Memory operations currently take 30% of execution time. A new widget called a cache speeds up...

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Question

Memory operations currently take 30% of execution time. A new widget called a cache speeds up...

Memory operations currently take 30% of execution time. A new widget called a cache speeds up 80% memory operations by a factor of 4. A second new widget called an L2 cache speeds up 1/2 of the remaining 20% by a factor of 2. What is the total speed up?

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Answer 1

Answer #1

Answer is as follows :

The method is to solve by formula =

Speedup = 1 / [(1 - non-speedup portion) + (Speed up portion 1)/speedup1 + (speed up portion 2)/speedup2 + ...]

and result is =

Speedup = 1 / [0.7 + 0.3*0.8/4 + 0.3*0.2*0.5/2 + 0.3*0.2*0.5] = 1.2422

Explanation :

Memory operations = 30% = 30/100 = 0.3

80% of memory operations = 0.3 * 80 % = (0.3 * 80) / 100 = 0.24, this now takes 0.06 seconds, 0.18 seconds saved

20% of the memory operations = 0.3 * 20 % = (0.3*20) / 100 = 0.06

Half of the remaining 20% = 0.03, this now takes 0.015 seconds, 0.015 seconds saved , 0.195 seconds saved

What had taken 1 seconds,now takes 0.805 seconds.

1/0.79=1.2422 // 0.79 is calculated by denomenator of formula..

So total Speed up for given query is = 1.2422

if there is any query please ask in comments...

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Answer 2

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Memory operations currently take 30% of execution time. A new widget called a cache speeds up...

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