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

JDMurray

The block size is calculated by subtracting the netmask value of the "interesting octet" from 256. So in your example, /26 = 255.255.255.192 = 256 - 192 = 64 subnet block size. You can also just memorize that /26 = 255.255.255.192 = 64 block size.

Digitaljunkie

Ahhh that's where I was messing up thanks JD got it now.........

Getting the questions right now so ....practice practice practice...

Thanks again JD

Dj.

JDMurray

No problem.

I should point out the other way of calculating block size, which is the way you were going, is to subtract the CIDR from the next largest octet value and raise two to that power.

For example, for /26:

1. 32 is the next octet size up from 26
2. 32 - 26 = 6
3. 2^6 = 64

It's good to know the math, but just remembering that a /26 subnet is always a 64 IP addresses in length is usually easier.

Digitaljunkie

Yea that is the way i'm calculating the block size.

Perfect thanks again for the info JD

Dj.

Digitaljunkie

boredgamelad wrote: »

There is a maximum of 32 bits... 22 + 8 doesn't equal 32, so one or both of those numbers has to be wrong.

172.17.0.0 / 255.255.254.0 = 172.17.0.0/23

Default class B: /16

23 - 16 = 7

2^7 = 128 subnets.

32 - 23 = 9

(2 ^ 9) - 2 = 510 hosts.

I just need to check a rule here with the host example,

32 - 23 = 9

(2 ^ 9) - 2 = 510 hosts.

While the CiDER is 23 you are not using the next boundary which is 24. Can someone explain why?
As in when do you decide to move to the next boundary to get the correct answer...?

Is it because 24 is in the next boundary set so the last boundary you can use is 22 ?

Thanks Dj................

alphagolf

Digitaljunkie wrote: »

I just need to check a rule here with the host example,

32 - 23 = 9

(2 ^ 9) - 2 = 510 hosts.

While the CiDER is 23 you are not using the next boundary which is 24. Can someone explain why?
As in when do you decide to move to the next boundary to get the correct answer...?

Is it because 24 is in the next boundary set so the last boundary you can use is 22 ?

Thanks Dj................

When answering questions such as: "Question: How many subnets and hosts per subnet can you get from the network 172.17.0.0 255.255.254.0?"..........

.........When you get to working out the hosts numbers you always use 32 as the "boundary" number.
When working out the number of subnets use the boundary of the "interesting octet" (as you have been doing)

255.255.255.11111111

Imagine counting from right to left in the last octet. 32, 31, 30, etc. 32 is always your max boundary number when working with hosts.

How many subnets and hosts per subnet can you get from the network 172.17.0.0 255.255.254.0?

I talk myself through it in my head as follows:
What they've given me (as a CIDR slash number) minus default slash for the class they've given me:
23 - 16 (default class

= 7.
2^7 = 128 subnets.

Because I now gotta find out the hosts I use 32 as my max boundary number:
32 minus CIDR slash number they've given me:
32 - 23 = 9
2^9 = 512
512 -2 = 510 hosts.

HTH

Also...just wanted to say thanks to the OP. I found this thread a number of years ago and spent a few hours going over it one afternoon with practice questions from subnettingquestions.com and have never looked back!

Digitaljunkie

That clears it up thanks a million alphagolf. And yes I also have to thank the OP. I tried 5 books and countless videos and really had no clue till I found this thread.

LordFlashart got my attention when I got to the line " 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."

"Hail LordFlashart"

Thanks again alphagolf.......Dj

Mell

I registered here just to thank all the people who are helping us with our preparation. Again, thank you very much, folks!

hs_parmar

unitedfan1 wrote: »

Hi guys, kinda new at all of this....i`m currently doing CCNA1 and i have a question i was hoping you could help me out with.

Question : ABC has aquired a class B address 172.16.0.0 ,the Company needs to create a subnetting scheme to provide the following:

36 subnets with at least 100 hosts
24 " " " " " " " " 255 hosts
10 " " " " " " " " 50 hosts

NB : It is not necessary to create a WAN connection

a)How many subnets are needed for this network
b)What is the minimum number of bits that can be borrowed
c)What is the subnetmask for this network in dotted decimal, binary and slash format
d)How many usable subnets are there
e)How many usable hosts are there per subnet

I know its a bit long but it would be great if you guys were to walk me through this one

thank you

hey unitedway1: i dont know if i am right in this..
but this is my approach to this question.

