Member Posts: 10 ■□□□□□□□□□
Hi Everyone, I am having the hardest time trying to get subnettting to "click". I have watched videos, read tutorials, and I still cant seem to figure it out! I am decent at binary, I can figure out the subnet mask no problem. But when it comes down to finding the first host to the last host, broadcast address, ect ect... I get completly confused. I am not good at math and I knew this would be a challenge, but I have really applied my self and I cant seem to get it.
Any help would be appreciated.. Thanks

Heres a question that we had in class that I cant figure out, but maybe if you could explain it, I can get it to click?

Given a subnet of 140.125.80.0 with a sunet mask of 255.255.252.0 what is the address of the nxt higher subnet?

Given a subnet mask of 255.255.255.0 with a class b address, how many subnets would be available?

• Member Posts: 18 ■□□□□□□□□□
Hey Mate,

I know you said that you watched the Video's but did you watch the CBT Nuggets CCNA video on subnetting? Fantastic video, described it really well. Theres tables for subnetting in the CCENT Certification book from Cisco Press that's pretty good aswell.
• Member Posts: 441
I'm at work so I don't really have time to go through and explain the answers but I can recommend a couple things. first look at this thread: http://www.techexams.net/forums/ccna-ccent/61908-subnetting-resources.html

Also there are tons of threads with very good explanations for subnetting questions especially in the ccent/ccna forum do a search and read through some.

Watch the Professor messer net+ videos on subnetting. After all that you should be getting a good idea then watch this video. It is my favorite method and if I've been practicing I can do any subnetting question within seconds using it. Creating Your Own Mental Subnet Calculator
• Member Posts: 4,214 ■■■■■■■■□□
Given a subnet of 140.125.80.0 with a sunet mask of 255.255.252.0 what is the address of the nxt higher subnet?

This is really easy. Look at the subnet mask, 255.255.252.0. I am going to rewrite this to 255.255.11111100.0. All I did was change the 3rd octet from Decimal to Binary and left everything else as it is. To determine how broad a subnet is, look at the last 1 and then find its value. (128 64 32 16 8 4 2 1) So that last 1 has a value of 4, which makes 4 your counter.

So lets write out the network addresses...
140.125.80.0 (You were given this)
140.125.84.0 (We added 4 to the 3rd octet since that is where the change occurs in the subnet mask)
140.125.88.0
140.125.92.0

We could continue, but it is just going to keep going up by 4. Remember that you can't have a value larger than 255, so when the time comes you would eventually increment the second octet by 1 and start over at 0 in the 3rd. (Much like when you get to 10 you increase the 10's spot by one and then start back at 0 in the ones position)

Another thing to point out that is not part of this question is that the addresses above are just the network addresses. The range of that subnet would be from the network address to 1 below the next network address. So that first range is
140.125.80.0 to 140.125.83.255

Given a subnet mask of 255.255.255.0 with a class b address, how many subnets would be available?
This is also pretty easy, you just need to remember a formula. (2^n)-2 where n is the number of bits used for the network address. (That is just a fancy way of saying the number of 1's in the subnet mask. 0's are used for the hosts)

The class B part is important because it means that 16 of your 1's are already accounted for. This leaves you with 8 left (255.255.255 is 24 1's and we subtracted the first 16). So your answer is 254.

The first thing to understand is that /29 is a subnet mask. What it means is that there are 29 1's. 11111111.11111111.11111111.11111000 or 255.255.255.248 or 255.255.255.11111000 if you write it the way I like to look at it.

The counter on this is going to be 8 since that is the position value of the last 1. 201.x.x.x is a class C address. Lets start writing down the network addresses,
201.222.10.0
201.222.10.8
201.222.10.16
201.222.10.24
201.222.10.32
201.222.10.40
201.222.10.48
201.222.10.56
201.222.10.64
201.222.10.72

I'm going to stop here since 69 is going to fall between the 64 and 72.

Subnet mask: 255.255.255.248 (converted from binary to decimal)
Class: C (gotta know the ranges of class A, B, and C addresses)

I'm not an expert subnetter, and as you can see I have to slowly work through the problem. Let me know if I made any mistakes!
Decide what to be and go be it.
• Member Posts: 10 ■□□□□□□□□□
Devilsbane, between you and the thread Subnetting made easy by @LordFlasheart, i am starting to get it..

I had a few more questions and I was wondering if you could do this one

15.16.193.6/21 ...I know that 24 - 21 = 3 2^3 = 8, but how do I know where I start the 0 and start going in incremnets of 8?

