GATE CS 2014 Set 2 Q.19 (CS Q.9): Memory Hierarchy: Cache, Main Memory and Secondary Storage

A 4-way set-associative cache memory unit with a capacity of 16 KB is built using a block size of 8 words. The word length is 32 bits. The size of the physical address space is 4 GB. The number of bits for the TAG field is _____.

Your answer

Official answer and solution

Official answer: 20
From: Official GATE 2014 answer key

Solution

Decompose the physical address into tag, set index, and block offset fields using cache capacity and set associativity.

Physical Address Space:
- Size = $4\text{ GB} = 2^{32}$ bytes.
- Physical address length = $32$ bits.
Block Offset (Byte Offset):
- Block size = $8$ words.
- Word size = $32$ bits = $4$ bytes.
- Block size in bytes = $8 \times 4 = 32$ bytes $= 2^5$ bytes.
- Block offset = $5$ bits.
Set Index Bits:
- Total cache capacity = $16\text{ KB} = 16 \times 1024$ bytes $= 2^{14}$ bytes.
- Total number of cache blocks: $\text{Lines} = \frac{2^{14}}{2^5} = 2^9 = 512\text{ blocks}.$
- The cache is 4-way set-associative ( $4$ blocks per set): $\text{Number of sets} = \frac{512}{4} = 128 = 2^7\text{ sets}.$
- Set index = $7$ bits.
Tag Field Bits:
$\text{Tag} = \text{Total Address Bits} - (\text{Set Index Bits} + \text{Block Offset Bits})$ $\text{Tag} = 32 - (7 + 5) = 32 - 12 = 20\text{ bits}.$

Final answer

Common trap

Treating block size as 8 bytes instead of 8 words (32 bytes).

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

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 2
Question: 19 of 65, printed as CS Q.9
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source