Subnetting Made Easy

captobvious

Satcom wrote: »

thanks for the write up I hate to even think about sub-netting when I think about study. I would rather learn how routing protocols work and how routers talk to each other than subnet. Back to the books

bookmarked this thread for future reference going to get heavy on subnetting!

You have to learn to crawl before you can run. I still credit this post with helping me get over the subnet "hump".

thomas130

is there an easy to work this one out

class a network 10.0.0.0 create a subnet mask for the 600 subnets. THen Identify the 100th subnet

I can work out doing the subnet easy but is there an easier way to finding out the 100th subnet rather than writing each subnet down like this

10.0.0.0
10.0.64.0
10.0.128.0
10.0.192.0

It would take forever

cleanwithit

I'm kinda confused. How do you know which octet to subtract from?

Like this:

"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."

How do you know which boundary to use, 8, 16, 24, 32 to subtract?

billscott92787

cleanwithit wrote: »

I'm kinda confused. How do you know which octet to subtract from?

Like this:

"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."

How do you know which boundary to use, 8, 16, 24, 32 to subtract?

Look at it this way. You have a /29 mask. This means that there are 29 "on" bits in the subnet mask.

11111111.11111111.11111111.11111 | 000

So, since you have a /29 you see the split in the fourth octect. That's where you know how to determine the subnet blocked and the next available block.

Subnet block: 256 - 248 = 8, this means that each subnet block size is 8

192.168.12.0 /29
192.168.12.8 /29
192.168.12.16 / 29
192.168.12.24 /29
192.168.12.32 /29
192.168.12.40 /29
192.168.12.48 /29
192.168.12.56 /29
192.168.12.64 /29
192.168.12.72 /29
192.168.12.80 /29
192.168.12.88 /29
192.168.12.96 /29
192.168.12.104 /29
and so on.

miller811

cleanwithit wrote: »

I'm kinda confused. How do you know which octet to subtract from?

Like this:

"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."

How do you know which boundary to use, 8, 16, 24, 32 to subtract?

you are almost there
since the mask is 29 = 3 the block size is 8 as you have stated.
the first subnet is 192.168.12.0 hosts 192.168.12.1 - 192.168.12.6 broacast 192.168.12.7
2nd subnet 192.168.12.8 hosts 192.168.12.9 - 192.168.12.14 broadcast 192.168.12.15
etc....
the address you were given is 198.168.12.78 /29

multiplying by 8 you would realize 80 is subnet, and one less is 72....
192.168.12.72 subnet
hosts 192.168.12.73-192.168.12.78
broadcast 192.168.12.79

hope that helps

ukman2003

Hi guys. New to subnetting, and having a few problems. I found out how to get the network number through this great method, but I'm still having trouble.

I need to figure out :

If the IP (Host) valid?

What is the Network (Subnet) Number? ( I think this is covered by using the method above)

What is the Broadcast Number?

What is the valid IP (Host) Range?

What is the subnet mask? If given dotted decimal convert to the prefix length and vice versa?

The first one is :

192.168.10.23 / 25


The second one is :

192.168.1.253
255.255.255.240


I'm just looking for a little help.
I've gotten the basic concept, but that's about it.


Thanks guys.

miller811

ukman2003 wrote: »

Hi guys. New to subnetting, and having a few problems. I found out how to get the network number through this great method, but I'm still having trouble.

I need to figure out :

If the IP (Host) valid?

What is the Network (Subnet) Number? ( I think this is covered by using the method above)

What is the Broadcast Number?

What is the valid IP (Host) Range?

What is the subnet mask? If given dotted decimal convert to the prefix length and vice versa?

The first one is :

192.168.10.23 / 25


The second one is :

192.168.1.253
255.255.255.240


I'm just looking for a little help.
I've gotten the basic concept, but that's about it.


