Spanning Tree Protocol (Part 2) Frame Behavior and BPU's
Spanning Tree Protocol (part 2) Frame Behavior and BPU's For the second part of our Spanning Tree Protocol lesson, we look at frame behavior and introduce you to BPUs fields and define why they are important.
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29 hours 18 minutes
Spanning Tree Protocol (part 2) Frame Behavior and BPU's For the second part of our Spanning Tree Protocol lesson, we look at frame behavior and introduce you to BPUs fields and define why they are important. We'll diagram and explain what a Bridge ID is, how it correlates to a MAC address and explain what a priority field is on a switch, and what function the lowest bridge ID has.
when each of these switches powers up or when Cisco switches power up the immediately start sending out frames to each other out off all ports called bpd. Use or bridge protocol data units. Now, these bpd use have four field in in them. That our off importance to us S C. CNN candidates,
our route bridge I d.
Route, path cost. Send a bridge. I D. And center port I d.
This these fields, I will explain as we go along through the spanning tree process. But if you notice these fields are quite similar to the spanning tree decision process, the only pretty much difference being that the word lowest is in front.
Now, before I move on, I have to explain what a bridge I d actually is. A bridge. I d for a switch is a combination of a priority field plus your Mac address.
So you're Mac. Address is 48 bits, and your priority is 16 bits 16 to the power to
so 16 to the power to is 65 5 30
Okay. The priority field can range between
zero and 65 5 35 which zero inclusive makes 65 5 36
The priority feel for each to school device. Each Cisco switch. It's set to
So every switch has a bridge i d. With the priority feel set to 32 768 How did they come up with? 32 768 Simple. 65 5 36 Divided by two smack in the middle,
a rude bridge is elected according to the Louis Bridge I. D. So the switch with the Louis Bridge I D becomes route.
So now, if every switches
priority field is set to 32 768
and that is sane, the switch as a tiebreaker with the Louis Mac address becomes the route.
So now, as I said previously, that when each switch powers up, they start sending people used to each other every two seconds.
Now it's which initially announces that it's the route it's announces itself as the route. By setting the route Bridge I D field in the bpd, you equal to the sender bridge. I d feel so the
Each switch is basically saying the route bridge, for example, Swiss air saying the route bridge is a and I am a hence I am the route.
Now when a switch receives a B p. D. You from another switch with a lower bridge i d or a better bridge I d in spanning tree. A lower number is always better. So when a switch receives a BP do from another switch with the lower bridge, i d. It stops sending its own BP to use
and start relaying those superior bpd use.
Given enough time,
the switch with the Louis Bridge i D. Or in this case since the Bridge I. D Field is Priority and Mac address and the priority field is set to 32 768 for all switches, the switch with the lowest Mac address becomes the Route bridge.
So all other switches in a layer to topology will relay the root bridges bpd use
and not produce their own BP dues.
So in this case, which has a Mac address off a switch, B has a touch, too. Has a Mac address of B B B B. So it's three has a Mac address off CC CC, and Switch four has a Mac address of dddd.
In this case, everybody starts off announcing themselves as rude bridge by sending B p to use with the Route Bridge I. D Field equal to the sender Bridge I D Field.
Eventually, as you see in hacks, is a lower number than see, then be and then be
eventually in this topology switch, A will become route and continue sending bpd use,
and everybody else will be relaying the bpd. You switch one sentence.
So in the fourth step decision process, the first step is the Louis Bridge I D.
And using just the first step, we have elected the Route Bridge, so the route Bridge election occurs. According to the Louis Bridge I d. Now
again, in this case since the Bridge I. D Field consists of Priority and Mac address and everybody's priority is the same, which is 32 768 The tiebreaker becomes the Louis Mac address, and the switch with the Louis Mac address becomes the Route bridge.
In this case, it's which one becomes the Route bridge.
The next step.
Step two in the spanning tree conversions process
from spanning tree to go from beginning to end. That's what conversions is
is a root port election per non route switch,
so I'm going to erase this.
The route port election on Lee happens on non route switches,
which one of these in anthropology are non roots, which is switched to switch four and switch three.
Which one is your route? Bridge?
route toward election happens according to route path cost and what is root path costs
Well spanning tree designates
path costs. You can think of it as distance for each link speed. For Lynn speeds of 10 megabits per second, the cost is 100 for ling speeds of 100 megabits per second. The cost is 19 and fooling speeds of 1000
megabits per second.
