Consider a machine with byte addressable memory of 2^32 bytes divided into blocks of size 32 bytes. Assume a 2-set associative cache having 512 cache lines is used with this machine. The size of tag field in bits is _____
(A) 18
(B) 16
(C) 19
(D) 21


Answer: (C)

Explanation: Total address space = 32 bit
Block size (W.O.) = 5 bit
No. of bit to represent Set offset (S.O.) = log2(512) – log2(2) = 9 – 1 = 8.
Therefore, number of Tag bits are = 32 – 8 – 5 = 19 bits.

Option (C) is correct.

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