We are aware that 64 KB equals 64K words, or 2 to the power of 14 words (2 to the power of 1 word = 2 to the power of 14 blocks).
Each block has 32 bits of data, a 32-14-2 bit tag, and a valid bit. Therefore, for a 64-KB cache, the overall cache size is (2 to the power 14)*49=784 bits, or 98KB. Since there are 64 bytes per line and a cache line's size equals a main memory block's size, the block offset is 6 bits. Associative 2-way cache stores two lines as a single set. The remaining 18 bits for the tag are calculated as follows: number of sets = total cache lines/2 = (512KB/64)/2 = 212, translating to 12 bits for the set.
Learn more about block here-
https://brainly.com/question/17354891
#SPJ4
