Computer networks miscellaneous

Computer networks miscellaneous

Easy Questions

Computer Networks

Computer networks miscellaneous
  1. Which one of the following is TRUE about the interior gateway routing protocols - Routing Information Protocol (RIP) and Open Shortest Path First (OSPF)?








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    RIP uses distance vector routing and OSPF was link state rating.

    Correct Option: A

    RIP uses distance vector routing and OSPF was link state rating.

  1. Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 byte long and the transmission time for such a packet is 50 ms Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 ms. What is the maximum achievable through put in this communication?








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    Now, Transmission time for 1 packet = 50 μs
    Transmission time for 5 packet = 5 × 50 μs = 250 μs
    Propagation delay = 200 μs
    Total time = Transmission time + Propagation Delay
    = 250 + 200 = 450 μs = 450 × 10– 6 μs
    Finally, Maximum Achievable Throughput

    =
    Size of window
    		 bps
    Total time

    =
    5000
    		 = 11.11 × 10-6 bps
    450 × 10-6

    Correct Option: B

    Now, Transmission time for 1 packet = 50 μs
    Transmission time for 5 packet = 5 × 50 μs = 250 μs
    Propagation delay = 200 μs
    Total time = Transmission time + Propagation Delay
    = 250 + 200 = 450 μs = 450 × 10– 6 μs
    Finally, Maximum Achievable Throughput

    =
    Size of window
    		 bps
    Total time

    =
    5000
    		 = 11.11 × 10-6 bps
    450 × 10-6

  1. A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is








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    In the first back off race, there are two conditions 0 and 1 In the second back off race there are four conditions 0, 1, 2 and 3 Hence, winning probability of A can be calculated as
    1/2 × 3/4 + 1/2 × 1/2
    = 3/8 + 1/4
    = 5/8 = 0.625

    Correct Option: B

    In the first back off race, there are two conditions 0 and 1 In the second back off race there are four conditions 0, 1, 2 and 3 Hence, winning probability of A can be calculated as
    1/2 × 3/4 + 1/2 × 1/2
    = 3/8 + 1/4
    = 5/8 = 0.625

  1. Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48 bit jamming signal is 46.4 ms. The minimum frame size is








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    We are given with Jamming Signal = 48 bit
    Round trip propagation delay = 46.4 ps
    Now, To get actual round propagation time, subtract transmission time of 48 bit We get actual round propagation time = 51.2 µs.
    Now, the frame length must be such that to its transmission time must be more than 51.2 µs, so minimum frame length = 51.2 * 10–6 × 10 × 10–6 = 512 bits.

    Correct Option: D

    We are given with Jamming Signal = 48 bit
    Round trip propagation delay = 46.4 ps
    Now, To get actual round propagation time, subtract transmission time of 48 bit We get actual round propagation time = 51.2 µs.
    Now, the frame length must be such that to its transmission time must be more than 51.2 µs, so minimum frame length = 51.2 * 10–6 × 10 × 10–6 = 512 bits.

  1. In a packet switching network, packets are routed from source to destination along a single path having two intermediate nodes. If the message size is 24 byte and each packet contains a header of 3 byte, then the optimum packet size is








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    Packet switched network message = 24 byte
    Header size = 3
    Case 1: If packet size = 4
    Then data = 4 – 3 = 1 byte only
    So require 24 packets
    Case 2: Packet size 6
    Data = 6 – 3 = 3
    Require 8 pakets.
    Case 3:
    Packet size = 7
    Data = 7 – 3 = 4
    Require 4
    24 = 4 packets.
    Case 4: Packet size = 9
    Data = 9 – 3 = 6
    Require 6 = 24 = 4 packets.
    So min requirement is in case 4.

    Correct Option: D

    Packet switched network message = 24 byte
    Header size = 3
    Case 1: If packet size = 4
    Then data = 4 – 3 = 1 byte only
    So require 24 packets
    Case 2: Packet size 6
    Data = 6 – 3 = 3
    Require 8 pakets.
    Case 3:
    Packet size = 7
    Data = 7 – 3 = 4
    Require 4
    24 = 4 packets.
    Case 4: Packet size = 9
    Data = 9 – 3 = 6
    Require 6 = 24 = 4 packets.
    So min requirement is in case 4.