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Cache Size is 64KB, Block size is 32B and the cache is Two-Way Set Associative. For a 32-bit physical address, give the division between Block Offset, Index and Tag.
Answer:
if block size is 32B,then block offset = 2^5 so 5 bits, index has equation, cache size/(block size*associative) so 64K/(32*2) = 1K, 2^10, so index = 10 bits,now tag = physical address-block offset-index, which is 32-5-10 = 17bits
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