Free CCNA | Spanning Tree Protocol (Part 1) | Day 20 | CCNA 200-301 Complete Course

Welcome to Jeremy’s IT Lab. This is a free, complete course for the CCNA. If you like these videos, please subscribe to follow along with the series.

Also, please like and leave a comment, and share the video to help spread this free series of videos. Thanks for your help. In this video we will move on to another important topic for the CCNA, and a topic that is very important for network engineers understand.

That is STP, Spanning Tree Protocol. The CCNA Exam Topics list mentions Rapid spanning tree, an updated and superior version of STP. However, to understand rapid STP, I think its important to understand classic STP first.

So, first we will talk about STP, which will probably be over two separate days since there is a lot to cover, and then in a future video I will teach you about rapid STP. Let’s take a look at what we’ll cover in this video. First, I’ll talk about redundancy in networks, and why it's so important.

Then I will introduce STP, Spanning Tree Protocol. I will introduce its purpose, the problem it solves, etc. Also, remember to watch until the end of today’s quiz.

I will once again feature a bonus question from Boson ExSim for CCNA, Boson’s set of practice exams for the CCNA. If you’re looking for a set of practice exams to get ready for the real thing, Boson is without a doubt the best there is. I used ExSim for my CCNA and CCNP, so I feel very confident about recommending them to you.

If you want to grab a copy of Boson ExSim, please follow the link in the video description. Okay let’s get started. First, just a few points about redundancy in networks.

Redundancy is an essential part of network design. A network that is not redundant is simply not acceptable. Modern networks are expected to run 24 hours a day, 7 days a week, 365 days a year.

Even a short downtime can be disastrous for a business. Imagine if Amazon’s network went down for an hour, that would certainly be bad for business. If one network component fails, you must ensure that other components take over with little or no downtime.

Finally, as much as possible, you must implement redundancy at every possible point in the network. As network engineers, we are responsible for business-critical infrastructure, so we have to make sure that that infrastructure is resilient to failures as much as possible. First off, here is a poorly designed network.

There are many points of failure here which could cut off connectivity. For example if this connection is cut due to a hardware failure, this entire network loses connectivity to the Internet. Or, if this connection is cut off due to a hardware failure, these hosts lose connectivity within the LAN, and out to the Internet.

Okay, those are just two examples, let’s look at a better network design. This network here is a much better design. If this PC wants to reach the Internet, it might use this path in a normal situation.

However, even if this router has a hardware failure and goes down completely, the PC can reach the Internet via this or another alternate path. Perhaps traffic to this other PC in the LAN usually follows this path to the destination. What if this switch fails?

That’s not a problem, because this alternate path is available. So, I think you can see the benefit of designing redundant networks. However, you may be asking, what if this switch fails?

Well, if that is the case, all hosts connected to this switch would lose connectivity. Unfortunately, Most PCs only have a single network interface card (NIC), so they can only be plugged into a single switch. However, important servers typically have multiple NICs, so they can be plugged into multiple switches for redundancy.

We will cover many protocols that are used to enable network redundancy throughout this course, and spanning tree is one of them. Spanning tree is a Layer 2 protocol by the way, it enables redundant layer 2 networks, so within the LAN here, not routing out to the Internet and between networks at Layer 3. I just showed you the benefits of a redundant LAN, having multiple paths between these switches provides alternate paths if one connection fails.

However, without spanning tree, there is a MAJOR problem here which can destroy your network. So, where is the problem? Let me introduce the concept of ‘broadcast storms’.

I’ll use a simplified network topology to demonstrate the issue. PC1 is 10. 0.

0. 1, PC2 is 10. 0.

0. 2, and PC3 is 10. 0.

0. 3. You already know what a switch does with a broadcast frame or an unknown unicast frame.

Let’s say, for example, PC1 wants to send some traffic to PC2. To do that, it needs to know PC2’s MAC address. So, let’s say PC1 sends an ARP request frame, which is a broadcast frame, it uses the broadcast MAC address of all F's as its Layer2 address.

When SW1 receives the frame, what will it do? As I said, you already know what a switch does with broadcast and unknown unicast frames. It will flood it out of all interfaces, except the one it was received on.

So, SW2 and SW3 both receive a copy of the frame. They then do the same thing, they flood it out all interfaces except the one it was received on. So, PC2 receives the ARP request and will reply with a unicast ARP reply.

