Cache Mapping

Question

In a 2 level hierarchy, the cache has an access time of 15 ns and the main memory has an access time of 110 ns, the hit rate of the cache is 90%. If the block size of the cache is 16 Bytes, then average memory access time including miss penalty is?

Answer 1

Cache access time  = 15 ns

Main memory access time =  110 ns

Block size = 16 bytes

Block access time = 16 * 110 = 1760 ns

Miss penalty means after access the memory we have to include the data in cache

Average memory access time = Hit rate of cache * cache access time + (1 – hit rate)(cache access  + block access + from memory again put on  cache

= 0.9 * 15 + 0.1* (15+1760+15)

= 192.5 ns