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