All good? Actually no, NOT all good. Although PC2 received the ARP request and sent its reply, these broadcast frames still remain on the network.

I’ve cleaned up the arrows so you can see this easier. As I just said, PC2 received the ARP request and sent the reply, but what about these broadcast frames in the network? The switches will continue flooding them .

So, what will happen after this? SW1 just received two broadcast frames, on two different interfaces. It will once again flood them.

Let me clean that up again. SW2 and SW3 both just received broadcast frames, what will they do? They will flood them.

So, I think you get the point. This will continue FOREVER. Do you remember the TTL, or time to live, field of the IP header?

It is used to prevent infinite loops at Layer 3. But the Ethernet header doesn’t have a TTL field. These broadcast frames will loop around the network indefinitely.

If enough of these looped broadcasts accumulate in the network, the network will be too congested for legitimate traffic to use the network. This is called a broadcast storm. Eventually your network will look like this, so full of looping broadcast frames that no regular traffic can pass through your network.

The red arrows represent the clock-wise loop between the three switches, and the purple arrows the counter-clockwise loop. However, network congestion isn’t the only problem. Each time a frame arrives on a switchport, the switch uses the source MAC address field to ‘learn’ the MAC address and update its MAC address table.

When the frames with the same source MAC address repeatedly arrive on different interfaces, the switch is continuously updating the interface in its MAC address table. This is known as MAC Address Flapping. So, how can we design a network with redundant paths that doesn’t result in Layer 2 loops?

Well, Spanning Tree Protocol is one answer to this problem. So let’s take a look at Spanning Tree protocol What we now call ‘classic spanning tree protocol’ is an industry standard protocol, IEEE 802. 1D.

This is the type of STP we will focus on in today’s video, we will focus on the newer Rapid STP later. Because it is so important to prevent Layer 2 loops, switches from ALL vendors run STP by default. So, you won’t only find STP on Cisco switches.

STP prevents Layer 2 loops by placing redundant ports in a blocking state, essentially disabling the interface. These interfaces act as backups that can enter a forwarding state if an active interface, meaning an interface that is currently forwarding, fails. Interfaces in a forwarding state behave normally.

They send and receive all normal traffic. However, Interfaces in a blocking state only send or receive STP messages (called BPDUs, or Bridge Protocol Data Units), and some other specific traffic. Before going more in depth , let me talk about that word ‘bridge’.

I told you about Ethernet hubs in a previous video. Hubs were used before switches were invented, and instead of learning MAC addresses to forward frames to the correct destination, they simply flooded frames out of all interfaces. But actually, before switches, there was another kind of device called a bridge.

You don’t need to know about bridges for the CCNA, they are an old technology, but they’re like a transitional stage between the hub and the switch. However, the reason I’m telling you about bridges is that Spanning Tree Protocol still uses the term ‘bridge’. However, when we use the term ‘bridge’, we really mean ‘switch’.

Bridges are not used in modern networks. So, in this lecture, and really any time I talk about STP, you’ll hear me use the term ‘bridge’, but really it means switch. So, if we look at this topology again, perhaps these interfaces are in a forwarding state, while this one interface on SW3 is in a blocking state, effectively disabling the connection between SW2 and SW3.

So effectively its like that link doesn’t exist, and this is our topology. If PC1 sends that same ARP request broadcast frame, it will be flooded like this, no more loops. However, if at some point another interface fails, perhaps this one.

. . The switches will automatically adjust the topology, and the broadcast frame would be flooded like this, again no loops.

So, that is just a basic outline of the purpose of Spanning Tree Protocol. Now let’s go a little deeper into how spanning tree protocol works. By selecting which ports are forwarding and which ports are blocking, STP creates a single path to and from each point in the network.

This prevents Layer 2 loops. There is a set process that STP uses to determine which ports should be forwarding and which should be blocking. That process is what we will cover next.

STP-enabled switches send Hello BPDUs out of all interfaces, the default timer is 2 seconds, so the switch will send a Hello BPDU out of every interface, once every 2 seconds. If a switch receives a Hello BPDU on an interface, it knows that interface is connected to another switch, because routers, PCs, etc. do not use STP, so they do not send Hello BPDUs.

So, back to our topology here, these switches will send BPDUs out of each interface, like this . They use these BPDUs to advertise themselves to other switches, and to learn about other switches. Now, what exactly are these BPDUs used for?

