GATE | GATE CS 2021 | Set 1 | Question 32
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.
Explanation: Given, byte-addressable primary memory of size 232 bytes = 32 bits of MM width.
Cache block size = 64 bytes = 26 bytes = 6 bits in mem block width.
Number of cache block (#lines) = Cache size / Cache block size = 32 KB / 64 bytes = 512 = 29 = 9 bits in LO.
Hence, Tag bits are = 32 – 9 – 6 = 17 bits.
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