172. 16. 0 . 0
255.255. 0 . 0

172. 16.NNNNNNHH.HHHHHHHH /22 - 64 SUBNETS AND 1024-2 = 1022 HOSTS
255.255.252.0

172. 16. 64. 0 /21 - 32 SUBNETS AND 2048-2 = 2046 HOSTS
255.255.248. 0

172. 16. 72. 0 /20 - 16 SUBNETS AND 4096-2 = 4094 HOSTS
255.255.240. 0

Can somebody please correct me if i'm wrong in this. Much thanks in advance.

boredgamelad

The question you just responded to was posted in 2009.

Denholm

Everyone keeps saying they understand this easily, well i wish i did . i'm doing my N+ soon and for the love of all that's good and pure i can't figure out subnetting.....

Denholm

then again in all honesty i have no networking experience, i'm literally learning everything from professor messer vids for the N10-005 and practice exams. I'm 'stuck' at a 'wall' so to speak, when doing practice exams that involve 100+ questions i keep scoring in the 70% range never in 60% range which is good but i cant go above 80% and i've done alot of N+ exams it's the same all around. Dont know if it's cause iv'e only been studying for 3 weeks and as i said no experience at all. any advice would be greatly apprciated

m900

What is the address range for the 11th subnet in the network 10.0.0.0/16 ? Please explain me how to do this using original poster's method.

mendysue

I'm a student too, so hopefully someone can tell you if I'm right.

24-16= 8 , 2 to the 8th power is 256 and that is your block size. I think your subnets would look like this:

0. 10.0.0.0
10.0.255.0
1. 10.0.1.0
10.0.1.255
2. 10.0.2.0
10.0.2.255
3. 10.0.3.0
10.0.3.255
4. 10.0.4.0
10.0.4.255
5. 10.0.5.0
10.0.5.255
6. 10.0.6.0
10.0.6.255
7. 10.0.7.0
10.0.7.255
8. 10.0.8.0
10.0.8.255
9. 10.0.9.0
10.0.9.255
10. 10.0.10.0
10.0.10.255
11. 10.0.11.0
10.0.11.255
12. 10.0.12.0
10.0.12.255

ph1x10n

m900 wrote: »

What is the address range for the 11th subnet in the network 10.0.0.0/16 ? Please explain me how to do this using original poster's method.

255.255.0.0 (nnnnnnnn.ssssssss.hhhhhhhh.hhhhhhhh) /16

Your entire 2nd octet is used for the subnet, giving you 256 subnets, from 0 as your first all the way to 255 as your last
1st subnet: 10.0.0.0 (broadcast of 10.0.255.255)
2nd subnet: 10.1.0.0 (broadcast of 10.1.255.255)
And it continues on like this for 254 more subnets.
Does that make sense?

steveyeung

m900 wrote: »

What is the address range for the 11th subnet in the network 10.0.0.0/16 ? Please explain me how to do this using original poster's method.

a class A network subnetted with the /16 mask
10.0.0.0 is the first subnet, so
10.10.0.0 is the 11th subnet,
10.11.0.0 is the 12th subnet

so the host range is 10.10.0.1 to 10.10.255.254

mannygo

Ok, I have a question here that I am having trouble answering:

1) Assuming the 172.16.0.0 network, create a subnet mask where 500 hosts are required per network
2) Calculate the host range for the 5th network
3) Assuming the 172.17.0.0 network, create a subnet mask where 1000 subnets are required
4) Calculate the host range for the 6th network

Exaples on how to would be great, not just the answer.

Thanks in advance !!!

Clevernamehere

Let's see if I get this right. For 1-2 you need 500 hosts. 2 to the 9th power is 512 so that covers the requirement. 32 bits in a subnet mask so 32-9=23. /24 is the end of the third octet so 24-23=1 and, again the powers of 2, 2 to the first power is obviously 2. So your counting subnets by 2. 0,2,4,6,8. 8 would be the 5th network. Since all subnets are being counted in 2's the host range is 172.16.8.1-172.16.9.254.