First of all, switches use one field in the STP BPDU, the Bridge ID field, to elect a root bridge for the network. The switch with the lowest Bridge ID becomes the root bridge. I’ll talk about the bridge ID in the next slide.

ALL ports on the root bridge are put in a forwarding state, and other switches in the topology must have a path to reach the root bridge. So, as I mentioned previously STP puts ports in either a blocking or forwarding state, to avoid Layer 2 loops in the network. However, as I just said, on the root bridge, all ports are forwarding, and all other switches must have a path to reach the root bridge.

Traditionally, the bridge ID field of the spanning tree BPDU looked like this. There is a bridge priority field, which is 16 bits in length, and then there is the MAC address of the switch, which as you already know is 48 bits in length. The default bridge priority is 32768 on all switches, so by default the MAC address is used as the tie-breaker.

As I said before, the switch with the lowest bridge ID becomes the root bridge, so therefore by default the switch with the lowest MAC address becomes the root bridge. So here’s that topology once again, and I’ve written the priority and MAC address for each switch. As you know MAC addresses are 12 hexadecimal digits, but I’ve shortened them to three.

I’ve also added port lights for the interfaces, to show if they are forwarding or blocking. The G0/2 interface on each switch is connected to a PC, so because it isn’t receiving any BPDUs, it knows it is safe to go into forwarding mode, there is no risk of creating a Layer 2 loop, so these port lights are all green. Now, all three switches have the default priority of 32768, so in order to know which one will be the root bridge we will have to compare the 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 to 11, and C is equal to 12, so SW1 has the lowest MAC address.

Therefore, SW1 will become the root bridge of this network. All ports on the root bridge become designated ports, in a forwarding state. So, that is the traditional bridge ID.

However, the bridge ID was actually updated to look like this. In reality, the bridge priority has been updated to be made of two parts, the bridge priority which is 4 bits, and the ‘extended system ID’, which is just the VLAN ID, which is 12 bits, because as you know a VLAN number is 12 bits in length. Why include a VLAN ID in the bridge priority?

Well, Cisco switches use a version of STP called PVST, which stands for Per-VLAN Spanning Tree. PVST runs a separate STP ‘instance’ in each VLAN, so in each VLAN different interfaces can be forwarding or blocking. One interface could be forwarding in VLAN1, but blocking in VLAN2, for example.

By adding the VLAN ID into the bridge priority, the switch will have a different bridge ID in each VLAN. Here’s a deeper look at the bridge priority field. You may have wondered why 32768 is the default bridge priority.

Well, it’s because this total field is 16 bits in length, and the most significant bit is set to 1 by default. Therefore, the default bridge priority WAS 32768. However, with the addition of the extended-system ID, adding the VLAN ID number to the bridge priority, that changed.

So, the default VLAN ID is 1, therefore the bridge priority in total actually ISN’T 32768, it’s 32769. In the default VLAN of 1, the default bridge priority is actually 32769, which is 32768 + 1. Now, here’s a question, If you want to increase the switch’s bridge priority without changing VLAN numbers, what is the minimum unit of increase/decrease?

Let me explain what I mean in the next slide. The bridge priority + extended system ID is a single field of the bridge ID, however the extended system ID is set and cannot be changed because it is determined by the VLAN ID. Therefore, you can only change the total bridge priority (that is, the bridge priority + extended system ID) in units of 4096, the value of the least significant bit of the bridge priority portion.

Let me demonstrate. Currently, the bridge priority here is 32769. Let’s reduce it to make this switch the root bridge.

If I want to reduce it just a little, I can reduce it to 28673, which is 16384 plus 8192 plus 4096 plus 1. I could reduce it more, of course, but the point is this: the STP bridge priority can only be changed in units of 4096. So, the valid values you can configure are listed here, starting from 0 and increasing in units of 4096.

The extended system ID will then be added to this number to make the total bridge priority value. So let’s look at this topology again. We’ll just be looking at the STP topology for a single VLAN, VLAN1, so the priority for each switch is 32769.

But if there are multiple VLANs, say VLAN1, VLAN2, and VLAN3 in this network, the priority would be 32770 for VLAN2, and 32771 for VLAN3, etc. We could also change the bridge priority on the switches for a specific VLAN, so for example SW1 is the root bridge in VLAN1, SW2 could be the root bridge in VLAN2, and SW3 could be the root bridge in VLAN3. I’ll talk about how you can do that in the next video, I just want to let you know some of the possibilities.

