Memory Hierarchy Q.80

80Consider a machine with a byte addressable main memory of 232 divided into blocks of size 32 bytes. Assume that a direct mapped cache having 512 cache lines is used with this machine. The size of the tag field in bits is ................ .

Option : A
Explanation : Total address space = 32 bit
offset bit (B) = 5 bit
No. of bit to represent line no (L) = 9.
Tag bit + B + L = 32
= X + 5 + 9 = 32
= X = 18

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