Subnetting Made Easy

Hi all,

I've received an email from one of your members asking me to post up my technique for subnetting as links to external blogs are not allowed due to forum rules. I know that he benefited from it and he wishes to help out others so here goes:

First of all I need you to get rid of all of the negative thoughts surrounding subnetting. Put down all of the books that you have read about the subject and navigate away from other sites claiming to provide an easy way to subnet. This technique requires no charts, just simply the know-how to work with the powers of 2.

We need to start with the fundamentals of IP addressing. An IP address is made up of 32 bits, split into 4 octets (oct = 8, yes?). Some bits are reserved for identifying the network and the other bits are left to identify the host.

There are 3 main classes of IP address that we are concerned with.

Class A Range 0 - 127 in the first octet (0 and 127 are reserved)
Class B Range 128 - 191 in the first octet
Class C Range 192 - 223 in the first octet

Below shows you how, for each class, the address is split in terms of network (N) and host (H) portions.

NNNNNNNN . HHHHHHHH . HHHHHHHH . HHHHHHHH Class A
NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B
NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C

At each dot I like to think that there is a boundary, therefore there are boundaries after bits 8, 16, 24, and 32. This is an important concept to remember.

We will now look at typical questions that you may see on subnetting. More often than not they ask what a host range is for a specific address or which subnet a certain address is located on. I shall run through examples of each, for each class of IP address.

What subnet does 192.168.12.78/29 belong to?

You may wonder where to begin. Well to start with let's find the next boundary of this address.

Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8.

We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-

192.168.12.0
192.168.12.8
192.168.12.16
192.168.12.24
192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64
192.168.12.72
192.168.12.80
.............etc

Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.

What subnet does 172.16.116.4/19 sit on?

Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2^5 = 32.

We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-

172.16.0.0
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc

Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet. Easy eh?

What subnet does 10.34.67.234/12 sit on?

Our mask is 12. Our next boundary is 16. Therefore 16 - 12 = 4. 2^4 = 16 which gives us our block size.

We have borrowed from the second octet as bit 12 sits in the second octet so we count up the block size in that octet. The subnets are:-

10.0.0.0
10.16.0.0
10.32.0.0
10.48.0.0
.............etc

Our address is 10.34.67.234 which must sit on the 10.32.0.0 subnet.

Hopefully the penny is starting to drop and you are slapping the side of your head realising that you were a fool to think that subnetting was hard. We will now change the type of question so that we have to give a particular host range of a subnet.

What is the valid host range of of the 4th subnet of 192.168.10.0/28?

Easy as pie! The block size is 16 since 32 - 28 = 4 and 2^4 = 16. We need to count up in the block size in the last octet as bit 28 is in the last octet.

192.168.10.0
192.168.10.16
192.168.10.32
192.168.10.48
192.168.10.64
.................etc

Therefore the 4th subnet is 192.168.10.48 and the host range must be 192.168.10.49 to 192.168.10.62, remembering that the subnet and broadcast address cannot be used.

What is the valid host range of the 1st subnet of 172.16.0.0/17?

/17 tells us that the block size is 2^(24-17) = 2^7 = 128. We are borrowing in the 3rd octet as bit 17 is in the 3rd octet. Our subnets are:-

172.16.0.0
172.16.128.0

The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254. You must remember not to include the subnet address (172.16.0.0) and the broadcast address (172.16.127.255).

What is the valid host range of the 7th subnet of address 10.0.0.0/14?

The block size is 4, from 16 - 14 = 2 then 22 = 4. We are borrowing in the second octet so count in the block size from 0 seven times to get the seventh subnet.

The seventh subnet is 10.24.0.0. Our valid host range must be 10.24.0.1 to 10.27.255.254 again remembering not to include our subnet (10.24.0.0) and the broadcast address (10.27.255.255).

What if they give me the subnet mask in dotted decimal?

If you're lucky and they give you a mask in dotted decimal format then you should have an even easier time. All you need again is your block size.

Let's say they have given a mask of 255.255.255.248 and you wish to know the block size. Here's the technique:

1. Starting from the left of the mask find which is the first octet to NOT have 255 in it.
2. Subtract the number in that octet from 256 to get your block size e.g. above it is 256 - 248 = block size of 8.
3. Count up from zero in your block size in the octet identified in step 1 as you have learned above (the example above would be in the last octet).

Another example is a mask of 255.255.192.0 - you would simply count up in 256 - 192 = 64 in the third octet.

One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the second octet.

What now?

Now it's time to go and pick up those books again and go straight to the practice questions, completely by-passing any of their techniques. Use my method and you will be laughing!

