next, we're going to discuss what the sub net Mass does
Now, just like your home address
You have your street address
and then your house number. So let's go with 21 jump street.
The street address is Jump Street, and the house number is 21
Your I P address has two portions.
It has a network portion which corresponds to the street address,
and then it has a host portion
which corresponds to the house address.
What? The subnet mask Does it separate
the stream portion of the address from the house portion of the address?
So a sub net mass separates the network address
from the host address I have drown out in binary
And four AQ tats for the sub net mask.
And the I P we're gonna be converting to binary is 210 10 10
and the sub net mass being the Class C Summit mass. Since 200 falls in the Class C range, it is a number between 1 92 and 2 23 So the seven mass has to be 2552552550
If the Ark *** are not clearly visible on camera, I have drawn out
a bigger octet on top. Here. Each one of these fields between the dots looks like this,
with the name numbers going from one than Double 248 16 32 64 1 28
So let's convert the I pee into Binary
200. How do I get 200?
plus eight is 200. Sorry, 1 92 plus eight is 200
then The second octet is 10 which is simply eight and two.
So eight plus two is 10.
The third octet is also 10 which is eight plus two
and the fourth octet. I'm gonna be off camera here for a second. I didn't have enough space to draw this out larger than that. Eight and two
and the rest are zeros. Now let's look at the subject mask.
The first octet is 255 How do I get to 55? Once again, all the bits are turned on.
So 1 28 64 plus 32 plus 16 plus eight plus one plus two plus one
The second octet is also 255
So is the third octet on the fourth ock tent is zero, which means all zeros. This is our sub net masterworks
So bit value 1 28 We're looking at
so for any bit in the i p address.
If the corresponding bit
in the sub net mask is on,
then that bit in the I P address belongs to the network portion of the address or the street portion of your address.
So 1 28 belongs to the network portion because the correspondent bit in the sub net mask is on.
How about this? 32 bit
the course the bit in the I. P address
is off. But that doesn't matter.
If the corresponding bet in the sub net mascots turned on then this dim, it belongs to the network. So it does not matter whether bit in the i P addresses on with the value one or off with the value zero as long as the corresponding bit in the sub net mask is on. That *** belongs to the network.
So this first hole are Tet has on bits in the sub net mask. The corresponding bits for the I P in the sub net mask are all on
this first octet. The whole thing belongs to the network.
So does the second octet
because the corresponding bit for the i p
the corresponding bits in the sub net mass. For this I p
address, the corresponding bits are on. So the second octet also belongs to the network.
And sure does the third doctor
the 4th 4th octet for the bits in the I p. The corresponding bits in the sub net mask are all off.
So that portion off the I P address belongs to the host portion.
the 1st 2nd and third octet have the bits in the sub net mask as on
so the 1st 2nd and third octave belong to the network.
the last octet all belongs to the host address or four host addresses.
So after this point, I want to ask you, give me just the street name.
Give me just the network address.
You would say, or I would expect you to say, I'm gonna erase this tough. The network address is 200 dot Tenn 200.0.10. In the computer world, we can't just leave a field empty,
we simply put a zero here.
So the street number
for this I p address just the street number or the network address is 210 10 0 And again, this is slash 24 or 25525525508 bits plus eight bits plus eight bits
is 24 bits, hence its last 24.
Now, the first valid address on this street or the first house on this street is simply
If I were to ask you, how many total houses can you have on this street? Or, in other words, what is the last address on this street?
as you have seen, if an octet is full,
so you take any octet and turn all the bits on. That's the maximum number you can derive from it, which is 255 So the last address
on the street is going to be 200
200 dot Tenn dot Tenn 22.214.171.124
Making sure these dots are visible for you guys. So
the rule in all networks being
is that you cannot use
the first address and the last address in any network for your PC, you cannot assign a PC or any machine the first address and the last address in a network.
Why? Well, because the first address, the 0.0 address is the network address itself. So if you ask for my home address and I told you I live on Jump Street, would you be able to get to my house? No. You would just be able to get to my street.
So assigning this address through machine is not allowed because it is just the street address
or the network address.
The last address in any network
Any message sent to this address will be received by all houses on that street, or any message sent through the dot to 55 address will be received by all hosts
that my valid range of addresses and I'm going to erase this 10 here. We don't really need it anymore.
