Let's do one more exercise.
Let's go ahead and change the Mask.
We're going to make it 255.0.0.0 for our new Mask.
So this is going to be a zero and a zero,
and we're going to have all these zeroed out here as well.
So now if this is what our computer showed us,
now my question is what is the actual network address
that this computer is on?
How many bits is it?
How long is it?
And how many host bits are available?
So go ahead and pause, make your decisions,
and then let's compare notes after you resume.
[MUSIC PLAYING]
Thanks for taking a moment to do that.
So here's one way we can approach this.
We simply look at the Mask first.
That's the very first clue, and we're going to say, OK.
This Mask says that the first eight
bits are being used for the network,
and if all the rest of the bits are zeros,
that's the dividing line.
And the cool thing is about a Mask, it won't jump.
This network Mask or subnet Mask doesn't jump.
They're going to be contiguous bits moving from left to right.
So a Mask will never say 110111.
It's going to be contiguous.
So because the dividing line is right here,
our very first octet is the network.
Everything after that is zeroed out from the network address
perspective.
And that's eight, because it's the first 8
bits representing the network.
And now we have the remaining 24 bits
of that IP address, which are now representing
host addresses on network 192.
So let's do one more example to help reinforce the concept.
Let's say we have a Macintosh or a Linux
or a Windows box sitting on a network,
and it's a happy, happy computer.
You want to know why it's so happy?
Because it's got an IP address.
It's got the IP address of 10.25.3.99.
That's its IP address, and guess what it also has?
It also has a Mask.
That Mask in decimal is 255.255.0.0.
What I would love for you to do right now is tell me
what is the actual network that this computer is sitting on?
What is the host address for this computer?
And what is the number of bits that
are currently being used by that network for network addressing?
And how many bits are left over for host
addressing on that network?
So take a moment.
Go ahead and pause me right now.
Take as much time as you need.
And once you've calculated what the actual network address is
that this computer is sitting on, go ahead and resume me
and then we'll compare notes.
Awesome.
Welcome back.
So in this example of 10.25.3.99,
the very first thing we're going to look at
isn't the IP address at all.
It's the Mask, because the Mask is controlling
which portion of that IP address is the network
and which portion is available for host addresses.
And the Mask is saying, I'm consuming--
I'll use green for this--
I am consuming the first octet for network.
So we know that the 10 is part of the network address
that the computer is sitting on.
We also know based on the Mask that this second octet is also
consumed and being allocated for the purpose of network
addressing, and that's it.
The third and fourth octet, the Mask isn't taking anything.
So we can draw our dividing line right there.
This computer is sitting on network 10.25.
And then we simply zero out the bits
that host addresses are going to have,
0.0 as we describe the network.
And that is a total of 8 bits here and 8 bits here,
a total of 16 bits being used for the network address.
Now that also implies that we have 16 bits left over that
are not being used for the network address portion
that we can then use for individual host
assignments on that network.
So this computer with the IP address of 10.25.3.99,
the first two numbers are the actual network right here,
and the last two numbers of 3.99 represent the actual host
portion or this guy's unique identifier
on that network regarding IP addressing.
And as we mentioned there are 16 bits
being used for the network, and there's also 16 bits
available for host addressing.
I hope this has been informative for you,
and I'd like to thank you for viewing.