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  • Answer with detailed steps?

    qwerty123

    qwerty123

    @qwerty123-ZuiVHD
    Updated: Oct 22, 2024
    Views: 883
    In an enhancement of a design of a CPU, the speed of a floating point until has been increased by 20% and the speed of a fixed point unit has been increased by 10%. What is the overall speedup achieved if the ratio of the number of floating point operations to the number of fixed point operations is 2:3 and the floating point operation used to take twice the time taken by the fixed point operation in the original design?
    (A) 1.155 (B) 1.185 (C) 1.255 (D) 1.285
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  • sookie

    Member • Jun 4, 2012

    Let say the total no. of instructions to be executed is 100
    By the ratio of 2:3
    Floating point instructions N(Ft): 40
    Fixed point instruction N(Fd): 60

    Before scenario
    Time taken to process a floating point instruction T(Ft):2 units
    Time taken to process a fixed point instruction T(Fd): 1 unit

    Total time to execute T(B)=N(Ft) * T(Ft) + N(Fd) * T(Fd)=40*2+60*1=140 units

    After scenario
    When speed of floating point processing is increased by 20% means time to execute a floating instruction in reduced by 20%
    Before T(Ft) was 2 units
    Now time taken to process a floating point instruction T(Ft):2 - 20% of 2 =1.6 units (After)

    Similarly speed of fixed point processing is increased by 10% hence
    Before T(Fd) was 1 units
    Now time taken to process a fixed point instruction T(Fd): 1 - 10% of 1=0.9 units (After)

    Total time to execute T(A)= 40*1.6+60*0.9=118units

    So Speed up gained is T(B)/T(A) =140/118=1.186 (approx)
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