GATE CS 2017 Set 1 Q25 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

To find the local miss rate of the $L2$ cache, we need to determine the ratio of $L2$ misses to $L2$ accesses.

1. Calculate $L2$ Accesses per Instruction

Every memory access first goes to the $L1$ cache. If it misses in $L1$, it is then directed to the $L2$ cache. Therefore, the number of accesses to the $L2$ cache is equal to the number of misses in the $L1$ cache.

• Memory accesses per instruction: $1.4$
• $L1$ miss rate ($MR_{L1}$): $0.1$

$$\text{Accesses to } L2 \text{ per instruction} = \text{Misses in } L1 \text{ per instruction}$$ $$\text{Accesses to } L2 = 1.4 \times 0.1 = 0.14 \text{ accesses/instruction}$$

2. Calculate $L2$ Misses per Instruction

We are given that the $L2$ cache experiences $7$ misses per $1000$ instructions.

$$\text{Misses in } L2 \text{ per instruction} = \frac{7}{1000} = 0.007 \text{ misses/instruction}$$

3. Calculate $L2$ Miss Rate

The local miss rate of the $L2$ cache ($MR_{L2}$) is the fraction of $L2$ accesses that result in a miss:

$$MR_{L2} = \frac{\text{Misses in } L2 \text{ per instruction}}{\text{Accesses to } L2 \text{ per instruction}}$$ $$MR_{L2} = \frac{0.007}{0.14} = \frac{7}{140} = 0.05$$

Thus, the miss rate of the $L2$ cache is 0.05.