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Here’s one more example using the binary method

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

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If a PC had an address of 172.16.129.1/17 or 172.16.129.1 255.255.128.0

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

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

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are used for network or subnet and the remaining 15 bits

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are used as the host portion of the address.

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

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

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

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

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

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

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or subnet portion of the address.

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

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

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

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

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how to convert decimal into binary and vice versa.

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

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

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

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

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

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

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

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

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

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

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So the subnet that this host 172.16.129.1 resides on is 172.16.128.0

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

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

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and that would give you 172.16.128.1

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

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except for the last bit which is set to binary 0

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so that would give you 172.16.255.254

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

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

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

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which gives you a value of 255.

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

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which gives you a decimal equivalent of 254.

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To work out the broadcast address, fill the host portion

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

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

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which is 255 so the broadcast address is 172.16.255.255

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So in summary, host 172.16.129.1 is on subnet 172.16.128.0

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

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in the subnet is 172.16.255.254 and the broadcast address is 172.16.255.255

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

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

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

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

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

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

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

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This method is reliant in you remembering tables and methods

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rather than relying on binary.

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

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

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

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

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You should be quite comfortable to write out this table

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from memory before attempting any subnetting question.

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

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

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

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So I’m not gonna cover it again here.

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

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the top value which will give you the second value.

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So 256 minus 128 gives you 128

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

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

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

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to work out the values in the second line.

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

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

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

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

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

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

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with the subnet mask is not equal to 255

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

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

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

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That means that in this octet there is a split

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between the subnet and the host portions.

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

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

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

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

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

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

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The 16 lets us now the increment of the networks.

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

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

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so 256 less 240 gives you 16.

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

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

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

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So in step 3 we worked out where 35 fits

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in the range of networks worked out in step 2

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

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so our networks are in multiples of 16.

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

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

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

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

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So 35 sits somewhere between 32 and 48 and thus

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we know that 172.16.35.123 is on network 172.16.32.0

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

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

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

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in step 3 gets rounded down to the nearest value.

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

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So the 3rd octet is equivalent to 32.

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

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So you now know that 172.16.35.123 is on network 172.16

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because the blue portion or network portion remains the same

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

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

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