Consider a computer system with a byte-addressable primary memory of size $2^{32}$ bytes. Assume the computer system has a direct-mapped cache of size 32 KB ($1\text{ KB} = 2^{10}$ bytes), and each cache block is of size 64 bytes. The size of the tag field is ________ bits.

Physical Address bits:

• Primary memory size = $2^{32}$ bytes.

• Physical address space requires $\log_2(2^{32}) = 32$ bits.

Block Offset (Word Offset) bits:

• Cache block (line) size = $64\text{ bytes} = 2^6$ bytes.

• Number of block offset bits = $\log_2(2^6) = 6$ bits.

Cache Line (Index) bits:

• Total cache size = $32\text{ KB} = 32 \times 2^{10}\text{ bytes} = 2^{15}$ bytes.

• Number of cache lines = $\frac{\text{Cache Size}}{\text{Block Size}} = \frac{2^{15}}{2^6} = 2^9 = 512$ lines.

• Number of index bits = $\log_2(2^9) = 9$ bits.

Tag bits:

• In a direct-mapped cache:

  $$\text{Tag bits} = \text{Address bits} - (\text{Index bits} + \text{Block offset bits})$$

  $$\text{Tag bits} = 32 - (9 + 6) = 32 - 15 = 17\text{ bits}$$