MSTP Reflection VectorIEEE 802.1 March 2005 Atlanta 1 MSTP Reflection Vector Norman Finn, Cisco Systems

MSTP Reflection Vector IEEE 802.1 March 2005 Atlanta 1

MSTP Reflection Vector

Norman Finn,

Cisco Systems


• The use of a separate spanning tree per source node to garner the advantages of IP-style routing at Layer 2 is not new. However, it has not proven easy to apply that idea to 802.1 bridges. (The problem solved by the Reflection Vector, below, is one that is easy to miss.) An incremental advance to the 802.1S Multiple Spanning Tree Protocol enables 802.1 bridges to offer optimal across Layer 2 networks.

Abstract


Symmetrical Spanning Trees


• 9 Bridges A-I, connected as shown.

Letters are Bridge IDs. Lower letter (A) is “better” than higher letter (D).

Numbers are path costs.

Each pair of bridges agrees on link’s path cost.

Spanning Tree Uses Sub-Optimal Paths

(Well, it does now.)

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• Bridge A is the Root Bridge.

• Bridge E breaks the two spanning tree loops by blocking the marked ports.

• Path from E to G is E-F-I-A-B-D-G.

• This clearly qualifies as “sub-optimal.”

Spanning Tree Uses Sub-Optimal Paths

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• Instead of 1 spanning tree, we create 9 spanning trees.

Each bridge is the root of its own spanning tree instance (STI).

802.1S (MSTP) supports 64 STIs trivially, 4k with modest effort.

But, we should be able to do better!

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• Whenever Bridge A sends a frame, it uses VLAN A, which is attached to STI A.

• Of course, the STI with A as the root is the optimal path is always straight away from A, along the least-cost path.


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• Similarly, traffic originating from Bridge D uses VLAN D on STI D, and thus takes the optimal path from D to its destination.


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• All STIs use the same FID.

What about MAC address learning?

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2DA


• MAC address learning absolutely depends on symmetrical paths.

• You can learn the source MAC address X in a from X to Y only if the path from X to Y is identical to the path from Y to X.

• That’s what spanning tree is all about.

Can anything go wrong? YES!


Asymmetrical Spanning Trees

• For example, suppose the VLANs rooted on Bridges A and E are blocked as shown.

• The path from E to A is E-H-C-B-A.

• The path from A to E is A-I-F-E.

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• If A learns E’s address and transmits a frame for E towards B, the frame will hit the blocked port H-E and not get to E.

• Basically, nothing works.

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• But, if the two spanning trees are symmetrical, then A-B-C-H-E is used for both directions, and MAC address learning works just fine.

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• Root of each MSTP BPDU says, “I did (not) chose this route on this STI.”

Needs one bit vector of other STIs for each STI

Relayed from each root down the trail.

• Among equal-cost root ports, I must select one receiving a lower “did” indication.

There may be more than one, e.g. E receiving A.

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• This is how the 9 STIs would be created if we do nothing.

• Many STI pairs are compatible, e.g. A-I, B-F, and D-H.

• But some are not, e.g. A-E, D-F, and B-E.

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• This layout would be perfect.

• But, how do we get there?

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• Because link costs can be configured, as well as computed from the link speed, the two bridges on the two ends of the link can disagree on the link cost used in the STP algorithm. This makes the symmetrization much more difficult.

Coordinating path costs


• So, carry a little extra in MSTP to ensure that both bridges use the same link cost!

The bridge advertises its link configured costs in obMSTP BPDUs.

All bridges on a given LAN use the link costs advertised by the CSTI Designated Bridge.

The CSTI does not play this game.

• One link cost parameter is required for each STI.

Coordinating path costs


Identifying the core of the problem

• When can STP break loops asymmetrically?

• A ring with an odd number of nodes cannot have a problem; routes are always symmetrical.

• A ring with an even number of nodes has two equal paths from one side to the other, and so can have asymmetrical paths.

OddOKC D

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EvenBad

Cost = 2

Cost = 2

Cost= 3

Cost= 3

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“X” == “pointopposite root on ring”


Identifying the core of the problem

• The addition of link costs changes the definition of “odd” and “even”, but not the nature of the problem.

