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Data Communication and Computer Networks - Functions of OSI and TCP/IP Layers

Avatto > > UGC NET COMPUTER SCIENCE > > PRACTICE QUESTIONS > > Data Communication and Computer Networks > > Functions of OSI and TCP/IP Layers

1. In a data link protocol, the frame delimiter flag is given by 0111. Assuming that bit stuffing is employed, the transmitter sends the data sequence 01110110 as

  • Option : D
  • Explanation : Delimiter flag: 0111
    Data sequence: 01110110
    So, for a flag of 4 bits we will compare data sequence with pattern of 3 bits i.e. 011. Data sequence : 0 1 1 1 0 1 1 0
    The underlined bits are found matched, hence, 0 is stuffed.
    Thus resulting in the data sequence as 0110101100 which is option (d).
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2. A bit-stuffing based framing protocol uses an 8- bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is

  • Option : B
  • Explanation : The given 8-bit delimiter pattern is 01111110.
    The output bit-string after stuffing is 01111100101.
    The above bold underline bit is stuffed bit.
    So input bit-string must be 0111110101.
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3. A computer on a 10 Mbps network is regulated by a token bucket. The token bucket is filled at a rate of 2 Mbps. It is initially filled to capacity with 16 Megabits. What is the maximum duration for which the computer can transmit at the full 10 Mbps?

  • Option : B
  • Explanation : Let duration is t sec.
    = 16 + 2t = 10 * t
    = t = 2 sec
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4. A host is connected to a Department network which is par t of a University network. The University network, in turn, is part of the Internet . The largest network in which the Ethernet address of the host is unique is

  • Option : D
  • Explanation : Ethernet address is basically the MAC address, which is supposed to be unique to a NIC. Thus it is unique over the Internet.
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5. A router has two full-duplex Ethernet interfaces each operating at 100 Mb/s. Ethernet frames are at least 84 bytes long (including the Preamble and the Inter-Packet-Gap).
The maximum packet processing time at the router for wirespeed forwarding to be possible is (in micro-seconds)

  • Option : B
  • Explanation : Minimum transmission time
    = (84 bytes) / (2*100 bps)
    = (84 * 8 bits) / (2*100 bps)
    = 3.36 seconds
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