Consider a small two-way set-associative cache memory, consisting of four blocks. For choosing the block to be replaced, use the least recently used (LRU) scheme. The number of cache misses for the following sequence of block addresses is:
$8, 12, 0, 12, 8$.
We have 4 blocks and 2 blocks in a set
=> there are 2 sets. So blocks will go to sets as follows:
Since the lowest bit of block address is used for indexing into the set, so 8, 12 and 0 first miss in cache with 0 replacing 8 (there are two slots in each set due to 2-way set) and then 12 hits in cache and 8 again misses. So totally 4 misses.
This one ....
Sir I have ordered GO PDF on 16 Aug 2018 still ...