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

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

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

 

---

@endurance1

@rupesh17

Thank you guys. I did the same mistake. Thank you for correcting

---

These are simple aptitude questions, no COA concepts will be used here

Answer 1

SpeedUp $=\dfrac{\text{Original time taken}}{\text{new time taken}}$

Let $x$ be the time for a fixed point operation,

Original time taken $=\dfrac{(3x + 2\times 2x)}{5} =\dfrac{7x}{5}$
New time taken $=\dfrac{\left(\dfrac{3x}{1.1} + \dfrac{4x}{1.2}\right)}{5}=\dfrac{8x}{1.32\times 5}$
So, SpeedUp $=\dfrac{7\times 1.32}{8} =1.155$

Correct Answer: $A$

Comments

(the floating point operation used to take twice the time taken by the fixed point operation in the original design), in this statement it is written floating is twice of fixed, meaning (2*3x) instead of (2*2x), can someone clear this doubt

---

floating point op take twice the time taken by the fixed point op. It means if fixed point is taking x time then floating point takes 2x time. And we multiplied this time with number of operations which is given in question.

---

I’m following a different approach and am getting an answer not in the options,

Floating point Speed (original) = x

Fixed point Speed (original) = y

Floating point Speed (new) = 1.2x

Floating point Speed (new) = 1.1y

Total operations being either of float or fixed type, 2:3 gives 40% and 60% float and fix point operations

Since a floating pt operation took double time than fixed in original, its speed should be half of what a fix pt operation has
Or
2x=y

Speed up (S) = New Speed / Old Speed :

[0.4(1.2x)+0.6(1.1y)]/[0.4x+0.6y]

Putting 2x = y and solving gives S = 1.8x/1.6x

= 1.125

What is coming different here?

Answer 2

You can tabulate down the given data from question.

	Speed in Original Design	Speed in New Design	# of Operations given(2:3)
Floating Point (FP) operations	100	120	2x
Fixed Point (XP) operations	200	220	3x

Here, Speed = #operations per unit time (Just like speed = distance/time)

Took 100,200 for FP and XP for easy understanding.

Say 5x is total number of operations, in which 2x FP and 3x XP operations

Speed up = $\frac{Time in Original Design}{Time  in New Design}$

=>$\frac{2x/100 + 3x/200}{2x/120 + 3x/220}$ (Since we know time = #operations/speed)

we get Speed up = 1.155, So Option (A)

Hope this solution was easy to understand

Comments

when amount of floating point and fixed point data are same then you can compare speed of floating point and fixed point ,but here fixed point is 3x and floating point is 2x hence you cannot compare speed directly….

---

just perfect smooth !!! thank youuu

---

Amazing Explanation !!

Answer 3

• $k$ = no of times one floating point operation takes more time than one fixed point operation.
• $x:y$ = ratio of no of fixed point operations to that of floating point operations.
• $a$ = % increase in speed of floating point unit in fraction {i.e. 20% means $a$ = $0.2$}
• $b$ = % increase in speed of fixed point unit

Then Net speedup 

$\begin{align*} S &= \frac{\text{total time in old design}}{\text{total time in new design}} \\ & = \frac{ky+x}{\frac{ky}{1+a}+\frac{x}{1+b}} \end{align*}$

Putting all these values S = $1.155$.

NOTE: That is not a standard formula. It has been derived , by taking total time  = $T$ and total operations = $N$

Comments

https://gateoverflow.in/101176/floating-point

which is also explained by debashih sir :)

Answer 4

Suppose we have 20 floating instructn and 30 fixed.

Initially Speed of Floating instruction is 20 and fixed is 10,

so total time=20*20+30*10=700

After increasing float speed 20%=speed(float)=24

10% increase in fixs speed=speed(fix)=11

So total time= 24*20+30*11=480+330=810

Speed up=810/700=1.1571

Comments

why is it not 700/810???

---

Because speed up of new =Old time taken/new time taken

---

This solution is wrong.

It says "initially Speed of Floating instruction is 20 and fixed is 10,"

Speed of floating point is double than fixed here but this is not given,they have given that time of floating point is double.Although this is giving correct answer but approach is wrong.@Sushmita you were right.It should be 710/800 and it gives wrong answer

---

Speed has  increased by 20% (since speed has increased ) hence it will take lesser time and it should be 16 for floating point operations after the enhancement. 

and based on above argument it should be 70/59=1.186

pls correct me if my approach is wrong. Thanks in advance.

Answer 5

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

Comments

Here, the following statements are wrong - 
float takes 20% less time  = 2*80/100 =1.6
fixed take 10% less time   = 1*90/100= .9

---

@mv_ind why if speed increses then time decreses

---

Speed increases 20% does not mean time decreases 20%.

(By that logic, if speed increases 100% then time will reduce by 100%, which is illogical)

---

ys bro got it , my bad, thanks

Answer 6

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