Answer:
a. 100%
b. 50%
c. 66.6%
Explanation:
a.
The memory references D6, 58, D8, 9A, DE and 5C all results in misses and the miss ratio is 100 (that is , 6 misses/ 6 references*100 = 100%)
b.
If fully associative cache is used, there will be a 3 cache misses with miss ratio of (3/6*100) = 50%.
c.
With a 2-way set-associative cache, 4 memory reference misses will occur, with a miss ratio of (4/6*100)= 66.6% misses.
A. string? maybee its a string of units
It's often Linux since Linux is free, but it could be Windows.
Answer:
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