All language subtitles for Free-CCNA-Spanning-Tree-Protocol-Part-1-Day-20-CCNA-200-301-Complete-Course_en

This is a free, complete course for the CCNA.

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In this video we will move on to another important\n

for network engineers understand.

That is STP, Spanning Tree Protocol.

The CCNA Exam Topics list mentions Rapid spanning\n

However, to understand rapid STP, I think\n

So, first we will talk about STP, which will\n

and then in a future video I will teach you about rapid\nSTP.

Let’s take a look at what we’ll cover\nin this video.

First, I’ll talk about redundancy in networks,\n

Then I will introduce STP, Spanning Tree Protocol.

I will introduce its purpose, the problem\nit solves, etc.

Also, remember to watch until the end of today’s\nquiz.

I will once again feature a bonus question\n

If you’re looking for a set of practice\n

is without a doubt the best there is.

I used ExSim for my CCNA and CCNP, so I feel\n

If you want to grab a copy of Boson ExSim,\n

First, just a few points about redundancy\nin networks.

Redundancy is an essential part of network\ndesign.

A network that is not redundant is simply\nnot acceptable.

Modern networks are expected to run 24 hours\n

Even a short downtime can be disastrous for\na business.

Imagine if Amazon’s network went down for\n

If one network component fails, you must ensure\n

Finally, as much as possible, you must implement\n

As network engineers, we are responsible for\n

to make sure that that infrastructure is resilient\n

First off, here is a poorly designed network.

There are many points of failure here which\n

For example if this connection is cut\n

loses connectivity to the Internet.

Or, if this connection is cut off due to a\n

within the LAN, and out to the Internet.

Okay, those are just two examples, let’s\n

This network here is a much better design.

If this PC wants to reach the Internet, it\n

However, even if this router has a hardware\n

reach the Internet via this or another alternate\npath.

Perhaps traffic to this other PC in the LAN\n

That’s not a problem, because this alternate\npath is available.

So, I think you can see the benefit of designing\n

However, you may be asking, what if this switch\nfails?

Well, if that is the case, all hosts connected\n

Unfortunately, Most PCs only have a single\n

only be plugged into a single switch.

However, important servers typically have\n

multiple switches for redundancy.

We will cover many protocols that are used\n

course, and spanning tree is one of them.

Spanning tree is a Layer 2 protocol by the\n

so within the LAN here, not routing out to\n

I just showed you the benefits of a redundant\n

provides alternate paths if one connection\nfails.

However, without spanning tree, there is a\n

Let me introduce the concept of ‘broadcast\nstorms’.

I’ll use a simplified network topology to\ndemonstrate the issue.

PC1 is 10.0.0.1, PC2 is 10.0.0.2, and PC3\nis 10.0.0.3.

You already know what a switch does with a\n

Let’s say, for example, PC1 wants to send some traffic\nto PC2.

To do that, it needs to know PC2’s MAC address.

So, let’s say PC1 sends an ARP request frame,\n

MAC address of all F's as its Layer2\naddress.

When SW1 receives the frame, what will it\ndo? As I said

you already know what a switch does with broadcast\n

It will flood it out of all interfaces, except\n

So, SW2 and SW3 both receive a copy of the\nframe.

They then do the same thing, they flood it\n

So, PC2 receives the ARP request and will\n

Although PC2 received the ARP request and\n

I’ve cleaned up the arrows so you can see\nthis easier.

As I just said, PC2 received the ARP request\n

The switches will continue flooding them . So,\n

SW1 just received two broadcast frames, on\n

SW2 and SW3 both just received broadcast frames,\n

So, I think you get the point.

Do you remember the TTL, or time to live,\n

It is used to prevent infinite loops at Layer\n3.

But the Ethernet header doesn’t have a TTL\nfield.

These broadcast frames will loop around the\n

If enough of these looped broadcasts accumulate\n

for legitimate traffic to use the network.

This is called a broadcast storm.

Eventually your network will look like this,\n

regular traffic can pass through your network.

The red arrows represent the clock-wise loop\n

arrows the counter-clockwise loop.

However, network congestion isn’t the only\nproblem.

Each time a frame arrives on a switchport,\n

to ‘learn’ the MAC address and update\nits MAC address table.

When the frames with the same source MAC address\n

the switch is continuously updating the interface\n

This is known as MAC Address Flapping.

So, how can we design a network with redundant\n

Well, Spanning Tree Protocol is one answer\nto this problem.

So let’s take a look at Spanning Tree protocol\n

is an industry standard protocol, IEEE 802.1D.

This is the type of STP we will focus on in\n

Because it is so important to prevent Layer\n

So, you won’t only find STP on Cisco switches.

STP prevents Layer 2 loops by placing redundant\n

These interfaces act as backups that can enter\n

meaning an interface that is currently forwarding,\nfails.

