Functions of OSI and TCP/IP Layers Q.14

  • Option : A
  • Explanation : For 1st packet,
    128.75.43.16 BITWISE-AND 255.255.255.0
    = 128.75.43.0
    128.75.43.16 BITWISE-AND 255.255.255.128
    = 128.75.43.0
    Now, since both these subnet masks are producing the same Network ID, hence The one with greater number of ones will be selected, and the packet will be forwarded there. Hence packet 1 will be forwarded to Eth1.
    For 2nd packet,
    192.12.17.10 BITWISE-AND 255.255.255.0
    = 192.12.17.0
    {Does not match with any of the network addresses}
    192.12.17.10 BITWISE-AND 255.255.255.128
    = 192.12.17.0
    {Does not match with any of the network addresses}
    192.12.17.10 BITWISE-AND 255.255.255.255
    =192.12.17.10
    {Does not match with any of the network addresses}
    Hence, default interface must be selected for packet 2, i.e Interface Eth2.
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