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Thread: Subnetting Question

  1. #1

    Subnetting Question

    Suppose the administrator of a class B IP network wishes to divide the network to create subnets with up to 240 hosts on each subnet. Then what would be a suitable net/host address boundary and a subnet mask.

    My Solution:

    240 host = log2(240) = 8

    therefore there are 32 bits in ip address and we need the 8 bits to accommodate 240 hosts.

    Therefore 32-8 = 24. Therefore we need the first 24 bits for the network park, which suggest the
    host/net boundary of /24 is suitable.

    This means that the first 24 bits are all 1s and therefore we get the following:
    11111111 11111111 11111111 = 255.255.255.

    Therefore our subnet mask becomes 255.255.255.0 with a boundary of /24 on class B IP network.

    HI Can anybody check my solution and please let me know whether it is right or wrong.

    Thank you

  2. #2
    You can use at least 254 different networks, like so:

    network 172.16.1.0/24
    broadcast 172.16.1.255

    network 172.16.2.0/24
    broadcast 172.16.2.255

    etc, etc.

    In other words, you are on the right track.
    arrogance breeds ignorance

    Screaming Electron, Full of BSD Goodness

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