Thanks guys.

practice.......you can figure out all of the answers by working through the methods listed in the thread

192.168.10.23 / 25 = 255.255.255.128 1 bit is borrowed...
128
64
32
16
8
4
2
1

that determines the subnets possible
256/128 = 2 subnets .0 and .128
256/64 = 4 subnets .0 and .64 and .128 and .192
256/32 = 8 subnets .0 and .32 and .64 and .96 and .128 etc
256/16 = 16 subnets .0 and .16 and .32 and .48 and .64 etc
256 /8 = 32 subnets .0 and .8 and .16 and .24 and .32 etc
256/4 - 64 subnets .0 and .4 and .8 and .12 and .16 etc


192.168.10.0 is the subnet
valid host range is 192.168.10.1 - 192.168.10.126
broadcast address 192.168.10.127
so your address falls into the range so yes it is valid
the second subnet starts at 192.168.10.128
this allowed them to take a network and divide it... subnetting:)


192.168.1.253 255.255.255.240 = /28 = 16 bit address blocks
subnet 192.168.1.240
valid host range 192.168.1.241 -192.168.1.254
broadcast 192.168.1.255
so yes it is valid host because it is in the valid range

in this case 192.168.1.0 was subnetted as follows
1 subnet 192.168.1.0 /28
2 subnet 192.168.1.16 /28
3 subnet 192.168.1.32 /28
4 subnet 192.168.1.48 /28
5 subnet 192.168.1.64 /28
6 subnet 192.168.1.80 /28
etc
your example was the last subnet available with the /28
find the subnet size by examining the mask
one ip above the subnet it the first available host address
one ip address below the next subnet is the broadcast
one ip address below the broadcast is the last host address

Bosefus

Really good guide.

Thank you, will defiantly pass this on.

Assassin2005

well this was a great tut to learn subnetting.

i have a problem

Question :-
What is the broadcast address of the network 10.4.64.0 255.255.240.0?
ANSWER : 10.4.79.255

but the problem is how to find out broadcast address when there are 4096 subnets and 4094 host. because in exam you have to do it quickly without wasting time. is there a trick to find out quickly first valid host, last valid host and broadcast address.

miller811

Assassin2005 wrote: »

well this was a great tut to learn subnetting.

i have a problem

Question :-
What is the broadcast address of the network 10.4.64.0 255.255.240.0?
ANSWER : 10.4.79.255

but the problem is how to find out broadcast address when there are 4096 subnets and 4094 host. because in exam you have to do it quickly without wasting time. is there a trick to find out quickly first valid host, last valid host and broadcast address.

1. find the subnet size by examining the mask
2. one ip address above your network address is the first available host address
3. one ip address below the next subnet is the broadcast (use the subnet size to find the boundary)
4. one ip address below the broadcast is the last host address

Firemarshalbill.com

Try my whitepaper on subnetting, I tried to make it as easy and understandable as possible

http://www.linkedin.com/in/firemarshalbill

and if you are on linkedin send me a link request

georgemc

thomas130 wrote: »

is there an easy to work this one out

class a network 10.0.0.0 create a subnet mask for the 600 subnets. THen Identify the 100th subnet

I can work out doing the subnet easy but is there an easier way to finding out the 100th subnet rather than writing each subnet down like this

10.0.0.0
10.0.64.0
10.0.128.0
10.0.192.0

It would take forever

You've almost answered it yourself...

For every 4 increments in your third octet you 2nd octet will increase by 1.

so....100 / 4 = 25 increments of your second octet will get you to 100


10.0.0.0 - 10.24.255.255 is your first 100 subnets (2nd octet 0 thru 24)

The last subnet in that range (the 100th) should be 10.24.192.0


It's very late and I'm completely out of it right now...so if I'm completely wrong and off-base with this, forgive me

I hope this helps...


George

subnetaggravation

Everytime I think I finally think I get subnetting I try a few questions and get them wrong. My blood pressure cant take much more of this. Here are a few of my main concerns:

1. What if my mask matches the next boundary? For instance, 172.21.197.0/24 well 24-24 = 0 and 2 to the power of zero .... well I think you can see that this is not going to work.

2. I am still baffled by which octet you are supposed to be working with and when. I followed the tutorial here to a tee when doing questions on subnettingquestions.com and ALWAYS turned out wrong. For instance it would seem I would use the 3rd octect since that is where the 24th bit is but I get the answer wrong all the time with this logic.