Happy subnetting!

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Comments

nimrod.sixty9

That cleared it up very well, thank you all. I didnt realize subtract from class for subnet bits. I can do the math and the numbers, but this one confused me:

Question: What valid host range is the IP address 172.17.247.48/23 a part of?

Answer: 172.17.246.1 through to 172.17.247.254

I got 2 Subnets and 510 Hosts...
With that how do I figure out the above .246 and .247?

Also, how heavy in subnetting in the Net+ exam?

Jas21

There isn't any subnetting in the Network + exam (certainly wasn't when I took it a few years back!)

jdfriesen

nimrod.sixty9 wrote: »

I can do the math and the numbers, but this one confused me:

Question: What valid host range is the IP address 172.17.247.48/23 a part of?

Answer: 172.17.246.1 through to 172.17.247.254

I got 2 Subnets and 510 Hosts...
With that how do I figure out the above .246 and .247?

In this question you've got to figure out the block size. There are a couple ways to do that. The way I do it is 256 - the mask of the last non-zero octet. In that case, /23 = 255.255.254.0, so they last non zero octet is 254, so 256-254=2 as the block size.

With the block size known, you've got to then find the start of the current subnet. With a block size of 2, it's pretty easy, it's going to be 246, so the first valid address is 172.17.246.1 (.0 isn't allowed as that's the network address). The next valid subnet given a block size of 2 would be 172.17.248.0/23, so that means the last allowable address is going to be 172.17.248.254 (.255 is the broadcast address).

To subnet quickly, you've got to be able to quickly count in multiples of 2, 4, 8, 16, 32, 64 and 128, which enables you to find the start of each subnet quickly.

j-man

jdfriesen wrote: »

In this question you've got to figure out the block size. There are a couple ways to do that. The way I do it is 256 - the mask of the last non-zero octet. In that case, /23 = 255.255.254.0, so they last non zero octet is 254, so 256-254=2 as the block size.

Forget the subtraction stuff.

128 192 224 240 248 252 254 255 - last non-zero of SM

128 64 32 16 8 4 2 1 - block size

That's the block size based on the final non zero octet of the subnet mask.

While it's important to be able to come up with the numbers quickly, it's also extremely important to know why and how those numbers are found.

Learn both the binary and decimal ways to manually calculate the subnet, first host, last host and broadcast and it will start to make sense.

*edit* jdfriesen - the last bit wasn't directed at you. It's a general statement to anyone that wants to learn to subnet.

binaryhat

I have been doing subnetting in class and I don't understand how you get this answer:

[FONT=Arial, sans-serif]Which network prefix will implement the IP addressing scheme for 2 LANs one with hosts 172.22.0.62 and 172.22.0.37 and the other with hosts 172.22.0.94 and 172.22.0.75. [/FONT]

/27 is the answer but I don't see y

Futura

binaryhat wrote: »

I have been doing subnetting in class and I don't understand how you get this answer:

[FONT=Arial, sans-serif]Which network prefix will implement the IP addressing scheme for 2 LANs one with hosts 172.22.0.62 and 172.22.0.37 and the other with hosts 172.22.0.94 and 172.22.0.75. [/FONT]

/27 is the answer but I don't see y

27 bit mask is 255.255.255.224, so interesting octet is the last one.

224 looks like this

11100000
3 subnet bits and 5 host bits.
block size is 32, therefore the ranges are

172.22.0.0 - 31
172.22.0.32 - 63
172.22.0.64 - 95
172.22.0.96 - 127
and so on

This means that the addresses above fall in two subnets.

Hope this make sense, please speak up if i have got this wrong? I'm learning too.

okplaya

binaryhat wrote: »

I have been doing subnetting in class and I don't understand how you get this answer:

[FONT=Arial, sans-serif]Which network prefix will implement the IP addressing scheme for 2 LANs one with hosts 172.22.0.62 and 172.22.0.37 and the other with hosts 172.22.0.94 and 172.22.0.75. [/FONT]

/27 is the answer but I don't see y

I can see why they chose a /27, but a /26 could be used as well based off of that information.

Priston

IPv6 subnetting
I have a /8 address
FD00::0
I want 987954853651556842951 subnets
70 bits gives me 1180591620717411303424 subnets
so I use /78
which leaves me 50 bits for host 1125899906842624

FD00::0 - FD00::3:FFFF:FFFF:FFFF
FD00::4:0:0 - FD00::7:FFFF:FFFF:FFFF
FD00::8:0:0 - FD00::B:FFFF:FFFF:FFFF
Is this right?

