# Network Packet Overhead Calculator

> See how many packets and bytes on the wire it takes to send data over Ethernet with IPv4 or IPv6 and TCP or UDP, and the resulting protocol efficiency.

Category: IT, networking & security · Applies to: Worldwide · Last updated: 2026-10-29
Interactive calculator: https://calculators.geekmidst.com/calculators/network-packet-overhead-calculator

Every packet carries headers: Ethernet (14 bytes + 4 FCS, plus 8 preamble and 12 inter-frame gap on the wire), IPv4 (20) or IPv6 (40), and TCP (20 + options) or UDP (8). This shows how much of the link actually carries your data.

Also called: TCP/IP overhead calculator, Ethernet frame efficiency calculator.

## Inputs
- Application data to send (unit bytes; default 1460)
- MTU (unit bytes; default 1500)
- IP version (options: IPv4, IPv6; default v4)
- Transport (options: TCP, UDP; default tcp)
- TCP options (unit bytes; default 0; optional)
- VLAN tag (options: None, 802.1Q (+4 bytes); default none)

## Results
- Efficiency (data ÷ bytes on the wire)
- Packets
- Max data per packet (MSS)
- Ethernet frame bytes
- Bytes on the wire
- Overhead

## Formula

```
MSS = MTU − IP header − TCP/UDP header;  Packets = ⌈Data ÷ MSS⌉;  Frame = 14 (+4 VLAN) + IP + L4 headers + data + 4 FCS, at least 64 bytes;  Wire = Frames + 20 bytes per packet (preamble, SFD and inter-frame gap)
```

Where:
- MTU: largest IP packet the link carries
- MSS: largest data chunk per packet

## Worked example
1. MSS = 1,500 − 20 (IPv4) − 20 (TCP) = 1,460 bytes, so one packet.
2. Frame = 14 + 1,500 + 4 = 1,518 bytes; on the wire add 8 + 12 = 1,538 bytes.
3. Efficiency = 1,460 ÷ 1,538 = 94.93%.

**Result:** A full-size TCP/IPv4 segment on standard Ethernet is 94.93% efficient, so a 1 Gbit/s link carries at most about 949 Mbit/s of TCP data.

## What the result means

Small packets are very inefficient: one byte of UDP data still occupies 84 bytes of wire time. Larger MTUs (jumbo frames) raise efficiency only if every hop supports them.

## Assumptions and limits
- Data larger than one packet is split into full-size packets plus a final partial packet.
- For UDP, the application sends separate datagrams rather than relying on IP fragmentation.
- TCP acknowledgements, retransmissions and tunnel headers are not included.

## Questions

### Why is the minimum frame 64 bytes?

Ethernet requires at least 64 bytes from destination address to FCS so collisions can be detected; shorter frames are padded.

### What MSS does IPv6 give on a 1,500-byte MTU?

1,500 − 40 − 20 = 1,440 bytes, 20 less than IPv4 because the IPv6 header is twice as long.

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Source: GM¹X CalcVerse — https://calculators.geekmidst.com/calculators/network-packet-overhead-calculator