• Member Posts: 4,214 ■■■■■■■■□□
15.16.193.6/21 ...I know that 24 - 21 = 3 2^3 = 8, but how do I know where I start the 0 and start going in incremnets of 8?

15.16.193.6 with a mask of 255.255.248.0 or 255.255.11111000.0

As you said, 15.x.x.x is a class A address. When you buy a class A address, what you are buying is 15. anything. So you would start with 15.0.0.0 (If you were buying a class B address then you would be starting with something like 130.15.x.x)

Counter is 8.
15.0.8.0
15.0.16.0
15.0.24.0
15.0.32.0
.........
15.0.240.0
15.0.248.0
15.1.0.0
15.1.8.0

As you can guess, this is going on for a long time. Why is that? Lets look at that formula I gave you earlier. (2^13)-2. I listed out 8 of the subnets, buy I could list out 8,190 of them!

The first network address will be 15.0.0.0
The range will be from 15.0.0.0 to 15.0.7.255
15.0.7.255 would be the broadcast. (subtracting 1 from the next network address. Since you can't go negative then you need to borrow from the higher up octet. So what I did was rewrite 15.0.8.0 to 15.0.7.256 and then I can take 1 away from 256 and make it 255)

The second network address is 15.0.8.0
The range is from 15.0.8.0 to 15.0.15.255

The third network address is 15.0.24.0
The range is from 15.0.24.0 to 15.0.31.255

See the pattern? You will always increment in the octet where the 0's begin in the subnet mask. For example, the above mask it is in the 3rd octet. If your mask was 255.255.255.0 then it would be the 4th octet.
Decide what to be and go be it.
• Member Posts: 10 ■□□□□□□□□□
but why did you start in the 3rd octet? and what is the subnet address?

Sorry its the most fustrating because I have the concept I just cant seem to apply it properly
• Member Posts: 4,214 ■■■■■■■■□□
I started in the 3rd octet because of the subnet mask, which is 255.255.248.0 .
11111111.11111111.11111000.00000000 written in binary.We don't care about the first or second octet because it is just all 1's. How boring. We don't care about the 4th octet because it is just all 0's. The change happens in the 3rd octet, specifically the octet where the first 0 appears, so that is where you are going to be incrementing your numbers. That value of the last 1 is 8 and it is in the 3rd octet.

Remember what these 1's and 0's mean. 1's are used for the network portion of the address. 0's represent what is available for the hosts to use. A subnet mask of 255.0.0.0 has few bits used for the network portion, which means that there are few subnets but many hosts can be on each one. A mask of 255.255.255.240 on the other hand is going to have many subnets but only a few hosts available to be on each one.
Decide what to be and go be it.
• Banned Posts: 703
Chris,

We started in the third octet because of the mask

Remember a Class A address will always have it's first bit 0
Class B will always be 01
and Class C is 110

Remember the default classes 255.0.0.0 (/ 255.255.0.0 (/16) 255.255.255.0 (/24) - this is most important in doing subnetting on non-class C addresses

so 15.16.193.6/21 is a Class A address which has a default mask of 255.0.0.0

our /21 tells us that our mask is 255.255.248.0 (8+8+5=21)

Now remember since we're in the 3rd octet doing our subnetting we need to change the math a little

11111111.11111111.11111000.00000000

our hosts are 2^11
and our sub netis: 2^13

The reason it is 2^13 is because our default class for a class A address is 255.0.0.0 since we borrowed 13 bits to make our sub net to put our 2^11 hosts in the amount of sub nets grow exponentially.

Also since you have the mask in place if you take the IP address

15.16.193.6
255.255.248.0

248 = 11111000
193= 11000001
You just AND them together which basically states 1+1 = 1 1+0=0 and 0+0 = 0

This IP is part of 11000000 or 192.0 network or 15.16.192.0 respectively.

Hope this helps

Forgot to say this - your block size is still 8 you're just subtracting 21 from 24 if we used the entire 32bits it'd be 32-21 = 11 (which is your # of subnets) the 8 is just the "count" so we have
15.16.192.0
15.16.200.0

so our first is 15.16.192.1 through 15.16.199.254 and .255 as broadcast
• Senior Member Mod Posts: 6,926 Mod
I struggled a bit with subnetting mainly because I saw so many different methods and none of them stuck. Imagine trying to drive to a town you have no idea where it is and 10 people give you 10 different routes, none of them evident to you. Like you, I understood the basics but had some trouble figuring out which octect was the "interesting" one, etc. Two things helped me overcome my problem:

