Memory Management is a core OS topic and one of the most repeated in ECET. Questions from Paging and Segmentation are simple, often direct, and scoring.
📘 Concept Notes
🔹 Paging
- Paging is a memory management technique where process memory is divided into fixed-size pages and physical memory is divided into frames.
- Pages are mapped to frames using a Page Table.
- Avoids external fragmentation but may have internal fragmentation.
Formula – Logical Address Breakdown
If:
- Logical address size =
bits - Page size =
bytes
Then:
- Page number =
bits - Page offset =
bits
Physical Address
🔹 Example – Paging
- Logical address =
bits - Page size =
→ Offset = 
bits
→ Page number = 
bits
So logical address = 
🔹 Segmentation
- Segmentation divides memory into variable-sized segments based on logical divisions of a program (code, data, stack, etc.).
- Each segment has:
- Segment Number
- Segment Base Address
- Segment Limit
Formula – Physical Address in Segmentation
If logical address = (segment number, offset):
Condition: 
🔹 Comparison Table
| Feature | Paging | Segmentation |
|---|---|---|
| Division | Fixed-size pages | Variable-size segments |
| Address parts | Page no. + Offset | Segment no. + Offset |
| Fragmentation | Internal | External |
| Basis | Physical memory division | Logical division of program |
🔟 10 Expected MCQs – ECET 2026
Q1. Paging divides memory into:
A) Segments
B) Frames & Pages
C) Blocks & Files
D) Tables
Q2. Segmentation divides memory based on:
A) Fixed-size blocks
B) Logical divisions
C) Cache levels
D) Virtual memory
Q3. In paging, the mapping between page and frame is stored in:
A) Segment Table
B) Page Table
C) TLB
D) Cache
Q4. In paging, if logical address size = 12 bits, page size = 
→ page number bits = ?
A) 4
B) 8
C) 12
D) 16
Q5. In segmentation, physical address = ?
A) Segment Number + Offset
B) Base + Offset
C) Limit + Offset
D) Page No. + Offset
Q6. Which problem exists in paging?
A) External fragmentation
B) Internal fragmentation
C) Both A & B
D) None
Q7. Which problem exists in segmentation?
A) External fragmentation
B) Internal fragmentation
C) Both A & B
D) None
Q8. If page size = 
and logical address = 
bits → offset bits = ?
A) 8
B) 10
C) 12
D) 14
Q9. Which technique allows logical division of program (code, data, stack)?
A) Paging
B) Segmentation
C) Swapping
D) Fragmentation
Q10. Advantage of paging is:
A) No external fragmentation
B) No internal fragmentation
C) Faster access
D) Both A & B
✅ Answer Key
| Q.No | Answer |
|---|---|
| Q1 | B |
| Q2 | B |
| Q3 | B |
| Q4 | B |
| Q5 | B |
| Q6 | B |
| Q7 | A |
| Q8 | B |
| Q9 | B |
| Q10 | A |
🧠 Explanations
- Q1 → B: Paging divides memory into pages & frames.
- Q2 → B: Segmentation is based on logical program structure.
- Q3 → B: Page Table stores mapping.
- Q4 → B:
bits → page number. - Q5 → B: Segmentation uses base + offset.
- Q6 → B: Paging causes internal fragmentation.
- Q7 → A: Segmentation causes external fragmentation.
- Q8 → B: Page size
→ 10 offset bits. - Q9 → B: Segmentation allows logical division.
- Q10 → A: Paging removes external fragmentation.
🎯 Why Practice Matters
- Paging & Segmentation questions are compulsory in OS section of ECET.
- Simple formulas & definitions → easy 2–3 marks.
- Also important for interviews and advanced computer architecture.
