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Here is another example, ABC limited have been allocated subnet 10.128.192.0/18

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for several offices in the USA

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Paul a network administrator once again needs to split the subnet

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into smaller subnets, Paul requires 30 subnets with as many hosts

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as possible on each subnet, and once again he asked you for your help.

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You need to decide which formula to use.

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Please note we’ve been ask for networks or subnets

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so we need to use the formula 2 to the n and not the formula 2 to the n - 2.

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And we need to remember to count the bits

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from the left hand side to the right hand side.

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So on step 2 you need to work out the number of bits required to cover the number

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of hosts or in this case the number of networks that we’ve been asked for.

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Paul is asked for 30 subnets so we will require 5 bits because

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using the formula 2 to the n and substituting n with 5 will give you 32.

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So we will actually end up having 32 subnets rather than just 30.

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So we now know that we need to steal 5 bits from the host portion of the address

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and allocate that to the network portion

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because 5 binary bits are required to give us 32 networks.

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The 3rd step is to convert the host portion of the original network into binary.

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so the original network we were given was 10.128.192.0/18 or 10.128.192.0

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with the mask of 255.255.192.0 now 255 tells us that the first octet is network

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the 2nd 255 tells us that the 2nd octet is network

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however in the third octet, the octet is not fully populated with binary 1's.

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So in the 3rd octet there’s a split between network and host.

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The last octet is filled with binary 0's so that entire octet is host.

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Converting 192 into binary gives us 2 binary 1's followed by 6 binary 0's

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0 in decimal converted to binary give us 8 binary 0's.
.

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So we have converted the 3rd octet where we have both network and host bits

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and the last octet into binary and we have drawn a line separating the network

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and the host portion of the address.

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How do we know that we need to draw the line here?

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because we have 18 bits in the network mask

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The first octet is 8 bits, the 2nd octet is 8 bits,

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8 plus 8 is 16, plus 2 gives us 18 so this line indicates

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the separation between network and host.

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Now the original network once again is 10.128.192.0/18 or

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could be written as 255.255.192.0 in dotted decimal notation.

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So once again the network portion is 10.128 the network host portion is 192

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and the host portion is 0. We are going to take 5 bits from the host portion

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and allocate that to the subnet, so the network portion is 10.129

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and then on the 3rd octet is the first 2 bits are the network

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and we count 5 bits from the left hand side to the right hand side

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so 12345 and we draw a line here indicating that this 5 bits are subnet

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and all bits to the right of the second line are host.

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So we have now stolen 5 bits from the host portion and allocated

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that to the subnet portion of the address.

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So we need to work out what the new subnet mask is.

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It’s equal to the number of bits in the network and subnet portion of the address.

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So it’s equal to this portion of the address

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plus the extra 5 bits allocated to the subnet portion.

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Just to remind you once again 1 octet is 8 bits.

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so the first octet is 8 bits, the 2nd octet is 8 bits

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so that gives you a total of 16 bits.

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We’ve got 2 bits in the 3rd octet which are part of the network plus

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5 additional bits which have been allocated to subnet, so that gives us 7 bits.

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so the total number of bits in the network subnet portion is equal

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to 8 plus 8 plus 2 plus 5 which equals 23 bits

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you could also work this backward once again,  there are 32 bits in an IPv4 address

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and notice in the host portion there are 8 bits in the last octet

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allocated to host plus 1 bit in the 3rd octet

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so 1 plus 8 equals 9, 32 less 9 gives you 23.

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Either method is fine, the result is the same 23 bits have now been

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allocated to network and subnet where's before only 18 bits were allocated.

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So now it's possible to work out to new subnet.

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Once again, to work out the subnet go through the various

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binary combinations for the subnet portion of the address.

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So this portion in green mark the subnet

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so the first network or subnet is equal to 10.128.

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a 2 binary bits which part of the original network plus 5 additional

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binary bits which would now allocated to subnet.

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So once again the subnet mask is /23 which can be written

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in dotted decimal notation as 255.255.254.0

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To work out the first subnet, fill the subnet portion of the address

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with 0's and populate the host portion of the address with 0s.

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please note this 2 binary 1's. the 5 green binary 0's that are part of the subnet

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and the 1 red binary 0 that’s part of the host portion

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all form part of the same subnet.

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So 11 followed by 6 binary 0's equals 192 in decimal.

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To work out the 2nd network or subnet, we go through binary combination.

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The next binary combination is 4 binary 0's followed by binary 1

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taking the whole octet into account that equals 194 in decimal.

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Please note the host portion is always set to binary 0's.

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So the last octet is once again 0.

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So the second network or subnet is 10.128.194.0

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Now you probably already guess what the 3rd one is gonna be

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because we're going up in multiples of 2.

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But if we go to the whole process again getting the next binary value

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would be 3 binary 0's followed by binary 1 followed by binary 0.

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And converting that whole octet back into decimal will give us 196.

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So we know that we're going in multiples of 2,

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so the first 1 is 192, then 194, then 196, then 198, then 200, 202, 204, etc.

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all the way up to the last subnet.

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To work out the last subnet, fill the subnet portion of the address with binary 1's

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so we end up having 10.128 followed by 7 binary 1's,

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followed by binary 0 in the 3rd octet.

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7 binary 1's followed by binary 0 in an octet is equal to 254.

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The last octet is once again equal to 0.

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So the last subnet is 10.128.254.0 with the /23 mask

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or it can be written as 10.128.254.0 with the mask of 255.255.254.0

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I hope that’s helped you learn how to subnet based on a requirement

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first specific number of hosts or specific number of networks.

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So what have we covered?

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We look at the reason for subnetting, subnetting is very important for this course

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and it’s important that you have a good understanding of subnetting.

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so we spent time looking at the binary method and the quick method

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for determining the subnet address, broadcast address

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first host address and last host address for a given IP address.

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I also showed you how to create multiple subnets

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based on specific host or network requirements.