For 3-4 you need 1000 subnets. 172 is a class b address with its octect ending at /16. With the powers of two we know 2 to the 10th is 1,024 which meets the requirements. 16+10=26. 32-26=6. 2 to the 6th power is 64. So the first network is 172.17.0.0. Then 172.17.0.64. Counting by 64 until you get to the 6th network 172.17.1.64. So the host range would be 172.17.1.64-172.17.1.126.

I'm pretty confident on the number of hosts but not on the subnets. Please don't be afraid to tell me how wrong I am here.

mannygo

What about the broadcast and network bits?

TheNewITGuy

mannygo wrote: »

Ok, I have a question here that I am having trouble answering:

1) Assuming the 172.16.0.0 network, create a subnet mask where 500 hosts are required per network
2) Calculate the host range for the 5th network
3) Assuming the 172.17.0.0 network, create a subnet mask where 1000 subnets are required
4) Calculate the host range for the 6th network

Exaples on how to would be great, not just the answer.

Thanks in advance !!!

172.16.0.0 is by default a /16 network which contains 2^16 hosts - how many bits give us 500 hosts per subnet?

we know that 2^9 = 512 hosts so we can just say ok, we need 9 bits for hosts

11111111.11111111.11111110.00000000 = 23 bits (we could just have done 32-9)

so our network is now 172.16.0.0/23 or 255.255.254.0

256-254 is 2 and is our block size

example:
172.16.0.0 - 172.16.1.255
172.16.2.0 - 172.16.3.255
172.16.4.0 - 172.16.5.255
172.16.6.0 - 172.16.7.255
172.16.8.0 - 172.16.9.255

same goes for 172.17.0.0 well we know that 2^9 is 512 then 2^10 is 1024 so we just change ths subnet mask for 252.0 and block size is now 4 and a /22 and just count

mannygo

TheNewITGuy wrote: »

172.16.0.0 is by default a /16 network which contains 2^16 hosts - how many bits give us 500 hosts per subnet?

we know that 2^9 = 512 hosts so we can just say ok, we need 9 bits for hosts

11111111.11111111.11111110.00000000 = 23 bits (we could just have done 32-9)

so our network is now 172.16.0.0/23 or 255.255.254.0

256-254 is 2 and is our block size

example:
172.16.0.0 - 172.16.1.255
172.16.2.0 - 172.16.3.255
172.16.4.0 - 172.16.5.255
172.16.6.0 - 172.16.7.255
172.16.8.0 - 172.16.9.255

same goes for 172.17.0.0 well we know that 2^9 is 512 then 2^10 is 1024 so we just change ths subnet mask for 252.0 and block size is now 4 and a /22 and just count

Great explanation to both!!!

It seems so difficult but it gets easier now

Clevernamehere

11111111.11111111.11111110.00000000 = 23 bits (we could just have done 32-9)

same goes for 172.17.0.0 well we know that 2^9 is 512 then 2^10 is 1024 so we just change ths subnet mask for 252.0 and block size is now 4 and a /22 and just count

But if you did that you would have 10 host bits. 11111111.11111111.11111100.00000000. The question asks for 1000 subnets. Since you have a class B network the bits are 11111111.11111111.00000000.00000000. 2 to the 10th is 1024 so you would need to add 10 subnet bits. 11111111.11111111.11111111.11000000. Which is /26 or 255.255.255.192. Again let me know if I'm doing this horribly wrong.

Magic Johnson

Clevernamehere wrote: »

But if you did that you would have 10 host bits. 11111111.11111111.11111100.00000000. The question asks for 1000 subnets. Since you have a class B network the bits are 11111111.11111111.00000000.00000000. 2 to the 10th is 1024 so you would need to add 10 subnet bits. 11111111.11111111.11111111.11000000. Which is /26 or 255.255.255.192. Again let me know if I'm doing this horribly wrong.

Yeah you are right. You are sacrificing host bits for subnet bits. You couldn't have 1000 subnets and 500 hosts on the same network because there isn't enough bits. So you'd have 62 hosts (64-2) per subnet, and there would be 1024 subnets.