3. I am still 100% confusedon how to answer questions when given the mask in decimal form.

I very much appreciate all of your great information but I cant help but get frustrated when I constantly get this wrong over and over and over again.

Firemarshalbill.com

OK as you can see everyone on the posting does it a little different, I think it depends on how your brain is wired.

I assume when you say decimal format you mean something like 255.255.255.240. Remember that when we are talking mask we are starting at the left with 1's. each group between the periods represents 8 bits hence 255 = 11111111 because all 8 bits are 1's. When you see that value change as in my example here 240 that is the field we are going ot work in. the way to add up the value to determine the 1's is like this

on a scrap piece of paper at the top right this (these are the 2 raised to the powers)

128 64 32 16 8 4 2 1

then you are going to subtract these values from the 240 until you reach 0 and put a 1 in the column if you can subtract.

so you get

128 64 32 16 8 4 2 1
1 1 1 1 0 0 0 0

because
240 - 128 = 112

112 - 64 = 48

48 - 31 = 16

16 - 16 = 0

so we now see that the first 4 bits are ones and the last 4 are 0's.
to give us the / (cidr) value we add up the one's. so for 255.255.255.240 we get 8+8+8+4 = 28 or as we normally write it /28

Good luck I hope this helps

Picker

subnetaggravation wrote: »

Everytime I think I finally think I get subnetting I try a few questions and get them wrong. My blood pressure cant take much more of this. Here are a few of my main concerns:

1. What if my mask matches the next boundary? For instance, 172.21.197.0/24 well 24-24 = 0 and 2 to the power of zero .... well I think you can see that this is not going to work.

2. I am still baffled by which octet you are supposed to be working with and when. I followed the tutorial here to a tee when doing questions on subnettingquestions.com and ALWAYS turned out wrong. For instance it would seem I would use the 3rd octect since that is where the 24th bit is but I get the answer wrong all the time with this logic.

3. I am still 100% confusedon how to answer questions when given the mask in decimal form.

I very much appreciate all of your great information but I cant help but get frustrated when I constantly get this wrong over and over and over again.

If im right this is one of the most easyiest subnet. 172.21.0.0 is the network address. 255.255.255.0 is the subnet mask.

Subnet = 256

Host = 254

Valid subnet - 1.0 all the way to 255.

Remember that this is a class B address by looking at the first octect. therefore the first two octect dont change at all. the third is the subnet which in this case is all 1's = 256 the host is all 0's = 256 - 2 = 254

subnetaggravation

Picker wrote: »

If im right this is one of the most easyiest subnet. 172.21.0.0 is the network address. 255.255.255.0 is the subnet mask.

Subnet = 256

Host = 254

Valid subnet - 1.0 all the way to 255.

Remember that this is a class B address by looking at the first octect. therefore the first two octect dont change at all. the third is the subnet which in this case is all 1's = 256 the host is all 0's = 256 - 2 = 254

Ok I think you answered my question about which octet to manipulate, so a class A address we would manipulate the second octect, class B the 3rd and class C the 4th?

subnetaggravation

lol well I couldve swore Cisco standard practice did not allow subnet zero but I guess I was wrong, no wonder I was off most of the time.

Geetar28

Dude, I signed up for the forum just to THANK YOU for posting this!!!!

I was able to understand subnetting from the long route used in the Bryant method...but I knew that it took too long to answer the questions. It took me a bit to streamline things but now I'm able to answer EVERY question at subnettingquestions.com with ease.

I write out the powers of 2: 2 4 8 16 32 64 128 256 etc...up to 65,536 and I write out the possible masks i.e. 128 192 224 240 248 252 254 255. This allows me to answer most every question within just a few seconds (many I can even do in my head now!!)

My only regret is not finding this easy method before drudgging through all the other garbage out there.

albanga

I cannot thank you enough for this! I thought i would never understand sub netting and then found this! I must admit it took me a whole day before it started to click. As time went by it started making a little bit more sense and then a little more and then BAM!!!

I can now go onto subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online and answer all the questions in my head. Phew!!

