# Optional fragmentation · example 2

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/practice/packet-support/fragment-2

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/practice/packet-support--fragment-2/

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



$$
P=T-H
$$





$$
P_{\max}=8\left\lfloor\frac{\mathrm{MTU}-H}{8}\right\rfloor
$$





$$
\mathrm{offset}=\mathrm{preceding\ payload\ bytes}/8
$$



Subtract 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.

#### What this result means

Only payload positions determine offsets; repeating a header does not move the offset of later original data.