Question

g A CPU is equipped with a cache; Accessing a word takes 20 clock cycles if the data is not in the cache and 5 clock cycles if the data is in the cache. What is the effective memory access time in clock cycles if the hit ratio is 90%

Answer 1

Answer:

6.5

Explanation:

The formula to calculate the efficiency is;

= (cache-click-cycle x hit ratio) + ( memory-clock-cycle x 1 - hit ratio)

= (5 x 0.9) + ( 20 x 0.1)

= 4.5 + 2

= 6.5

Answer 2

In this exercise we have to use the knowledge of efficiency, in this way, using the data already informed, we can say that the result of this is:

6.5 effective memory

In this exercise, it was informed that some conditions must be established, that is:

• Hit radio: 90%

• Memory: 20

• Cache: 5

The formula to calculate the efficiency is;

$efficiency= ((cache-click-cycle)(hit \ ratio)) + ( (memory-clock-cycle)(1 - hit \ ratio))$

Substituting the known values ​​in the formula above, we find that:

$= (5)(0.9) + ( 20)(0.1)\\= 4.5 + 2\\= 6.5$

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