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Return to GATE2004 IT q50.

In an enhancement of a design of a CPU, the speed of a floating point until has been increased by <math>20%</math> and the speed of a fixed point unit has been increased by <math>10%</math>. What is the overall speedup achieved if the ratio of the number of floating point operations to the number of fixed point operations is <math>2:3</math> 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

Solution[edit]

Speed up = Original time taken/ new time taken
Let $x$ be the time for a fixed point operation
Original time taken = (3$x$ + 2*2$x$)/5 = 7$x$/5
New time taken = ((3$x$/1.1) + (4$x$/1.2))/5 = 8$x$/1.32*5
So, speed up = 7*1.32/8 = 1.155




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