Awesome, thank you!!!
Pat
1 Like
Hey Rene,
As per my understanding, F0/1 should be blocked since it has the lower port no in the below case.
Please suggest.
Hello Aamir
When selecting which port to block on a switch, the following four parameters that are found within received BPDUs. Specifically, one BPDU is considered superior to another if it has:
- The lower root bridge ID
- The lower path cost to the Root
- The lower sending Bridge ID
- The lower sending port ID
These parameters are checked in order, and the next is only checked if the previous is a tie. So in this case, SW2 will receive BPDUs from the ROOT on both links. Both root bridge IDs are the same, both path costs are the same, and both sending Bridge IDs are the same. The last thing to check is the sending port ID.
Fa0/1 on SW2 receives a BPDU with the port ID of Fa0/1 from SW1. Similarly, Fa0/2 on SW2 receives a BPDU with the port ID of Fa0/2 from SW1. The port that receives a BPDU with the lowest port ID is the one chosen to forward traffic. Therefore Fa0/2 on SW2 becomes blocked. Does that make sense?
I hope this has been helpful!
Laz
hello Lazaros,
Thanks for the explanation. I appreciate that.
Amir
1 Like
Hi Rene,
Somebody maybe already ask this question. I just curious about this statement below,
‘SW1 will forward this broadcast frame on all it interfaces, except the interface where it received the frame on’
is that means SW2 will received two ARP packets at the same time on its fa0/0 and fa0/1 interfaces, or is that only one packet at a time?
Thank You
Hello Christevand
This statement simply indicates the fundamental operation of a switch. If we were to assume that Spanning Tree Protocol (STP) is not functioning at all, then the answer to your question is that yes, SW2 will receive ARP requests on both Fa0/0 and Fa0/1.
This however, as further described in the lesson, would cause problems, because SW2 would then take the ARP request it got on Fa0/0 and it would send it out of Fa0/1, and it would take the ARP request it got on Fa0/1 and send it out of Fa0/0. This results in a layer 2 loop, and a broadcast storm will take place, causing both switches to be overwhelmed, resulting in massive network slowdowns.
This example shows why STP is necessary. If STP is enabled, as it should be, this will not happen. STP will cause one of the two links between the two switches to be in a blocking state, thus only one of the two interfaces (Fa0/0 or Fa0/1) on SW2 will receive the ARP request. Which port that will be depends on the STP configuration. This resolves the layer 2 loop problem. Does that make sense?
I hope this has been helpful!
Laz
Hi Laz,
This is now make sense. Thank you for answering, much appreciate it!
Evand
1 Like
Hi Rene
I am not understanding how this topology would cause a switching loop. When H1 sends and arp request, SW1 will broadcast the frame out of all its ports except the source port. When Sw2 gets the broadcasts from both the gig1 and gig2 interfaces, the switch would check its MAC address table against the destination MAC address of the ETH Frame and send both frames to H2. Im not understanding how the frame would again be broadcasted out of SW2. Please help me understand where my thinking is wrong here.
Hello Pancratius
ARP requests use a destination MAC address of FF:FF:FF:FF:FF:FF. This is a broadcast address. When a frame with this address as a destination is received on a switch, it will be sent out of all of the switch’s ports except the one it received it on. So the MAC address table does not play any role when such a frame is received.
So if such a frame is received on Fa0/0 of SW2 it will be sent out of all ports including Fa0/1, and will be sent back to SW1. Similarly, it is also received on Fa0/1 of SW2 and it will be sent back to SW1 via Fa0/0. SW1 will receive these broadcasts and will rebroadcast them to all ports including those going back to SW2, and so on, resulting in a broadcast storm due to a Layer 2 loop. Does that make sense?
For more information about ARP, take a look at this lesson:
I hope this has been helpful!
Laz
Hi Lazaros
This does help a lot. But I have one more question. When the broadcast of FF:FF:FF:FF:FF:FF arrives at SW2, it will eventually get to H2 and H2 would reply to the ARP request. The SW2 should then add H2s mac address to the table wouldnt it? I know this would not stop the broadcast storm but just wanted clarity on this regard.
Thanks
Hello Pancratius
Yes that is exactly correct. Because the ARP reply from H2 contains H2’s MAC address in the source MAC address field of the frame, the switch will place that MAC address into the MAC address table, and it will correspond with the port number on which H2 is connected to the switch.
This is true of any frame that arrives on a switch’s port, the source MAC address in the header will be added to the MAC address table.
I hope this has been helpful!
Laz