Explore BrainMass

Explore BrainMass

    Format of Memory Address for Various Cache Organizations

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    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)

    © BrainMass Inc. brainmass.com October 10, 2019, 12:41 am ad1c9bdddf

    Solution Preview

    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 Summary

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