Computer networks miscellaneous

Computer networks miscellaneous

Easy Questions

Computer Networks

Computer networks miscellaneous
  1. An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are








  1. View Hint View Answer Discuss in Forum


    Correct Option: A


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








  1. View Hint View Answer Discuss in Forum

    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?








  1. View Hint View Answer Discuss in Forum

    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 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?








  1. View Hint View Answer Discuss in Forum

    Let the duration of transmission be x s.
    Transmission Network of Computer by y s
    Network initially filled with p Megabits
    Initial rate of transmission be q per second
    Now, from the above, we can easily formulate
    y + qx = xy ⇒ y = x (y – q)
    ⇒ 16 = x(10 – 2) ⇒ x = 2

    Correct Option: B

    Let the duration of transmission be x s.
    Transmission Network of Computer by y s
    Network initially filled with p Megabits
    Initial rate of transmission be q per second
    Now, from the above, we can easily formulate
    y + qx = xy ⇒ y = x (y – q)
    ⇒ 16 = x(10 – 2) ⇒ x = 2

  1. In a token ring network, the transmission speed is 107 bps and the propagation speed is 200 m/ms. The 1-bit delay in this network is equivalent to








  1. View Hint View Answer Discuss in Forum

    107 bits require time = 1 s
    ⇒ 1 Bit require = 1/107 s (unitary method)

    ⇒
    1
    		 = 10-6 / 10 s =
    1
    		 ms
    10 × 10610

    Now, 1 ms = 200 m cable
    ⇒ 1/10 ms = 20 m cable

    Correct Option: C

    107 bits require time = 1 s
    ⇒ 1 Bit require = 1/107 s (unitary method)

    ⇒
    1
    		 = 10-6 / 10 s =
    1
    		 ms
    10 × 10610

    Now, 1 ms = 200 m cable
    ⇒ 1/10 ms = 20 m cable