The valid range of addresses that are assign a ble to PCs
is 0.12 dot 254 This is your valid range. The valid range of addresses could also be achieved using a formula you couldn't use the Formula Two to the power and minus two where n equals number
host bits. How many holes bits do I have
in this network? 210 10 0
Well, since the mask is 2552552550 the first octet belongs to the network.
The second octet belongs to the network. The third octet belongs to the network. The four talk Ted, The last act. It belongs to
in this last are kept eight.
So two of the power and minus two. Where N is the number of host bits? You could say two to the power eight minus two, which is 2 56 minus two, which is 2 54 Which gives you the same number we got here. The valid ranges
0.1 through dot to 54
254 total valid addresses. Let's we'll get another example. What if we have a Class B address? So 1 50 is a number between
1 28 and 1 91 which makes it a Class B address.
So let's say we have one
100 dot Tenn 100.0.10
and the mask is 255.255
If I were to ask you
what the street name is, or just the network address, this is an I. P address. This is a host address.
So if I were to ask you just for the network address, I would expect you to answer
Why? 1 51 100 0.0? Because the first octet belongs to the network and the second octave belongs to the network. The third and fourth thought that belonged to host addresses.
So we just set them to zero for the street name or the network name. Why this last 16 here? Because eight bits turned on in the subnet mask here, plus a bit turned on in the subnet mask in the second act.
So eight plus eight is 16. If I were to ask you for the first valid address here.
You would say 1 51 100 0.1 Why? 0.1? Well, the octet to the right. The right most active has to be full
to the architect to the left of it.
So this would go 0.10 dot 20.30 that for all the way through
0.255 Then you would have won
1.11 dot 21.3 all the way through 1.255 Then you would have 2.2 dot one and so on and so forth.
Well, I'll just erase this.
The last possible address would be 1 50.100 dot 255.255 Both the architects are full
off course. This last address would be your broadcast address. Again, any
data sent to this address will be received by all hosts on the network.
So my first valid address was 1 51 a 10 dot won. My last valid address would be 1 51 100.255 dot 254 This would be 0.1 through 255.254 would be my valid
range if I were to use that two to the power and minus two formula
and the minus two. Because you can't use the first and the last address the last address being the broadcast address
and the first address being the network of the street address itself. If I were to come up with how many total addresses are between 0.1 and 255.254 I could use the Formula
Two to the power and minus two minus two because we take away the first and the last address.
So in this case, the value of N
well, how many host bits do we have?
The first octave belongs to the network because all the bits underneath in the sub net master turned on
the second octave belongs again to the network. That is all the bits in the summit mascot turned on
the third and fourth octet belonged to host addresses. So this is eight bits plus eight bits 16 bits. True to the power 16 minus two.
So two to the power 16 iss, I believe 65 5 36 minus two, which is 65 5 34 Now let's do a class example. I'm gonna pick a class A I p 10
This the mask for Class A. Since the first octet, the value is between one and 1 26
My subject mask is 255
0.0 dot 0.0. So if I were to ask you just for the street name just for the network address, you should say
eight life slash eight. Because on Lee, eight bits total are turned on in the subnet mask,
meaning the first, the whole first doctor. Now, the first valid address in this case would be 10.0 dot 0.1.
Then it would go to $10.0.23 dot for all the way to $10 0 as you're about to 55
then it would go $10.0 dot to 0.0.2 dot 12.2553 dot and so on and so forth. When this number gets to 255 Then you will add one over here.
The last possible address
for a Class A would be
I p Class A would be 10.255 dot 255 dot
255 which, of course, is your broadcast. The first address, of course, is your network address.
It's first address and this last address are unusable. The first address being the street address or the network address itself and the last address being the broadcast. Again, any message sent to this address will be received by all hosts. All
devices on the network.
The last valid address would be 10.255 dot 255.254
$10? $0? You're not 1 to 10 to 55255.254 would be my valid
These are assigned herbal addresses that I can provide true machines to computers. Two PCs
Now. If you were to ask me what what is the total range of addresses between that one and dot to 54?
Let's see if I remember so this would be again to the parlor and minus two
minus two. Once again, we take away the first and the last address and we say minus two.
So this is going to be true to the power. The first
our 10 belongs to the network, so Well, that's gone. The 2nd 3rd and fourth octet belongs to host addresses. So this is to the power 24
minus two. And I believe to the power 24 is 16777216 minus two, which yields 16777214 addresses.
And that concludes the class full I p address ing portion off this class
next up, we're going to get into class less. I p Addressing