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

The answer is "1.24"

Step-by-step explanation:

Formula:

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

Operations in the memory = 30%

= 0.3

80% Memory
`= 0.3 * 80\%`

`=`
`((0.3 * 80) )/(100)`

`= 0.24`

Now it takes 0.06 seconds, saved 0.18 seconds.

20% of the memory operations:

`= 0.3 * 20\% \\\\= ((0.3 * 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* (0.8)/(4)} + 0.3*0.2*(0.5)/(2) + 0.3*0.2*0.5`

`Speedup = 1.2422`