I'm trying to figure out what is the lowest layer of the OSI model that can be used to extract a bit rate (through a packet sniffer). I'm not sure if it's possible through only the physical layer, or it's necessary to have information about the link layer.


To answer your question as intended, the network card pass to the packet sniffer only valid frames. That means it has already filtered corrupted frame and this is a function of the data link layer (OSI layer 2).

In fact all the physical layer functions (and some of the data link layer functions) are performed by the NIC itself and not by the operating system, so using a standard packet sniffer you can't see what happens at the physical layer. You would need specialized devices to do that.

But actually the answer should be "none", because the OSI model is just that, a model. It is not used in real life.

What is actually used is the TCP/IP model, in which the Network Access is a single layer that encompass all roles that are separated in two layers in the OSI model: physical and data link.

Note that despite this, even experienced people well aware of the TCP/IP model still split the Network access layer in two, and refer to the IP layer as layer 3 but strictly speaking this terminology is incorrect, and IP should be referred as layer 2

But since most networking courses begin with the OSI model instead of the TCP/IP model, this would cause many confusion.

  • Well put. I usually refer to the lower four (OSI) layers using the OSI model and the rest is just application layer like in the TCP/IP / DoD / IETF model - causes less confusion and is precise enough.
    – Zac67
    Mar 26 '18 at 17:09
  • Thank you very much. I am new to networking, and studied the 5 layer model (The top layers were just combined into the application layer). I didn't think of it like that.
    – Tom Dawn
    Mar 26 '18 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.