172.16.0.0-63
172.16.0.64-127
172.16.0.128-191
172.16.0.192-255

172.16.1.0-63

ETC

Did thenewitguy possibly misread and think it was for 500 hosts then 1000 hosts?

avaismughal

Given that the subnet mask for PCs above is 255.255.240.0, identify host portion of the IP address 146.191.111.129 (express as a decimal number).

mohammadzzz

hi

i am struggling with IPv6,any help would be apperciated on subnetting,prefix etc

LoukasV

Please help me understand the method(way of thinking) of solving this. My exams is in only one day!!! Reference binary codes for help

Thank you

xnx

Why book your exam if you're not ready??

/25 - .128 ----> Goes up in ( 256 - 128 )
/26 - .192 ----> Goes up in (256 -192)
/27 - .224
/28 - .240
/29 - .248
/30 - .252

You can work the rest out yourself

LoukasV

Thank you for the quick reply..
i can understand how to solve 2nd and 4th line of the array
but not 1st and 3rd..
how can you calculate network and subnet mask when given only the range?

Network:150.140.142.0

Subnet?

Range:150.140.142.0-150.140.142.63
Network: ?

Subnet?

Range:150.140.140.0-150.140.140.255

mohammadzzz

hi

any help on this question. A router is connected to following networks 10.0.0.0,
10.0.1.0,
10.0.2.0, and
10.0.3.0

What is the most appropriate summarization for these routes?

A. 10.0.0.0 /21B. 10.0.0.0 /22C. 10.0.0.0 /23

D. 10.0.0.0 /24

why not /21.Please can some one explained logic behind the correct answer of /22 mask.My idea is 24-21=3 so 2^3 =8,still should cover all the subnets .while the correct answer is /22 which gives us 2^2 =4,i am confused....any help plz.
Thanks

mella060

mohammadzzz wrote: »

hi

any help on this question. A router is connected to following networks 10.0.0.0,
10.0.1.0,
10.0.2.0, and
10.0.3.0

What is the most appropriate summarization for these routes?
A. 10.0.0.0 /21B. 10.0.0.0 /22C. 10.0.0.0 /23
D. 10.0.0.0 /24

why not /21.Please can some one explained logic behind the correct answer of /22 mask.My idea is 24-21=3 so 2^3 =8,still should cover all the subnets .while the correct answer is /22 which gives us 2^2 =4,i am confused....any help plz.
Thanks

So assume the networks in your example are...

10.0.0.0/24
10.0.1.0/24
10.0.2.0/24
10.0.3.0/24

Step 1 - Re-list the subnet ID's (numbers) and prefix lenghs in decimal and calculate the subnet broadcast addresses.

10.0.0.0/24 - broadcast address is 10.0.0.255
10.0.1.0/24 - broadcast address is 10.0.1.255
10.0.2.0/24 - broadcast address is 10.0.2.255
10.0.3.0/24 - broadcast address is 10.0.3.255

Step 2 - Note the low and high end of the range of addresses for all combined subnets by noting the numerically lowest subnet ID and the numerically highest subnet broadcast address. So in your example, 10.0.0.0 would be the lowest subnet ID and 10.0.3.255 would be the highest broadcast address.

Step 3 - Pick the shortest prefix length of all subnets to be summarized (subordinate) subnets, and then subtract 1. So in your example, all the subnets have a prefix length of /24, so subtract 1 from 24...24-1 = 23. So calculate the range of addresses you would get with a prefix length of /23.

10.0.0.0/23 - 10.0.0.1 - 10.0.1.255...doesn't match all the addresses to be summarized. So subtract 1 from 23 and see if that includes all the subnets and their addresses to be summarized.

10.0.0.0/22 - range 10.0.0.1 - 10.0.3.255...This exactly matches the range of addresses shown above so you have found the subnet and mask to use in the best summary route: 10.0.0.0/22

Answer A 10.0.0.0/21 would also work but it is not the best summary route in this example. The best summary route is the one that includes all the subnets to be summarized, but with as few other addresses as possible.