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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?
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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
 
'''(A) 1.155''' (B) 1.185 (C) 1.255 (D) 1.285

Revision as of 22:01, 27 February 2014

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

Solution

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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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

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




blog comments powered by Disqus