The cost is four in Oct Apology.
If you look all these links a f f f f f, which means all these links are fast eat in it.
So the path costs for all these links is 19.
Now, remember, at this point, switch a switch. One with Mac address is your route bridge. So switch one is the only one sending bpd use.
remember this rule
the path costs inside off a bpd you the route path costs inside off a BP do is implemented when a VP Du enters a switch.
Not when it leaves the switch.
Now, if you think of this cost as distance,
when a B P D you leave, switch one port F 01 The route path cost inside of that BP do, and the route path cost is the distance or the cost from a local switch. Let's say switch for to the route bridge
so the distance from the root to the root has to be zero. Correct
the distance to yourself from yourself
So a BBDO leaves
switch one port as zero slash one, heading out to switch to and another bebe do leaves switch 10 slash two heading out to switch. Three.
Now these be produced. When they leave switch one. Have a cost of zero. A route path cost off. Zero.
When this BB do leaving switch one port F zero slash one
enters switched to F zeros last won. The cost is implemented to 19.
Now we're looking at switch to At this point, we're just looking from the point of view of sweets, too.
The other baby do that left switch one F zero slash to as its cost implemented to 19 0 plus 19 is 19
at 33 zeros. Last two.
Now that BP do transfers a Swiss tree and leave switched three as heroes, last one still, with the cost of 19 remember the only increments route path costs when a BP do you enters the switch, not when it exits the switch. Then the BBDO enters switch for F zero slashed one,
and another 19 is added to the route path cost. Since this is a fascinating it late
and 19 plus 19 is 38
so the cost at zero slash one switch for F zero slash one is 30.
The same people do
leaves switch for F zero slash to an F zeros last three with
a different centre port i d. So the sender port on switch for zeros. Last two is as it was last two and the center port over here, the last three. So it's the same Bebe, do you? Nothing else has changed except the center port I D Field
inside of the baby. Do you
this b B D u transfers is this link and enter switched to poor F zero slash two and switch to port of zeros last three and the cost is implemented by 19 again. So 38 plus 1957
57 here on 57 away here
switch to now looks at the cost,
which is 19 on port at zero slash one
and looks at the cost at Zito's last two and half years Last three, which is 57 the Louis route path cost on Swiss to is on port as heroes Last one hence f zero slash one on switch to
is elected. I'm gonna race that 19 as Route
So I'm gonna go ahead and raise Thies numbers because we're gonna work through the rest of the switches.
Now, once again,
let's look at Swiss Street. A similar process gonna happen. That happened for switch to
a B P. D. You leave stretch one F zero slashed to with the cost of zero with the root path cost of zero
when it enters Swiss three f zero slash to the root path, cost is incremental. Goodbye path cost, which is 19 and the BP do now has a cost of 19 as it sits here on sweets too. Yeah, zero slash, too.
Now again another BP do leave switch one F zero slash one with the cost of zero with the root path. Cost of zero and enter switched to zero slash won, and the cost is implemented to 19.
The same BBDO transfers to switch to and the cost is not implemented when it leaves. Switch to F zero slash two and zeros last three.
The BB do enters switch for F zero slash to an F zero slash three with the different center port idea because the center Port I d from switch to will be at zero slash two and after his last three.
And the cost on that BP do is implemented once again
on poor, too, and poor tree off switch for
then that BP do leaves switch for heading up two word switch three leave switch for zero slash one
and enter switch 30 slash won, and the cost is implemented to 57.
Switch three compares to cost at zero slash two, which is 19 and F zero slash one, which is 57 since switch trees costs at zero slash two. Is lore remember the decision process? The second step, says Lewis, Route path costs
switch three F zero slashed to becomes your
Once again, I am going to erase
and then we're going to work through switch for looking at switch for now.
Ah, BP do leaves
switch one poor to zero slash two with the root pat Cost of zero and enter. Switched to be, as you know, slashed too. And the cost is implemented by path cost,
which is 19.
So the cost is 19 over here.
And then the bpd you transfers his switch three leaves, Swiss three of zero slash one and the BP do you enter? Switch for zero slash won and the cost is implemented to 38.
The other baby do the route sent out
left port port F zero slash one with the root pat cost of zero and enter switched to zero slash won, and the cost is implemented to 19.
BP do transfers his switch to and it is converted to two separate bpd use. It's actually the same BP. Do you the only difference being that the sender port I d field, in fact, of the BP do you now stays port to for the bpd, leaving poor too. And port three for the bpd. You leaving port three.
