Consider a byte-addressable computer with 24-bit addresses, a cache capable of storing a total of 64KB of data, and blocks of 32 bytes. Show the format of a 24-bit memory address for:
A. Direct Mapped Cache: word, block, tag (number of bits in decimal)
B. Fully Associative Cache: word, tag (number of bits in decimal)
C. 4-way set Associative Cache: word, set, tag (number of bits in decimal)
Given that it is a byte-addressable computer with 24-bit addresses.
Size of cache (Sc) = 64 KB or 64*1024 bytes or 2^16 bytes
Size of cache block (Sb) = 32 bytes or 2^5 bytes
Number of cache blocks (Nb) = Sc/Sb = 2^16/2^5 = 2^11
A. In case of direct ...
Solution shows relevant computations as to how the size of different fields of memory address is obtained in each case.