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Thread: Subnetting help

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  1. #1

    Subnetting help


    Right, im going over subnets again after doing them last year, i thought i had remembered it until the Cisco testerererer cd hit me with this: (I have put this in code tags as it allows spacing)

    Code:
            Given the IP address 172.156.100.100 & a subnet mask of 255.255.255.224 what is the broadcast address:
    
            A:172.156.100.127
    
            B: 172.156.100.255
    
            C: 172.156.32
    
            D: 172.156.100.128
     
    So I have done my working like this:
    
    IP = 172.156.100.100 /27
    
    .                            | N | N | N | S |  H  |
    Broadcast:                               011 11111    = 172.156.100.127
    
    But the correct answer is D: 172.156.100.128??
    
    I have tried to work through the example given again using ANDing:
    
    Subnet mask = 225.225.225.224
    IP                 = 172.156.100.100
    
    Only showing 'The intresting Octet':
    
    Subent =    224   =   11100000
    IP     =    100   =   01100100
                           ------------
    Network address =     01100000 = 96
    
    Therefore;
    
    Broadcast       =     01111111 = 127
    Could the Cisco CD have a little error, or a i doing this the wrong way?

    I have worked through another example from the same CD:

    What is the broadcast for IP subnet 192.168.100.32, given a mask of 255.255.255.244.

    So, 192.168.100.32 /27

    Code:
            | S  |    H    |
    224      111  00000
             001  00000    - Network
             001  11111    - Broadcast = 192.168.100.63
    Which, according to the CD is correct. Is the CD wrong? This is confusing me a lil.

    Thanks much

  2. #2
    May be you are right.There will be some mistake in cisco cd.

    Put your IP address and Subnet Mask.The result will be the same.

    http://www.subnetonline.com/subcalc/subnet8.html

  3. #3
    Linux Newbie
    Join Date
    Mar 2005
    Location
    Romania
    Posts
    186
    Yes, the CD is wrong. The broadcast address for 172.156.100.100/27 is 172.156.100.127... 172.156.100.128 is the network address for the following subnet.
    You can only be young once. But you can always be immature.

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