Question

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

Answer 1

Answer:

The answer is "1.24"

Explanation:

Formula:

$\bold {Speedup= \frac{1}{(1 - non-speedup-portion) +(Speed up portion 1)}+\frac{(speed up-portion \ 2)}{ speedup\ 2}+ ....}$Given value:

Operations in the memory = 30%

                                              = 0.3  

80% Memory $= 0.3 \times 80\%$

                       $=$ $\frac{(0.3 \times 80) }{100}$

                       $= 0.24$  

Now it takes 0.06 seconds, saved 0.18 seconds.

20% of the memory operations:

$= 0.3 \times 20\% \\\\= \frac{(0.3 \times 20)}{100} \\\\= 0.06$

Half of the other 20% = 0.03  

It takes 0.015 seconds now, saved 0.015 seconds, saved 0.195 seconds.  

It takes 0.805 seconds in the 1 second.

$Speedup = \frac{1} {0.7 + 0.3\times \frac{0.8}{4}} + 0.3\times0.2\times\frac{0.5}{2} + 0.3\times0.2\times0.5$

$Speedup = 1.2422$