• If the “point opposite me on this loop” (X) is a link, I’m OK, and if a bridge, I may be in trouble.

“Even”bad

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“Odd”OK

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1Cost = 3

Cost = 3

Cost = 3

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Cost = 3

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“X” == “pointopposite root on ring”

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• It has been observed often, in IEEE 802.1, that when a bridge has equal root path costs to choose from, any decision it makes for its Root Port is perfectly compatible with the spanning tree algorithms.

But, Root Port selection can be changed!


Problems are always visible at both ends

• As long as the link costs are forced to be symmetrical:

• If any Bridge “X” has an equal-cost root port choice on STI Y, then Bridge Y has an equal-cost root port choice on STI X!

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The Reflection Vector


• If both ends know about the problem, they should be able to do something about it!

• So, we add, for each STI in the MSTP BPDU, a Reflection Vector containing one bit of information about each of the other STIs.

Aha!



• A slightly different diagram lets us flesh out the details of the Reflection Vector.

• Looking at exactly two STIs, rooted at Bridges A and I, we see that Bridges B and E will each want to block one port for one of the two STIs.

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• Let’s look at the two Roots’ BPDUs, and see what each STI’s Reflection Vector says about the other’s STI.

• The box is the color of the BPDU, the letter inside shows what the vector says about the other Root’s STI.

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• When the Root Bridge initiates the Reflection Vector, all of the other STIs’ bits are Yes.

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• A bit in a Reflection Vector received from the Root Port is passed unchanged through all of the receiver’s Designated Ports that are not blocked on that bit’s STI.

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• When the Reflection Vector from the Root Port is passed through a port that is blocked on any STI, that STI’s bit is set to “No” in the Reflection Vector, whatever its former value.

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• This rule applies to the ports on the Root Bridge, as well.

• Assuming E is a Root Bridge in the above diagram, the A bit of its Reflection Vector are “No” on the port(s) that block STI A.

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• Whenever a bridge receives a “No” bit from any other STI on its own STI’s Root Port, it knows that it is not in synch with that other STI, and that something needs to be done.

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• If the bridge has an Alternate Port that:

1. Has the same Root Path Cost as the Root Port; and

2. Is receiving a “Yes” from the STI in question;

• And if the Root ID comparison so indicates;

• Then the bridge selects as the Root Port any Alternate Port that has the same Root Path Cost, but a “Yes” for that other STI.

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• Bridges A and I have no Alternate Ports of equal cost to their Root Ports, so cannot do anything.

• Root Bridge A is superior to Root Bridge I, so Bridge B does nothing.

• Only Bridge E meets all the criteria. It selects a new Root Port from among its Alternate Ports.

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• However, Bridge E does not advertise its decision!

• It continues to advertise information about what it would do if left to itself; otherwise, a chain of dependencies could slow convergence even further.

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Convergence Times


Convergence times

• In general, spanning trees converge by:

Various bridges’ claims to being the Root Bridge propagate outward in expanding “claim domains” from the best bridges.

As the “claim domains” collide, the best bridge wins, until one bridge’s “claim domain” covers the whole network.

• All STIs converge simultaneously; they don’t depend on each other.


Convergence times

• So, A’s claim reaches E at the same time I’s claim reaches B.

• A’s Reflection Vector, however, cannot be computed until it knows I’s claim.

• Therefore, an extra trip across the network may be required for A’s Reflection Vector to reach E.

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Worst case convergence time:

• A and B die at the same time. (Note that the letters have changed.)

• Wave 1: D’s (C’s) claim to being root propagates to C (D).

• Wave 2: C’s (D’s) better claim to being root propagates back to D (C).

• Wave 3: The Reflection Vectors propagate.

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Drat!

• So, this plan can converge more slowly than standard MSTP, because of the extra wave of information.

• Bottom line: Convergence is slower.


Documents

MSTP Reflection VectorIEEE 802.1 March 2005 Atlanta 1 MSTP Reflection Vector Norman Finn, Cisco Systems