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STP lower 10Gb below 1Gb weight

pashdown1
Level 1
Level 1

A

| \

|  >C

| /

B

C is connected to A&B with 2Gb port channels.

A&B are connected to each other with a 10Gb link.

Right now, "show mac-address-table" show the mac addresses for C reachable directly over the link on B.  However doing the same on A does not point to C, but rather B as the source of the mac addresses.  I'd rather use the link between A&B as backup in case the link between A&C fails.  I presume that the STP would weight this accordingly, with the weight of 2Gb+10Gb > 1Gb, but it isn't.  Is there something I'm missing, or is there something I need to adjust to behave as desired?

8 Replies 8

cadet alain
VIP Alumni
VIP Alumni

Hi,

If  switch C was the root bridge then the Port-channel interfaces on A and B would be forwarding and whichever one had the lowest BID would be designated bridge and its 10 Gig interface would be forwarding and the other end would be blocking.

Can you provide output of sh spanning-tree  on A and B.

Regards.

Alain

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Peter Paluch
Cisco Employee
Cisco Employee

Hi Pete,

I am afraid that the STP you are running is using an old table of interface costs, and is unable to distinguish between the two alternatives. Please check the show spanning-tree command output to see what costs have been assigned to different interfaces. It may be necessary to manually modify the interface costs, or possibly move to MSTP that in Cisco implementation uses a more progressive interface cost evaluation method that allows for far greater speeds than the original 802.1D STP.

Best regards,

Peter

Here is "show spanning-tree" on the particular vlan and ports in question for both A & B:

A:

#show spanning-tree vlan 20

VLAN0020

  Spanning tree enabled protocol rstp

  Root ID    Priority    8212

             Address     0019.0732.e000

             Cost        2

             Port        1025 (TenGigabitEthernet9/1)

             Hello Time   2 sec  Max Age 20 sec  Forward Delay 15 sec

  Bridge ID  Priority    28692  (priority 28672 sys-id-ext 20)

             Address     0019.07f1.1c00

             Hello Time   2 sec  Max Age 20 sec  Forward Delay 15 sec

             Aging Time 14440

Interface           Role Sts Cost      Prio.Nbr Type

------------------- ---- --- --------- -------- --------------------------------

Te9/1               Root FWD 2         112.1025 P2p

Po12                Desg FWD 3         128.1668 P2p

B:

#show spanning-tree vlan 20

VLAN0020

  Spanning tree enabled protocol rstp

  Root ID    Priority    8212

             Address     0019.0732.e000

             This bridge is the root

             Hello Time   2 sec  Max Age 20 sec  Forward Delay 15 sec

  Bridge ID  Priority    8212   (priority 8192 sys-id-ext 20)

             Address     0019.0732.e000

             Hello Time   2 sec  Max Age 20 sec  Forward Delay 15 sec

             Aging Time 14440

Interface           Role Sts Cost      Prio.Nbr Type

------------------- ---- --- --------- -------- --------------------------------

Te9/1               Desg FWD 2         128.1025 P2p

Po12                Desg FWD 3         128.1669 P2p

Hi Pete,

The switch B is currently acting as the root bridge - it is visible from the output you have posted. Try making the switch C a root bridge using the command

spanning-tree vlan 1-4094 priority 4096

The network should then converge to the form you want it to have.

Best regards,

Peter

Will B then route through A to get to C, or go directly to C still?

Hi Pete,

With C being the root switch, both A and B will use their direct links to reach C. They won't go through each other to reach C.

Best regards,

Peter

A&B act as a core for many other switches.  Setting C as the root may cause some unexpected topology problems for me.  What I wish is for A&B to share the roll of root.

Hello Pete,

While you cannot have A and B share the root role in a single VLAN (each spanning tree has exactly one root bridge only), what you request is possible. I suggest you increase the port cost of the Te9/1 interface on switch A to, say, 10:

SwitchA:

interface Te9/1

  spanning-tree cost 10

This will make A select a new root port - the Po12. Thus, the spanning tree will converge on the topology

B (root) --- C --- A

Best regards,

Peter

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