So, here in VLAN1, SW1 is the root bridge. All interfaces on the root bridge are designated ports, and designated ports are in a forwarding state. Designated port is one of the port roles in spanning tree.

There are a couple other port roles. I will introduce those in a minute. Okay just a few more points about the root bridge.

When a switch is powered on, it assumes it is the root bridge. It will only give up its position if it receives a ‘superior’ BPDU, and superior means a BPDU from a switch with a lower bridge ID. Once the topology has converged and all switches agree on the root bridge, only the root bridge sends BPDUs.

The reason all switches send BPDUs at first is because they all think they are the root bridge. Other switches in the network will forward BPDUs from the root bridge, but they will not generate their own original BPDUs. Before moving on, let’s see if you understand by doing a few practice questions.

In this network of 4 switches, which will become the root bridge? Pause the video now to think about the answer. Okay, did you find the root bridge?

The answer is SW3. Both SW1 and SW3 have the same priority, 12289, but SW3’s MAC address is lower. The first half, 014A 38 is the same, but the next digit is F for SW1, but 2 for SW3, so SW3’s MAC address is lower.

Let’s do another practice question. Okay here it is. Which switch will become the root bridge in this case?

Pause the video to think about your answer. Okay, the answer is SW4. It has the lowest priority of the 4 switches, 4097.

Okay, now let’s move on. So far we have covered the first step of spanning-tree’s process of creating loop-free Layer 2 LANs. Step 1: the switch with the lowest bridge ID is elected as the root bridge.

All ports on the root bridge are designated ports, so they are in a forwarding state. It’s important that this is the first step that spanning tree takes, because the rest of the steps depend on knowing which switch is the root bridge. Now let’s go on to step 2.

All other switches will select ONE of its ports to be its ‘root port’. So, that means there is one root port on each switch in the network, EXCEPT on the root bridge. The interface with the lowest root cost will be the root port.

Root ports are also in a forwarding state. Now let’s talk about what that ‘root cost’ is. Each interface has an associated spanning tree ‘cost’.

A regular Ethernet interface, with a speed of 10 megabits per second, has a cost of 100. Fastethernet, 100 megabits per second, has a cost of 19. Gigabit ethernet has a cost of 4, and 10 gigabit ethernet has a cost of 2.

Make sure you remember these path costs for the exam. Of course, there will be flashcards in the deck for this video, so use those to help you remember. So, these are gigabit ethernet ports, so they all have a cost of 4.

The root cost is the total cost of the outgoing interfaces along the path to the root bridge. SW1 is the root bridge, so it has a cost of 0 on all interfaces. They are gigabit ethernet interfaces, but you don’t count the cost of the receiving interface, just the sending, the outgoing interface.

So, SW1 advertises its root cost of 0 in its BPDUs. SW2 will receive the BPDU and add the cost of its outgoing interface, G0/1, which is 4, when it floods those BPDUs out of its interfaces. SW3 will do the same.

So, which port do you think SW2 will choose as its root port? Here is its logic. It was advertised a cost of 0 on its G0/1 interface, however the cost of its interface is 4, therefore the total root cost via G0/1 is 4.

It was advertised a cost of 4 on G0/0, from SW3. However its interface also has a cost of 4, so the total root cost via G0/0 is 8. 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 cost of 8 via G0/1, so it will select G0/0 as its root port. In this case, the ports directly across from each root port are the root bridge, so they are already designated ports.

However, keep in mind that the port connected to another switch’s root port MUST be designated. Because the root port is the switch’s path to the root bridge, another switch must not block it. Okay, so I’ve updated our spanning-tree summary here.

First, one switch is elected as the root bridge. All ports on the root bridge are designated ports. There is only one step in selecting the root bridge, that is the switch with the lowest bridge ID.

Next, each remaining switch will select ONE of its interfaces to be its root port, which is also in a forwarding state. Ports across from, ports connected to, the root port are always designated ports. The first criteria for root port selection is the port with the lowest root cost.

However, what if a switch has multiple ports with the same root cost? In that case, the interface connected to the neighbor with the lowest bridge ID will be selected as the root port. Let’s see an example.

