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Calcrivo

MTU Calculator

Calculate how many fragments a payload requires for a given MTU, and the resulting overhead.

Inputs

bytes
bytes

Standard Ethernet MTU is 1500 bytes; jumbo frames are typically 9000 bytes

bytes

IPv4 (20B) + TCP (20B) headers by default

Fragments Required

4

Usable Payload per Packet

1,460bytes

Total Bytes on Wire

4,660bytes

Header Overhead

3.56%

Step by step

  1. Values used

    Payload Size = 4,500 bytes; MTU = 1,500 bytes; Header Overhead per Packet = 40 bytes

  2. Fragment count

    fragments = ceil(payload_size / (MTU − header_overhead))

  3. Fragments Required

    = 4

  4. Usable Payload per Packet

    = 1,460 bytes

  5. Total Bytes on Wire

    = 4,660 bytes

  6. Header Overhead

    = 3.56

How it works

The MTU (Maximum Transmission Unit) is the largest packet size a network link can carry in a single frame — 1500 bytes for standard Ethernet, or up to 9000 bytes for 'jumbo frames' on networks that support them. If a payload plus its headers exceeds the MTU, it must be split into multiple fragments, each carrying its own header overhead, which reduces the useful data ratio and adds reassembly work at the destination. Path MTU Discovery is the mechanism modern stacks use to avoid fragmentation by finding the smallest MTU along a route.

Formula

Fragment count

fragments = ceil(payload_size / (MTU − header_overhead))

S
Payload size in bytes
h
Header overhead per packet (e.g. IPv4+TCP = 40 bytes)

Frequently Asked Questions

What is the standard MTU for Ethernet?

1500 bytes is the standard Ethernet MTU, defined by the original Ethernet II frame format. This has remained the default across most LANs and the public internet for decades.

What are jumbo frames and why use them?

Jumbo frames raise the MTU to typically 9000 bytes, reducing the number of packets (and thus header overhead and CPU interrupts) needed to move the same amount of data. They're common in data center and storage networks but must be supported end-to-end to be useful.

What happens if a packet is larger than the path MTU?

Routers either fragment the packet into smaller pieces (adding overhead and reassembly work) or, if the Don't Fragment (DF) bit is set, drop the packet and return an ICMP 'Fragmentation Needed' message so the sender can reduce its packet size.

Why does fragmentation increase overhead?

Every fragment needs its own copy of the packet headers (IP, and sometimes TCP), so splitting one large packet into several smaller ones multiplies the header bytes sent, reducing the fraction of the transmission that's actual payload data.

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