IP Grouping

Discussion in 'MCSA' started by RogueIT, Feb 25, 2009.

  1. RogueIT

    RogueIT Guest

    While answering a subnetting question I got this one wrong. Can someone
    explain the answer to me?

    TIA
    Rogue.
    ******************************************
    Question:
    Suppose that your network has an Internet Assignted Numbers Authority
    (IANA)-assigned Class C CIDR block of eight addresses. Which subnet mask
    groups these IP address into a single subnet.

    255.255.240.0
    255.255.248.0
    255.255.255.0
    255.255.255.248
     
    RogueIT, Feb 25, 2009
    #1
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  2. The question is, arguably, poorly written.

    The apparently obvious answer is to the question: What subnet provides for
    eight hosts, and that answer is 255.255.255.248.

    But I suspect the intent of this question is to ask: Suppose that your
    organization has an Internet Assigned Numbers Authority (IANA)-assigned
    block of eight Class C CIDR networks. Which subnet mask groups these eight
    networks into a single supernet.

    And the answer to *that* question is: 255.255.248.0


    --
    Lawrence Garvin, M.S., MCITP(x2), MCTS(x5), MCP(x7), MCBMSP
    Principal/CTO, Onsite Technology Solutions, Houston, Texas
    Microsoft MVP - Software Distribution (2005-2009)

    MS WSUS Website: http://www.microsoft.com/wsus
    My Websites: http://www.onsitechsolutions.com;
    http://wsusinfo.onsitechsolutions.com
    My MVP Profile: http://mvp.support.microsoft.com/profile/Lawrence.Garvin
     
    Lawrence Garvin \(MVP\), Feb 25, 2009
    #2
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  3. RogueIT

    RogueIT Guest

    And the answer to *that* question is: 255.255.248.0

    That is the correct answer but how did you come to get 248.0?
     
    RogueIT, Feb 25, 2009
    #3
  4. Consider this block of contiguous "Class C CIDR" networks:

    192.168.64.0
    192.168.65.0
    192.168.66.0
    192.168.67.0
    192.168.68.0
    192.168.69.0
    192.168.70.0
    192.168.71.0

    any one of those networks would have a subnet mask of 255.255.255.0 -- a
    standard Class C network.

    increasing the fourth octet pushes us into subnets-of-one-Class-C territory,
    and that's not where we want to go, we need to make the adjustment the other
    direct, in the third octet, to combine individual networks into a
    supernetwork.

    Using simple brute force methodology and working backwards:

    255.255.255.0 = 1 Class C network (e.g. 192.168.64.0 with 256-2 hosts in
    the network)
    255.255.254.0 = 2 Class C networks (192.168.64.0 and 192.168.65.0, with
    512-2 hosts in the supernet)
    255.255.252.0 = 4 Class C networks (192.168.64.0 thru 192.168.67.0, with
    1024-2 hosts in the supernet)
    255.255.248.0 = 8 Class C networks (192.168.64.0 thru 192.168.71.0, with
    2048-2 hosts in the supernet)

    Or, you can also use some basic arithmetic...

    Just as 256 - 8 = subnet mask of 248 for 8 *hosts* in a single Class C
    network,
    so does 256 - 8 = subnet mask of 248 one octet up, to get 8 *networks*.




    --
    Lawrence Garvin, M.S., MCITP(x2), MCTS(x5), MCP(x7), MCBMSP
    Principal/CTO, Onsite Technology Solutions, Houston, Texas
    Microsoft MVP - Software Distribution (2005-2009)

    MS WSUS Website: http://www.microsoft.com/wsus
    My Websites: http://www.onsitechsolutions.com;
    http://wsusinfo.onsitechsolutions.com
    My MVP Profile: http://mvp.support.microsoft.com/profile/Lawrence.Garvin
     
    Lawrence Garvin \(MVP\), Feb 25, 2009
    #4
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