GATE IT 2004 | Question: 50

Question

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?

• A. $1.155$
• B. $1.185$
• C. $1.255$
• D. $1.285$

Comments

i tried doing this in the normal way by increasing the speed by 20% and 10% the answer i got was 1.186.

However what i think is happening is i am reducing time with respect to speed linearly. Meaning if speed is increased by 20% i am assuming that time must also decrease by 20% but it’s not correct the change in %tage of time will be 17% instead of 20 and by doing that way i am able to arrive at 1.155.

Am i thinking in the right direction ?

---

@endurance1 yes ...It won’t change linearly ..say 100km with speed of 10kmph take 10 hrs ….now speed got increased by 20%  ..i.e. 10+2=12kmph..time will be 100/12=8.33hrs….not 8 hrs…

in this case if old time say  1 sec  then new time will be  will be  1/1.2 =0.833 sec and  1/1.1=0.909sec..

---

If in doubt from where did 1.1 came in the denominator, watch this: 

https://www.youtube.com/watch?v=YJeNHLRpvWw&ab_channel=DEWSAcademy

 

Answer 1

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