Given a word addressable machine, with following memory configuration.
- 64 cache lines divided into 4 sets.
- main memory contains 4K blocks of 128 words each.
How does the main memory address format look?
Since 64 cache lines are divided into 4 sets, so the number of lines in each set = 64/4 = 16
Since each cache line can hold one memory block, so size of each line = 128 words
There are 4K i.e. 4096 blocks in main memory. To ...
Solution describes the computation of 19 bits address format, used to address a word in the main memory.