Question

Consider a processor that generates 32-bit virtual addresses. It uses a direct mapped cache whose size is 32 KByte and has 32 Byte blocks. What is the number of bits needed for cache addressing?

• A. 5
• B. 10
• C. 15
• D. 20

Hint:

Cache addressing bits are calculated as: Index bits + Block offset bits.

Answer 1

The Correct Option is C

Step 1: Understanding cache parameters.
Cache size = 32 KB = $2^{15}$ bytes
Block size = 32 bytes = $2^5$ bytes
Step 2: Calculating number of cache blocks.
\[ \text{Number of blocks} = \frac{2^{15}}{2^5} = 2^{10} \]
Step 3: Calculating block offset bits.
Block size is $2^5$ bytes, so block offset requires $5$ bits.
Step 4: Calculating index bits.
Number of cache blocks is $2^{10}$, so index bits required are $10$.
Step 5: Total cache addressing bits.
Cache addressing bits = Index bits + Block offset bits
\[ 10 + 5 = 15 \]
Step 6: Final conclusion.
Thus, the number of bits needed for cache addressing is 15.