1. A computer system exploits “direct mapping” cache. If cache has 64 blocks of
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1. A computer system exploits “direct mapping” cache. If cache has 64 blocks of 16 words, then how many bits are there in the “tag” field of the 15-bit address for this architecture? 1. A computer system exploits “direct mapping” cache. If cache has 64 blocks of 16 words, then how many bits are there in the “tag” field of the 15-bit address for this architecture? 1. A computer system exploits “direct mapping” cache. If cache has 64 blocks of 16 words, then how many bits are there in the “tag” field of the 15-bit address for this architecture?Explanation / Answer
Given number of cache blocks = 64-byte blocks.
and data cache is 16 KB in size.
Now we are calculating cache blocks...
16 KB/16 = 16 * 1024 byte/16 byte = 256 cache blocks...
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Now total sets = blocks divided by sets = 256 / 4 = 64 sets
therefore 2^6 and also 2^6 = 64 possible sets...
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Given 15 bit address architectire
Assume x as number of bits to tag field.
15 = x + 6 + 6
therefore x = 15 - 12 = 3 bits for tag field.
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