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# Format of Memory Address for Various Cache Organizations

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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)

#### Solution Preview

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 Summary

Solution shows relevant computations as to how the size of different fields of memory address is obtained in each case.

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