Computer Networks · TEE · practice
Optional fragmentation · example 2
IP packet details · supporting practice · optional practice
Optional extension · no matching calculation in the two supplied papers
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Practice question. An unfragmented IPv4 packet is 2000 bytes including a 20-byte header. It crosses a link of MTU 620 bytes. DF=0 and every fragment repeats the same header. Find each fragment’s payload length, total length, offset and MF flag. [3 practice marks]
Answer
Given
Total=2000; H=20; MTU=620 bytes; DF=0.
To find
All fragments and their header-related values.
Formula and why
LaTeX: P=T-H
LaTeX: P_{\max}=8\left\lfloor\frac{\mathrm{MTU}-H}{8}\right\rfloor
LaTeX: \mathrm{offset}=\mathrm{preceding\ payload\ bytes}/8Subtract the header before splitting the original payload. Round non-final fragment payload capacity down to a multiple of eight.
Step-by-step solution
Step 1. Original payload
2000−20=1980 bytes.
Step 2. Non-final payload capacity
MTU−H=620−20=600. Floor((600)/8)×8=600 bytes.
Step 3. Allocate each fragment
| Fragment | Payload bytes | Total bytes | Starting byte | Offset /8 | MF |
|---|---|---|---|---|---|
| 1 | 600 | 620 | 0 | 0 | 1 |
| 2 | 600 | 620 | 600 | 75 | 1 |
| 3 | 600 | 620 | 1200 | 150 | 1 |
| 4 | 180 | 200 | 1800 | 225 | 0 |
Step 4. Verify no payload is lost
600+600+600+180=1980 bytes. Every total length is at most 620.
Final answer: 4 fragments; lengths, offsets and MF are listed in the table.