Now that part is over i guess its time to get WAN's down pat

But thank you again for introducing me to this concept. I LOVE IT!! I even just showed my manager a few questions and answered them all in about 8 seconds and he was utterly impressed

miller811

here is my favorite site to test your skills

IP Subnet Practice

click the new problem button and then solve

with practice, you will be able to subnet in your head

jonfc3s

Why isnt this a sticky? Great thread!

Jon

IA-Daigakusei

I believe LordFlasheart`s post deserves a sticky. That was the best post teaching how to do subnets that I have ever seen! Good job!

baraev

I have no words to describe how EASY you made for me subnetting! After using your method I am indeed laughing each time I have to subnet when I remind my self how easy you have made it for me! God bless you man

daveccna

This thread was helpful to me, while I had 'got' subnetting I was bogged down in slow methods. This is damned fast, I think knowing the bog standard theory and then learning a quick technique which you know makes sense due to your grounding is great.

Rep flashheart.

daveccna

daveccna wrote: »

This thread was helpful to me, while I had 'got' subnetting I was bogged down in slow methods. This is damned fast, I think knowing the bog standard theory and then learning a quick technique which you know makes sense due to your grounding is great.

Rep flashheart.

Seriously it's awesome, I can now do the questions on the subnetting questions internet site in less than 5 seconds in most cases.

Nzastudios

I got stuck with these - Can someone give me a clear breakdown of how these answers were gotten pls
What is the last valid host on the subnetwork 192.168.39.0/28?
Answer: 192.168.39.14

What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
Answer: 172.19.179.254

What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
Answer: 10.5.176.1

miller811

Nzastudios wrote: »

I got stuck with these - Can someone give me a clear breakdown of how these answers were gotten pls
What is the last valid host on the subnetwork 192.168.39.0/28?
Answer: 192.168.39.14

What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
Answer: 172.19.179.254

What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
Answer: 10.5.176.1

I]What is the last valid host on the subnetwork 192.168.39.0/28?
[/I]Answer: 192.168.39.14

since the mask is /28 the block size if 16 = 2 to the 4th or 2X2X2X2
192.168.39.0 is the network address
192.168.39.1 is the first valid host address
192.168.39.15 would be the broadcast address
which makes 192.168.39.14 as the last useable address in the subnet
192.168.39.16 would start the next network segment with a /28 mask

What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
Answer: 172.19.179.254
block size is 4 =2X2 since 6 bits are used for subnetting
172.19.176.0 is the network address
172.19.179.255 would be the broadcast address
172.19.179.254 would be the last address
172.19.180.0 would be the next network address with a /252
the key here is you are subnetting the 3rd octet and not the 4th

What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
Answer: 10.5.176.1[/QUOTE]

240 mask means you are borrowing the first 4 bits of the third octet. Like the first example your block size is 16, Like the second question the subnetting is taking place in the third octet

10.5.0.0 1st subnet
10.5.16.0 2nd subnet
10.5.32.0 3rd subnet
10.5.48.0 4th subnet
10.5.64.0 5th subnet
10.5.80.0 6th subnet
10.5.96.0 7th subnet
10.5.112.0 8th subnet
10.5.128.0 9th subnet
10.5.144.0 10th subnet
10.5.160.0 11th subnet
10.5.176.0 12th subnet

on the 12 subnet
10.5.176.1 would be the first valid host address
10.5.191.255 would be the broadcast address
10.5.191.254 would be the last host address

10.5.192.0 13th subnet
etc

Hope that helps

The keys that help me figure this out is figure out the block size
from the block size figure out your network address
figure out the next network subnet from the block size
one address above your network address is the first host
one address below the next network segment is your broadcast address
one address below your broadcast address is the last host address

Nzastudios

I am actually starting to get this together in my head and its pretty motivating lol

It all came together for me (for those who are still baffled like I was for a longgggg time - From example below)

What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
Answer: 10.5.176.1[/QUOTE]

240 mask means you are borrowing the first 4 bits of the third octet. Like the first example your block size is 16, Like the second question the subnetting is taking place in the third octet

