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In an enhancement of a design of a CPU, the speed of a floating point unit has been increased by $\text{20%}$ and the speed of a fixed point unit has been increased by $\text{10%}$. 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?

  1. $1.155$
  2. $1.185$
  3. $1.255$
  4. $1.285$
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Answer is (B)

Say in old system:               20 floating point operation and 30 fixed point operations

                                                 say floating point takes 2 units and fixed point takes 1 unit

Total time in old system=  20*2 +30 *1= 70 units

In Enhanced system         floating point takes = 2* 0.8= 1.6 units

                                               fixed point takes= 1*0.9= 0.9 units

Total time = 20*1.6 +30 *0.9 = 59 units

Speedup= 70/59= 1.1864

                                           

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Explanation:
(1.27)
Let total operation be 100 and total time taken is t = 100 sec.
Floating point operations are =610×100=60
Fixed point operations are =410×100=40
Let time taken by floating point is t1 and time taken by fixed point is t2. So t1+t2=100 … (1)
Time taken by 60 floating point operation = t1
1 floating operation = 160
Similarly 1 fixed point operation takes 240
∵ 160=2240
2t1 = 6t2
From equation 1 and 2
1=6008  2=2008
After speed-up 1′=11.3=6001.3×8=6000104
2′=21.2=1001.2×8=200096
1′=1′+2′=6000104+200096
2=′×1
2=100[6000104+200096]×1=10057.69+20.83×1
2=10078.52×1=1.27 1
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Let us assume number of operation =100;

So number of floating point operation=100*2/5=40

So number of fixed point operation=60

Lets time taken by floating point operation is t1 and for fixed point operation it take t2

So t1+t2=100(assume).

since 60 floating point operation take t1 time

then only one=t1/40

and fixed point=t2/60

and it is given that::
t1/40=2*t2/60

3t1=4t2;

since t1+t2=100

t2=300/7

t1=400/7

Now time after speed up let us assume t1' and t2'

t1'=400/3*1.1=121.22(Approx)

t'2=300/7*1.2=35.72

Tt=t'1+t'2=156.94

S2=100/156.94*S1

S2=0.64*S1
Answer:

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