Question

Consider a computer system with a byte-addressable primary memory of size \(2^{32}\) bytes.
Assume a direct-mapped cache of size 32 KB and cache block size of 64 bytes. The size of the tag field is \(\underline{\hspace{2cm}}\) bits.

Hint:

Tag bits = Address bits − (Index bits + Offset bits).

Answer 1

Correct Answer: 17

\[ \text{Main memory size} = 2^{32} $\Rightarrow$ \text{Address bits} = 32 \] \[ \text{Cache size} = 32\,\text{KB} = 2^{15} \text{ bytes} \] \[ \text{Block size} = 64 = 2^6 $\Rightarrow$ \text{Offset bits} = 6 \] \[ \text{Number of blocks} = \frac{2^{15}}{2^6} = 2^9 $\Rightarrow$ \text{Index bits} = 9 \] \[ \text{Tag bits} = 32 - (9 + 6) = 17 \] Final Answer: \[ \boxed{17} \]