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5 August, 20:40

For a bus with width=32Bits, speed=33MHz, what is the theoretical throughput in MiBytes?

a. 32*33 * (10^6) / (2^20)

b. 32/8*33 * (2^20) / (10^6)

c. 32/8*33 * (10^6) / (2^20)

d. 8*33 * (10^6) / (2^20)

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Answers (1)
  1. 5 August, 21:09
    0
    Option C

    Explanation:

    Given:

    - Clock speed f = 33 MHz

    - The width of the bus w = 32 bits

    Find:

    what is the theoretical throughput in MiBytes?

    Solution:

    - First step is to convert the width of the bus to bytes as follows:

    Bytes = 32 bits * (1 Bytes / 8 bits)

    Bytes = 32 / 8

    - Second step is to evaluate the time of the cycle:

    Time period of clock T = 1 / f

    Time period of clock T = 1 s / (33*10^6)

    T = (1 / 33*10^6)

    - Third step is to formulate the number of byte:

    Number of byte = Bytes * T

    = (32 / 8*33*10^6)

    - Fourth step is to convert to Mi bytes:

    Mibytes = Number of byte / 2^20

    Mibytes = (32 / 8*33*10^6 * 2^20)

    - The correct option is C
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