GATE CS 2017 Set 2 Q.53 (CS Q.53): Memory Hierarchy: Cache, Main Memory and Secondary Storage

Consider a machine with a byte addressable main memory of $2^{32}$ bytes divided into blocks of size 32 bytes. Assume that a direct mapped cache having 512 cache lines is used with this machine. The size of the tag field in bits is

Your answer

Official answer and solution

Official answer: 18
From: Official GATE 2017 answer key

Solution

Decompose the physical address into Tag, Cache Line Index, and Block Offset fields.

Given parameters:

Byte-addressable main memory size $= 2^{32}$ bytes $\implies$ Physical Address $= 32$ bits.
Cache block size $= 32$ bytes $= 2^5$ bytes $\implies$ Block Offset $= 5$ bits.
Number of cache lines in direct-mapped cache $= 512 = 2^9 \implies$ Cache Index $= 9$ bits.

In a direct-mapped cache, the physical address is partitioned as:

\text{Physical Address (32 bits)} = \text{Tag} + \text{Index} + \text{Offset}

\text{Tag} = 32 - (\text{Index} + \text{Offset}) = 32 - (9 + 5) = 32 - 14 = 18 \text{ bits}

Thus, the size of the tag field is 18 bits.

Final answer

Common trap

Confusing line index with word offset or forgetting that cache size in lines gives $2^9$ directly.

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

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 53 of 65, printed as CS Q.53
Section: Computer Science and Information Technology

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