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?

The answer is "1.24"

Explanation:

Formula:

Given value:

Operations in the memory = 30%

                                              = 0.3  

80% Memory

Now it takes 0.06 seconds, saved 0.18 seconds.

20% of the memory operations:

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.