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Thread: Fun subnetting

  1. Senior Member Devilsbane's Avatar
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    #1

    Default Fun subnetting

    You are the network administrator for the small branch office of a large corporation. Your network consists of 85 Windows XP Professional computers and 10 Windows Server 2003 computers. You have been assigned the IP address of 204.29.99.0/24. You are not allowed to use a different IP address range. However, subnetting is permitted. All three subnets connect to a central router. An additional subnet will be added in the future.
    You must divide the IP addresses, reserving the maximum number of IP addresses for the future subnet. Subnet1 contains 10 computers, Subnet2 contains 25 computers, and Subnet3 contains 60 computers.
    How should each subnet's network IP address be configured?

    A.
    Subnet1 - 204.29.99.192/27
    Subnet2 - 204.29.99.128/26
    Subnet3 - 204.29.99.0/25


    B.
    Subnet1 - 204.29.99.240/29
    Subnet2 - 204.29.99.224/28
    Subnet3 - 204.29.99.192/27


    C
    Subnet1 - 204.29.99.224/28
    Subnet2 - 204.29.99.192/27
    Subnet3 - 204.29.99.128/26


    D
    Subnet1 - 204.29.99.192/27
    Subnet2 - 204.29.99.192/27
    Subnet3 - 204.29.99.192/27


    I am curious to hear what your answers are, I will provide the "correct answer" along with my thought process later.

    Thanks
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    #2
    You simply need to select the minimum number of bits that allows that many hosts.

    10 = 4 (/28 )
    25 = 5 (/27)
    60 = 6 (/26)

    The answer is obvious with those choices. If you have a question that has multiple answers with the same subnets, check the ranges and find the one that doesn't have any overlap or other oddities.
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  4. Senior Member Devilsbane's Avatar
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    #3
    You know I did that, and came up with 28 for the first one, but then went back and changed it to 27 for some reason. I guess I just can't count...

    And yes, the answer is C.
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  5. Senior Member knownhero's Avatar
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    #4
    Quote Originally Posted by dynamik View Post
    You simply need to select the minimum number of bits that allows that many hosts.

    10 = 4 (/28 )
    25 = 5 (/27)
    60 = 6 (/26)

    The answer is obvious with those choices. If you have a question that has multiple answers with the same subnets, check the ranges and find the one that doesn't have any overlap or other oddities.
    Thats what I did. I just remembered that its doubles when taking 1 of the mask length:

    For example:

    /32 = 1
    /31 = 2
    /30 = 4
    /29 = 8
    /28 = 16

    And worked it out that way.
    70-410 [x] 70-411 [x] 70-462[x] 70-331[x] 70-332[x]
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    #5
    Quote Originally Posted by knownhero View Post
    Thats what I did. I just remembered that its doubles when taking 1 of the mask length:

    For example:

    /32 = 1
    /31 = 2
    /30 = 4
    /29 = 8
    /28 = 16

    And worked it out that way.
    I did it similar to this, also
    No longer work in IT. Play around with stuff sometimes still and fix stuff for friends and relatives.
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    #6
    Quote Originally Posted by knownhero View Post
    Thats what I did. I just remembered that its doubles when taking 1 of the mask length:
    Yep. With binary, if you remember 1 bit = 2 values, 5 bits = 32 values and 10 bits = 1024 values, you can easily double/halve your way to any other one (that'll you'll likely be tested on, anyway).
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    #7
    It works for IPv6 subnetting also.
    No longer work in IT. Play around with stuff sometimes still and fix stuff for friends and relatives.
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