GATE CS 2021 Set 1 Q.32 (CS Q.22): Memory Hierarchy: Cache, Main Memory and Secondary Storage

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 $32\text{ KB}$ ( $1\text{ KB} = 2^{10}\text{ bytes}$ ), and each cache block is of size $64\text{ bytes}$ .

The size of the tag field is ___________ bits.

Your answer

Official answer and solution

Official answer: 17
From: Official GATE 2021 answer key

Solution

Decompose the 32-bit physical address into Tag, Line/Index, and Block Offset fields for a direct-mapped cache.

Given parameters:

Byte-addressable memory of size $2^{32}$ bytes $\implies$ Physical Address size is 32 bits.
Cache block (line) size $= 64\text{ bytes} = 2^6$ bytes $\implies$ Block offset field $= 6$ bits.
Total cache size $= 32\text{ KB} = 32 \times 2^{10} = 2^{15}$ bytes.

In a direct-mapped cache, the number of cache blocks (lines) is:

\text{Number of lines} = \frac{\text{Cache size}}{\text{Block size}} = \frac{2^{15}}{2^6} = 2^9 = 512

Thus, the Line Index field requires $\log_2(2^9) = 9$ bits.

The tag field size is obtained by subtracting the index and offset bits from the total address width:

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

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2021 Computer Science and Information Technology (CS, Set-1)
Question: 32 of 65, printed as CS Q.22
Section: Computer Science and Information Technology

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