All language subtitles for Free CCNA - Subnetting (Part 1) - Day 13 - CCNA 200-301 Complete Course

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In this we will be talking about ‘subnetting’.
This is a very big topic for the CCNA, but

not just for the test, it’s an essential
skill for a network engineer. Many people

have trouble understanding subnetting, but
let me assure you, it is NOT difficult. Subnetting

is very simple if you take it step-by-step.
So, I’m going to split subnetting into 2

, or maybe even 3 videos so we can take our
time to really understand subnetting without

getting lost. Now, because subnetting is such
an important topic and many people have trouble

with it, there are already plenty of subnetting
videos on youtube. Of course feel free to

check out those videos too, there are some
different tricks and techniques people teach

that can speed up the subnetting process.
I, however, will simply outline the basic

steps involved in subnetting, I will avoid over-complicating the topic. My end

goal for these videos is that you understand
and can do subnetting. So let’s get started.

So, what will we cover in this video? Only
a couple things. First is C I D R, pronounced

CIDR, which stands for classless inter-domain
routing. What exactly is that? Well, remember

I introduced the IPv4 address classes, such
as class A, B, and C? Well, CIDR throws all

that away and lets us be more flexible with
our IPv4 networks. Then, of course, we’ll

cover the process of subnetting, taking it
step-by-step so you don’t get lost.

Now, before I get into CIDR, let’s review
these IPv4 address classes, so we can then

understand the need for classLESS IPv4 addressing.
There are five classes of IPv4 addresses,

A, B, C, D, and E. Class A addresses have
a first octet beginning with 0, and the rest

of the bits can either be 0 or 1. This leads
to a decimal range for the first octet of

0 to 127. Remember, an IPv4 address is 32
bits, so there are 4 octets, 4 groups of 8

bits, in an IPv4 address. This makes the class
A address range from 0.0.0.0 through 127.255.255.255.

Now, remember there are some special and reserved
addresses in these ranges that can’t be

used for normal IP addresses on a device,
but for this video we’ll just include all

of them in Class A. Class B addresses have
a first octet beginning with 1 0 , and the

other 6 bits can be either 0 or 1. This gives
a range for the first octet of 128 through

191. The address range for class B is 128.0.0.0
through 191.255.255.255. Class C addresses

have the first three bits set to 1 1 0, and
the others can be either 0 or 1. If you write

that range in decimal that is 192 through
223. The address range is therefore 192.0.0.0

through 223.255.255.255. Class D addresses
begin with 1 1 1 0 in binary, which gives

a range of 224 through 239 for the first octet
of the address. This means that the address range

for class D is 224.0.0.0 through 239.255.255.255.
Finally, class E address begin with 1 1 1 1

in binary, so the first octet range is 240
through 255, and therefore the address range is 240.0.0.0

through 255.255.255.255.

However, only the class A, B and C addresses
can be assigned to a device as an IP address,

as classes D and E have special purposes I
mentioned in the IPv4 addressing videos. Class

A addresses have a /8 prefix length, meaning
the first octet identifies the network and

the other three octets are used for individual
hosts within the network. Class B addresses

have a /16 prefix length, so the first two
octets identify the network, and the last

two octets identify individual hosts within
that network. Class C addresses have a prefix

length of /24, so the first three octets are
used to identify the network, and only the

last octet is used to identify individual
hosts within that network.

The different prefix lengths give different
characteristics to these classes. As you can

see, there are few class A networks available,
only 128, actually less than that because

some are reserved, like the 127.0.0.0/8 range,
which you may remember is used for loopback

addresses. Because only the first octet of
a class A address is used for the network

ID, there are three whole octets available
for addresses within each class A network,

so there are 16 million, 777 thousand, 216
addresses in each class A network. That is

2 to the power of 24, because there are 3
octets, 3 times 8 equals 24 bits. Class B

addresses are different, there more class
B networks, 16,384, but fewer addresses per

network, 65,536, which is still very many
addresses of course. Finally, there are very

many class C networks, 2 million 97 thousand
152 networks, but only 256 addresses per network.

So, how does a company get their own network
to use? Well, IP addresses are assigned to

companies or organizations by a non-profit
American corporation called the IANA, the

Internet Assigned Numbers Authority. The IANA
assigns IPv4 address and networks to companies

based on their size. For example, a very large
company might receive a Class A or Class B

network, remember there are lots of available
addresses to use for hosts in each class A

and class B network, while a small company
might receive a class C network, because there

are fewer addresses in each class C network,
only 256. However, this system led to many

wasted IP addresses, so multiple methods of
improving this system have been created. Let

me give you an example of how this strict
system of addresses can waste IP addresses.

