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SET 5


  Question 1

In Ethernet when Manchester encoding is used, the bit rate is:

A : Half the baud rate.
B : Twice the baud rate.
C : Same as the baud rate.
D : None of the above


  •  
    .

     Correct answer is :A

     Solution :
      In Manchester encoding, the bitrate is half of the baud rate.

  •   Question 2

    Which one of the following uses UDP as the transport protocol?

    A : HTTP
    B : Telnet
    C : DNS
    D : SMTP


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    .

     Correct answer is :C

     Solution :
      DNS uses UDP as the transport protocol

  •   Question 3

    There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that ONLY one station transmits in a given time slot?

    A : (1-p)n-1
    B : np(1-p)n-1
    C : p(1-p)n-1
    D : 1-(1-p)n-1


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     Correct answer is :B

     Solution :
      The probability that a particular station transmits and no body else transmits = p*(1-p)^(n-1)
    The probability that any station can transmit = n*(probability that a particular station transmits) = n*p*(1-p)^(n-1).

  •   Question 4

    In a token ring network the transmission speed is 10^7 bps and the propagation speed is 200 metres/micro second. The 1-bit delay in this network is equivalent to:

    A : 500 metres of cable.
    B : 200 metres of cable.
    C : 20 metres of cable.
    D : 50 metres of cable.


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     Correct answer is :C

     Solution :
      Transmission delay for 1 bit t = 1/(10^7) = 0.1 micro seconds. 200 meters can be traveled in 1 micro second. Therefore, in 0.1 micro seconds, 20 meters can be traveled.

  •   Question 5

    The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?

    A : 62 subnets and 262142 hosts
    B : 64 subnets and 262142 hosts
    C : 62 subnets and 1022 hosts
    D : 64 subnets and 1024 hosts


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    .

     Correct answer is :C

     Solution :
      Maximum number of subnets = 2^6-2 =62.
    Note that 2 is subtracted from 2^6. The RFC 950 specification reserves the subnet values consisting of all zeros (see above) and all ones (broadcast), reducing the number of available subnets by two.
    Maximum number of hosts is 2^10-2 = 1022.
    2 is subtracted for Number of hosts is also. The address with all bits as 1 is reserved as broadcast address and address with all host id bits as 0 is used as network address of subnet.
    In genera

  •   Question 6

    The message 11001001 is to be transmitted using the CRC polynomial x^3 + 1 to protect it from errors. The message that should be transmitted is:

    A : 11001001000
    B : 11001001011
    C : 11001010
    D : 110010010011


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    .

     Correct answer is :B

     Solution :
      The polynomial x^3+1 corresponds to divisor is 1001.
    11001001 000  <--- input right padded by 3 bits
    1001          <--- divisor
    01011001 000  <---- XOR of the above 2
     1001         <--- divisor
    00010001 000
       1001
    00000011 000
          10 01
    00000001 010
           1 001
    00000000 011 <------- remainder (3 bits)

    After dividing the given message 11001001 by 1001, we get the remainder as 011 which is the CRC. The transmitted data is, message + CRC which is 11001001 011.

  •   Question 7

    The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocolis used, is:



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


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     Correct answer is :C

     Solution :
      Distance between stations = L KM
    Propogation delay per KM = t seconds
    Total propagation delay = Lt seconds
    Frame size = k bits
    Channel capacity = R bits/second
    Transmission Time = k/R
    Let n be the window size.
    UtiliZation = n/(1+2a) where a = Propagation time / transmission time
    = n/[1 + 2LtR/k]
    = nk/(2LtR+k)
    For maximum utilization: nk = 2LtR + k
    Therefore, n = (2LtR+k)/k


  •   Question 8

    Match the following:



    A : P 2 Q 1 R 3 S 5
    B : P 1 Q 4 R 2 S 3
    C : P 1 Q 4 R 2 S 5
    D : P 2 Q 4 R 1 S 3


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     Correct answer is :B

     Solution :
      SMTP is an application layer protocol used for e-mail transmission.
    TCP is a core transport layer protocol.
    BGP is a network layer protocol backing the core routing decisions on the Internet
    PPP is a data link layer protocol commonly used in establishing a direct connection between two networking nodes.

