Given a computing system with two levels of cache (L1 and L2) and a main memory. The first level (L1) cache access time is 1 nanosecond (ns) and the “hit rate” for L1 cache is 90% while the processor is accessing the data from L1 cache. Whereas, for the second level (L2) cache, the “hit rate” is 80% and the “miss penalty” for transferring data from L2 cache to L1 cache is 10 ns. The “miss penalty” for the data to be transferred from main memory to L2 cache is 100 ns. Then the average memory access time in this system in nanoseconds is _________ . (rounded off to one decimal place)

To calculate the Average Memory Access Time ($T_{\text{avg}}$) for a hierarchical memory system with two cache levels:

Given data:

• L1 cache access time, $T_1 = 1\text{ ns}$

• L1 hit rate, $H_1 = 90\% = 0.9$

• L1 miss rate, $(1 - H_1) = 1 - 0.9 = 0.1$

• L2 hit rate, $H_2 = 80\% = 0.8$

• L2 miss rate, $(1 - H_2) = 1 - 0.8 = 0.2$

• L1 miss penalty (transfer time from L2 to L1), $T_{\text{penalty, L2}} = 10\text{ ns}$

• L2 miss penalty (transfer time from Main Memory to L2), $T_{\text{penalty, MM}} = 100\text{ ns}$

The average access time is given by:

Where the miss penalty at L1 is:

Now substituting back into $T_{\text{avg}}$:

Thus, the average memory access time is 4.0 ns.