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SET 6


  Question 1

Packets of the same session may be routed through different paths in

A : TCP, but not UDP
B : TCP and UDP
C : UDP, but not TCP
D : Neither TCP, nor UDP


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     Correct answer is :B

     Solution :
      Packet is the Network layer Protocol Data Unit (PDU). TCP and UDP are Transport layer protocols. Packets of same session may be routed through different routes. Most networks donít use static routing, but use some form of adaptive routing where the paths used to route two packets for same session may be different due to congestion on some link, or some other reason.

  •   Question 2

    The address resolution protocol (ARP) is used for

    A : Finding the IP address from the DNS
    B : Finding the IP address of the default gateway
    C : Finding the IP address that corresponds to a MAC address
    D : Finding the MAC address that corresponds to an IP address


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     Correct answer is :D

     Solution :
      Address Resolution Protocol (ARP) is a request and reply protocol used to find MAC address from IP address.

  •   Question 3

    The maximum window size for data transmission using the selective reject protocol with n-bit frame sequence numbers is:

    A : 2n
    B : 2n-1
    C : 2n-1
    D : 2n-2


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     Correct answer is :B

     Solution :
      In Selective Reject (or Selective Repeat), maximum size of window must be half of the maximum sequence number.

  •   Question 4

    In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge-routing?

    A : For shortest path routing between LANs
    B : For avoiding loops in the routing paths
    C : For fault tolerance
    D : For minimizing collisions


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     Correct answer is :B

     Solution :
      The main idea for using Spanning Trees is to avoid loops.

  •   Question 5

    An organization has a class B network and wishes to form subnets for 64 departments. The subnet mask would be

    A : 255.255.0.0
    B : 255.255.64.0
    C : 255.255.128.0
    D : 255.255.252.0


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     Correct answer is :D

     Solution :
      The size of network ID is 16 bit in class B networks. So bits after 16th bit must be used to create 64 departments. Total 6 bits are needed to identify 64 different departments. Therefore, subnet mask will be 255.255.252.0.

  •   Question 6

    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 bytes and each packet contains a header of 3 bytes, then the optimum packet size is:

    A : 4
    B : 5
    C : 6
    D : 7


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     Correct answer is :D


  •   Question 7

    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

    A : 94
    B : 416
    C : 464
    D : 512


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     Correct answer is :C

     Solution :
      Transmission Speed = 10Mbps.
    Round trip propagation delay = 46.4 ms
    The minimum frame size = (Round Trip Propagation Delay) * (Transmission Speed) = 10*(10^6)*46.4*(10^-3) = 464 * 10^3 = 464 Kbit

  •   Question 8

    For which one of the following reasons does Internet Protocol (IP) use the timeto- live (TTL) field in the IP datagram header

    A : Ensure packets reach destination within that time
    B : Discard packets that reach later than that time
    C : Prevent packets from looping indefinitely
    D : Limit the time for which a packet gets queued in intermediate routers


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     Correct answer is :C

     Solution :
      Time to live (TTL) or hop limit is a mechanism that limits the lifespan or lifetime of data in a computer or network. TTL may be implemented as a counter or timestamp attached to or embedded in the data. Once the prescribed event count or timespan has elapsed, data is discarded. In computer networking, TTL prevents a data packet from circulating indefinitely.

  •   Question 9

    Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use?

    A : 20
    B : 40
    C : 160
    D : 320


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     Correct answer is :B

     Solution :
      Round Trip propagation delay = 80ms
    Frame size = 32*8 bits
    Bandwidth = 128kbps
    Transmission Time = 32*8/(128) ms = 2 ms
    Let n be the window size.
    UtiliZation = n/(1+2a) where a = Propagation time / transmission time
    = n/(1+80/2)
    For maximum utilization: n = 41 which is close to option (B)

  •   Question 10

    Two computers C1 and C2 are configured as follows. C1 has IP address 203.197.2.53 and netmask 255.255.128.0. C2 has IP address 203.197.75.201 and netmask 255.255.192.0. which one of the following statements is true?

    A : C1 and C2 both assume they are on the same network
    B : C2 assumes C1 is on same network, but C1 assumes C2 is on a different network
    C : C1 assumes C2 is on same network, but C2 assumes C1 is on a different network
    D : C1 and C2 both assume they are on different networks


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     Correct answer is :C

     Solution :
     
    Network Id of C1 = bitwise '&' of IP of C1 and subnet mask of C1 
                     = (203.197.2.53) & (255.255.128.0) 
                     = 203.197.0.0
    C1 sees network ID of C2 as bitwise '&' of IP of C2 and subnet mask of C1 
                    = (203.197.75.201) & (255.255.128.0) 
                    = 203.197.0.0
    which is same as Network Id of C1.
    
    Network Id of C2 = bitwise '&' of IP of C2 and subnet mask of C2
                     = (203.197.75.201) & (255.255.192.0) 
                     = 2

  •   Question 11

    Station A needs to send a message consisting of 9 packets to Station B using a sliding window (window size 3) and go-back-n error control strategy. All packets are ready and immediately available for transmission. If every 5th packet that A transmits gets lost (but no acks from B ever get lost), then what is the number of packets that A will transmit for sending the message to B?

    A : 12
    B : 14
    C : 16
    D : 18


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     Correct answer is :C

     Solution :
      Total 16 packets are sent. See following table for sequence of events. Since go-back-n error control strategy is used, all packets after a lost packet are sent again.
    Sender      Receiver
     1            
     2            1
     3            2
     4            3
     5            4
     6           
     7            6
                  7             
              [Timeout for 5] 
    
     5
     6            5
     7            6
     8
     9           
                  8
                  9
              [Timeout for 7]
    
     7       

  •   Question 12

    Consider the diagram shown below where a number of LANs are connected by (transparent) bridges. In order to avoid packets looping through circuits in the graph, the bridges organize themselves in a spanning tree. First, the root bridge is identified as the bridge with the least serial number. Next, the root sends out (one or more) data units to enable the setting up of the spanning tree of shortest paths from the root bridge to each bridge. Each bridge identifies a port (the root port) through which it will forward frames to the root bridge. Port conflicts are always resolved in favour of the port with the lower index value. When there is a possibility of multiple bridges forwarding to the same LAN (but not through the root port), ties are broken as follows: bridges closest to the root get preference and between such bridges, the one with the lowest serial number is preferred.

    For the given connection of LANs by bridges, which one of the following choices represents the depth first traversal of the spanning tree of bridges?




    A : B1, B5, B3, B4, B2
    B : B1, B3, B5, B2, B4
    C : B1, B5, B2, B3, B4
    D : B1, B3, B4, B5, B2


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     Correct answer is :A


  •   Question 13

    Consider the data given in above question. Consider the correct spanning tree for the previous question. Let host H1 send out a broadcast ping packet. Which of the following options represents the correct forwarding table on B3?



    A : A
    B : B
    C : C
    D : D


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     Correct answer is :A


  • MY REPORT
    TOTAL = 13
    ANSWERED =
    CORRECT / TOTAL = /13
    POSITIVE SCORE =
    NEGATIVE SCORE =
    FINAL SCORE =