These two bbd use enter switch four poor too. And port three. And the cost for both of these is implemented to 38.
At this point, Switch four compares to cost at port one port to and port three and discovers that the costs are the same. So it cannot elect the report yet.
Hence, switch four has no other choice but to move to the Turks. Step in the spanning three decision process. Which, being Louis Centerbridge I d
The sender bridge I d. That switch for is receiving on port F zero slash one is the bridge i d off switch three
with the Mac address C C C c and the priority being the same 32 768 And the bridge I d French four is receiving on poor too. And port three is B B B B from switch to
so the center port I d is the i d or the sender bridge. I d is the i d off the sending switch,
which is with the priority 32 768 on switch to and Mac address B B B B.
Switch four compares the sender ridge I d on port to port three and port four. I mean, import one
since the center bridge i d is lore. Since B B B B is lower than C C C C switch four now starts looking at port to and port three to elect a route port.
Now the sender bridge i d on port to is B B B B is the bridge I d off switch to
and the sender bridge I d on poor tree is also B B B B.
So switch for now still can't decide whether poor to or port three is going to be the root port.
Hence switch four has to move to the last step in the spanning tree decision process, which is the Louis Sender port I d.
So the sender Port i D, which is the idea of descending switch on port to a switch
switch. Four is too.
And the center port I d on port three off switch four is three since two is lower than three
f zero slash too.
On switch four becomes your
So at this point, we have concluded the second step in the spanning tree conversions process, which was elect a route board per non route switch. So which two has a route port, which is F zero slash one which four has a route port, which is F zero slash two and switch tree has the report, which is F zero slash, too.
The last step in the spanning tree convergence process is to elect one designated port per segment.
So, as I mentioned earlier, there are five segments in this topology between switch one and switch to between switch one and switch three between switch three and switch for this link between Swiss story and switch for and the two links between
switch four and switch to
At this time,
when we were electing
the route port, our view of the topology was based on. We were sitting on each switch and looking at how BP to use enter our switch.
Now, we're gonna take a view of the apology by sitting on this segment.
So there are two ports in the on the segment between switch one and switch to which being switched one port one and switch to Port one.
The designated port on each segment is also elected according to the lowest route path cost.
So let's look at switch one first
on the segment between switch one and switch to
the route path. Cost on switch one is zero
F zero slash one is zero since the distance fronts which one to switch one is zero
and then the BBD leaves at zero slash one. Switch one and enter switch to at zero slash won, and the cost is implemented to 19.
So between zero and 19 0 wins since
since the Lewis route path cost wins.
Port one switch one becomes the designated port.
So I'm going to put
a D on port one for switch one
now on the segment between switch one and switched three. The route path cost at switch one as yours. Last two is zero and the route path cost at switch three. Ab zero slash two is 19. So between zero at switch one and 19 at switched three, switch one as heroes last two becomes your designated port.
Always remember, all ports on a rude bridge will become designated port, no matter what segment, no matter who they're connected to on every segment, the cost from the root to itself is zero. So all ports on a rude bridge will become designated ports.
Now let's look at the segment between switch three and switch for
in the BP. Do the route pot costs inside of the bpd you as it leaves Which one and transfer sir. Switch three and then leaves 33 f zero slash one is still 19.
When this BB do you enter switch 40 slash one. The cost is implemented to 38 so 19 had switched three of zero slash one and 38 at switch 40 slash one.
So between 19 and 38 switch three F zero slash one Becomes your designated pork.
Now let's look at the last two segments
between switch for and switch. Switch to and switch for
the route path. Cost at switch to F zero slash two and zero slash three is still 19. It was zero as a BB. Do you left as you go slash one switch one. The route it got implemented to 19. It's switched to zero slash one, and when the BB do leaves which two,
as you know slash two and zero slash three.
The cost is still 19. Remember, the cost is only implemented when a BP do enters a switch.
So when this VP Du travels, when these two bpd use travel and enter switch four zeros last two and zero slash three, the cost is implemented from 19 to 38. So the cost is 19 at zero slash to an F zero slash three off sweets, too.
And the cost is 38 at zero slash two and zeros last three off switch for
so as zero slash too,
and F zeros last three off switch to becomes
Now are we done with the spanning tree convergence process? Yes, we are.
At this point. We got to see what happens to the sports. Are they put in blocking out? They put in forging now can before traffic. So let's get to that