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Direct Mapped Cache - Memory Blocks

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Suppose that a computer using direct-mapped cache has 2^24 words of main memory and a cache of 64 blocks, where each cache block contains 16 words.

A. How many blocks of main memory are there?

B. What is the format of a memory address as seen by the cache? That is, what are the sizes of the tag, block and word fields?

C. To which cache block will the memory reference (20DB63)base16 map?

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Solution Summary

This solution shows brief and relevant computations for number of blocks of main memory, number of bits needed to represent the word, block, and tag field, and which cache block the memory reference will apply to.

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Size of main memory (Sm) = 2^24 words
Size of cache block (Sc) = 16 or 2^4 words
Number of blocks in cache (Nc) = 64

A. Number of blocks of main memory (Nm) = Sm/Sc = 2^24 / 2^4 = 2^20 blocks

B. Number of bits needed to represent word field = 4 (because there are 2^4 words / cache block)
Number of bits ...

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