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How many total bits are required for a direct- mapped cache with 256 KB of data and 8- word block size, assuming a 32-bit address?

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Answer:

it's direct mapped (associativity=1), it holds 16k of data, and our blocksize is 64 bytes. so we have 16k/64 = (16*1024)/64 = 256 sets. log2(256) = 8, so we need 8 bits for the index.

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