Computer networks miscellaneous

Computer networks miscellaneous

Easy Questions

Computer Networks

Computer networks miscellaneous
  1. An IP router implementing Classless Inter-domain Routing (CIDR) receives a packet with address 131.23.151.76. The router’s routing table has the following entries :

    The identifier of the output interface on which this packet will be forwarded is ______.








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    Address 131.23.151.76
    Coming to I field of given routing table.
    ⇒ 131.16.0.0/12
    131.0001 0111.151.76
    131.0001 0000.0.0 (mask bit = 12)
    131.16.0.0 (matched)
    Coming to II field of given routing table.
    ⇒ 131.28.0.0/14
    131.0001 0111.151.76
    131.0001 0100.0.0 (mask bit = 14)
    131.20.0.0 (not matched)
    Coming to III field of given routing table
    ⇒ 131.19.0.0/16
    131.0001 0111.151.76
    131.0001 0111.0.0 (mask bit = 16)
    131.23.0.0 (not matched)
    Coming to IV field of given routing table
    ⇒ 131.22.0.0/15
    131.0001 0111.151.76
    131.0001 0110.0.0 (mask bit = 15)
    131.22.0.0 (matched)
    ∵ I and IV entries are matched, we pickup longest mask bit
    ∴ Output interface identifies is 1.

    Correct Option: B

    Address 131.23.151.76
    Coming to I field of given routing table.
    ⇒ 131.16.0.0/12
    131.0001 0111.151.76
    131.0001 0000.0.0 (mask bit = 12)
    131.16.0.0 (matched)
    Coming to II field of given routing table.
    ⇒ 131.28.0.0/14
    131.0001 0111.151.76
    131.0001 0100.0.0 (mask bit = 14)
    131.20.0.0 (not matched)
    Coming to III field of given routing table
    ⇒ 131.19.0.0/16
    131.0001 0111.151.76
    131.0001 0111.0.0 (mask bit = 16)
    131.23.0.0 (not matched)
    Coming to IV field of given routing table
    ⇒ 131.22.0.0/15
    131.0001 0111.151.76
    131.0001 0110.0.0 (mask bit = 15)
    131.22.0.0 (matched)
    ∵ I and IV entries are matched, we pickup longest mask bit
    ∴ Output interface identifies is 1.

  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. 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. 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








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    Correct Option: A


Direction: Consider a network with 6 routers R1 to R6 connected with links having weights as shown in the following diagram.

  1. Suppose the weights of all unused links in the previous question are changed to 2 and the distance vector algorithm is used again until all routing tables stabilize. How many links will now remain unused?








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    As in the previous question and not used. So, we give them weight = 2, by applying same method only one link not used.

    Correct Option: B

    As in the previous question and not used. So, we give them weight = 2, by applying same method only one link not used.