  •   Question 9

    What is the maximum size of data that the application layer can pass on to the TCP layer below?

    A : Any size
    B : 216 bytes - size of TCP header
    C : 216 bytes
    D : 1500 bytes


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     Correct answer is :A

     Solution :
      The default TCP Maximum Segment Size is 536. Where a host wishes to set the maximum segment size to a value other than the default, the maximum segment size is specified as a TCP option, initially in the TCP SYN packet during the TCP handshake. Because the maximum segment size parameter is controlled by a TCP option, a host can change the value in any later segment.

  •   Question 10

    Which of the following system calls results in the sending of SYN packets?

    A : socket
    B : bind
    C : listen
    D : connect


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     Correct answer is :D

     Solution :
      socket() creates a new socket of a certain socket type, identified by an integer number, and allocates system resources to it.
    bind() is typically used on the server side, and associates a socket with a socket address structure, i.e. a specified local port number and IP address.
    listen() is used on the server side, and causes a bound TCP socket to enter listening state.
    connect() is used on the client side, and assigns a free local port number to a socket. In case of a TCP socket, it it causes an attempt to establish a new TCP connection. When connect() is called by client, following three way handshake happens to establish the connection in TCP. 1) The client requests a connection by sending a SYN (synchronize) message to the server. 2) The server acknowledges this request by sending SYN-ACK back to the client. 3) The client responds with an ACK, and the connection is established.

  •   Question 11

    In the slow start phase of the TCP congestion control algorithm, the size of the congestion window

    A : does not increase
    B : increases linearly
    C : increases quadractically
    D : increases exponentially


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     Correct answer is :D

     Solution :
      Although the name is slow start, during the slow start phase, window size is increased by the number of segments acknowledged, which means window size grows exponentially. This happens until either an acknowledgment is not received for some segment or a predetermined threshold value is reached.

  •   Question 12

    If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?

    A : 1022
    B : 1023
    C : 2046
    D : 2047


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     Correct answer is :C

     Solution :
      The binary representation of subnet mask is 11111111.11111111.11111000.00000000. There are 21 bits set in subnet. So 11 (32-21) bits are left for host ids. Total possible values of host ids is 2^11 = 2048. Out of these 2048 values, 2 addresses are reserved. The address with all bits as 1 is reserved as broadcast address and address with all host id bits as 0 is used as network address of subnet. In general, the number of addresses usable for addressing specific hosts in each network is always 2

  •   Question 13

    A computer on a 10Mbps network is regulated by a token bucket. The token bucket is filled at a rate of 2Mbps. It is initially filled to capacity with 16Megabits. What is the maximum duration for which the computer can transmit at the full 10Mbps?

    A : 1.6 seconds
    B : 2 seconds
    C : 5 seconds
    D : 8 seconds


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     Correct answer is :B

     Solution :
      New tokens are added at the rate of r bytes/sec which is
    2Mbps in the given question.
    Capacity of the token bucket (b) = 16 Mbits
    Maximum possible transmission rate (M) = 10Mbps
    So the maximum burst time = b/(M-r) = 16/(10-2) = 2 seconds

  •   Question 14

    A client process P needs to make a TCP connection to a server process S. Consider the following situation: the server process S executes a socket(), a bind() and a listen() system call in that order, following which it is preempted. Subsequently, the client process P executes a socket() system call followed by connect() system call to connect to the server process S. The server process has not executed any accept() system call. Which one of the following events could take place?

    A : connect () system call returns successfully
    B : connect () system call blocks
    C : connect () system call returns an error
    D : connect () system call results in a core dump


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     Correct answer is :C

     Solution :
      Since accept() call is not executed then connect () gets no response for a time stamp to wait & then return no response server error.

  • MY REPORT
    TOTAL = 14
    ANSWERED =
    CORRECT / TOTAL = /14
    POSITIVE SCORE =
    NEGATIVE SCORE =
    FINAL SCORE =