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A satellite link with bandwidth 20 kbps has a propagation time of 300 ...

  • Q. A satellite link with bandwidth 20 kbps has a propagation time of 300 ms. The transmitter employs Go-back-3 ARQ scheme and each frame is assumed to be of 500 bytes. What is the possible throughput?
  • filter_dramaExplanation
    Answer is : C

    The answer to the question is 15 Kbps

    • It uses the sliding window protocol for the transmission of data. 
    • The question takes into consideration the variant of sliding window protocol namely GO BACK N ARQ. In this protocol, the sender can have up to N packets unacknowledged that are still remaining in the pipeline.
    • The receiver only sends cumulative acknowledgments. In case of encountering an error, the sender has to resend all the data frames following the error.

    According to the question:

    • The data rate of the link is 20 Kbps and the propagation delay = 300 ms

    So, the time required to transmit 500 bytes long data will be given by

    • Transmission Time \( t = {Number \,of \,bits \,to\, be\, transmitted \over data\, rate\, of\, the\, link }\)

                                                 = \((500~* ~8 ~bits) \over 20 Kbps\) =  200 ms

    • Now, the propagation delay is given as d = 300 ms

    So the efficiency of the link is given by:


    Where N = window size

    • \(E = {3 ~* ~200ms \over (200~ms~+~2~*~300ms)}\) = 0.75

    So, the maximum data rate attainable = 0.75 * 20 Kbps = 15 Kbps
    So, the answer will be 15Kbps.


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Similar Questions

  • 1. Consider two database relations R and S having 3 tuples in R and 2 tuples in S. What is the maximum number of tuples that could appear in the natural join of R and S?
  • filter_dramaExplanation
    Answer is : D


    If there are two relations A and B with m and n number of tuples respectively. Then maximum number of tuples in their natural join will be m*n. Maximum tuples will be when one attribute will contain all same values as the other relation having same attribute.


    Given: two relations R and S having tuples 3 and 2 respectively. 

    So, their natural join will contain maximum of 3 * 2= 6 tuples.


    Consider table R (a, b) as ;

    a b
    1 3
    1 5


    Relation S(a, c) as :

    a c
    1 2
    1 4
    1 6

    Here, the natural join of R and S will be :

    a b c
    1 3 2
    1 3 4
    1 3 6
    1 5 2
    1 5 4
    1 5 6
    So, it contains total of 6 tuples.
  • 2. Which one of the following protocols is NOT used to resolve one form of address to another one? 
  • filter_dramaExplanation
    Answer is : C

    For, this we must know what each of these protocol does.


    DNS stands for domain name server. DNS converts domains to IP address. These are internet’s equivalent of a phone book.


    ARP stands for address resolution protocol.  This is a protocol uses to convert IP address into MAC address. This protocol operates below the network layer as part of interface between OSI network and link layer.


    RARP stands for reverse address resolution protocol. This is used to convert a particular MAC address into a corresponding IP address.


    DHCP stands for dynamic host configuration protocol. It enables a server to automatically assign an IP address to a computer from a defined range of numbers. It is not used to resolve one address form to another.

  • 3. A computer network uses polynomials over GF (2) for error checking with 8 bits as information bits and uses x3 +x +1 as the generator polynomial to generate the check bits. In this network. The message 01011011 is transmitted as
  • filter_dramaExplanation
    Answer is : C


    Append no. of 0’s equal to the degree of the generator polynomial to the original data and then divide the message with the generator polynomial. Remainder obtained from this will be added with the original message and transmitted as CRC.


    Generator is given as x3 +x +1

    Degree of the generator is 3.

    So, 3 0’s is added to the data and new data will be 01011011000 and it is divided by 1011.


    Here remainder is 101. Append this to the original message. The message will become 01011011101. So, the original message 01011011 will be transmitted as 01011011101.

  • 4. Which of the following is not vulnerable to replay attack?
  • filter_dramaExplanation
    Answer is : B

    A one-time password (OTP), also known as a one-time pin, is a password that is valid for only one login session or transaction, on a computer system or other digital device.

    The most important advantage that is addressed by OTPs is that they are not vulnerable to replay attacks. This means that a potential intruder who manages to record an OTP that was already used to log into a service or to conduct a transaction will not be able to misuse it since it will no longer be valid.
  • 5. The protocol that indicates when a transaction may lock and unlock each of the data items is called as __________
  • filter_dramaExplanation
    Answer is : A
    The protocol that indicates when a transaction may lock and unlock each of the data items is called as locking protocol. Locking protocols restrict the number of schedules.

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