So, here are two routers. As you can see,
R1 has three networks connected to it here.

Remember that routers are used to connect different
networks, so each of these links are separate

Layer 3 networks, different IP networks. R2
also has three networks connected here. Perhaps

each of these networks will have a few switches,
with many end hosts such as PCs and servers

connected to these switches. However, there
is one more network here. That’s this network

connecting these two routers. This is known
as a ‘point-to-point’ network, meaning

that its a network connecting two points,
in this case R1 and R2. For example, this

might be a connection between offices in different
cities, let’s say San francisco and new

york.

So, because this is a point-to-point network,
we don’t need a large address block, so

let’s use a class C network, 203.0.113.0/24.
Because this is a class C network, there are

256 addresses in the network. Minus 1 for
the network address, 203.0.113.0, minus one

for the broadcast address, 203.0.113.255,
minus one for R1’s address, which I’ll

assign as 203.0.113.1, and minus 1 for R2’s
address, which I’ll assign as 203.0.113.2.

That’s a total of 4 addresses used, and
252 addresses WASTED. Clearly, this is not

an ideal system.

Before introducing CIDR, here’s another
quick example of address waste. A company,

company X, needs IP addressing for 5000 end
hosts. This is a problem. Why? A class C network

does not provide enough addresses, so a class
B network must be assigned. Because a class

B network allows for about 65,000 addresses,
this results in about 60,000 addresses being

wasted.

When the Internet was first created, the creators
did not predict that the Internet would become

as large as it is today. This resulted in
wasted address space like the examples I showed

you, and there are many more examples that
I could show you. The total IPv4 address space

includes over 4 billion addresses, and that
seemed like a huge number of addresses when

IPv4 was created, but now address space exhaustion
is a big problem, there's not enough addresses. One way to solve, or remedy this is

CIDR. The IETF (Internet Engineering Task
Force) introduced CIDR in 1993 to replace

the ‘classful’ addressing system.

With CIDR, the requirements of ‘class A
address must use a /8 network mask, class

B must use /16, and class C must use /24’
were removed. This allowed larger networks

to be split into smaller networks, allowing
greater efficiency. These smaller networks

are called ‘subnetworks’ or ‘subnets’.
Let’s look at an example of splitting a

larger network into a smaller network so you
can see how it works.

Here’s the same point-to-point network we
looked at before. Previously, it was assigned

the 203.0.113.0/24 network space, but that
resulted in lots of wasted addresses. Let’s

write this out in binary. Here’s the binary,
with the dotted decimal underneath. Now, the

prefix length is /24, so here’s the network
mask, also known as the subnet mask, 255.255.255.0.

Remember, all ‘1s’ in the subnet mask
indicate that the same bit in the address

is the network portion. In this case, I made
the network portion blue, and the host portion

is red. Well, how many host bits are there? 8, because
it’s one octet. So, how many potential hosts, or how

many usable addresses are there? Well, the
formula is this. 2 to the power of 8, minus

2, equals 254 usable addresses. What is the
8? Well, it’s the number of host bits, which is

8 in this case. And why minus 2? Those are
the network address and broadcast address,

we can’t assign them to a device so we have
to remove them from the number of usable addresses.

So, we have 254 usable addresses, but we only
need two, one for R1 and one for R2.

However, CIDR allows us to assign different
prefix lengths, it doesn’t have to be /24.

Let’s get some practice calculating the
number of hosts with different prefix

lengths. 203.0.113.0/25. 203.0.113.0/26, 203.0.113.0/27,
/28, /29, /30, /31, and finally /32. I’ve

put /31 and /32 in red because they’re a
little bit special, you’ll see when you

try to calculate it. So, pause the video here
and try to calculate how many usable address

are in each network...okay, let’s check
out the answers.

So, here is 203.0.113.0, but this time with
a /25 mask. Notice that the network portion

of the address has extended into the first
bit of the last octet, and the mask

in dotted decimal is now written as 255.255.255.128.
I changed the color of the extra bit to purple,

but it is part of the network portion, the blue
part. If you don’t remember how to convert

from binary to dotted decimal, make sure you
review that, it’s very important for subnetting.

Now there are 7 bits in the host portion of
the address, so the number of usable addresses

is 2 to the power of 7, minus 2, which equals
126. Once again, we only need 2 addresses,

one for R1 and one for R2, so we will be wasting
124 addresses. That’s better than wasting

252 addresses with a /24 prefix length, but
still its wasteful.

