I had with messrs J. Smith and J. Daniel earlier in the evening :biggrin
    I've not done much classless subnetting in my studies so far so I'm still at the stage of having to work them out from first principles. Doing this would have been useful for the OP as it would have indicated to him that his assumption of /21 (despite how the question was worded) was wrong.
     
    Certifications: A+, N+,MCDST,MCTS(680), MCP(270, 271, 272), ITILv3F, CCENT
    WIP: Knuckling down at my new job
  8. zebulebu

    zebulebu Terabyte Poster

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    Yeah - they confuse the hell out of me too!
     
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  9. albertc30

    albertc30 Kilobyte Poster

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    Thanks everybody for all the comments. I must say that some of the questions on the netspace challenge are sometime quite wrong as one question I got last week about how many pins do a SATA cable has when I answered 7 apparently the correct answer was 10.

    Anyways, about my initial question, that is how it was asked, 3 bits borrowed from the network portion so I just went backwards with the bits borrowed and ended up with a /21 hence the increments of 8. I always have done subneting as you borrow bits from the host portion of the address but I guess this question confused me.

    Thanks everybody for all comments.

    Cheers
     
    Certifications: CCNA
    WIP: 220-701 - A+

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