Performance

Ethernet Frame & PPS

Pick a frame size and link speed to get on-wire bytes, packets per second at line rate, efficiency and real goodput. Handy for sizing firewalls, routers and NICs.

redzilla.cl — eth

Packets per second

—

at line rate

On-wire bytes

—

frame + preamble + IFG

Efficiency

—

payload / wire

Wire usage

Useful payload —

—

PPS —

Wire —

Efficiency —

Goodput —

Frame breakdown

Field	Bytes	Notes

How it is computed · overhead, PPS and goodput

1. Every L2 frame (dst 6 + src 6 + ethertype 2 + payload + FCS 4) also carries preamble+SFD (8 B) and the IFG (12 B): +20 B that are not data but still occupy the wire.

2. On-wire bytes: wire = frame + 20. At line rate, PPS = speed / (wire × 8).

3. The payload is the useful data (frame − 18: minus header and FCS; with a VLAN the frame already includes those 4 B). Efficiency: payload / wire.

4. Real goodput: PPS × payload × 8. It is the "application" data bandwidth after removing all link-layer overhead.

Reference: 64 B @1G = 1,488,095 pps · 1518 B @10G = 812,743 pps · efficiency 1518 B = 97.5%.

Runs locally in your browser · no sign-up · nothing leaves your browser

How it works

The calculator takes an Ethernet frame size (as a full L2 frame or as payload) and the link speed (from 10 Mbps to 100 GbE) and returns the bytes it actually occupies on the wire, the packets per second (PPS) at line rate, the efficiency and the real data goodput. It is the key calculation for sizing firewalls, routers and NICs, whose limit is usually PPS rather than Gbps.

Besides the frame itself (14 B header, payload and 4 B FCS, per IEEE 802.3), every transmission consumes 8 B of preamble+SFD and 12 B of inter-frame gap: 20 extra bytes that are not data but occupy the wire. Hence wire = frame + 20, PPS = speed / (wire × 8), efficiency is payload / wire and goodput is PPS × payload × 8. The optional 802.1Q tag adds 4 B, and frames from 64 B up to 9018 B jumbo are supported.

Example: minimum 64 B frame on a 1 GbE link

Bytes on the wire: 64 + 8 (preamble+SFD) + 12 (IFG) = 84 B = 672 bits.
PPS at line rate: 10⁹ / 672 ≈ 1,488,095 packets per second.
Useful payload: 64 − 18 = 46 B, efficiency 46/84 ≈ 54.8%: the real goodput is about 548 Mbps out of the 1000 the link offers.

Frequently asked questions

How many packets per second does a 1 Gbps or 10 Gbps link carry?

It depends on the frame size. With minimum 64 B frames, 1 GbE carries 1,488,095 pps and 10 GbE reaches 14.88 million pps. With maximum 1518 B frames, the figures drop to 81,274 pps and 812,743 pps respectively. The minimum-frame value is the one used to benchmark routers and firewalls in the worst case.

Why does Ethernet efficiency never reach 100%?

Because every frame drags fixed overhead: a 14 B header, 4 B FCS, 8 B preamble+SFD and 12 B inter-frame gap, 38 B in total that are not data. With 1518 B frames the efficiency is about 97.5%, but with 64 B frames it falls to 54.8%: more than half of the wire goes to overhead. That is why small-packet traffic is so demanding.

What do I gain with 9000-byte jumbo frames?

Less overhead per transported byte and far fewer packets to process: a 9018 B jumbo frame reaches 99.6% efficiency and cuts PPS about 6 times compared with 1518 B, offloading the CPU in storage (iSCSI, NFS) and virtualization. The condition is that every device along the path supports the same MTU; if one hop does not, you get fragmentation or drops.

Does the VLAN tag change the maximum frame size?

Yes: the 802.1Q tag adds 4 B between the source MAC and the EtherType, so the standard maximum frame grows from 1518 to 1522 B (IEEE 802.3ac) and the minimum from 64 to 68 B. The calculator adds it automatically when you tick the box, both in the on-wire bytes and in the PPS.

Was this tool useful?

Disclaimer

We take great care to keep every tool accurate and review it thoroughly; even so, we can't guarantee it is free of errors or take responsibility for how the results are used. We recommend double-checking anything critical.

Found an error? Let us know →