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

Step 1: Determine total address size.
Main memory size is 2³² bytes, so the number of address bits required is:

Address bits = 32

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Step 2: Analyze cache parameters.
Cache size = 32 KB = 32 × 1024 bytes = 2¹⁵ bytes

Block size = 64 bytes = 2⁶ bytes

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Step 3: Calculate offset and index bits.
Offset bits = log₂(block size) = 6

Number of cache blocks = 2¹⁵ / 2⁶ = 2⁹

Index bits = log₂(number of blocks) = 9

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Step 4: Compute tag bits.
Tag bits = Total address bits − (Index bits + Offset bits)

Tag bits = 32 − (9 + 6) = 17

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Final Answer:

17