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

Welcome to Jeremy’s IT Lab.

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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.

Thank you for watching.

Please subscribe to the channel, like the
video, leave a comment, and share the video

with anyone else studying for the CCNA.

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That's all for now.