Interfaces in a forwarding state behave normally.

They send and receive all normal traffic.

However, Interfaces in a blocking state only\n

or Bridge Protocol Data Units), and some other\nspecific traffic.

Before going more in depth , let me talk about\n

I told you about Ethernet hubs in a previous\nvideo.

Hubs were used before switches were invented,\n

frames to the correct destination, they simply\n

But actually, before switches, there was another\n

You don’t need to know about bridges for\n

they’re like a transitional stage between\n

However, the reason I’m telling you about\n

However, when we use the term ‘bridge’,\nwe really mean ‘switch’.

Bridges are not used in modern networks.

So, in this lecture, and really any time I\n

‘bridge’, but really it means switch.

So, if we look at this topology again, perhaps\n

while this one interface on SW3 is in a blocking\n

So effectively its like that link doesn’t\n

If PC1 sends that same ARP request broadcast\n

However, if at some point another interface\n

The switches will automatically adjust the\n

flooded like this, again no loops.

So, that is just a basic outline of the purpose\n

Now let’s go a little deeper into how spanning tree protocol

By selecting which ports are forwarding and\n

path to and from each point in the network.

There is a set process that STP uses to determine\n

That process is what we will cover next.

STP-enabled switches send Hello BPDUs out\n

seconds, so the switch will send a Hello BPDU\n

If a switch receives a Hello BPDU on an interface,\n

switch, because routers, PCs, etc. do not use STP,\n

So, back to our topology here, these switches\n

this . They use these BPDUs to advertise themselves\n

Now, what exactly are these BPDUs used for?

First of all, switches use one field in the\n

The switch with the lowest Bridge ID becomes\nthe root bridge.

I’ll talk about the bridge ID in the next\nslide.

ALL ports on the root bridge are put in a\n

topology must have a path to reach the root\nbridge.

So, as I mentioned previously STP puts ports in either a blocking

state, to avoid Layer 2 loops in the network.

However, as I just said, on the root bridge, all ports are\n

a path to reach the root bridge.

Traditionally, the bridge ID field of the\n

There is a bridge priority field, which is\n

address of the switch, which as you already\n

The default bridge priority is 32768 on all\n

As I said before, the switch with the lowest\n

by default the switch with the lowest MAC\n

So here’s that topology once again, and I’ve\n

As you know MAC addresses are 12 hexadecimal\n

I’ve also added port lights for the interfaces,\n

The G0/2 interface on each switch is connected\n

BPDUs, it knows it is safe to go into forwarding\n

so these port lights are all green.

Now, all three switches have the default priority\n

be the root bridge we will have to compare\nthe MAC addresses.

Remember, the LOWEST bridge ID wins.

Which of these MAC addresses is the lowest?

Well, hexadecimal A is equal to 10, B is equal\n

Therefore, SW1 will become the root bridge\nof this network.

All ports on the root bridge become designated\n

So, that is the traditional bridge ID.

However, the bridge ID was actually updated\nto look like this.

In reality, the bridge priority has been updated\n

which is 4 bits, and the ‘extended system\n

12 bits, because as you know a VLAN number\nis 12 bits in length.

Why include a VLAN ID in the bridge priority?

Well, Cisco switches use a version of STP\n

PVST runs a separate STP ‘instance’ in\n

One interface could be forwarding in VLAN1,\n

By adding the VLAN ID into the bridge priority,\n

Here’s a deeper look at the bridge priority\nfield.

You may have wondered why 32768 is the default\nbridge priority.

Well, it’s because this total field is 16 bits\n

Therefore, the default bridge priority WAS\n32768.

However, with the addition of the extended-system\n

So, the default VLAN ID is 1, therefore the\n

In the default VLAN of 1, the default bridge\n

Now, here’s a question, If you want to increase\n

VLAN numbers, what is the minimum unit of\nincrease/decrease?

Let me explain what I mean in the next slide.

The bridge priority + extended system ID is\n

extended system ID is set and cannot be changed because it is

Therefore, you can only change the total bridge\n

system ID) in units of 4096, the value of\n

Currently, the bridge priority here is 32769.

Let’s reduce it to make this switch the\nroot bridge.

If I want to reduce it just a little, I can\n

I could reduce it more, of course, but the\n

only be changed in units of 4096.

So, the valid values you can configure are listed\n

The extended system ID will then be added\n

So let’s look at this topology again.

We’ll just be looking at the STP topology\n

But if there are multiple VLANs, say VLAN1,\n

would be 32770 for VLAN2, and 32771 for VLAN3, etc.

We could also change the bridge priority on\n

SW1 is the root bridge in VLAN1, SW2 could be the\n

I’ll talk about how you can do that in the next video,\n

So, here in VLAN1, SW1 is the root bridge.

