Pipe-lining with floating point operation..!!

Question

Suppose that a task makes extensive use of floating point operations with 40% of the time is consumed by floating point operations. With a new hardware design, the floating point module is speeded up by a factor of 4. What is the overall speedup?

A. 1.05

B. 1.42

C. 2.5

D. 4

Please explain the little bit problem also.

Comments

Hemant try this  

if you still have doubt then please ask .

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40% floating point means we can do parallalism and (1-40%) we have to do serial execution .

so now Find speed up ie.[(1-f) +f/speed up factor ] ^-1

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@anu thank you..!! Link is useful. :)

Answer 1

speed up= execution time before enhancement / execution time after enhancement

execution time before enhancement =1

execution time after enhancement = 0.6 + 0.4/K  =0.6 +0.1=0.7

speedup=1/0.7=1.42

Comments

I didnt understand that how you calculate execution time after enhancement. Please help.

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@Sakshi Jaiswal Basically there is an Amdhal's law which states that If $T$ is total execution time of a program using single processor. Then, speedup achieved with $N$ processors is $\frac{T}{T*(1-f)+ T*\frac{f}{N}}$ where $f$ is a fraction of parallelizable unit/program.

Here, floating point module which constitutes fraction of $0.4$ has been speeded up which is equivalent to parallelization with $4$ modules. Now you can calculate.

Answer 2

without pipline total time=x

with 40% part done in 1/4th time we get total time =.4x/4+.6x

so speedup=x/(.4x/4+.6x)=1/.7=1.42

Answer 3

let t is the total time for executing the task 0.6t ( time for general operations) + 0.4t (time for floating point operation)

since floating point operations are speed up by 4 times the time taken for its execution decreases by 4 i.e 0.4t/4 = 0.1t

now total time for executing the task = 0.6t ( time for general operations) + 0.1t (time for floating point operation)

hence speed up = (0.6t + 0.4t)/ (0.6t + 0.1t) = t/0.7t = 1.42

Answer 4

Initially :

Let time taken by FPU = x.

This means 40% of time = x.

This means 60% of time by other units = 1.5 x. 

This means total time taken = (1+1.5)x  = 2.5 x

 

After Improvement: 

Time taken by FPU = $\frac{1}{4}x = 0.25x$

Time taken by other units = 1.5 x

Total time taken = 1.75 x

 

Speedup = $\frac{2.5x}{1.75x} = 1.42$