Explore BrainMass

Explore BrainMass

    Numerical: direct mapped cache and byte addressable memory

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

    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.

    © BrainMass Inc. brainmass.com October 9, 2019, 8:17 pm ad1c9bdddf

    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 ...

    Solution Summary

    Solution shows the computations in sufficient details that should help you in solving similar problems.