All interfaces on the root bridge are designated\n

Designated port is one of the port roles in\nspanning tree.

There are a couple other port roles.

I will introduce those in a minute.

Okay just a few more points about the root\nbridge.

When a switch is powered on, it assumes it\nis the root bridge.

It will only give up its position if it receives\n

a BPDU from a switch with a lower bridge ID.

Once the topology has converged and all switches\n

The reason all switches send BPDUs at first\n

Other switches in the network will forward BPDUs from the root

Before moving on, let’s see if you understand\n

In this network of 4 switches, which will\n

Pause the video now to think about the answer.

Okay, did you find the root bridge?

Both SW1 and SW3 have the same priority, 12289,\n

The first half, 014A 38 is the same, but the\n

Let’s do another practice question.

Which switch will become the root bridge in\nthis case?

Pause the video to think about your answer.

It has the lowest priority of the 4 switches,\n4097.

So far we have covered the first step of spanning-tree’s\n

Step 1: the switch with the lowest bridge\n

All ports on the root bridge are designated\n

It’s important that this is the first step\n

of the steps depend on knowing which switch\nis the root bridge.

All other switches will select ONE of its\n

So, that means there is one root port on each\n

The interface with the lowest root cost will\nbe the root port.

Root ports are also in a forwarding state.

Now let’s talk about what that ‘root cost’\nis.

Each interface has an associated spanning\ntree ‘cost’.

A regular Ethernet interface, with a speed\n

Fastethernet, 100 megabits per second, has\na cost of 19.

Gigabit ethernet has a cost of 4, and 10 gigabit\n

Make sure you remember these path costs for\nthe exam.

Of course, there will be flashcards in the\n

So, these are gigabit ethernet ports,\n

The root cost is the total cost of the outgoing\n

SW1 is the root bridge, so it has a cost of\n0 on all interfaces.

They are gigabit ethernet interfaces, but\n

interface, just the sending, the outgoing interface.

So, SW1 advertises its root cost of 0 in its\nBPDUs.

SW2 will receive the BPDU and add the cost\n

4, when it floods those BPDUs out of its interfaces.

So, which port do you think SW2 will choose\nas its root port?

It was advertised a cost of 0 on its G0/1\n

is 4, therefore the total root cost via G0/1\nis 4.

It was advertised a cost of 4 on G0/0, from\nSW3.

However its interface also has a cost of 4,\n

So, it will select G0/1 as the root port.

SW3’s logic follows the same process.

It has a total cost of 4 via G0/0, and a total\n

In this case, the ports directly across from\n

However, keep in mind that the port connected\n

Because the root port is the switch’s path\n

Okay, so I’ve updated our spanning-tree\nsummary here.

First, one switch is elected as the root bridge.

All ports on the root bridge are designated\nports.

There is only one step in selecting the root\n

Next, each remaining switch will select ONE\n

is also in a forwarding state.

Ports across from, ports connected to, the root port are always

The first criteria for root port selection\n

However, what if a switch has multiple ports\n

In that case, the interface connected to the\n

Okay, let’s practice that with a\nquiz, actually.

First, which switch will become the root bridge?

Pause the video to think about the answer.

Okay, the answer is SW2, because it has the lowest\npriority.

So, SW2’s ports are all designated.

Now, which ports will become root ports?

All interfaces are gigabit ethernet, so all\nhave a cost of 4.

Remember, if there is a tie in root cost,\n

to the neighbor with the lowest bridge ID.

So, pause the video here to think about your\n

Okay, on SW1 and SW4, the answer is obvious, SW1’s\n

Via G0/0 it has a cost of 8, 4 plus 4.

Via G0/1 it has the same, a cost of 8, 4 plus\n4.

So, we have to use the tiebreaker, which neighbor\n

It’s SW1, the priorities are the same, but\n

So, G0/0 is selected as the root port, and\n

So, this is the process so far.

HOWEVER, there is ONE more tiebreaker that\n

What if two switches have two connections\n

neighbor bridge ID are the same?

Then we get to the final tie-breaker, the\n

neighbor switch with the lowest port ID will\n

Okay, let me briefly explain port ID.

So, here is the output of the command SHOW\n

in a future video when we look at spanning\ntree configuration.

I just want to show you this section, this\n

Notice the column title is Prio dot number.

So, each port has a default priority of 128,\n

2 for G0/1, etc on this switch.

So, the STP port ID equals the port priority\n

Similar to the bridge ID, where the MAC address\n

tie, in this case the port number is used\n

I won’t explain the port ID in more depth\n

about it or change it, so you can just look\nat the port number.

For example, G0/0 is lower than G1/0, or G0/3\n

So, one more quiz to practice that.

Now there are two connections between SW1\nand SW3.

Which port will SW3 select as the root port?

Pause the video to think about your answer.

