The width of the physical address on a machine is 40 bits. The width of the…
2016
The width of the physical address on a machine is 40 bits. The width of the tag field in a 512 KB 8-way set associative cache is _________ bits.
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Correct answer: 24
Key idea: split the physical address into tag, index, and block offset. Tag bits = physical address width - index bits - offset bits.
Assume a 64-byte block (common default). Block offset bits = log2(64) = 6.
Number of sets = cache size / (associativity × block size) = 512 KB / (8 × 64 B) = 524,288 / 512 = 1,024 sets.
Index bits = log2(1,024) = 10.
Tag bits = 40 (physical address width) − 10 (index) − 6 (offset) = 24 bits.
Note: If the block size differs from 64 bytes, recompute the block offset and index accordingly; the procedure above still applies.
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