Answer with detailed steps?

Replies

• 

  sookie

  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)