subnetting

Discussion in 'Hardware' started by Creativeballance, Jan 30, 2008.

  1. Creativeballance

    Creativeballance Private First Class

    Couple questions, as I'm just learning about it. Why do you need to subnet?
    I understand that you do it when you need to subdivide your network, but why?

    Also, I'm looking at http://support.microsoft.com/default.aspx?scid=KB;en-us;164015&
    and I'm just not understanding why you can't use certain addresses, "Remember, again, that binary host addresses with all ones or all zeros are invalid, so you cannot use addresses with the last octet of 0, 63, 64, 127, 128, 191, 192, or 255." Can someone maybe expain it a bit differently than what is explained in the microsoft article?

    Thanks!
     
  2. PEBKAC

    PEBKAC Private First Class

    I am by no means a subnetting expert, so someone else may be able to explain this better than I can, but I'll take a shot at answering your question... In the example Microsoft provides, they are using a network of 192.168.123.0/24 (class C) and breaking down that network into smaller networks (subnets). You may want to subnet a network for a number of reasons--different physical locations, network security, etc. For example: Maybe you place all your servers on one subnet and then have your workstations at other offices on other subnets. Then you can firewall your servers from the workstations--if the network(s) containing your workstations gets owned, then there's added protection for your servers. That's just one example.

    As the article describes, octets are actually in binary. For the last octet of a class C, you are looking at 00000000 through 11111111 (or 0 through 255 in decimal).

    For the subnet examples they provide...

    192.168.123.1-62
    192.168.123.65-126
    192.168.123.129-190
    192.168.123.193-254

    0, 64, 128, & 192 are the first addresses of each network and 63, 127, 191, and 255 are the last addresses of each network (broadcast).

    Hope that helps.
     
  3. Creativeballance

    Creativeballance Private First Class

    Ok, so we've got these 4 networks we're borrowing 2 bits, and they've got:
    A) 00000000 = .0
    B) 01000000 = .64
    C) 10000000 = .128
    D)11000000 = .192
    So, A) 0 is invalid so the range begins at 1, but shouldn't it run to .63, with .64 as the broadcast? If it does end at .62 like you said, what happened to .63?
    I'm trying to get this straight in my mind.
     
  4. lbmest

    lbmest MajorGeek

    You are close but your assumption that the first number is invalid is incorrect.

    A) 00000000 = .0 (Network address unable to be assigned to a host.)
    B) 01000000 = .64 (Network address unable to be assigned to a host.)
    C) 10000000 = .128 (Network address unable to be assigned to a host.)
    D) 11000000 = .192 (Network address unable to be assigned to a host.)

    E) .1 - .62 (Valid range of host addresses.)
    F) .65 - .126 (Valid range of host addresses.)
    G) .129 - .190 (Valid range of host addresses.)
    H) .193 - .254 (Valid range of host addresses.)

    I) 00111111 = .63 (Broadcast address for network unable to be assigned to a host.)
    J) 01111111 = .127 (Broadcast address for network unable to be assigned to a host.)
    K) 10111111 = .191 (Broadcast address for network unable to be assigned to a host.)
    L) 11111111 = .255 (Broadcast address for network unable to be assigned to a host.)

    Does this make sense?
     
  5. Creativeballance

    Creativeballance Private First Class

    Ok, yeah it does. Thanks!
    Now, in practical sense...
     

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