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− | In an enhancement of a design of a CPU, the speed of a floating point until has been increased by | + | 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 |
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
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
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
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