A computer has a memory hierarchy consisting of two-level cache (L1 and L2) and a main memory. If the processor needs to access data from memory, it first looks into L1 cache. If the data is not found in L1 cache, it goes to L2 cache. If it fails to get the data from L2 cache, it goes to main memory, where the data is definitely available. Hit rates and access times of various memory units are shown in the figure. The average memory access time in nanoseconds ($ns$) is ________. (rounded off to two decimal places)

To find the Average Memory Access Time ($AMAT$):

Given data from the hierarchy:

• L1 Cache:

  • Hit rate ($H_1$) = $95\% = 0.95$

  • Miss rate ($M_1$) = $1 - 0.95 = 0.05$

  • Access time ($T_1$) = $10\text{ ns}$

• L2 Cache:

  • Hit rate ($H_2$) = $85\% = 0.85$

  • Miss rate ($M_2$) = $1 - 0.85 = 0.15$

  • Access time (including L1 cache miss penalty) ($T_2$) = $20\text{ ns}$

• Main Memory:

  • Access time (including L1 and L2 cache miss penalty) ($T_m$) = $200\text{ ns}$

Since the access times for L2 and Main Memory already include the miss penalties of previous levels (hierarchical access):

$$\begin{aligned} AMAT &= H_1 \times T_1 + M_1 \times \left( H_2 \times T_2 + M_2 \times T_m \right) \\ &= 0.95 \times 10 + 0.05 \times \left( 0.85 \times 20 + 0.15 \times 200 \right) \\ &= 9.5 + 0.05 \times (17 + 30) \\ &= 9.5 + 0.05 \times 47 \\ &= 9.5 + 2.35 \\ &= 11.85\text{ ns} \end{aligned}$$

Hence, the average memory access time is 11.85 $\text{ns}$ [range: 11.83 to 11.87].