d
Let say the total no. of instructions to be executed is 100 By the ratio of 2:3 Floating point instructions N(Ft): 40 Fixed point instruction N(Fd): 60
Before scenario Time taken to process a floating point instruction T(Ft):2 units Time taken to process a fixed point instruction T(Fd): 1 unit
Total time to execute T(B)=N(Ft) * T(Ft) + N(Fd) * T(Fd)=40*2+60*1=140 units
After scenario When speed of floating point processing is increased by 20% means time to execute a floating instruction in reduced by 20% Before T(Ft) was 2 units Now time taken to process a floating point instruction T(Ft):2 - 20% of 2 =1.6 units (After)
Similarly speed of fixed point processing is increased by 10% hence Before T(Fd) was 1 units Now time taken to process a fixed point instruction T(Fd): 1 - 10% of 1=0.9 units (After)
Total time to execute T(A)= 40*1.6+60*0.9=118units
So Speed up gained is T(B)/T(A) =140/118=1.186 (approx)
ankitkhare72
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