Okay, let’s practice that with a quiz, 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 priority. So, SW2’s ports are all designated. Now, which ports will become root ports?

All interfaces are gigabit ethernet, so all have a cost of 4. Remember, if there is a tie in root cost, the switch will select the interface connected to the neighbor with the lowest bridge ID. So, pause the video here to think about your answer, which ports will be selected as root ports, one on each switch.

Okay, on SW1 and SW4, the answer is obvious, SW1’s G0/0 and SW4’s G0/1 have a cost of 4, so they are selected. How about SW3? 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 4. So, we have to use the tiebreaker, which neighbor switch has the lowest bridge ID, SW1 or SW4? It’s SW1, the priorities are the same, but SW1’s MAC address is lower.

So, G0/0 is selected as the root port, and SW1’s G0/1 becomes designated. So, this is the process so far. HOWEVER, there is ONE more tiebreaker that might be needed to select the root port.

What if two switches have two connections between them, so both the root cost and the neighbor bridge ID are the same? Then we get to the final tie-breaker, the interface connected to the interface on the neighbor switch with the lowest port ID will become the root port. Okay, let me briefly explain port ID.

So, here is the output of the command SHOW SPANNING-TREE, we’ll talk about it more in a future video when we look at spanning tree configuration. I just want to show you this section, this lists the spanning tree port ID of each interface on the switch. Notice the column title is Prio dot number.

So, each port has a default priority of 128, and then a unique port number, 1 for G0/0, 2 for G0/1, etc on this switch. So, the STP port ID equals the port priority plus the port number. Similar to the bridge ID, where the MAC address is used as a tiebreaker if the priorities tie, in this case the port number is used as a tiebreaker if the priorities tie.

I won’t explain the port ID in more depth than this, usually you don’t need to worry about it or change it, so you can just look at the port number. For example, G0/0 is lower than G1/0, or G0/3 is lower than G1/2. So, one more quiz to practice that.

Now there are two connections between SW1 and 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 to a lower port ID on the neighbor switch, SW1. This is an important point. The NEIGHBOR switch’s port ID is used to break the tie, not the local switch’s port ID.

That’s why G0/2 was selected over G0/0, because G0/0 is connected to a higher port ID on SW1. So, SW1’s G0/1 interface is a designated port, because it is connected to SW3’s root port. Okay, so this is our process so far.

But, it’s not complete. We still haven’t blocked any ports, and we need to block some ports to prevent Layer 2 loops. So, let’s return to our previous topology.

All that’s left is this connection between SW2 and SW3. So far, all of our ports are in a forwarding state, both root ports and designated ports are always in a forwarding state. So, to prevent loops do we block both of these ports?

SW2’s G0/0 and SW3’s G0/1? Actually no, here’s an important rule to remember: every collision domain has a single spanning tree designated port. Remember, unlike old Ethernet hubs, which we don’t use anymore, when we use switches, each link is a separate collision domain.

This collision domain between SW1 and SW2 has one designated port, SW1’s G0/0. This connection between SW1 and SW3 has one, SW1’s G0/1. And the connections with the PCs are all designated ports in the forwarding state, because the PCs don’t participate in spanning tree.

So, we need one designated port on the connection between SW2 and SW3. How do we determine which port will be designated, in a forwarding state? The switch with the lowest root cost will make its port designated.

However, in this case both switches have the same root cost, 4 for SW2 via its G0/1 interface and 4 for SW3 via its G0/0 interface. So, for the tie-breaker we compare the bridge ID. SW2 has the lower bridge ID, so its G0/0 interface will be designated.

Finally, the other switch will make its port non-designated, which means it is in a blocking state. So SW3’s G0/1 is non-designated, it blocks the port to prevent Layer 2 loops. So, here is the process for selecting the different port roles and states in spanning tree.

One switch is selected as the root bridge, the switch with the lowest bridge ID. Then, each remaining switch selects ONE of its interfaces to be a root port. The interface with the lowest root cost is selected, if that’s a tie the interface connecting to a neighboring switch with the lowest bridge ID is selected, if that’s a tie also, the interface connected to the lowest port ID on the NEIGHBOR switch is selected.

Then finally, each remaining collision domain will select ONE interface to be a designated port, and the other port will be non-designated. The interface on the switch with the lowest root cost will be designated, if that’s a tie the interface on the switch with the lowest bridge ID will be designated, and then the other interface will be a non-designated port, in a blocking state. There are still many important things left to explain regarding spanning-tree, I will cover those in part 2, before moving on to another type of spanning tree, called rapid spanning tree.

