Question

Consider a memory system with a cache access and mapped time of 110 ns and a memory access time of 210 ns (assume the memory access time includes the time to check the cache, page table, and access time). If the effective-access time is 10% greater than the cache access and mapped time, what is the value of hit ratio H? [Show the procedure]

Answer 1

Answer:

memory

Explanation:

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Answer 2

Answer:

0.89

Explanation:

Effective-access time(EAT) = cache access time * 1.1 (i.e. 10% greater)

So, EAT = 110 * 1.1 = 121

Let x = Hit Ratio

Now, EAT = x*cache access time + (1-x)*memory access time

121 = 110*x + 210(1-x)

Solving the above equation, we get, x = 0.89