Radiant9

This has really made subnetting SO much easier. I have had charts and formulas and such all jumbled up in my brain. This makes it SO much easier. Im gonna practice this until I can answer questions practically instantly.

Again, Thank you for sharing this!

518

I did a few questions from subnettingquestions website. One of them:

What valid host range is the IP address 10.180.194.157 255.255.240.0 a part of?

Answer: 10.180.192.1 through to 10.180.207.254

I did this to get the answer:

32-28=4
2^4 = 16 – 31
32 - 47
48 - 63
64 - 79
80 - 95
96 - 111
112 - 127
128 - 143
144 - 159
160 - 175
176 – 191
192 – 207
208 - 254

Given that on CCENT exam, a candidate should only spend 1min or < for each question. And most books suggested that a candidate must be able to answer subnetting questions within 15-30secs. Was it possible for the question above to get it in 30secs or less? I mean, I can probably remember the numbers for the first 5 range, but not all the way down to 192-207 range

Am I that really slow??? I understand that I need to keep practicing, and that's what I'm doing.

Thanks.

chmorin

518 wrote: »

I did a few questions from subnettingquestions website. One of them:

What valid host range is the IP address 10.180.194.157 255.255.240.0 a part of?

Answer: 10.180.192.1 through to 10.180.207.254

I did this to get the answer:

32-28=4
2^4 = 16 – 31
32 - 47
48 - 63
64 - 79
80 - 95
96 - 111
112 - 127
128 - 143
144 - 159
160 - 175
176 – 191
192 – 207
208 - 254

Given that on CCENT exam, a candidate should only spend 1min or < for each question. And most books suggested that a candidate must be able to answer subnetting questions within 15-30secs. Was it possible for the question above to get it in 30secs or less? I mean, I can probably remember the numbers for the first 5 range, but not all the way down to 192-207 range Am I that really slow??? I understand that I need to keep practicing, and that's what I'm doing.

Thanks.

You are doing it the hard way. You want to do Binary ANDing.

10.180.194.157 = 00001010.10110100.11000010.10011101
255.255.240.0 = 11111111.11111111.11110000.00000000
AND=======================================
00001010.10110100.11000000.00000000

This give you the network address:10.180.192.0. Then use the information you already know how to gather to finish up the rest of the range.

There may be an even faster way, but I can't remember.

518

chmorin wrote: »

You are doing it the hard way. You want to do Binary ANDing.

10.180.194.157 = 00001010.10110100.11000010.10011101
255.255.240.0 = 11111111.11111111.11110000.00000000
AND=======================================
00001010.10110100.11000000.00000000

This give you the network address:10.180.192.0. Then use the information you already know how to gather to finish up the rest of the range.

There may be an even faster way, but I can't remember.

Thanks. But, it took me 1+ minute just writing those binary numbers. My main concern was HOW to quickly get the answer. Yes, ANDing is easier. But, it still took longer than 1-min to get the answer.

Unless, subnettingquestions website was only concerned whether a person learned the concept, or not.

chopsticks

okplaya wrote: »

I can see why they chose a /27, but a /26 could be used as well based off of that information.

Agree with your opinion.

MickQ

okplaya wrote: »

Originally Posted by binaryhat
I have been doing subnetting in class and I don't understand how you get this answer:
[FONT=Arial, sans-serif]
Which network prefix will implement the IP addressing scheme for 2 LANs one with hosts 172.22.0.62 and 172.22.0.37 and the other with hosts 172.22.0.94 and 172.22.0.75. [/FONT]/27 is the answer but I don't see y

I can see why they chose a /27, but a /26 could be used as well based off of that information.

It could be, but consider that you're looking for an address range which holds the hosts 172.22.0.37 and 172.22.0.62

/26 gives you 62 usable host addresses (.1 to .62).
Shame to be wasting half of that when you can use a /27 for 30 usable host addresses, instead (.33 to .62).

Don't forget the old subnet zero rule. It basically says that when you're subnetting into classless subnets that you're not able to use the first subnet (and also last subnet - "all ones subnet").
What does that mean for us here? Well, if you're subnetting with /27, you won't be able to use address ranges .0 - .31, and .224 - .255

As far as I know (someone correct me if I'm wrong), you should consider the subnet zero rule to be in place unless specifically told that you can use the first and last subnets, or it shows the "ip subnet-zero" command.

Of course, we could also make it a /24, but that would include the other addresses (.75 and .94) in the same range as the first couple.

miller811

518 wrote: »

I did a few questions from subnettingquestions website. One of them:

What valid host range is the IP address 10.180.194.157 255.255.240.0 a part of?

