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  1. #1
    Junior Member
    Join Date
    Nov 2008
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    another subnetting question

     
    i do not understand how to do this problem in subnetting

    12.0.0.0 break up into 25 network


    i know that you convert to binary
    128 64 32 16 8 4 2 1
    0 0 0 1 1 0 0 1

    which will give you 5 network bits

    you determine the interval by

    255.0.0.0 =11111111.00000000.00000000
    .11111000.00000000

    which will give you an interval of 8


    12.0.0.0-12.7.255.255


    my question is how do you get the 12.7.255.255 ?
    12.8.0.0-
    Last edited by huntert; March 1st, 2009 at 02:51 AM.

  2. #2
    Junior Member
    Join Date
    Mar 2009
    Posts
    1

    Reply to 12.0.0.0

    First you look at "how many sub networks you need"?
    in this case you need 25
    so
    2 to the power of what gives you 25 or more but not less
    2 to the power of 5
    2^5=32 (number of subnets)
    you cannot use 2^4 cause it is only 16 which is less than 25
    than you can use your default subnet mask for class A (cause your IP is 12.0.0.0 -class A)
    and write it in binary:
    here
    255.0.0.0 -decimal
    11111111.00000000.00000000.00000000 - binary
    so you need to borrow from hosts (0's) 5 bits
    so you will get
    11111111.11111000.00000000.00000000
    now you can look at the second octet of the subnet mask
    it goes like that -128-64-32-16-8-4-2-1
    so you can take the last 1 in your subnet mask and compare it to that decimal representation
    it is 5 from the left = 8
    so your subnet mask are:
    12.0.0.0 ------12.7.255.255-this is last one before next one (broadcast)
    12.8.0.0
    12.16.0.0
    12.24.0.0
    12.32.0.0
    and so on.................
    I hope that was helpful
    and I hope I am right.

  3. #3
    Junior Member
    Join Date
    Nov 2008
    Posts
    17
    thanks

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