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--> align:middle line:84%
Here�s one more example using the binary method
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--> align:middle line:84%
and after this I�m gonna show you the quick method.
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--> align:middle line:84%
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02:52:16,129 --> 02:52:16,129
If a PC had an address of .1/17 or .1 255.255.128.0
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--> align:middle line:84%
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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--> align:middle line:84%
are used as the host portion of the address.
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--> align:middle line:84%
So 172.16.129.1/17 means that the split takes place in the 3rd octet.
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--> align:middle line:84%
The reason why once again is the first octet is 8 bits in size, the second octet
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--> align:middle line:84%
is 8 bits in size so that gives us 16 bits, 17 bits in the network or subnet
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--> align:middle line:84%
means that the split between subnet and host is in the 3rd octet.
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--> align:middle line:84%
So once again you need to convert the 3rd and 4th octet into binary.
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--> align:middle line:84%
There�s no need to convert the 1st 2 octet as they are part of the network
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--> align:middle line:84%
or subnet portion of the address.
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--> align:middle line:84%
You only need to convert the host portion of the address into binary.
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--> align:middle line:84%
So in binary 1 followed by 6 binary 0's followed by 1 equals 129 in decimal,
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--> align:middle line:84%
7 0's followed by binary 1 is the binary equivalent of 1 in decimal.
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--> align:middle line:84%
Once again refer to the binary section of this course if you're not sure
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--> align:middle line:84%
how to convert decimal into binary and vice versa.
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--> align:middle line:84%
So once again 172.16.1 is the network or subnet portion of the address
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--> align:middle line:84%
and the remaining bits of the host portion of the address.
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--> align:middle line:84%
So to work out the network or subnet portion of the address
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--> align:middle line:84%
you need to fill the host portion of an address with binary 0's.
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--> align:middle line:84%
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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--> align:middle line:84%
So in the 3rd octet we have 1 binary 1 followed by 7 binary 0's
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--> align:middle line:84%
which give us the equivalent decimal value of 128, the 4th octet is filled
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--> align:middle line:84%
with binary 0's which will give us the equivalent decimal value of 0.
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--> align:middle line:84%
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So the subnet that this host .1 resides on is .0
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--> align:middle line:84%
To work out the first host in the same subnet, you need to fill the host portion
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--> align:middle line:84%
with binary 0's except for the last bit which is set to binary 1.
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--> align:middle line:84%
and that would give you 172.16.128.1
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--> align:middle line:84%
To work out the last host, you fill the host portion of the address with binary 1s
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--> align:middle line:84%
except for the last bit which is set to binary 0
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--> align:middle line:84%
so that would give you 172.16.255.254
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--> align:middle line:84%
Now just to make sure that you understand this, notice the 3rd octet is filled
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--> align:middle line:84%
with binary 1's, there is a single red binary 1 followed by 7 green binary 1's.
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--> align:middle line:84%
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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--> align:middle line:84%
The 4th octet is filled with 7 binary 1's, followed by binary 0
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--> align:middle line:84%
which gives you a decimal equivalent of 254.
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--> align:middle line:84%
To work out the broadcast address, fill the host portion
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--> align:middle line:84%
of the address with binary 1's, so that would give you 172.16
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--> align:middle line:84%
8 binary 1's in the 3rd octet which is 255 and 8 binary 1's on the 4th octet
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--> align:middle line:84%
which is 255 so the broadcast address is 172.16.255.255
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--> align:middle line:84%
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So in summary, host .1 is on subnet .0
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--> align:middle line:84%
The first host in the subnet is 172.16.128.1, the last host
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--> align:middle line:84%
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in the subnet is .254 and the broadcast address is .255
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--> align:middle line:84%
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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--> align:middle line:84%
Now that we�ve seen the binary method, let me show ou the quick method
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--> align:middle line:84%
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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--> align:middle line:84%
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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--> align:middle line:84%
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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--> align:middle line:84%
So remember it's 128 64 32 16 8 4 2 and 1
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--> align:middle line:84%
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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--> align:middle line:84%
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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--> align:middle line:84%
256 minus 64 gives you 192 and so forth and so on,
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--> align:middle line:84%
as an example 256 minus 32 gives you 224, 256 minus 1 gives you 255
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--> align:middle line:84%
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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--> align:middle line:84%
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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--> align:middle line:84%
equivalent 255.255.240.0 the first thing you need to work out is, why is the
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--> align:middle line:84%
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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--> align:middle line:84%
with the subnet mask is not equal to 255
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--> align:middle line:84%
So in this example, once again we have an address 172.16.35.123
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--> align:middle line:84%
and the subnet mask is 255.255.240.0, so in the 3rd octet
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--> align:middle line:84%
the subnet mask is not equal to 255 but is equal to a value of 240.
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--> align:middle line:84%
That means that in this octet there is a split
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--> align:middle line:84%
between the subnet and the host portions.
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--> align:middle line:84%
So the 1st 2 octets are network or subnet the last octet is host
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--> align:middle line:84%
but in the third octet there is a split between subnet and host.
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--> align:middle line:84%
Step 2 is to subtract that subnet mask value that is now 255 from 256.
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--> align:middle line:84%
So 256 less 240 would give you 16, what 16 tells us is that network are
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--> align:middle line:84%
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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--> align:middle line:84%
The 16 lets us now the increment of the networks.
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--> align:middle line:84%
Now the table I showed you in step 1 will allow you very quickly and easily
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--> align:middle line:84%
to work this out, so in the third octet we have a value of 240,
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--> align:middle line:84%
so 256 less 240 gives you 16.
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--> align:middle line:84%
So remember in the 3rd octet the subnet mask was 240, 256 less 240 gives us 16
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--> align:middle line:84%
notice in the IP address the 3rd octet value is 35
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--> align:middle line:84%
so part of 35 is network and part of 35 is host.
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--> align:middle line:84%
So in step 3 we worked out where 35 fits
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--> align:middle line:84%
in the range of networks worked out in step 2
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--> align:middle line:84%
Now in step 2 we worked out that 256 less 240 is 16
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--> align:middle line:84%
so our networks are in multiples of 16.
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--> align:middle line:84%
So just start at 0 and go until you pass the value in the question.
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--> align:middle line:84%
So as an example, the first network would be 0 in the 3rd octet,
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--> align:middle line:84%
the 2nd network would be 16 on the 3rd octet
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--> align:middle line:84%
the 3rd one would be 32 and the 4th one would be 48.
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--> align:middle line:84%
So 35 sits somewhere between 32 and 48 and thus
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--> align:middle line:84%
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we know that is on network
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--> align:middle line:84%
The way you work that out is to leave the network portion of the address the same.
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--> align:middle line:84%
In other words, this blue portion the first 2 octets remains the same,
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--> align:middle line:84%
the subnet or host octet that lies between 32 and 48 as per our calculation
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--> align:middle line:84%
in step 3 gets rounded down to the nearest value.
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--> align:middle line:84%
so 35 is between 32 and 48, and rounding 35 down we get 32.
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--> align:middle line:84%
So the 3rd octet is equivalent to 32.
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--> align:middle line:84%
Lastly the host portion of the address is just set to 0.
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--> align:middle line:84%
So you now know that 172.16.35.123 is on network 172.16
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--> align:middle line:84%
because the blue portion or network portion remains the same
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--> align:middle line:84%
35 is rounded down to 32 because the subnet host portion lies
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--> align:middle line:84%
between 32 and 48 and the host portion is just set to 0, in other words 172.16.32.0
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--> align:middle line:84%
It's as simple as that to work out the subnet that our hosts resides on.
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