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    Numerical: direct mapped cache and byte addressable memory

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    Given, a byte addressable main memory of 2^16 bytes and block size of 8 bytes, using direct mapped cache consisting of 32 lines.

    a. How many total memory bytes can be stored in the cache?
    b. How is a 16 bit memory address divided into tag, line number and byte number?
    c. Into what line would byte with the following address be stored?
    0001 0001 0001 1011.

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    https://brainmass.com/computer-science/memory/numerical-direct-mapped-cache-byte-addressable-memory-145891

    Solution Preview

    a. total memory bytes that can be stored in the cache = number of cache lines * size of each cache line
    = 32 * 8 (because each cache line holds one memory block)
    = 256 ...

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