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In an enhancement of a design of a CPU, the speed of a floating point unit has been increased by 30% and the speed of a fixed point unit has been increased by $20$%. The overall speedup achieved if the ratio of the number of fixed point operations to floating point operations is $4:6$ and the floating point operation used to take twice the time taken by fixed point operation in the original design(upto $2$ decimal places is __________.

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Speed up = (Original time taken)/ New time taken
Let x be the time for a fixed point operation
Original time taken = (4x + 6*2x)/10(since total instruction is 4+6= 10) = 16x/10
New time taken = ((4x/1.2(since increse 20% so 120%)+(12x/1.3)(since increse 30% so 130%))/10

               =19.6x/1.56*10
So, Speed up = 16*1.56/19.6 = 1.27

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Speed up = (final execution time) / (initial execution time)

= (1.3*T(floating)*6 + 1.2*T(fixed) *4) / (6*T(floating) + 4*T(fixed)

Since, T(floating) = 2* T(fixed)    #given

So,

Speed up = (1.3*2*6 + 1.2*4) / (4*2 + 6)

=20.4/ 14

=1.45

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