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 ?

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@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..

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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

in old cpu

let us assume fixed operation takes=1 unit time

then float operation takes = 2 unit time

as the operation are given 2:3

lets consider 2 float and 3 fixed operation

total execution time= 2*2 +3*1=7

in new machine

float takes 20% less time  = 2*80/100 =1.6

fixed take 10% less time   = 1*90/100= .9

total execution time = 1.6*2 + .9*3 = 5.9

speedup= performance of new machine/ performance of old machine

performance is inversely proportional to execution time

so speedup= execution time of old / execution time of new

                = 7/5.9

                =1.185

Answer 2

you can also do this without Amdahl’s law, but i was not willing to give up on Amdahl’s law

so after a lot of time, i solved it like this

 

Answer 3

Original time

Floating Point : 2y

Fixed point :  y

Floating point and fixed point instruction are in 2:3 ratio.

For simplicity assume there are 2 floating point and 3 fixed point instruction

Not original time 4y + 3y =7y             -------------------1

Now speed of floating point increases by 20%. we need to convert this 'SPEED' into 'TIME' because we need to calculate ratio of original time and new time.

speed increased by 20% means , it becomes 120/100 of original.

And not time becomes Inverse of this i.e. 100/120 (Newtons formulae applicable to even Computer science.)

Now time for one floating point instruction becomes = 100*2y / 120 -A

simillarly time for one fixed point becomes 100*y/110 -B

we have 2 Floating and 3 fixed point instruction . hence total time 2A+3B  = 800*y/132  ---------------2

Divide 1 by 2 gives answer A

Answer 4

	FLOATING	FIXED
OLD	1 instruction/unit time Conclusion: 1 instruction takes 1 unit time	2 instructions/unit time Conclusion: 1 instruction takes 0.5 unit time
NEW	1.2 instructions/unit time Conclusion: 1 instruction takes $\frac{1}{1.2}$ unit time	2.2 instructions/unit time Conclusion: 1 instruction takes $\frac{1}{2.2}$ unit time

 

$Old\: Time= 0.4x(1) + 0.6x(0.5)=0.7x$

$New\: Time= 0.4x(\frac{1}{1.2}) + 0.6x(\frac{1}{2.2})=0.606x$

$Speedup=\frac{0.7}{0.606}=1.155$