GATE SOLUTION 2025: Consider a memory system with 1M bytes of main memory and 16K bytes of cache

Question 36 GATE 2025 Operating Systems Explained | Cache Memory, Tag Bits & Direct Mapping

In this video, I solve and explain Question 36 from GATE 2025 Operating Systems step-by-step with complete intuition.

We cover:

Cache memory basics
Direct mapping
Tag bits
Index bits
Block offset
Byte addressable memory
Cache address structure

Instead of memorizing formulas, this video focuses on understanding why the address gets divided into:

Tag
Index
Offset

and how to calculate the total number of bits required to store all tag values.

Topics covered:

Main memory vs cache memory
Number of cache blocks
Memory addressing
Direct mapped cache
Address decomposition

If you're preparing for:

GATE CSE
Operating Systems exams
Computer Architecture
Placements & interviews

this video will help strengthen your low-level systems understanding.

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