Answer: 10.180.192.1 through to 10.180.207.254

I did this to get the answer:

32-28=4
2^4 = 16 – 31
32 - 47
48 - 63
64 - 79
80 - 95
96 - 111
112 - 127
128 - 143
144 - 159
160 - 175
176 – 191
192 – 207
208 - 254

Given that on CCENT exam, a candidate should only spend 1min or < for each question. And most books suggested that a candidate must be able to answer subnetting questions within 15-30secs. Was it possible for the question above to get it in 30secs or less? I mean, I can probably remember the numbers for the first 5 range, but not all the way down to 192-207 range Am I that really slow??? I understand that I need to keep practicing, and that's what I'm doing.

Thanks.

when I see this 255.255.240.0

I immediately notice the third octect is the interesting octet, (first non 255)
I then take 256 - the interesting octect
256-240 = 16
sixteen is the block size

now you need to find the one it fits in.

just randomly, I multiply be 10
160.... nope to small
add 16 = 176... nope still to small
192.... thats the one
so subnet is 10.180.192.0
one address above that is 10.180.192.1 = first valid host
go to the next subnet 192 + 16 = 208.
10.180.208.0

one address below that is the broadcast address 10.180.207.255
one address below that is the last valid host 10.180.207.254

hope that helps.

518

miller811 wrote: »

when I see this 255.255.240.0

I immediately notice the third octect is the interesting octet, (first non 255)
I then take 256 - the interesting octect
256-240 = 16
sixteen is the block size

now you need to find the one it fits in.

just randomly, I multiply be 10
160.... nope to small
add 16 = 176... nope still to small
192.... thats the one
so subnet is 10.180.192.0
one address above that is 10.180.192.1 = first valid host
go to the next subnet 192 + 16 = 208.
10.180.208.0

one address below that is the broadcast address 10.180.207.255
one address below that is the last valid host 10.180.207.254

hope that helps.

That's it! I could have jumped to x10 of the block size!

Thank you kindly!

gustavQ

Hi everyone.

I have a homework question that i don't know how to solve. I'm not asking to solve me the question, but some one to explain me how can I solve it.

I have this ip address space: 10.123.10.0/24
I want to segment this address space into the following segments:

1. Web servers,
2. Transactional systems,
3. Desktop computers.

The point is to divide this address space into these 3 segments. Provide the network address and the appropriate network mask for each of the segments.

Can anyone help me solve this?

ccnaomkar

mella060 wrote: »

VLSM is basically subnetting a subnet. You have to be able to subnet first. Once you have mastered subnetting then working out VLSM issues will be a lot easier.

Right. First master subnetting

hackmer

gustavQ wrote: »

Hi everyone.

I have a homework question that i don't know how to solve. I'm not asking to solve me the question, but some one to explain me how can I solve it.

I have this ip address space: 10.123.10.0/24
I want to segment this address space into the following segments:

1. Web servers,
2. Transactional systems,
3. Desktop computers.

The point is to divide this address space into these 3 segments. Provide the network address and the appropriate network mask for each of the segments.

Can anyone help me solve this?

How many IP addresses on each of your three segment you need ?

gustavQ

hackmer wrote: »

How many IP addresses on each of your three segment you need ?

I don't know. There is some kind of logic to the devision?

hackmer

First you must decide, how many IPs you want to have on each segment (for example, 10 web servers, 20 transactional systems, 500 desktop PCs etc.).
After that you can divide /24 into smaller subnets (/25 for 126 IPs, /28 for 14, etc.).

gustavQ

hackmer wrote: »

First you must decide, how many IPs you want to have on each segment (for example, 10 web servers, 20 transactional systems, 500 desktop PCs etc.).
After that you can divide /24 into smaller subnets (/25 for 126 IPs, /28 for 14, etc.).

But why the network mask has to be different? I'm new in this kind of theme.

chopsticks

gustavQ wrote: »

But why the network mask has to be different? I'm new in this kind of theme.

This is to satisfy the different networks and hosts requirement, and to be more effectively managing the given IP address spaces so as to cut down wasteful assignments.

gustavQ

chopsticks wrote: »

This is to satisfy the different networks and hosts requirement, and to be more effectively managing the given IP address spaces so as to cut down wasteful assignments.

Are you talking about VLSM?

So if I want 500 Desktop Computers, 10 Transaction systems and 20 Web Servers. What is the answer?

I'm not understanding nothing...

