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Editorial Team
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Asked: 2026-05-12T11:07:58+00:00

What is the overhead for PPP and Ethernet sending 5000 bytes? Frame size for

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What is the overhead for PPP and Ethernet sending 5000 bytes?

Frame size for Point-to-Point Protocol: 8 bytes
MTU: 500 bytes

Frame size for Ethernet: 18 bytes
MTU: 1500 bytes

Both sending 5000 bytes..

I know this is just a calculation, but I am not sure how to do it. I can’t find it anywhere. I would think that since a PPP frame takes 8 bytes and maximum transmission unit is 500 then it can send (500 – 8)bytes of information in one go. It sends 10 frames, resulting in 4920 bytes sent. Then sends the final (80+8)bytes with the last frame.

Similar for Ethernet. (1500 – 18)bytes with each frame. 3 frames sent means 4446 bytes sent. Sending (554+18)bytes in the last frame.

This obviously doesn’t answer the “overhead” question. Anyone have any ideas?

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1 Answer

  1. Editorial Team

    It really depends on how you define overhead. This answer will assume overhead is the number of bytes that you need to transmit in addition to the data itself.

    For Ethernet, assuming the 5000 byte payload is not encapsulated in an IP + TCP/UDP frame, you will have 18 bytes of overhead for every packet sent. That means each transmission with an MTU of 1500 will be able to hold 1482 bytes. To transmit 5000 bytes, this means 4 packets must be transmitted, which means an overhead of 72 bytes (18 * 4). Note that the overhead becomes bigger when you include things like the IP frame which contains a TCP frame.

    For PPP, as you’ve already shown, you can send 492 bytes per frame. Eleven frames means 88 bytes of overhead (11 * 8) – again, not including any additional protocol frames within the payload.

    In both these examples any protocols that build on top of these link layer protocols will contribute to overhead. For example, an Ethernet packet sent with an IPv4 frame which contains a UDP datagram will have an additional 28 bytes consumed by headers and not data (20 bytes in an IPv4 header and 8 in a UDP header, assuming no IP options). Considering the original Ethernet example, this means the amount of data per packet becomes 1454 bytes, which luckily still comes to 4 packets (the extra spills over into the smaller 4th packet), with 144 bytes of overhead.

    You can read more here (I find that page a little hard to read though).

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