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Assume CSMA/CD protocol. Find the least frame length in bytes for a 2 Mbps bit rate and 1.5km long network where propagation delay is 4.25 nano seconds per metre.
    Correct answer is '3.1875'. Can you explain this answer?
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    Assume CSMA/CD protocol. Find the least frame length in bytes for a 2 ...
    Minimum frame size for CSMA/CD = 2 X Propagation delay
    Minimum frame size = 2 X Propagation Delay per metre X Length of the network X Bit Rate
    Minimum frame size = 2 X 4.25 X 10– 9 X 1.5 X 103 X 2 X 106
    So, minimum frame size = 25.50/8 = 3.1875 bytes
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    Assume CSMA/CD protocol. Find the least frame length in bytes for a 2 ...
    CSMA/CD Protocol and Frame Length Calculation




    CSMA/CD Protocol:
    CSMA/CD stands for Carrier Sense Multiple Access with Collision Detection. It is a protocol used in Ethernet networks to avoid collisions between data packets sent simultaneously by two or more devices on the same network segment.




    Frame Length Calculation:
    The frame length is the length of the data packet transmitted over the network. It is calculated by adding the length of the data to the length of the header and trailer.




    Given Information:
    Bit rate = 2 Mbps
    Network length = 1.5 km
    Propagation delay = 4.25 ns/m




    Calculation:


    Propagation delay = Distance / Propagation speed
    Propagation speed = 2*10^8 m/s
    Propagation delay = 4.25 ns/m * 1.5*10^3 m = 6.375 µs




    The minimum frame length can be calculated as follows:
    Frame length = 2 * Propagation delay * Bit rate
    Frame length = 2 * 6.375 µs * 2 Mbps
    Frame length = 25.5 bits




    The minimum frame length in bytes can be calculated as follows:
    Frame length (bytes) = Frame length / 8
    Frame length (bytes) = 25.5 bits / 8
    Frame length (bytes) = 3.1875 bytes




    Answer:
    The least frame length in bytes for a 2 Mbps bit rate and 1.5km long network where propagation delay is 4.25 nano seconds per metre is 3.1875 bytes.
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    Assume CSMA/CD protocol. Find the least frame length in bytes for a 2 Mbps bit rate and 1.5km long network where propagation delay is 4.25 nano seconds per metre.Correct answer is '3.1875'. Can you explain this answer?
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