Original subtitles

84% Here is another example, ABC limited have been allocated subnet 10.128.192.0/18

84% for several offices in the USA

84% Paul a network administrator once again needs to split the subnet

84% into smaller subnets, Paul requires 30 subnets with as many hosts

84% as possible on each subnet, and once again he asked you for your help.

84% You need to decide which formula to use.

84% Please note we�ve been ask for networks or subnets

84% so we need to use the formula 2 to the n and not the formula 2 to the n - 2.

84% And we need to remember to count the bits 10

84% from the left hand side to the right hand side. 11

84% So on step 2 you need to work out the number of bits required to cover the number 12

84% of hosts or in this case the number of networks that we�ve been asked for. 13

84% Paul is asked for 30 subnets so we will require 5 bits because 14

84% using the formula 2 to the n and substituting n with 5 will give you 32. 15

84% So we will actually end up having 32 subnets rather than just 30. 16

84% So we now know that we need to steal 5 bits from the host portion of the address 17

84% and allocate that to the network portion 18

84% because 5 binary bits are required to give us 32 networks. 19

84% The 3rd step is to convert the host portion of the original network into binary. 20

84% so the original network we were given was 10.128.192.0/18 or 10.128.192.0 21

84% with the mask of 255.255.192.0 now 255 tells us that the first octet is network 22

84% the 2nd 255 tells us that the 2nd octet is network 23

84% however in the third octet, the octet is not fully populated with binary 1's. 24

84% So in the 3rd octet there�s a split between network and host. 25

84% The last octet is filled with binary 0's so that entire octet is host. 26

84% Converting 192 into binary gives us 2 binary 1's followed by 6 binary 0's 27

90% 0 in decimal converted to binary give us 8 binary 0's. 28

84% So we have converted the 3rd octet where we have both network and host bits 29

84% and the last octet into binary and we have drawn a line separating the network 30

84% and the host portion of the address. 31

84% How do we know that we need to draw the line here? 32

84% because we have 18 bits in the network mask 33

84% The first octet is 8 bits, the 2nd octet is 8 bits, 34

84% 8 plus 8 is 16, plus 2 gives us 18 so this line indicates 35

84% the separation between network and host. 36

84% Now the original network once again is 10.128.192.0/18 or 37

84% could be written as 255.255.192.0 in dotted decimal notation. 38

84% So once again the network portion is 10.128 the network host portion is 192 39

84% and the host portion is 0. We are going to take 5 bits from the host portion 40

84% and allocate that to the subnet, so the network portion is 10.129 41

84% and then on the 3rd octet is the first 2 bits are the network 42

84% and we count 5 bits from the left hand side to the right hand side 43

84% so 12345 and we draw a line here indicating that this 5 bits are subnet 44

84% and all bits to the right of the second line are host. 45

84% So we have now stolen 5 bits from the host portion and allocated 46

84% that to the subnet portion of the address. 47

84% So we need to work out what the new subnet mask is. 48

84% It�s equal to the number of bits in the network and subnet portion of the address. 49

84% So it�s equal to this portion of the address 50

84% plus the extra 5 bits allocated to the subnet portion. 51

84% Just to remind you once again 1 octet is 8 bits. 52

84% so the first octet is 8 bits, the 2nd octet is 8 bits 53

84% so that gives you a total of 16 bits. 54

84% We�ve got 2 bits in the 3rd octet which are part of the network plus 55

84% 5 additional bits which have been allocated to subnet, so that gives us 7 bits. 56

84% so the total number of bits in the network subnet portion is equal 57

84% to 8 plus 8 plus 2 plus 5 which equals 23 bits 58

84% you could also work this backward once again, there are 32 bits in an IPv4 address 59

84% and notice in the host portion there are 8 bits in the last octet 60

84% allocated to host plus 1 bit in the 3rd octet 61

84% so 1 plus 8 equals 9, 32 less 9 gives you 23. 62

84% Either method is fine, the result is the same 23 bits have now been 63

84% allocated to network and subnet where's before only 18 bits were allocated. 64

84% So now it's possible to work out to new subnet. 65

84% Once again, to work out the subnet go through the various 66

84% binary combinations for the subnet portion of the address. 67

84% So this portion in green mark the subnet 68

84% so the first network or subnet is equal to 10.128. 69

84% a 2 binary bits which part of the original network plus 5 additional 70

84% binary bits which would now allocated to subnet. 71

84% So once again the subnet mask is /23 which can be written 72

84% in dotted decimal notation as 255.255.254.0 73

84% To work out the first subnet, fill the subnet portion of the address 74

84% with 0's and populate the host portion of the address with 0s. 75

84% please note this 2 binary 1's. the 5 green binary 0's that are part of the subnet 76

84% and the 1 red binary 0 that�s part of the host portion 77

84% all form part of the same subnet. 78

84% So 11 followed by 6 binary 0's equals 192 in decimal. 79

84% To work out the 2nd network or subnet, we go through binary combination. 80

84% The next binary combination is 4 binary 0's followed by binary 1 81

84% taking the whole octet into account that equals 194 in decimal. 82

84% Please note the host portion is always set to binary 0's. 83

84% So the last octet is once again 0. 84

84% So the second network or subnet is 10.128.194.0 85

84% Now you probably already guess what the 3rd one is gonna be 86

84% because we're going up in multiples of 2. 87

84% But if we go to the whole process again getting the next binary value 88

84% would be 3 binary 0's followed by binary 1 followed by binary 0. 89

84% And converting that whole octet back into decimal will give us 196. 90

84% So we know that we're going in multiples of 2, 91

84% so the first 1 is 192, then 194, then 196, then 198, then 200, 202, 204, etc. 92

84% all the way up to the last subnet. 93

84% To work out the last subnet, fill the subnet portion of the address with binary 1's 94

84% so we end up having 10.128 followed by 7 binary 1's, 95

84% followed by binary 0 in the 3rd octet. 96

84% 7 binary 1's followed by binary 0 in an octet is equal to 254. 97

84% The last octet is once again equal to 0. 98

84% So the last subnet is 10.128.254.0 with the /23 mask 99

84% or it can be written as 10.128.254.0 with the mask of 255.255.254.0 100

84% I hope that�s helped you learn how to subnet based on a requirement 101

84% first specific number of hosts or specific number of networks. 102

84% So what have we covered? 103

84% We look at the reason for subnetting, subnetting is very important for this course 104

84% and it�s important that you have a good understanding of subnetting. 105

84% so we spent time looking at the binary method and the quick method 106

84% for determining the subnet address, broadcast address 107

84% first host address and last host address for a given IP address. 108

84% I also showed you how to create multiple subnets 109

84% based on specific host or network requirements.