- IP Subnet Practice

I went through that thread and for some reason it clicked in a very short time. I then used the San Bernardino practice to make sure I understood everything. The fact that their calculator gives you the answer but doesn't show anything else really made me stop and think how the heck they got that result.
• Member Posts: 10 ■□□□□□□□□□
drkat wrote: »
Chris,

We started in the third octet because of the mask

Remember a Class A address will always have it's first bit 0
Class B will always be 01
and Class C is 110

Remember the default classes 255.0.0.0 (/ 255.255.0.0 (/16) 255.255.255.0 (/24) - this is most important in doing subnetting on non-class C addresses

so 15.16.193.6/21 is a Class A address which has a default mask of 255.0.0.0

our /21 tells us that our mask is 255.255.248.0 (8+8+5=21)

Now remember since we're in the 3rd octet doing our subnetting we need to change the math a little

11111111.11111111.11111000.00000000

our hosts are 2^11
and our sub netis: 2^13

The reason it is 2^13 is because our default class for a class A address is 255.0.0.0 since we borrowed 13 bits to make our sub net to put our 2^11 hosts in the amount of sub nets grow exponentially.

Also since you have the mask in place if you take the IP address

15.16.193.6
255.255.248.0

248 = 11111000
193= 11000001
You just AND them together which basically states 1+1 = 1 1+0=0 and 0+0 = 0

This IP is part of 11000000 or 192.0 network or 15.16.192.0 respectively.

Hope this helps

It really does help and I thank all of you for all the time you are giving me to get this right.

So if the subnet is 15.16.192.0 what would the broadcast be? and what woudl the range be?
• Member Posts: 4,214 ■■■■■■■■□□
So if the subnet is 15.16.192.0 what would the broadcast be? and what woudl the range be?

Well the next subnet would start at 15.16.200.0. So the broadcast would be 15.16.199.255. The range of usable addresses would be 15.16.192.1 to 15.16.199.254 (notice how I removed the subnet address and the broadcast. You can't assign a host either of these addresses).
Decide what to be and go be it.
• Banned Posts: 703
Chris,

Forgot to say this - your block size is still 8 you're just subtracting 21 from 24 if we used the entire 32bits it'd be 32-21 = 11 (which is your # of subnet bits) the 8 is just the "count" so we have
15.16.192.0
15.16.200.0

so our first is 15.16.192.1 through 15.16.199.254 and .255 as broadcast

updated the post up top
• Banned Posts: 703
I'd just like to say getting subnetting to click is no easy challenge - for some it does, others it takes awhile.

It took me awhile especially with all the different methods out there. What I ended up doing was taking the methods out there and just taking the "best" approach from all of em (each method has to have something that makes sense) and just combined them.

I guess it also helps if you're working with it every day but just memorize it and you'll be good!
• Member Posts: 1,155 ■■■■■□□□□□
Get to know binary math and you'll figure out subnetting on your own

Cisco has a free binary math practice drill that plays out like a Tetris game so you can have some fun while you learn

The free Apple app is called Cisco Binary Game and you can also download it for free for PC on the Cisco site. They move it around a lot so I can't provide a link.

• Senior Member Mod Posts: 6,926 Mod
Netwurk wrote: »
Get to know binary math and you'll figure out subnetting on your own

Cisco has a free binary math practice drill that plays out like a Tetris game so you can have some fun while you learn

The free Apple app is called Cisco Binary Game and you can also download it for free for PC on the Cisco site. They move it around a lot so I can't provide a link.

Online Cisco Binary Game: https://learningnetwork.cisco.com/docs/DOC-1803
• Member Posts: 131
Devilsbane wrote: »

This is also pretty easy, you just need to remember a formula. (2^n)-2 where n is the number of bits used for the network address. (That is just a fancy way of saying the number of 1's in the subnet mask. 0's are used for the hosts)

The class B part is important because it means that 16 of your 1's are already accounted for. This leaves you with 8 left (255.255.255 is 24 1's and we subtracted the first 16). So your answer is 254.

Are there not 256 subnets available?
• Member Posts: 4,214 ■■■■■■■■□□
KenC wrote: »
Are there not 256 subnets available?

Depends who you ask. Modern routers can use all, but I generally still take the 2 away.
Decide what to be and go be it.
• Member Posts: 4,214 ■■■■■■■■□□
cyberguypr wrote: »

+1 for this. Definitely puts a fun spin on learning binary. I could only put up with 10 minutes though before I wanted to hurt small children.
Decide what to be and go be it.