How about a /26 prefix length? Notice that
it’s now written 255.255.255.192 in dotted

decimal, because two bits of the last octet
are now part of the network portion. Since

there are 6 host bits, there are now 62 usable
addresses in this network. If we were to use

a /26 network mask for the 203.0.113.0 network,
we would be wasting 60 addresses. Getting

better, but we can make this network even
smaller.

Now that you get the idea, let’s speed it
up. For a /27 prefix length, the mask is written

as 255.255.255.224 in dotted decimal. There
are now 5 host bits, so that means there are

30 usable addresses. As you can see, the address
space is getting smaller and smaller as we extend the

network mask.

For a /28 prefix length, the mask is written
as 255.255.255.240 in dotted decimal. There

are now only 4 host bits, so that means there
are 14 usable addresses. After assigning addresses

to R1 and R2 this would mean only 12 wasted
addresses. But we can make this address space

even smaller, to make our addressing even
more efficient.

If we use a /29 prefix length, the mask is
written as 255.255.255.248 in dotted decimal.

Now we have only 3 host bits, so that means
there are just 6 usable addresses. Again,

after we give R1 and R2 addresses there would
be only 4 wasted addresses.

If we use a /30 prefix length, the mask is
written as 255.255.255.252 in dotted decimal.

There are now only 2 host bits, so that means
2 usable addresses. So, this is perfect! There

are 4 total addresses, that's the network address,
the broadcast address, R1’s address, and

R2’s address. That means 0 wasted addresses!

Before moving on to /31 and /32 let me clarify
a little bit. So, instead of 203.0.113.0/24,

we will use 203.0.113.0/30, which is a subnet
of that larger class C network. 203.0.113.0/30

includes the address range of 203.0.113.0
through 203.0.113.3. Let me show you that

in binary. Here is 203.0.113.0 in binary,
the host portion all 0s. Here is 203.0.113.1,

203.0.113.2, and 203.0.113.3. These are the
4 addresses in the network, these two being

the two usable addresses which are assigned
to R1 and R2. So we took up 4 addresses with

this subnet, what about the other addresses
in the 203.0.113.0/24 range? The remaining

addresses in the address block, which are 203.0.113.4 – 203.0.113.255, are now available

to be used in other subnets! That’s the
magic of subnetting. Instead of using 203.0.113.0/24

and wasting 252 addresses, we can use /30 and
waste no addresses. Or, perhaps there is another

way to make this even more efficient? Let’s
look into it.

If we use a /31 prefix length, the mask is
written as 255.255.255.254 in dotted decimal.

There is now only 1 host bit, so that means...0
usable addresses. 2 to the power of 1 is 2,

minus 2 for the network and broadcast addresses,
means 0 addresses that we can assign to devices.

So, you used to not be able to use /31 network
prefixes because of this. HOWEVER, for a point

to point connection like this it actually
is possible to use a /31 mask. Let’s check

it out.

So here’s the 203.0.113.0/31 network, R1
is 203.0.113.0 and R2 is 203.0.113.1. The

203.0.113.0/31 network consists of the addresses
from 203.0.113.0 through 203.0.113.1...which

is actually only two addresses. Here they
are in binary. there’s 203.0.113.0, and

there’s 203.0.113.1. Normally this would
be a problem, because it leaves no usable

addresses after subtracting the network and
broadcast addresses, but for point-to-point

networks like this, a dedicated connection
like this between two routers, there is actually

no need for a network address or a broadcast
address. So, we can break the rules in this

case and assign the only two addresses in
this network to our routers. Note that, if

you try this configuration on a Cisco router,
you’ll get a warning like this, reminding

you to make sure that this is a point-to-point
link, but it is a totally valid configuration.

So, once again The remaining addresses in
the 203.0.113.0/24 address block, which are 203.0.113.2

– 255 are now available to be
used in other networks! But this time we’ve

saved even more addresses, using only 2 addresses
instead of 4 for this point-to-point connection.

People still do use /30 for point-to-point
connections at times, but /31 masks are totally

valid and more efficient than /30 so I recommend
this method!

But we still haven’t looked at the /32 mask.
A /32 mask is written as 255.255.255.255 in

dotted decimal, making the entire address
the network portion, there are no host bits.

If you calculate this using our formula, you
will get -1 usable addresses...clearly the

formula doesn’t work in this case. You won’t
be able to use a /32 mask in this case, and

you will probably never use a /32 mask to
configure an actual interface. However, there

are some uses for a /32 mask, for example
when you want to create a static route not

to a network, but just to one specific host,
you can use a /32 mask to specify that exact

host. Anyway, I’ll talk about that later
in the course, just know that /32 masks are

used at some points, but you don’t have
to worry about them for now.

