Question

Consider adding a fast second-level cache to a computer with only a single level cache. Suppose data can be accessed from the second level cache 12 times faster than it can be accessed from memory, and also suppose that the second level cache is used 30% of the time and the first level cache is used 70% of the time. What is the speedup of the system after adding the second-level cache, compared to the original system?

Answer 1

Let the time to access data from first cache be

Let the time to access data from the memory be

Then, time to access data from second cache will be

Average time to access data without second cache (time 1) = (hit probability at cache 1)* (hit time at cache 1) + (miss probability at cache 1)*(time to access memory)

=

Average time to access data with second cache (time 2)= (hit probability at cache 1)* (hit time at cache 1) + (miss probability at cache 1)*((hit probability at cache 2)*(hit time at cache 2) + (miss probability at cache 2)*(time to access memory))

=

0.70 * T1+0.30 * (0.30 * 12/12 +0.70 * T2)

$speedup=\frac{time 1 }{time 2}$

$= \frac{0.70*T1 + 0.30*T2}{0.70*T1 + 0.30*(0.30*T2/12 + 0.70*T2)}$

$= \frac{7*T1 + 3*T2}{7*T1 + 2.175*T2}$

$= \frac{1 + 0.43*\frac{T2}{T1}}{1 + 0.31*\frac{T2}{T1}}$

Where T2/T1 denotes how faster it is to access data from cache 1 as compared to memory.