10.5.0.0 1st subnet
10.5.16.0 2nd subnet
10.5.32.0 3rd subnet
10.5.48.0 4th subnet
10.5.64.0 5th subnet
10.5.80.0 6th subnet
10.5.96.0 7th subnet
10.5.112.0 8th subnet
10.5.128.0 9th subnet
10.5.144.0 10th subnet
10.5.160.0 11th subnet
10.5.176.0 12th subnet

on the 12 subnet
10.5.176.1 would be the first valid host address
10.5.191.255 would be the broadcast address
10.5.191.254 would be the last host address

10.5.192.0 13th subnet
etc

What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
Answer: 172.19.179.254
block size is 4 =2X2 since 6 bits are used for subnetting

Please confirm my thought process below is in the right track...

The 6 bits
128 64 32 16 8 4 2 1 =252 (128+64+32+16+8+4)
1 1 1 1 1 1 0 0 = 6 Bits >> Now find block 8 - 6 = 2, then 2 x block diff(2)=4

Now that I know its a block 4, I simply add the 4 to the 3rd octet in 172.19.176.0 (176+4=180)

Therefore I focus on the range
172.19.176.0
172.19.180.0

Then thats how you came about the answer? If my thinkin is correct I definitely finally understand Subnetting
172.19.179.255 would be the broadcast address
172.19.179.254 would be the last address

figure out the next network subnet from the block size
one address above your network address is the first host
one address below the next network segment is your broadcast address
one address below your broadcast address is the last host address

---

miller811 wrote: »

I]What is the last valid host on the subnetwork 192.168.39.0/28?
[/I]Answer: 192.168.39.14

since the mask is /28 the block size if 16 = 2 to the 4th or 2X2X2X2
192.168.39.0 is the network address
192.168.39.1 is the first valid host address
192.168.39.15 would be the broadcast address
which makes 192.168.39.14 as the last useable address in the subnet
192.168.39.16 would start the next network segment with a /28 mask

What is the last valid host on the subnetwork 172.19.176.0 255.255.252.0?
Answer: 172.19.179.254
block size is 4 =2X2 since 6 bits are used for subnetting
172.19.176.0 is the network address
172.19.179.255 would be the broadcast address
172.19.179.254 would be the last address
172.19.180.0 would be the next network address with a /252
the key here is you are subnetting the 3rd octet and not the 4th

What is the first valid host on the subnetwork that the node 10.5.178.10 255.255.240.0 belongs to?
Answer: 10.5.176.1

240 mask means you are borrowing the first 4 bits of the third octet. Like the first example your block size is 16, Like the second question the subnetting is taking place in the third octet

10.5.0.0 1st subnet
10.5.16.0 2nd subnet
10.5.32.0 3rd subnet
10.5.48.0 4th subnet
10.5.64.0 5th subnet
10.5.80.0 6th subnet
10.5.96.0 7th subnet
10.5.112.0 8th subnet
10.5.128.0 9th subnet
10.5.144.0 10th subnet
10.5.160.0 11th subnet
10.5.176.0 12th subnet

on the 12 subnet
10.5.176.1 would be the first valid host address
10.5.191.255 would be the broadcast address
10.5.191.254 would be the last host address

10.5.192.0 13th subnet
etc

Hope that helps

Thank you so much Lordflashheart and Miller811 !!!

If I get confused in the near future I will be back here again lol!

The keys that help me figure this out is figure out the block size
from the block size figure out your network address
figure out the next network subnet from the block size
one address above your network address is the first host
one address below the next network segment is your broadcast address
one address below your broadcast address is the last host address[/QUOTE]

JockVSJock

This thread is golden!!! Should be a sticky.

gouki2005

LordFlasheart wrote: »

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!

can you explain me both for example

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

i use the method so 16-14 = 2
now 2^2= 4

so size of the blocks is 4 for so

10.0.0.0
10.4.0.0
10.8.0.0
10.12.0.0
10.16.0.0
10.20.0.0
10.24.0.0
10.28.0.0

so i thought the valid range was 10.25.0.0 to 10.26.0.0 it works for class C