Question

Consider a machine with a byte addressable main memory of 2³² bytes 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 ______.

Answer 1

Data:

Memory size = 2³² bytes

Total address space = 32 bit

Block size = 32 bytes = 2⁵ byte

Cache lines = 512 = 2⁹

Formula:

Tag	No. of line	Block offset
(x bits)	(9 bits)	(5 bit)

In bits:

Total address space = tag + no. of lines + block offset

For direct mapped cache:

Calculation:

32 = x + 9 + 5

∴ x = 18

The size of tag field in bits = 18