Question

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?

Answer 1

Hey mate ^_^

Here is ua answer.... ⏬⏬


Amdahl’s law for overall speed up

Overall speed up=1/((1-f)+f/s)

Where f is fraction of processes, which are speeded up.

Floating point  operations are 40%.(given 2:3) And speed up is .2(20% given).

Overall speed up in floating point is given as

1/((1-.4)+(.4/.2))=1/(.6+2)=.38

Where as overall speed up  in fixed op is

1/(1-.6)+.1/.1=1/2.4=.41.

Let initially floating point takes 2 unit  and fixed operation takes 1 unit of execution.

Now after speed up floating point will take 2*.62 units

And fixed point will take .59units

Let we have  20 floating and 30 fixed operations. Initially its cost is  70 units.

After  overall speed up = 20*1.24+30*.59=42.5 units.

Overall speed achieved is 70/42.5 approx 1.62


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