GATE CS 2021 Set 2 Q.29 (CS Q.19): Memory Hierarchy: Cache, Main Memory and Secondary Storage

Consider a set-associative cache of size 2KB ( $1\text{ KB} = 2^{10}\text{ bytes}$ ) with cache block size of 64 bytes. Assume that the cache is byte-addressable and a 32-bit address is used for accessing the cache. If the width of the tag field is 22 bits, the associativity of the cache is ______.

Your answer

Official answer and solution

Official answer: 2
From: Official GATE 2021 answer key

Solution

Use the memory address format breakdown into Tag, Set Index, and Block Offset fields.

Given parameters:

Memory address width = 32 bits
Cache block size = 64 bytes $= 2^6$ bytes
Cache capacity = 2 KB $= 2048$ bytes
Tag width = 22 bits

Calculate the field bit widths:

\text{Block offset} = \log_2(64) = 6 \text{ bits}

\text{Set index bits} = \text{Address width} - (\text{Tag bits} + \text{Offset bits}) = 32 - (22 + 6) = 4 \text{ bits}

The number of sets in the cache is:

\text{Number of sets} = 2^4 = 16

The total number of cache lines (blocks) in the cache is:

\text{Total lines} = \frac{\text{Cache size}}{\text{Block size}} = \frac{2048 \text{ bytes}}{64 \text{ bytes}} = 32 \text{ lines}

The associativity $k$ (number of lines per set) is:

k = \frac{\text{Total lines}}{\text{Number of sets}} = \frac{32}{16} = 2

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 - Session 2 (CS2)
Question: 29 of 65, printed as CS Q.19
Section: Computer Science and Information Technology

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