I'm doing some research on switching algorithms using a Level 1 GSW-2476 switch and I have a problem regarding MAC learning. As far as I know, a switch stores the source MAC address of each incoming packet (regardless of the packet content) in its MAC address table. However, I cannot learn the switch using this simple rule! This is my experiment:

I have connected 4 ports of the switch to a hardware traffic generator. Using the traffic generator I send this packet from port 1:

E1 E1 E1 E1 E1 E1 13 13 13 13 13 13 08 00 55 55 55 ... 55 55

i.e. a destination MAC address of E1E1E1E1E1E1, a source MAC address of 131313131313, an Ethertype of 0800 and about 200 bytes of value 55 as the frame payload. Consequently, it is expected that the switch attaches 131313131313 to the port 1. However, when I send a packet with a destination MAX address of 131313131313 (such as the below packet) using one of the other ports, the switch sends the packet to all the ports!

13 13 13 13 13 13 14 14 14 14 14 14 08 00 55 55 55 ... 55 55

It should also be noted that the switch is in its reset factory settings.


A problem is that your MAC addresses have the I/G bit set to indicate that they are group addresses, not individual addresses. Group addresses are not allowed to be source addresses, only destination addresses, and frames with destination group addresses are sent to all interfaces.

If you are going to make up MAC addresses, you should set the U/L bit to indicate that the address is locally administered, and you can send to a group address, but not from a group address.

| improve this answer | |
  • Thank you @Ron-Maupin. As you mentioned, the problem was due to the I/G bit. I didn't know anything about this bit! – M.H Jun 8 '18 at 18:11
  • It really helps to get a good understanding of ethernet before you try to understand ethernet switching. You can get the IEEE 802 standards for free, and you should familiarize yourself with the ethernet (IEEE 802.3) and multi-protocol (IEEE 802.1) standards. – Ron Maupin Jun 9 '18 at 23:07

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.