Radiant9

This may help;

www.learntosubnet.com

hackmer

gustavQ wrote: »

Are you talking about VLSM?
So if I want 500 Desktop Computers, 10 Transaction systems and 20 Web Servers. What is the answer?
I'm not understanding nothing...

Let take one situation:
We want 10 web servers, 15 transactional servers and 56 desktops PCs.
With your 10.123.10.0 /24 you can do something like this:
For 10 host addresses use - "/28" (14 host addrs) - 255.255.255.240
for 15 addr - "/27" (30 host addr) - 255.255.255.224
for 56 addr - "/26" ( 62 host addr) - 255.255.255.192
-
/26 (255.255.255.192): 10.123.10.1 - 10.123.10.56
/27 (255.255.255.224): 10.123.10.65 - 10.123.10.94
/28 (255.255.255.240): 10.123.10.97 - 10.123.10.110
-
Am I right, guys?

chopsticks

gustavQ wrote: »

Are you talking about VLSM?

So if I want 500 Desktop Computers, 10 Transaction systems and 20 Web Servers. What is the answer?

I'm not understanding nothing...

Very sorry for the late reply as I am quite tired in the past few days (I am working full time and attending my IT evening classes for 5 days per week + 1 full-day class on weekend).

Yes Sir, you are right, I am talking about VLSM

If basing on your given requirements:

IP Address: 10.123.10.0/24

Desktop PC: 500
Transaction Systems: 10
Web servers: 20

Total host address needed: 530

Let's work out how many host address for the given IP Address and subnet mask:

10.123.10.0 (IP address)
255.255.255.0 (Subnet mask)

The ones in blue belong to the network portion, while the green one belongs to the host part.

Since the whole of the last octet belongs to the hosts and we know that it can be represented by eight bits, we can calculate the maximum hosts addresses that it can give:

( 2^8 ) - 2 = 256 - 2 = 254 valid host addresses

Obviously, with only 254 host addresses, this subnet mask does not cater enough addresses for your 530 hosts.

So, in this senario, you will need a subnet mask of at least 10 host bits to cater for 530 hosts ---> (2^10) - 2 = 1022 host addresses). Therefore, your subnet mask should be at least be a /22, or 255.255.252.0

Hope the above helps.

djyox

This is great. I printed it out and handed it out to everyone in my S6 shop.

Thanks very very much. Very helpful.

gustavQ

hackmer and chopsticks thank you very much for your help.

So if i want 100 Desktop computers, 20 Web servers and 10 transactional systems, the Network address and the network mask for each segment are:

For 100 Desktop Computers - /25, 124 hosts, that is, 255.255.255.128 Network Mask and 10.123.10.0 Network address.

For 20 web servers - /27, 30 hosts , that is, 255.255.255.224 Network mask and 10.123.10.0 Network address.

For 10 transaction systems /28, 14 hosts, that is, 255.255.255.240 Network mask and 10.123.10.0 Network address.

This is right?

miller811

gustavQ wrote: »

hackmer and chopsticks thank you very much for your help.

So if i want 100 Desktop computers, 20 Web servers and 10 transactional systems, the Network address and the network mask for each segment are:

For 100 Desktop Computers - /25, 124 hosts, that is, 255.255.255.128 Network Mask and 10.123.10.0 Network address.

For 20 web servers - /27, 30 hosts , that is, 255.255.255.224 Network mask and 10.123.10.0 Network address.

For 10 transaction systems /28, 14 hosts, that is, 255.255.255.240 Network mask and 10.123.10.0 Network address.

This is right?

right idea, but not applied correctly. You must apply the range after each segment. You always start with the largest to smallest, which you did correctly, but then come to the end of that defined range before you start the next subnet.

For 100 Desktop Computers - /25, 124 hosts, that is, 255.255.255.128 Network Mask and 10.123.10.0 Network address.

10.23.10.0 - 10.23.10.127

So for the next group you start at the next boundary
10.23.10.128 is the subnet, now assign the new mask to this portion.

" For 20 web servers - /27, 30 hosts , that is, 255.255.255.224 Network mask and 10.123.10.0 Network address".

so the /27 is 2 to the 5th which equals a 32 bit subnet so your next subnet is
10.23.10.128 -10.23.10.159

and the final subnet you start one above the last end point
10.23.10.160

"For 10 transaction systems /28, 14 hosts, that is, 255.255.255.240 Network mask and 10.123.10.0 Network address".

since this a 28 bit mask 2 to the 4th is 16 = your block size

so the range is 10.23.10.160 - 10.23.10.175

in this case each ip address gets used once and only once.
And you still have
10.23.10.176 - 10.23.10.255 available.