ISRO CS 2025 Q.70: solution and answer

Official answer: DSource: ISRO's final answer key of 15 May 2026

1. Optimal replacement evicts the page in memory whose next use lies farthest in the future, or that is never used again. Start with 3 empty frames.

2. References $6,0,5$ fill the empty frames: 3 faults. Frames $\{6,0,5\}$.

3. Reference $2$ is a fault (4th). Page $6$ is never used again, so evict it. Frames $\{2,0,5\}$. The next reference $0$ is a hit.

4. Reference $3$ is a fault (5th). Remaining references: $0,4,2,3,0,3,2,5,2$. Page $0$ is needed first, $2$ is needed 3rd, and $5$ is needed 8th, so evict $5$. Frames $\{2,0,3\}$. The next reference $0$ is a hit.

5. Reference $4$ is a fault (6th). Remaining references: $2,3,0,3,2,5,2$. The first uses are $2$, then $3$, then $0$, so evict $0$. Frames $\{2,4,3\}$. The next references $2$ and $3$ are hits.

6. Reference $0$ is a fault (7th). Remaining references: $3,2,5,2$. Page $4$ is never used again, so evict it. Frames $\{2,0,3\}$. The next references $3$ and $2$ are hits.

7. Reference $5$ is a fault (8th). The only remaining reference is $2$, so either $0$ or $3$ may be evicted; the count is unaffected. The final $2$ is a hit.

8. Total page faults $=3+1+1+1+1+1=8$.