The width of the physical address on a machine is 40 bits. The width of the tag field in a 512 KB 8-way set associative cache is _______ bits.

Let the cache parameters be:

• Cache size $C = 512\text{ KB} = 2^{19}\text{ bytes}$

• Set associativity $P = 8 = 2^3$

• Block size $= B = 2^b\text{ bytes}$ (where $b$ is the number of block offset bits)

The number of sets $S$ in the cache is:

Therefore, the number of bits required for the Set index field is:

A physical address is divided into three fields:

Substituting the given physical address width ($40\text{ bits}$):

Thus, the width of the tag field is 24 bits.