Consider a computer system with a byte-addressable primary memory of size 232…
2021
Consider a computer system with a byte-addressable primary memory of size 232 bytes. Assume the computer system has a direct-mapped cache of size 32 KB (1 KB = 210 bytes), and each cache block is of size 64 bytes.
The size of the tag field is ________ bits.
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Correct answer: 17
Key idea: the physical address is 32 bits because the memory size is 2^32 bytes.
Offset bits = log2(block size) = log2(64) = 6 bits.
Number of cache lines = cache size / block size = 32 KB / 64 B = 32768 / 64 = 512.
Index bits = log2(512) = 9 bits.
Tag bits = address bits - index bits - offset bits = 32 - 9 - 6 = 17 bits.
Answer: 17 bits.
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