In this network we've been told to subnet 1 9 2 1 6 8 2 1 0 slash 24 into 4 subnets.
In other words we need a subnet for site one site to the serial link and this serial link.
Now we haven't been told to use a slash of 30 mosque on the serial links.
We'll do that in a separate video so all we need to do after this point is take 1 9 2 1 6 8 wondered
0 slash 24 and subnet that into 4 subnets.
Now when we look at a network address such as 1 9 2 1 6 8 1 0 we need to determine which point is the
network and which part is the host portion.
Currently this portion is the network portion.
We can't change the network portion but we can manipulate and change the host portion of the address.
So in other words what we can do is change this lost octet.
The reason why we can't manipulate the first three octet is because that's part of the network 24.
Essentially means that we have 24 binary ones.
Now you don't have spaces in an octet like this.
I'm just doing it to make it easier to read but essentially 24 means that we have 24 ones into the subnet
mask so that's 24 binary ones which essentially equates to 255 255 255 is zero.
It's important that you know how to convert a decimal to binary this binary value equates to decimal
255 so the host portion of the address is once again that portion.
This portion is the network portion we can't manipulate the network portion but to the host portion
consisting of eight binary zeros can be manipulated.
Now when we subnet think of submitting kind of like stealing butts we're going to steal butts from the
host portion.
We need to work out how many bits are required for the number of subnets that we need in this example
we need four subnets.
Now when it comes to subletting there are two formulas that you need to know.
Two to the power of n and two to the power of n minus two.
This formula is used when we asked for subnets which is what we've been asked in this question.
This formula we need to use when we asked for a certain number of hosts.
So if we were asked to subnet this subnet 1 9 2 1 6 8 1 0 select 24 and create as many subnets as possible
each having 4 hosts we would use two to the power of n where n is the number of bits that are required
minus two.
So if we needed to get 4 hosts we would actually need two to the power of three minus two which would
be to to power of three which is eight minus two which means that we would be able to have six hosts
per subnet.
We need to subtract two because of broadcast and network address but we don't need that.
When asked for networks two to the power of two equals four which is what we need for the number of
subnets that we've been asked to create.
This is the number of bits that we stealing or using for the subnet portion.
This is the number of bits that we would keep for the host portion.
Now we're not using that formula so I can basically scratch that from our example.
So basically remove that to we are not using that formula.
In this example this is the formula that we need.
And once again that's because we've been off to four subnets we haven't been asked for hosts.
Now two to the power of one equals t that doesn't give us enough subnets due to the power of two gives
us four subnet two to the power of three gives us eight subnets two to the power of four would give
us sixteen subnets we don't need that many we only need four subnets.
So all we need to do is use two bits which means we can steal two bits from the host portion for our
subnet portion.
So the host portion is now only six bits in length the subnet portion is now two bits in length.
So write it like that.
Once again there's obviously no spaces here in binary but that hopefully just makes it easier to read.
So we're going to steal two bits from the host portion and allocate to that to the subnet portion.
So how many bits are part of the network and subnet.
We've got our 8 bits from our original example.
An additional eight but from our original example plus eight bits.
So that's sixteen plus one plus one.
So in other words this is now a slash twenty six subnet it's no longer slash 24 because we've stolen
bits from the host portion for the network portion.
So the first network that we have is 1 9 2 1 6 8 1 dot 0.
Look at to these 8 2 binary bits to make it easier to read.
Notice we've got eight binary bits I've just split them up with spaces here to make it easier to demonstrate
which party's subnet in which parties host.
But notice there are eight binary but say eight binary bits equates to zero in decimal.
So the network is 1 9 2 1 6 8 wondered 0 slash 26.
That's our first network.
Now what's the second network.
The second network is 1 and 2 1 6 8 1.
Dot and what we do now is we just go through the different binary options.
This is 0 0 this is 0 1.
This one would be 1 0 and this would be 1 1.
So those are the different binary options that we have.
So if we'll look at this network that second binary but to set to a one what does this equal to this
equals to sixty four in a decimal.
So the next network is 64.
What is is this equal.
It looks like this.
Which equals 128.
Now as soon as you've worked out this second subnet you can simply do addition by that number.
So zero plus 64 64 plus 64 is 128 plus 64 is 192.
But if you look at the binary it's 128 plus 64 which equals 192.
So there are are four subnets that we've been asked to work out in this example.
So we've got four subnets the first one is for side one the second one is for the link between right
of one an Internet router third one is for side 2 fourth one is for the link between right at 2 and
the Internet router so the first subnet would be 1 0 2 1 6 8 wondered slash 26.
And I'll zoom in here to make it clearer.
So that's our first subnet.
Per our calculations second subnet is 1 9 2 1 6 8 1 64 slash 26.
Third one is 1 9 2 1 6 8 1 128 slash 26.
And the last one is 1 9 2 1 6 8 1 192 flash 26 so those subnets are these subnets that we worked out
so we need to now configure the network with this information in the first step we need to work out
to the subnets but now we need to configure the route as per the instructions so as an example the lost
IP in the subnet should be configured on rather one second lost IP address should be configured on the
switch third lost IP address should be configured on the DHB server per these instructions we also need
to configure a DHB pool on the DHB server so I'm going to start with the subject first and get that
working and then I'll move to the other subnets.