Question

Suppose that we want to enhance the processor used for Web serving. The new processor is 10 times faster on computation in the Web serving application than the original processor. Assuming that the original processor is busy with computation 40% of the time and is waiting for I/O 60% of the time, what is the overall speedup gained by incorporating the enhancement

Answer 1

Answer:

The overall speedup gained by incorporating the enhancement is 1.563

Explanation:

Using Amdahl's Law;

The overall speedup  $(N) =\dfrac{1}{(1-P) + ( \dfrac{P}{n})}$

where;

P = Fraction Enhanced for Old Processor

n = speedup enhancement as a result of the new processor

P = 40% = 0.40  and n = 10

∴

The overall speedup (N) is:

$=\dfrac{1}{(1-0.4) + ( \dfrac{0.4}{10})}$

$=\dfrac{1}{(0.6) + ( 0.04)}$

$=\dfrac{1}{0.64}$

= 1.563

Thus, the overall speedup gained by incorporating the enhancement is 1.563