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GATE | GATE CS 2021 | Set 1 | Question 32

  • Last Updated : 24 May, 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.
(A) 17
(B) 18
(C) 15
(D) 9


Answer: (A)

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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