Suppose a unified cache has the following characteristics:

Question

Read/write penalty 1 cycle
Miss rate 3%
Load instructions (read data) 20%
Store instructions (write data) 5%
Miss penalty 20 cycles
What is the average access time?

Answer 1

We have to add up all the possibilities.
a. Access to cache (instruction read) (1 + 1) x 0.97 cycles = 1.94 cycles (note 1 + 1 because it’s a 1 cycle
penalty)
b. Cache miss (instruction read) 0.03 x (20 + 1) = 0.63 cycles
c. Load instruction hit 0.20 x 0.97 x (1 + 1) = 0.388 cycles
d. Load instruction miss 0.20 x 0.03 x (20 + 1) = 0.126
e. Store instruction hit 0.05 x 0.97 x (1 + 1) = 0.97 cycles
f. Store instruction miss 0.05 x 0.03 x (20 + 1) = 0.0315

Total time = 0.63 + 0.388 + 0.126 + 0.97 + 0.0315 = 2.1455 cycles
Note that this is a simplistic answer that does not take account of cache reloading.