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For a 16K-byte, direct-mapped cache, suppose the block size is 32 bytes, draw a cache diagram....

For a 16K-byte, direct-mapped cache, suppose the block size is 32 bytes, draw a cache diagram. Indicate the block size, number of blocks, and address field decomposition (block offset, index, and tag bit width) assuming a 32-bit memory address.

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IF YOU HAVE ANY DOBBT THEN COMMENT

LIKE IT IF YOU UNDERSTOOD

16k = 214 AS cache size = 14 Bits the cache. lu bits are used to be recognize 1464 cache line Block offset AS Block size is g

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