Original subtitles

Here�s one more example using the binary method

and after this I�m gonna show you the quick method.

If a PC had an address of .1/17 or .1 255.255.128.0

you would once again need to work out where the subnet and host portions are split.

In this example /17 means that 17 bits of the 32 bits IP address

are used for network or subnet and the remaining 15 bits

are used as the host portion of the address.

So 172.16.129.1/17 means that the split takes place in the 3rd octet.

The reason why once again is the first octet is 8 bits in size, the second octet 10

is 8 bits in size so that gives us 16 bits, 17 bits in the network or subnet 11

means that the split between subnet and host is in the 3rd octet. 12

So once again you need to convert the 3rd and 4th octet into binary. 13

There�s no need to convert the 1st 2 octet as they are part of the network 14

or subnet portion of the address. 15

You only need to convert the host portion of the address into binary. 16

So in binary 1 followed by 6 binary 0's followed by 1 equals 129 in decimal, 17

7 0's followed by binary 1 is the binary equivalent of 1 in decimal. 18

Once again refer to the binary section of this course if you're not sure 19

how to convert decimal into binary and vice versa. 20

So once again 172.16.1 is the network or subnet portion of the address 21

and the remaining bits of the host portion of the address. 22

So to work out the network or subnet portion of the address 23

you need to fill the host portion of an address with binary 0's. 24

So this green portion of the address needs to be filled with 0's and that will give 25

us the subnet which is 172.16.1, that binary 1 is part of the network address 26

followed by 7 0's, followed by 8 0's. 27

So in the 3rd octet we have 1 binary 1 followed by 7 binary 0's 28

which give us the equivalent decimal value of 128, the 4th octet is filled 29

with binary 0's which will give us the equivalent decimal value of 0. 30

So the subnet that this host .1 resides on is .0 31

To work out the first host in the same subnet, you need to fill the host portion 32

with binary 0's except for the last bit which is set to binary 1. 33

and that would give you 172.16.128.1 34

To work out the last host, you fill the host portion of the address with binary 1s 35

except for the last bit which is set to binary 0 36

so that would give you 172.16.255.254 37

Now just to make sure that you understand this, notice the 3rd octet is filled 38

with binary 1's, there is a single red binary 1 followed by 7 green binary 1's. 39

That however is the single octet, so there are 8 binary 1's 40

which gives you a value of 255. 41

The 4th octet is filled with 7 binary 1's, followed by binary 0 42

which gives you a decimal equivalent of 254. 43

To work out the broadcast address, fill the host portion 44

of the address with binary 1's, so that would give you 172.16 45

8 binary 1's in the 3rd octet which is 255 and 8 binary 1's on the 4th octet 46

which is 255 so the broadcast address is 172.16.255.255 47

So in summary, host .1 is on subnet .0 48

The first host in the subnet is 172.16.128.1, the last host 49

in the subnet is .254 and the broadcast address is .255 50

I hope those 3 examples have helped you learn the binary method to work out the 51

subnet, 1st host, last host and broadcast address when 52

presented with an IP address of a host and its subnet mask 53

Now that we�ve seen the binary method, let me show ou the quick method 54

which allows you very quickly to work out the answer to question like; what subnet 55

is this host on, what is the broadcast address, what is the first host 56

and last host in the same subnets as this specific host. 57

This method is reliant in you remembering tables and methods 58

rather than relying on binary. 59

So the first table to remember is the following; the values at the top of this 60

table like 128, 64 and so forth are the decimal equivalents for the binary values 61

such as 1 followed by 7 binary 0's is equal to 128, 3 binary 0's followed 62

by binary 1 followed by 4 binary 0's is equal to 16. 63

You should be quite comfortable to write out this table 64

from memory before attempting any subnetting question. 65

So remember it's 128 64 32 16 8 4 2 and 1 66

in the IP addressing section of this course I explained those values 67

in a lot of detail and explain how you get to those specific values. 68

So I�m not gonna cover it again here. 69

To work out the values in the second line of this table, just take 256 less 70

the top value which will give you the second value. 71

So 256 minus 128 gives you 128 72

256 minus 64 gives you 192 and so forth and so on, 73

as an example 256 minus 32 gives you 224, 256 minus 1 gives you 255 74

so you only need to remember the top values and then it�s very simple 75

to work out the values in the second line. 76

A lot of people just memorize the entire table for speed and efficiency 77

but once again write this table out before attempting any binary question. 78

So if you were given a host address of 172.16.35.123/20 or the decimal 79

equivalent 255.255.240.0 the first thing you need to work out is, why is the 80

subnet mask is not equal to 255 and secondly make a note of that octet, 81

in other words that the network and host portion both reside within that octet, 82

with the subnet mask is not equal to 255 83

So in this example, once again we have an address 172.16.35.123 84

and the subnet mask is 255.255.240.0, so in the 3rd octet 85

the subnet mask is not equal to 255 but is equal to a value of 240. 86

That means that in this octet there is a split 87

between the subnet and the host portions. 88

So the 1st 2 octets are network or subnet the last octet is host 89

but in the third octet there is a split between subnet and host. 90

Step 2 is to subtract that subnet mask value that is now 255 from 256. 91

So 256 less 240 would give you 16, what 16 tells us is that network are 92

incrementing in values of 16, so the first network would be 0 93

second one 16, third one 32, fourth one is 48 and so forth and so on. 94

The 16 lets us now the increment of the networks. 95

Now the table I showed you in step 1 will allow you very quickly and easily 96

to work this out, so in the third octet we have a value of 240, 97

so 256 less 240 gives you 16. 98

So remember in the 3rd octet the subnet mask was 240, 256 less 240 gives us 16 99

notice in the IP address the 3rd octet value is 35 100

so part of 35 is network and part of 35 is host. 101

So in step 3 we worked out where 35 fits 102

in the range of networks worked out in step 2 103

Now in step 2 we worked out that 256 less 240 is 16 104

so our networks are in multiples of 16. 105

So just start at 0 and go until you pass the value in the question. 106

So as an example, the first network would be 0 in the 3rd octet, 107

the 2nd network would be 16 on the 3rd octet 108

the 3rd one would be 32 and the 4th one would be 48. 109

So 35 sits somewhere between 32 and 48 and thus 110

we know that is on network 111

The way you work that out is to leave the network portion of the address the same. 112

In other words, this blue portion the first 2 octets remains the same, 113

the subnet or host octet that lies between 32 and 48 as per our calculation 114

in step 3 gets rounded down to the nearest value. 115

so 35 is between 32 and 48, and rounding 35 down we get 32. 116

So the 3rd octet is equivalent to 32. 117

Lastly the host portion of the address is just set to 0. 118

So you now know that 172.16.35.123 is on network 172.16 119

because the blue portion or network portion remains the same 120

35 is rounded down to 32 because the subnet host portion lies 121

between 32 and 48 and the host portion is just set to 0, in other words 172.16.32.0 122

It's as simple as that to work out the subnet that our hosts resides on.