Test by Bikram | Mock GATE | Test 1 | Question: 49

Question

Consider two machines, machine $A$ and machine $B$. Machine $B$ runs floating-point instructions $n$ times faster than machine $A$.
Consider a program that takes $100$ $seconds$ to run on machine $A$, and spends half  of its time in floating-point instructions. If speedup of performance is $1.714$ (speed up is calculated by dividing  the performance of  Machine $B$  by performance of machine $A$) then the value of $n$ is _________ (In case of fractional value of $n$ give nearest higher positive integer).

Comments

Speed Up of any system is calculated by the formula = {Time required by slower machine} / {Time required by faster machine}.

But here the definition of speed up given in question is just the reverse of that. Here Machine B is faster and machine A is slower.

If we took the actual formula then only we are getting 6 as answer. Otherwise the answer written by Amit is correct.

@Bikram sir

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Sir, in question the definition of speed up is not given in reverse. I mean they have used the term performance and which is inverse of time.

so, answer should be according to the formula and it is nearly 6 as you said.

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sir i just assumed program contain 100 inst solved i got 5.66 so 6

is this a valid approach to do fast calculation?

Answer 1

here is the solution

Comments

how  50 for other operations on Machine B

Answer 2

Bikram sir please provide the solution ,

Here it is saying that machine A is taking half of the time in floating point instruction . so if machine A spend t time in floating point instrcution then it complete the whole program in 2t time .

machine B is n time faster in processing floating point instructions as compared to A .
so as A takes t time in floating point instructions then B will take t/n time .

for rest of work both A and B will take t time

so speed up = total time taken by B/ total time taken by A

= [ t/n +t ] / [2t] =1.714

by solving n = 0.41186

correct me where i did mistake .

Comments

Can anyone please explain how you are getting 6. If possible please provide the solution.

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can you once give the solution for this?

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@neeraj

read my above comments, and just you need to calculate..

Answer 3

speed up is calculated by dividing  the performance of Machine B by performance of machine A

We know the more time a process takes, the worse is it's performance. So, $\frac{P_B}{P_A}=\frac{T_A}{T_B}$

Where P stands for performance, and T stands for time taken.

 

Machine B runs floating-point instructions n times faster than machine A.

A spends 50 seconds in FP operations. 50 seconds in other operations.

B will spend $\frac{50}{n}$ seconds in FP operations, and 50 seconds in other operations (assuming equal time taken).

 

As per the question

$\frac{T_A}{T_B}=1.714$

=> $\frac{100}{(50/n)+50}=1.714$

After simplifying

$0.286n=1.714$

=> $n=5.99$

=> $n=6$

Comments

are u shashin ?

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Uhm, no. Close enough, lol. I'm Jashan.