The answer is G0/2, because it is connected\n

The NEIGHBOR switch’s port ID is used to\n

That’s why G0/2 was selected over G0/0,\n

So, SW1’s G0/1 interface is a designated\n

Okay, so this is our process so far.

We still haven’t blocked any ports, and\n

So, let’s return to our previous topology.

All that’s left is this connection between\nSW2 and SW3.

So far, all of our ports are in a forwarding\n

are always in a forwarding state.

So, to prevent loops do we block both of these\nports?

Actually no, here’s an important rule to\n

spanning tree designated port.

Remember, unlike old Ethernet hubs, which\n

each link is a separate collision domain.

This collision domain between SW1 and SW2\n

This connection between SW1 and SW3 has one,\nSW1’s G0/1.

And the connections with the PCs are all designated\n

PCs don’t participate in spanning tree.

So, we need one designated port on the connection\n

How do we determine which port will be designated,\n

The switch with the lowest root cost will\n

However, in this case both switches have the\n

and 4 for SW3 via its G0/0 interface.

So, for the tie-breaker we compare the bridge\nID.

SW2 has the lower bridge ID, so its G0/0 interface\n

Finally, the other switch will make its port\n

So SW3’s G0/1 is non-designated, it blocks\n

So, here is the process for selecting the\n

One switch is selected as the root bridge,\n

Then, each remaining switch selects ONE of\n

The interface with the lowest root cost is\n

connecting to a neighboring switch with the\n

a tie also, the interface connected to the\n

Then finally, each remaining collision domain\n

port, and the other port will be non-designated.

The interface on the switch with the lowest\n

a tie the interface on the switch with the\n

the other interface will be a non-designated\n

There are still many important things left\n

cover those in part 2, before moving on to\n

We already did a few quiz questions throughout\n

questions to make sure you know the whole\nprocess.

If you get stuck, if you don't know the answer, go back to the

try to figure out the answers yourself.

I will also feature one question from Boson\n

Because there are still some important points\n

aren’t ready to answer the spanning tree\n

So, I will show you one question from Boson\n

Of course, if you have already studied spanning\n

Okay, let’s do a couple more practice questions\nfirst.

Identify the root bridge, and the role of\n

so which interfaces are root ports, which\n

Pause the video to think about your answer.

Okay, I hope you found the answer.

So, the root bridge is SW3, because the priority\n

These are the root ports, SW2 selected its\n

the lower-number interface on SW1, G0/0.

And these are the remaining connections.

In each case the interface on SW2 is non-designated,\n

Always remember to check that there is one\n

Do the same thing, but with this network topology.

Look carefully, some of these interfaces are\n

Pause the video to think about your answer.

SW4 is the root bridge because it has the\nlowest priority.

SW1 uses its G0/1 interface as the root port,\n

with a much higher spanning tree cost.

Finally, the remaining designated and non-designated\nports.

SW1’s F1/0 and F2/0 are non-designated because\n

is non-designated because SW4 IS the root\n

Okay, that’s all for the quiz, let’s take\na look at Boson ExSim.

Okay for today's Boson ExSim practice question\n

optional feature of spanning tree which I\n

So, I won't give the answer in this video,\n

give you the answer in the next lecture video,\nday 21.

Now, if you think you know the answer, if\n

feel free to let me know your answer in the\ncomments.

Or if you want to do some independent research\n

tree portfast' and do some reading, and then\n

know your answer in the comment section.

You want to decrease the amount of time that\n

Portfast is not configured on any of the switchports\n

portfast default' command from global configuration\nmode.

Which of the ports on Switch A will use portfast?

So A says 'No ports, because portfast cannot\n

B says 'All ports', C says 'All access ports',\n

Okay so as I said, this time we won't check\nthe answer.

Please wait for the next video to see the\n

Of course, if you think you know the answer,\n

If you want to get your own copy of ExSim,\n

the real thing, please follow my link in the\nvideo description.

These are by far the best practice exams out\nthere for the CCNA.

There will be supplementary materials for\nthis video.

There will be a review flashcard deck to use\n

Download the deck from the link in the description.

There will also be a packet tracer practice\nlab.

Please be sure to watch the practice lab,\n

process of figuring out a spanning tree topology,\n

commands which I didn’t have the time to\nshow in this video.

Before finishing today’s video I want to\n

Thank you to Joyce, Marek, Samil, Velvijaykum,\n

Boson Software, the creators of ExSim, Sidi,\n

Sorry if I pronounced your name incorrectly,\n

One of you is displaying as Channel failed\n

and I’ll see if YouTube can fix it.

This is the list of JCNP-level members at\n

up recently and your name isn’t on here\n

Please subscribe to the channel, like the\n

with anyone else studying for the CCNA.

If you want to leave a tip, check the links\nin the description.

I'm also a Brave verified publisher and accept\n