Computer networks miscellaneous

Computer networks miscellaneous

Easy Questions

Computer Networks

Computer networks miscellaneous
  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?








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    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








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    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

Direction: Consider a network with five nodes, N1 to N5, as shown below :

The network uses a distance vector routing protocol. Once the routes have stabilized, the distance vectors at different nodes are as following. Each distance vector is the distance of the best known path at that instance to nodes, N1 to N5, where the distance to itself is 0. Also, all links are symmetric and the cost is identical in both directions. In each round, all nodes exchange their distance vectors with their respective neighbours, Then all nodes update their distance vectors. In between two rounds, any change in cost of a link will cause the two incident nodes to change only that entry is their distance vectors.

  1. After the update in the previous question, the link N1 -N2 goes down. N2 will reflect this change immediately in its distance vector as cost ∞. After the next round of update, what will be the cost to N1 in the distance vector of N3?








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    First, as soon as N1 -N2 goes down, N2 and N1 both update that entry in their tables as infinity. So N2 at this moment will be N2 (inf,0,2,_,_). I have left blank because that details are not important.
    Now for N3 to get updated in the subsequent round it will get tables from N2 and N4 only. But first we need to find the N4 calculated in previous update. So in previous question N4 received updates from N3 and N5 which are N3 : (7, 6, 0, 2, 6),N5 : (4, 3, 6, 4, 0).
    Now this is very important as why N4 did not get updated tables from N3.
    So answer is that these tables were shared at the same moment and so in a particular round of update old values of all the tables are used and not the updated values.
    N3 was updates after it passed its old table to its neighbors as why would N4 wait for N3 to get updated first!!! So N4 will update its table (in prev question) to N4 (8,7,2,0,4).
    See here path to N1 exists via N5 and not via N3 bcoz when table was shared by N3 it contained path to N1 as 7 and N1 via N3 sums to 7 + 2 = 9. Now when N3 receives tables from N2(inf,0,_,_,_) and N4 (8,7,2,0,4). At first it will see its distance to N1 as "Inf" and NOT 3 because "inf" is the new distance with the same Next hop N2 (If next hop is same, new entry is updated even though it is larger than previous entry for the same next hop).
    But at the same time it sees distance to N1 from N4 as 8 and so updates with the value (N3 - N4 + N4 - N1 ) = (2 + 8) = 10. So N3-N1 distance in N3 (10,_,0,_,_) is 10.

    Correct Option: C

    First, as soon as N1 -N2 goes down, N2 and N1 both update that entry in their tables as infinity. So N2 at this moment will be N2 (inf,0,2,_,_). I have left blank because that details are not important.
    Now for N3 to get updated in the subsequent round it will get tables from N2 and N4 only. But first we need to find the N4 calculated in previous update. So in previous question N4 received updates from N3 and N5 which are N3 : (7, 6, 0, 2, 6),N5 : (4, 3, 6, 4, 0).
    Now this is very important as why N4 did not get updated tables from N3.
    So answer is that these tables were shared at the same moment and so in a particular round of update old values of all the tables are used and not the updated values.
    N3 was updates after it passed its old table to its neighbors as why would N4 wait for N3 to get updated first!!! So N4 will update its table (in prev question) to N4 (8,7,2,0,4).
    See here path to N1 exists via N5 and not via N3 bcoz when table was shared by N3 it contained path to N1 as 7 and N1 via N3 sums to 7 + 2 = 9. Now when N3 receives tables from N2(inf,0,_,_,_) and N4 (8,7,2,0,4). At first it will see its distance to N1 as "Inf" and NOT 3 because "inf" is the new distance with the same Next hop N2 (If next hop is same, new entry is updated even though it is larger than previous entry for the same next hop).
    But at the same time it sees distance to N1 from N4 as 8 and so updates with the value (N3 - N4 + N4 - N1 ) = (2 + 8) = 10. So N3-N1 distance in N3 (10,_,0,_,_) is 10.

  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.

  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.