Here’s a simple chart showing the dotted
decimal subnet masks, and their equivalent

in CIDR notation. That’s right, the way
of writing a prefix with a slash followed

by the prefix length, like /25, /26, etc.
is called CIDR notation, because it was introduced

with the CIDR system. Previously, only the
dotted decimal method was used. Note that

I’ve showed you only how to subnet a class
C network so far, but we will look at

class B and class A networks as well, with
prefix lengths like /17, /11, /9, etc.

I spent a lot of time on just that one
example, but I hope you can see the use of

subnetting, dividing a larger network into
smaller networks, called subnets. Instead

of using the whole 203.0.113.0/24 network
for the point to point connection, we can

use a /30 subnet and use only 4 addresses,
or even better use a /31 subnet and use only

2 addresses. I’ll give one more example
of subnetting before finishing up this video.

In the next video I’ll give you some practice
problems and walk you through them so you

can get some hands-on practice with subnetting.

So, here’s a scenario. There are 4 networks
connected to R1, with many hosts connected

to each switch. There are 45 hosts per network,
R1 needs an IP address in each network so

its address is included in the range. You
have received the 192.168.1.0/24 network,

and you must divide the network into four
subnets that can accommodate the number of

hosts required. First off, are there enough
addresses in the 192.168.1.0/24 network in

the first place? So, we need 45 hosts per network,
including R1, but also remember that each

network has a network and broadcast address,
so that’s plus 2, so we need 47 addresses

per subnet. 47 times 4 equals 188, so there’s
no problem in terms of the number of hosts.

192.168.1.0/24 is a class C network, so there
are 256 addresses, so we will be able to assign

4 subnets to accommodate all hosts, no problem.

Okay let’s see how we can calculate the
subnets we need to make. We need four equal

sized subnets with enough room for at least
45 hosts. Here, I’ve written out 192.168.1.0

with a /30 mask, 255.255.255.252. I skipped
/32 and /31, since these aren’t point to

point links, we can’t use /31, and definitely
cant use /32. Since there are 2 host bits,

the formula to determine the number of usable
addresses is 2 to the power of 2, minus 2.

2 to the power of 2 is 2 times 2, which is
4. So that means there are 2 usable addresses

in a /30 network. Clearly not enough room
to accommodate the 45 hosts we have.

How about if we use a /29 mask to make these subnets, can we fit
the 45 hosts we need? There are 3 host bits,

so the formula is 2 to the power of 3 minus
2. 2 to the power of 3 is 2 times 2 times

2, which is 8. Therefore there are 6 usable
addresses, not enough for 45 hosts.

How about if we use /28? There are 4 host
bits, so the formula is 2 to the power of 4

minus 2. 2 to the power of 4 is 2 times 2
times 2 times 2, which is 16. So, that means there are

14 usable addresses, once again not enough for 45 hosts.

How about /27? There are 5 host bits, so the
formula is 2 to the power of 5 minus 2. And 2 to

the power of 5 is 2 times 2 times 2 times
2 times 2, which equals 32. So that means

30 usable addresses, again not enough
for 45 hosts.

How about a /26 subnet mask? There are now 6
host bits, so the formula is 2 to the power

of 6 minus 2. 2 to the power of 6 is 2 times
2 times 2 times 2 times 2 times 2, which equals

64. That means there are 62 usable addresses.
So, it looks like we’ve found our number! /27

doesn’t provide enough address space. /26
provides more than we need, but we have to

go with /26. Unfortunately we can’t always
make subnets have exactly the number of addresses

you want. There might be some unused address
space. That’s actually fine, since its good

to have some room for growth anyway.

So I think this video has gone on long enough.
Instead of finishing this task in this video, I’ll make

it this week’s quiz. The first subnet (Subnet
1) is 192.168.1.0/26. What are the remaining

subnets? To help you out, here’s a hint. Find the broadcast address of Subnet

1. The next address after that is the network address
of Subnet 2. And then just repeat the process for Subnets

3 and 4. Post your answers in the comment
section, and I’ll also go over the answer

in the next video.

So, what did we cover in this video? We covered
CIDR, classless inter-domain routing, which

removes the rules of class A, B and C networks
and lets us be more flexible with network

addressing, according to the size of the network.
We also covered the process of subnetting,

but mostly just the basics. Hopefully you
understand the purpose of subnetting, and

know a little bit about how to do it. I’ll
clarify and expand upon many things in the

next video, but also feel free to ask any
questions you have in the comment section.

For today’s video there won’t be a practice
lab, that will be after I’ve finished explaining everything about

subnetting. There will be flashcards, however,
to help you review some of the things learned

in this video. You can download them from the link in the description.

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