OK so let's complete the submitting lab.
We were told of that previously network 1 9 2 1 6 8 1 0 slash 24 was broken up into four subnets to
support the various subnets in this lab.
So we have 190 16 had 1.0 slash 26 64 slash 26 192 26 and 128 26.
So going back to my calculations we had 1 9 2 1 6 8 1.0 slash 24.
We created 4 subnets.
We used this formula to get the subnets to remind you to to the power of n minus two is used when asked
to work out the number of bits required for a certain number of hosts with subnets.
We don't have to subtract two so two to the power of n was used.
Which gave us four subnets.
So the formula 2 to the N or two to the power of two was used we only had to steal two bits from the
host portion for subnets and that gave us our three subnets.
We've now been told to subnet the subnet further so in the lab we've been told to break up 1 8 2 1 6
8 1 64 slash 26 to support as many subnets as possible with 8 hosts per subnet
so this network 1 8 2 1 6 8 1 64 slash 26 needs to be broken up to support as many subnets as possible
with 8 hosts required per subnet so the formula that we're going to use is two to the power of n minus
two.
That gives us and the number of hosts that we require.
So if we work through that two to the power of one gives us two that will only give us zero hosts so
that's not enough to to the power of two minus two gives us two hosts also not enough due to the power
of three minus two would give a six two to power three is eight minus two gives us six not enough to
do the power of four with sixteen minus two gives us fourteen hosts so that is enough to tip power five
is thirty two minus two would give us thirty hosts and we could continue with that but we already know
that this is the number of hosts that we require
so we need four bits in the host portion which would give us sixteen combinations minus two one for
broadcast one for network gives us fourteen hosts so this is what our original subnet looks like
at the moment this is the network subnet portion and this is the host portion we only require four but
for the host portion so we can now move
two bits to the new subnet portion so these two bits which which I'll put in green are gonna be our
new subnet portion there are no spaces once again in an octet This is a full octet so 0 1 0 0 followed
by four zeros but I've just split it to make it easier to see who's portion is four bits subnet portion
is this so if we work out the subnets we have to go through the various binary combinations which would
look like that which you can also do if you want to save time is this is 128 This is 64 This is 32 this
is sixteen so you can just add sixteen to 64 to get to the next subnet so that would be eighty and and
just to verify that I've done it right eighty in decimal looks like this in binary nodes 0 1 0 1 followed
by four zeros that's what we've got there and then you can simply add sixteen to that and there would
be 96
96.
It looks like this in binary which is the same as what I've got here.
0 1 1 0 followed by four zeros and then you can simply add 16 to that again which should give us 112.
If my math is right.
Hundred twelve looks like that.
So zero followed by three one's followed by four zeros.
So those are new subnets.
We've got four subnets each supporting 14 hosts.
We only require eight hosts but we had to use an additional but because three bits wouldn't give us
enough hosts we'd only have six hosts.
So we've had to use four bits in binary to give us 16 less two which is 14 hosts.
We have four subnets because we have two bits to do the power of two gives us four subnets.
So we've created four new subnets from that single subnet so we could allocate the subnet to our new
network so this new network would now be 1 9 2 1 6 8 1
64 and it's no longer a slash 26.
So I need to update that.
It's eight plus eight.
Sixteen plus a binary bits 24 plus four.
It's 28.
Or if you prefer we originally had 26.
We've added an additional two bits to the subnet portion so slash 28.
So this is now slash 28.
Now before I configure the network I just want to make sure that I've done things right.
This network needs to change now.
That's what we used originally.
What we've been told to do is use the lost new subnet you got from 1 9 2 1 6 8 164 slash 26 and use
that with Slash 30 mosques.
So we need to take this subnet and then we need to subnet it again.
So this is the new subnet that we want to break up as a slash 30.
The reason why we use flash 30 is two to the power of two minus two gives us two hosts a Wayne link
such as a serial link only requires two IP addresses IP addresses on each side of the link so we can
steal two bits here for the subnet portion so that would be network.
This would be our new subnet and our who's portion would only consist of two bits.
So that would be the first network
to slash 30 mosques now
next network would look like that.
Just go through the winery combinations.
So in decimal that's a four so it'd be 16 20 24 and we could once again verify that by using a calculator.
So 1 1 2 looks like that in binary 1 1 6.
It looks like that which is correct 120 looks like that.
Notice 0 followed by 4 ones 0 followed by 4 ones followed by three zeros.
Again there's no gap here.
In an octet I'm only doing that to make it easier to read 124.
Looks like that and that's also correct.
So what I'll do is I'll firstly configure the wine links because they've changed now.
This when link needs to be 112 slash 30 hundred twelve slash 30.
So this writer needs to be configured with the first ip address in that subnet so show run and the serial
interface needs to be changed.
Interface serial 0 1 0 IP address is 1 9 2 1 6 8 1 1 1 3.
First IP address in that subnet it's a slash 30 mosque 0 SPF neighbor relationship has come down writers
require that they have the same subnet on both sides for them to form an SPF neighbor relationship show
run on this side.
This IP address needs to change
so interface serial 0 1 slash 0 IP address 1 9 2 1 6 8 1 1 1 4 slash 30 mosque paying 1 9 2 1 6 8 1
1 1 3 Paying works you can see that to the neighbor relationship has come up so that looks good.
So on Roorda one neighbor relationship has come up again show IP rot shows us default route via 1 1
4.
In other words this road is seeing this rather as its default gateway or or gateway of last resort so
paying Cisco to come can we still get to Cisco.
Can the strata IP domain look up IP name server should be called Triple H sipping Cisco dot com
maybe not a problem Packet Tracer let's see if the strata camping the DNS server yes it can.
Can it ping Cisco dot com name has resolved.
So Internet wrote a competing Cisco dot com.
What about right of one.
I was just impatient they notice it can ping Cisco dot com and can it ping Facebook dot com again.
It can so I can save the configuration I've now successfully updated the Internet router and route one
with the new subnet.
Next thing I need to do is configure a subnet on this link so this will be one on 2 1 6 8 1 and I'll
choose the next one which is 1 1 6 slash 30
so notice how I've taken one original subnet this one and I've broken it up into these three.
This one is being used now so I can't use that anymore because I've submitted that into these four subnets
but there's nothing stopping you taking one subnet and subdividing it and then subdividing it again
like I've done here.