Question: How many total bits are required for a direct mapped cache with 16KB of data and 4-word blocks assuming a 32-bit address? Please give a
How many total bits are required for a direct mapped cache with 16KB of data and 4-word blocks assuming a 32-bit address?
Please give a detail Explanation thanks
16KiB = 4096 (2^12 words) **EXPLAIN WHY 16KB = 4096
With Block size of 4 words (22 ) there are 1024 (210) blocks.
Each block has 4 x 32 or 128 bits of data plus a tag which is 32 10 2 2 bits, plus a valid bit
Thus total cache size is
2 10 x (4 x 32 + (32 10 2 - 2) + 1) = 210 x 147 = 147 KibiBits
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