gatebook COA test

Question

Consider a program being run on a processor. A modification in processor design caused 
30% of the program to speed up by ten times while three fourth of the remaining program 
has a speed up of 80 and 40% of the remaining part of the program performs poorer 
and looses its speed by 50%. The remaining program has a speedup of 1. The overall 
speedup of the program exact to two decimal places is:—

Comments

Arjun sir added the answer there, using Amdhal's law. You can check it there.

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where. plz share the link

Answer 1

Portion 1: $0.30$, Speed Up1 = $10$
Portion 2: $ \frac{3}{4}*0.70$, Speed Up2 = $80$
Portion 3: $\frac{1}{4}*0.70*0.40$, Speed Up3 = $\frac{1}{2}$
Portion 4: $\frac{1}{4}*0.70*0.60$, Speed Up4 = $1$

Use Amdal's law:
Final Speed Up = $\frac{1}{\frac{Portion1}{SpeedUp1}+\frac{Portion2}{SpeedUp2}+\frac{Portion3}{SpeedUp3}+\frac{Portion4}{SpeedUp4}}$
                          = $\frac{1}{\frac{0.30}{10}+\frac{\frac{3}{4}*0.70}{80}+\frac{\frac{1}{4}*0.70*0.40}{\frac{1}{2}}+\frac{\frac{1}{4}*0.70*0.60}{1}}$
                         =$\frac{1}{0.2815625} =3.55$ (Rounding Off)

Answer 2

Amdahl's law states that if a portion of a machinery is improved, then entire machine is improved by some extent.

The remaining portion has no contribution in the improvement of entire machinery.

In our case

30% of the program is done 10 times faster 

75% of the remaining 70% is now 80 times faster

25% of 40% of remaining program is now 50% slower

25% of 60% of remaining program has a speedup of 1 i.e no change.

 

speedup = $\frac{1}{\frac{0.3}{10}+\frac{\frac{3}{4}*0.7}{80}+\frac{\frac{1}{4}*0.7*0.4}{\frac{1}{2}}+\frac{\frac{1}{4}*0.7*0.6}{1}}$

=3.55

Answer 3

Amdahl's law states that if a portion of a machinery is improved, then entire machine is improved by some extent.

The remaining portion has no contribution in the improvement of entire machinery.

In our case

30% of the program is done 10 times faster 

75% of the remaining 70% is now 80 times faster

25% of 40% of remaining program is now 50% slower

25% of 60% of remaining program has a speedup of 1 i.e no change.

 

speedup = $\frac{1}{\frac{0.3}{10}+\frac{\frac{3}{4}*0.7}{80}+\frac{\frac{1}{4}*0.7*0.4}{\frac{1}{2}}+\frac{1}{4}*0.7*0.6}$

=3.55