We already did a few quiz questions throughout the video, but let’s do a few more practice questions to make sure you know the whole process. If you get stuck, if you don't know the answer, go back to the previous slide to remind yourself of the process, and try to figure out the answers yourself. I will also feature one question from Boson ExSim after the quiz, however today will be a little different.

Because there are still some important points to cover about spanning tree, we actually aren’t ready to answer the spanning tree questions on Boson ExSim. So, I will show you one question from Boson ExSim, and in the next video we will see the answer. Of course, if you have already studied spanning tree and already know the answer, please write it in the comment section.

Okay, let’s do a couple more practice questions first. Here is a network topology. Identify the root bridge, and the role of each interface on each switch in the network, so which interfaces are root ports, which are designated ports, and which are non-designated ports.

Pause the video to think about your answer. Okay, I hope you found the answer. So, the root bridge is SW3, because the priority is a tie and it has the lowest MAC address.

These are the root ports, SW2 selected its G0/2 interface because it is connected to the lower-number interface on SW1, G0/0. And these are the remaining connections. In each case the interface on SW2 is non-designated, because it has a higher root cost.

Always remember to check that there is one designated port for each connection, each collision domain. Okay, let’s do one more quiz. Do the same thing, but with this network topology.

Look carefully, some of these interfaces are fast ethernet interfaces, they have a spanning tree cost of 19, not 4. Pause the video to think about your answer. Okay, hopefully you solved it.

SW4 is the root bridge because it has the lowest priority. these are the root ports. SW1 uses its G0/1 interface as the root port, because it’s other two interfaces are fastethernet, with a much higher spanning tree cost.

Finally, the remaining designated and non-designated ports. SW1’s F1/0 and F2/0 are non-designated because SW2 has a lower root cost, and SW2’s G0/1 is non-designated because SW4 IS the root bridge, so its G0/1 interface must be designated. Okay, that’s all for the quiz, let’s take a look at Boson ExSim.

Okay for today's Boson ExSim practice question we're talking about PortFast, which is an optional feature of spanning tree which I haven't talked about in today's video. So, I won't give the answer in this video, I'll just read the question and then I'll give you the answer in the next lecture video, day 21. Now, if you think you know the answer, if you've already studied spanning tree please feel free to let me know your answer in the comments.

Or if you want to do some independent research of your own, perhaps type into Google 'spanning tree portfast' and do some reading, and then figure out the answer, again please let me know your answer in the comment section. So here is the question. You want to decrease the amount of time that it takes for switchports on Switch A to begin forwarding.

Portfast is not configured on any of the switchports on Switch A. You issue the 'spanning-tree portfast default' command from global configuration mode. Which of the ports on Switch A will use portfast?

Select the best answer. So A says 'No ports, because portfast cannot be enabled globally. ' B says 'All ports', C says 'All access ports', and D 'All trunk ports'.

Okay so as I said, this time we won't check the answer. Please wait for the next video to see the answer for this question. Of course, if you think you know the answer, let me know in the comment section.

If you want to get your own copy of ExSim, and I highly recommend you do before you take the real thing, please follow my link in the video description. These are by far the best practice exams out there for the CCNA. There will be supplementary materials for this video.

There will be a review flashcard deck to use with the software ‘Anki’. Download the deck from the link in the description. There will also be a packet tracer practice lab.

Please be sure to watch the practice lab, it will give you some more practice for this process of figuring out a spanning tree topology, but also I will introduce some valuable CLI commands which I didn’t have the time to show in this video. Before finishing today’s video I want to thank my JCNP-level channel members. Thank you to Joyce, Marek, Samil, Velvijaykum, C Mohd, Johan, Mark, Aleksa, Miguel, Yousif, Boson Software, the creators of ExSim, Sidi, Magrathea, Devin, Charlsetta, Lito, Yonatan, Mike, Aleksandr, and Vance.

Sorry if I pronounced your name incorrectly, but thank you so much for your support. One of you is displaying as Channel failed to load, if this is you please let me know and I’ll see if YouTube can fix it. This is the list of JCNP-level members at the time of recording by the way, May 10th, if you signed up recently and your name isn’t on here don’t worry, you’ll definitely be in the next video.

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