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Let's do one more exercise.

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Let's go ahead and
change the Mask.

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We're going to make it
255.0.0.0 for our new Mask.

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So this is going to
be a zero and a zero,

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and we're going to have all
these zeroed out here as well.

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So now if this is what
our computer showed us,

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now my question is what is
the actual network address

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that this computer is on?

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How many bits is it?

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How long is it?

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And how many host
bits are available?

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So go ahead and pause,
make your decisions,

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and then let's compare
notes after you resume.

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[MUSIC PLAYING]

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Thanks for taking a
moment to do that.

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So here's one way we
can approach this.

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We simply look at
the Mask first.

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That's the very first clue,
and we're going to say, OK.

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This Mask says that
the first eight

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bits are being used
for the network,

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and if all the rest
of the bits are zeros,

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that's the dividing line.

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And the cool thing is about
a Mask, it won't jump.

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This network Mask or
subnet Mask doesn't jump.

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They're going to be contiguous
bits moving from left to right.

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So a Mask will never say 110111.

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It's going to be contiguous.

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So because the dividing
line is right here,

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our very first octet
is the network.

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Everything after that is zeroed
out from the network address

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perspective.

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And that's eight,
because it's the first 8

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bits representing the network.

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And now we have the
remaining 24 bits

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of that IP address, which
are now representing

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host addresses on network 192.

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So let's do one more example
to help reinforce the concept.

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Let's say we have a
Macintosh or a Linux

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or a Windows box
sitting on a network,

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and it's a happy,
happy computer.

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You want to know
why it's so happy?

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Because it's got an IP address.

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It's got the IP
address of 10.25.3.99.

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That's its IP address, and
guess what it also has?

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It also has a Mask.

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That Mask in decimal
is 255.255.0.0.

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What I would love for you
to do right now is tell me

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what is the actual network that
this computer is sitting on?

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What is the host address
for this computer?

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And what is the
number of bits that

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are currently being used by that
network for network addressing?

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And how many bits are
left over for host

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addressing on that network?

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So take a moment.

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Go ahead and pause me right now.

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Take as much time as you need.

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And once you've calculated what
the actual network address is

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that this computer is sitting
on, go ahead and resume me

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and then we'll compare notes.

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Awesome.

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Welcome back.

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So in this example
of 10.25.3.99,

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the very first thing
we're going to look at

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isn't the IP address at all.

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It's the Mask, because
the Mask is controlling

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which portion of that IP
address is the network

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and which portion is
available for host addresses.

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And the Mask is
saying, I'm consuming--

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I'll use green for this--

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I am consuming the
first octet for network.

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So we know that the 10 is
part of the network address

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that the computer is sitting on.

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We also know based on the Mask
that this second octet is also

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consumed and being allocated
for the purpose of network

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addressing, and that's it.

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The third and fourth octet,
the Mask isn't taking anything.

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So we can draw our
dividing line right there.

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This computer is sitting
on network 10.25.

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And then we simply
zero out the bits

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that host addresses
are going to have,

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0.0 as we describe the network.

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And that is a total of 8
bits here and 8 bits here,

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a total of 16 bits being
used for the network address.

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Now that also implies that we
have 16 bits left over that

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are not being used for the
network address portion

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that we can then use
for individual host

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assignments on that network.

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So this computer with the
IP address of 10.25.3.99,

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the first two numbers are the
actual network right here,

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and the last two numbers of
3.99 represent the actual host

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portion or this guy's
unique identifier

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on that network
regarding IP addressing.

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And as we mentioned
there are 16 bits

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being used for the network,
and there's also 16 bits

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available for host addressing.

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I hope this has been
informative for you,

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and I'd like to thank
you for viewing.

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