Question
Given the cache access time is $200 \mathrm{~ns}$ and the memory access time is $400 \mathrm{~ns}$. If the effective access time is $20 \%$ greater than the cache access time, what is the hit ratio?(A) $80 \%$(B) $20 \%$(C) $40 \%$(D) $100 \%$
Step 1
The effective access time is $20 \%$ greater than the cache access time. We can write the effective access time as follows: \[ \text{Effective Access Time} = (1 - \text{Hit Ratio}) \times \text{Memory Access Time} + \text{Hit Ratio} \times \text{Cache Access Time} Show more…
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Consider an $L_{1}$ cache with an access time of $1 \mathrm{~ns}$ and a hit ratio of $H=0.95$. Suppose that we can change the cache design such that we increase $H$ to $0.98$, but increase access time to $1.5 \mathrm{~ns}$. Which of the following condition is met for this change to result in improved performance? (A) Next level memory access time must be less than $16.67$ (B) Next level memory access time must be greater than $16.67$ (C) Next level memory access time must be less than 50 (D) Next level memory access time must be greater than 50
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Cache and Main Memory, Secondary Storage
Consider a single-level cache with an access time of $2.5 \mathrm{~ns}$ and a line size of 64 bytes and a hit ratio of $H=$ 0.95. Main memory uses a block transfer capability that has a first-word (4 bytes) access time of $50 \mathrm{~ns}$ and an access time of 5 ns for each word thereafter. What is the access time when there is a cache miss? (A) $130 \mathrm{~ns}$ (B) $149.4 \mathrm{~ns}$ (C) $2.375 \mathrm{~ns}$ (D) $8.875 \mathrm{~ns}$
In a cache memory, cache line is 64 bytes. The main memory has latency of $32 \mathrm{~ns}$ and bandwidth of $1 \mathrm{~GB} / \mathrm{sec} .$ Then the time required to fetch the entire cache line from main memory is (A) $32 \mathrm{~ns}$ (B) $64 \mathrm{~ns}$ (C) $96 \mathrm{~ns}$ (D) $128 \mathrm{~ns}$
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