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A channel has 10 Kbps bit rate using stop and wait protocol and has propagation delay of 4 ms. For a frame with 400 bit, what will be the efficiency (in percentage)?
    Correct answer is between '82.5,83.5'. Can you explain this answer?
    Verified Answer
    A channel has 10 Kbps bit rate using stop and wait protocol and has pr...
    transmission time for 400 bits = 40ms
    utilization = 
     = 83%
    View all questions of this test
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    A channel has 10 Kbps bit rate using stop and wait protocol and has pr...
    Efficiency of a communication system is defined as the ratio of useful information transmitted to the total time taken. In this case, we are given a channel with a bit rate of 10 Kbps and a propagation delay of 4 ms. We need to calculate the efficiency for a frame with 400 bits.

    To calculate efficiency, we need to consider the time taken for transmission and the time taken for propagation.

    1. Time taken for transmission:
    - Given bit rate = 10 Kbps
    - Frame size = 400 bits
    - Time taken to transmit the frame = Frame size / Bit rate
    = 400 bits / 10 Kbps
    = 400 / 10,000 seconds
    = 0.04 seconds

    2. Time taken for propagation:
    - Given propagation delay = 4 ms
    - Time taken for propagation = Propagation delay / 1000
    = 4 ms / 1000
    = 0.004 seconds

    3. Total time taken:
    - Total time taken = Time taken for transmission + Time taken for propagation
    = 0.04 seconds + 0.004 seconds
    = 0.044 seconds

    4. Efficiency calculation:
    - Efficiency = (Useful information transmitted / Total time taken) * 100
    - Useful information transmitted = Frame size = 400 bits
    - Efficiency = (400 bits / 0.044 seconds) * 100
    = 9090.9091 bits/second * 100
    = 909,090.91 bits/second

    The correct answer lies between 82.5% and 83.5%. Let's calculate the efficiency within this range:

    - Lower limit: 82.5% of 909,090.91 bits/second
    = 0.825 * 909,090.91 bits/second
    = 750,000 bits/second

    - Upper limit: 83.5% of 909,090.91 bits/second
    = 0.835 * 909,090.91 bits/second
    = 759,090.91 bits/second

    Therefore, the efficiency lies between 750,000 bits/second and 759,090.91 bits/second, which can be rounded to 82.5% and 83.5% respectively.
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    A channel has 10 Kbps bit rate using stop and wait protocol and has propagation delay of 4 ms. For a frame with 400 bit, what will be the efficiency (in percentage)?Correct answer is between '82.5,83.5'. Can you explain this answer?
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    A channel has 10 Kbps bit rate using stop and wait protocol and has propagation delay of 4 ms. For a frame with 400 bit, what will be the efficiency (in percentage)?Correct answer is between '82.5,83.5'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A channel has 10 Kbps bit rate using stop and wait protocol and has propagation delay of 4 ms. For a frame with 400 bit, what will be the efficiency (in percentage)?Correct answer is between '82.5,83.5'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A channel has 10 Kbps bit rate using stop and wait protocol and has propagation delay of 4 ms. For a frame with 400 bit, what will be the efficiency (in percentage)?Correct answer is between '82.5,83.5'. Can you explain this answer?.
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