GATE | GATE-IT-2004 | Question 50
In an enhancement of a design of a CPU, the speed of a floating point unit 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?
Explanation: Speed Up = Time taken in original design / Time taken in enhanced design
In original design: Ratio of floating point operations to fixed point operations = 2:3 Therefore let floating point operations be 2n and fixed point operations be 3n. Ratio of time taken by floating point operation to fixed point operation =2:1 Therefore let time taken by floating point operation be 2t and by fixed point operation be t. Time taken by the original design = (2n * 2t) + (3n * t) = 7nt
In Enhanced design: As the speed of the floating point operation is increased by 20% (1.2 * original speed) time taken for a floating point operation would be 83.33% of the original time (original time/1.2)(This is because CPU speed(S) is inversely proportional to execution time (T) hence if speed becomes 1.2S time would become T/1.2 ) Similarly for a of fixed point operation speed is increased by 10% (1.1 * original speed), it means the time taken now would be 90.91% of the original time (original time / 1.1) taken in case of fixed point operation. Time taken by enhanced design= (2n * 2t /1.2) + (3n * t /1.1) = 6.06nt
Speed up= 7nt / 6.06nt = 1.155
This explanation has been contributed by Yashika Arora.
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