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    Set Associative Cache: Main Memory Address Format

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    Given a word addressable machine, with following memory configuration.

    - 64 cache lines divided into 4 sets.
    - main memory contains 4K blocks of 128 words each.

    How does the main memory address format look?

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    Solution Preview

    Since 64 cache lines are divided into 4 sets, so the number of lines in each set = 64/4 = 16
    Since each cache line can hold one memory block, so size of each line = 128 words

    There are 4K i.e. 4096 blocks in main memory. To ...

    Solution Summary

    The solution describes the computation of 19 bits address format, used to address a word in the main memory.