GATE CS 2016 Set 2 Q50 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

To find the smallest cache size required to ensure an average read latency of less than $6\text{ ms}$, we first formulate the average read latency equation.

Let:

• Latency to read from cache, $T_c = 1\text{ ms}$
• Latency to read from disk, $T_d = 10\text{ ms}$
• Miss rate of the cache = $m$

Since the cost of checking the cache is $1\text{ ms}$ and a disk access is required on a miss, the average read latency $T_{\text{avg}}$ is given by:

We require $T_{\text{avg}} < 6\text{ ms}$:

From the given miss rate graph, we observe the miss rates for different cache sizes:

• At $\text{Cache Size} = 10\text{ MB}$, $\text{Miss Rate } (m) = 0.8$
• At $\text{Cache Size} = 20\text{ MB}$, $\text{Miss Rate } (m) = 0.6$
• At $\text{Cache Size} = 30\text{ MB}$, $\text{Miss Rate } (m) = 0.4$

The smallest cache size that achieves a miss rate $m < 0.5$ is $30\text{ MB}$ (which gives $T_{\text{avg}} = 1 + 0.4 \times 10 = 5\text{ ms} < 6\text{ ms}$).