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Bronze

rstp weird scenario

hi every body!

I have this weird scenario:

sw2,sw3 are connected to each other by f0/2 on each switch,also sw2,sw3 are connected to root switch sw1 by fo/1 on each switch. They are connected in traingular fashion.Please assume the cost of each link being 1.

fo/1 ports on sw2 and sw3 are root ports,f0/2 port on sw2 is alternate(blocking)while sw3 f0/2 port is designated port.

Let say link between sw1 and sw2 (f0/1) goes down, and sw2 put its alternate port f0/2 in forwarding.

Now we change cost of the link between sw1 (f0/1) and sw2 (F0/1)from 1 to 50 and enable the link.

SW1 f0/1 port would be in designated(blocking) ,sends the bpdu with proposal bit set to sw2.

My question is what will sw2 do now? will sw2 ignore the proposal from sw1 because the bpdu is not superior?

if yes, then sw1 will continue to send bpdu with proposal bit set . Am i right?

when will sw1 will stop sending bpdu with proposal bit set to sw2?

thanks a lot!

2 ACCEPTED SOLUTIONS

Accepted Solutions
Hall of Fame Super Silver

Re: rstp weird scenario

Hello Sarah,

in STP and Rapid STP port cost is the cost for the incoming port that receives the BPDU.

If you change the cost on both ends of the link on sw2 it is the incoming cost that of its port that counts.

So the BPDU received by sw1 that is the root bridge is always superior with a cost to root of 0.

Actually all ports on the root bridge are placed in forwarding and there is no root port.

So the handshake is winned by sw1 port, the only difference is that the link will become the alternate port on sw2 for its higher path cost as calculated on sw2 in comparison to fas0/2 via sw3.

Hope to help

Giuseppe

Re: rstp weird scenario

As Guiseppe mentioned earlier, the cost configuration on the root bridge has no use. The cost is only used when electing the root port, and the root port has no root bridge.

Let's go back to when the link is coming up between sw1 (root) and sw2. Both ports send a bpdu to each other. The port on sw1 considers the bpdu it receives as worse and ignores it. The port on sw2 considers the bpdu it receives a better. However, it is not as good as the information it receives on f0/2, so f0/1 is elected as alternate.

Now, whether f0/1 on sw2 sends back an agreement to the designated port on sw1 depends on your software. The latest version should send back an agreement (in which case the designated port on the root bridge sw1 will go forwarding immediately), else the port will go forwarding slowing (2xforward-delay). In theory, it does not matter much as this p2p link is blocked on one side anyway.

Regards,

Francois

6 REPLIES
Hall of Fame Super Silver

Re: rstp weird scenario

Hello Sarah,

in STP and Rapid STP port cost is the cost for the incoming port that receives the BPDU.

If you change the cost on both ends of the link on sw2 it is the incoming cost that of its port that counts.

So the BPDU received by sw1 that is the root bridge is always superior with a cost to root of 0.

Actually all ports on the root bridge are placed in forwarding and there is no root port.

So the handshake is winned by sw1 port, the only difference is that the link will become the alternate port on sw2 for its higher path cost as calculated on sw2 in comparison to fas0/2 via sw3.

Hope to help

Giuseppe

Re: rstp weird scenario

I agree with Giuseppe. To give more details on the proposal bit, again, the designated port will propose as long as it is not forwarding. If f0/1 on sw2 does not send back and agreement, it will take 2xforward delay.

Actually, the latest RSTP standard specifies that even blocked port should send back an agreement. So when f0/1 on sw2 sends back the agreement, the port on sw1 goes to forwarding and stop proposing.

Regards,

Francois

Bronze

Re: rstp weird scenario

thanks a lot Francois and Guiseppe!please consider the old rstp specification .

Let me recap my scenario

When sw2 lost its root port(f0/1), it puts its f0/2 alternate port in forwarding mode right away. The cost on f0/1 on sw2 is changed to 50, same action is performed on sw1 port connected to sw2 f0/1.Now the link is brought up.Sw1 sends Bpdu on port connected to f0/1 of sw1 with proposal bit set.

sw1 receives bpdu with proposal bit set and concludes it is not superior bpdu.Will sw1 ignore this bpdu or it will send bpdu back with agreement bit set and designates the f0/1 port as alternate port? I know that f0/1 on sw1 becomes alternate port but will sw2 sends bpdu with agreement bit set so sw1 does not need to wait for 30 seconds to put the port in forwarding or sw2 won't send any agreement which will cause sw2 to wait for 30 seconds before it could port in forwarding?

thanks a lot!

Re: rstp weird scenario

As Guiseppe mentioned earlier, the cost configuration on the root bridge has no use. The cost is only used when electing the root port, and the root port has no root bridge.

Let's go back to when the link is coming up between sw1 (root) and sw2. Both ports send a bpdu to each other. The port on sw1 considers the bpdu it receives as worse and ignores it. The port on sw2 considers the bpdu it receives a better. However, it is not as good as the information it receives on f0/2, so f0/1 is elected as alternate.

Now, whether f0/1 on sw2 sends back an agreement to the designated port on sw1 depends on your software. The latest version should send back an agreement (in which case the designated port on the root bridge sw1 will go forwarding immediately), else the port will go forwarding slowing (2xforward-delay). In theory, it does not matter much as this p2p link is blocked on one side anyway.

Regards,

Francois

Bronze

Re: rstp weird scenario

Thanks Francois! I finally got it!

Bronze

Re: rstp weird scenario

Thanks Giuseppe! I finally got it!

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