Coursify

IPv4 Fragmentation with MTU 1500: Computing the Third Fragment’s Header Fields

IPv4 Fragmentation with MTU 1500: Computing the Third Fragment’s Header Fields

Verified Sources
Nov 4, 2026

When an IPv4 router forwards a datagram, it may need to fragment the packet if its size exceeds the outgoing link’s MTU.

For IPv4 fragmentation, three header fields determine how fragments are interpreted:

  • Identification
  • Flags
  • Fragment Offset

We’re given:

  • MTU = 15001500 bytes
  • Received IP packet size = 44044404 bytes
  • IP header length = 2020 bytes

So the original payload (data) size is: 4404−20=4384 bytes4404 - 20 = 4384\text{ bytes}

Assume the router fragments such that each fragment’s total size fits within MTU, and all fragments except the last have payload lengths that are multiples of 8 bytes (a requirement enforced by the use of 8-byte fragment offset units).

IPv4 Fragmentation (MTU, MF flag, Fragment Offset) Tutorial

Step 1: Determine maximum payload per fragment for MTU 1500

Each fragment carries its own IPv4 header of 20 bytes, so the maximum data per fragment is: 1500−20=1480 bytes1500 - 20 = 1480\text{ bytes}

However, the fragment offset counts in 8-byte units, so all non-final fragments should have data lengths that are multiples of 8. Since 14801480 is divisible by 8: 1480/8=1851480/8 = 185

Thus, the router will create fragments with:

  • Fragment 1 data = 14801480
  • Fragment 2 data = 14801480
  • Fragment 3 data = remaining payload

Compute the fragment boundaries and third fragment header fields

  1. 1
    Step 1

    Payload size is 43844384 bytes because the header is 2020 and total is 44044404.

  2. 2
    Step 2

    Max data per fragment is 1500−20=14801500 - 20 = 1480 bytes.

  3. 3
    Step 3

    After two full fragments, remaining data is 4384−1480−1480=14244384 - 1480 - 1480 = 1424 bytes, so fragment 3 carries 14241424 bytes.

  4. 4
    Step 4

    Fragment 3 is the last fragment, so MF = 0 (no more fragments follow).

  5. 5
    Step 5

    Fragment offset is in 8-byte units. Fragment 3 starts after fragment 1 and fragment 2 data: 1480+1480=29601480 + 1480 = 2960 bytes. Therefore offset = 2960/8=3702960/8 = 370.

  6. 6
    Step 6

    Identification is copied unchanged from the original datagram into all fragments. DF is typically cleared by the router when it performs fragmentation.

Answer: Fields in the header of the third IP fragment

Let the original datagram’s Identification be IDID (unknown from the problem statement). The router copies it into all fragments unchanged.

1) Flags

  • DF: The router fragments, so DF must be 0 in the sent fragments.
  • MF: Since this is fragment 3 (the last fragment), MF = 0.

So the flags bits are: MF = 0 and DF = 0 (with the reserved/other bit also 0 under standard usage).

2) Fragment Offset

Fragment 3 begins after 29602960 bytes of data from the original payload: 2960/8=3702960/8 = 370

So the fragment offset value is 370.

3) Total Length of third fragment (for completeness)

This isn’t explicitly requested in “relevant fields,” but is typically needed to sanity-check fragmentation:

  • Third fragment payload = 14241424 bytes
  • Third fragment header = 2020 bytes Total Length3=1424+20=1444 bytes\text{Total Length}_3 = 1424 + 20 = 1444\text{ bytes} And indeed 1444≤15001444 \le 1500.

4) Identification

Identification = original ID (copied unchanged into all fragments).

Key rule to remember

In IPv4, the fragment offset is measured in 8-byte blocks of the original payload, not including the IPv4 header.

Why your payload sizes must align

All fragments except the last must have data lengths that are multiples of 8 bytes; otherwise the offset values can’t represent their boundaries using 8-byte units.

Fragmentation decision flow in an IP router

Check MTU vs datagram size

1

If the datagram’s total size exceeds outgoing MTU and DF is not set, fragmentation is permitted."

Choose per-fragment payload size

2

Pick max data per fragment = MTU − header_len, then adjust to 8-byte alignment for non-final fragments."

Set header fields for each fragment

3

Copy Identification; set DF/MF; compute Fragment Offset in 8-byte units."

Emit fragments

4

Send fragments; receiver reassembles using Identification + offsets + MF/last-fragment behavior."

IPv4 fragmentation for payload 4384 bytes (MTU 1500, header 20 bytes)

Third fragment carries the remainder and therefore has MF = 0.

Quick mapping of computed values to header fields

IPv4 Fragmentation Mini-Deck

1 / 5
Question · Term

MTU 1500 with IPv4 header 20 bytes: max data per fragment?

Click to reveal
Answer · Definition

1480 bytes = 1500 − 20; for non-last fragments, ensure data is a multiple of 8.

Knowledge Check

Question 1 of 4
Q1Single choice

With MTU = 1500 and IPv4 header = 20 bytes, the maximum data size per fragment (before considering 8-byte alignment) is:

Explore Related Topics

1

Ethernet (IEEE 802.3) Header Formats and the Need for Padding

2

The Length of an IPv6 Address (IPv6 Addressing Fundamentals)

3

The checksum in IP must be recomputed at every router because of change in ____ fields. (i) TTL, Options, Identification Number, Offset (ii) TTL, Options, Data, Offset (iii) TTL, Options, Datagram Length, Offset (iv) TTL, Header Length, Offset, ToS

Browse all research articles