Subnetting Made Easy

Moki99

Thank you guys. Techexams is full of great people.

demonfurbie

tagged for later

KenC

Fantastic explanation of subnetting by OP (including their reply to follow-up questions). You have helped a great many people with this work, myself included. Sincere thanks, Kenneth.

luisYme

very cool - thanks a bunch!
this video helped me out, too:
Intro to IP Addressing (Part I) | Boss CBT | Video Tutorials for Network Professionals - CCNA CCENT

Moki99

Okay heres a few things that are confusing me. I'm sure it is super simple but I'm an idiot so please excuse me.

Question: How many subnets and hosts per subnet can you get from the network 172.16.0.0 255.255.255.224?
Answer: 2048 subnets and 30 hosts

- How is it 2048 subnets? The subnet mask is in the 4th octet so shouldn't it be only 8 subnets? I understand that it's taking the 8 bits from the 3rd octet and 3 bits from the 4th octet making it 2^11 = 2048 but why is it doing that considering the way the subnet mask is setup? I'm obviously missing a huge step here.

Question: What valid host range is the IP address 172.28.248.229/28 a part of?
Answer: 172.28.248.225 through to 172.28.248.238

- I'm wondering how to do questions like this in my head. I understand that the increment is 16 given the /28 CIDR. Now how do I quickly figure out that .225 is the start of the network(first host)? I can't sit there and write out all the network ranges during the test. Sometimes for questions like this I'll do some multiplication in my head but doing multiples of 16 takes too much time. Is there a better way?

I hope I made sense, let me know if you need me to clarify what I'm asking. I appreciate the help here so much.

alxx

binary maths also becomes very easy if you can learn the powers of 2

converting from binary to decimal is easy

11010101
128 64 32 16 8 4 2 1
2⁷ + 2⁶ + 2⁵ + 2⁴ + 2³ +2² +2¹ + 2⁰
1 + 1 + 0 + 1 + 0 + 1 + 0 + 1
=128 +64 +16 + 4 +1
=213

alxx

Moki99 which method and or books are you using for subnetting ?

Write out a table that helps you before the exam starts (suggested in a few books and videos as well as by lots of people here)


Don't worry about doing them quickly.
Just concentrate on understanding it first.

Moki99

I use Jeremy's method from CBT Nuggets. I do it a lot quicker than he does it in the video because I don't need to write out all the binary anymore.

alxx

>Question: How many subnets and hosts per subnet can you get from the network >172.16.0.0 255.255.255.224?
>Answer: 2048 subnets and 30 hosts

172.16.0.0 gives 16 bits for network id and 16 bits for host id

So for hosts thats /27 so 256 - 224 = 32 - subnet address - broadcast address = 30

subnets 172.16.0.0 255.255.255.224

255.255.255.224 = /27
so
for number of subnets

27 - 16 (network id) = 11 2¹¹ = 2048

alxx

>Question: What valid host range is the IP address 172.28.248.229/28 a part of?
>Answer: 172.28.248.225 through to 172.28.248.238


/28 = 255.255.255.240

256 - 240 = 16 or 32 -28 = 4 2⁴

so address range is a block of 16 so host name range is a block of 14

so address(subnet) must be on multiple of 16 , the next boundary is 240
so 240 -16 = 224

172.28.248.224
172.28.248.240
172.28.248.256

so hostrange is 225 to 238

ciscoman2012

Moki99 wrote: »

Okay heres a few things that are confusing me. I'm sure it is super simple but I'm an idiot so please excuse me.

Question: How many subnets and hosts per subnet can you get from the network 172.16.0.0 255.255.255.224?
Answer: 2048 subnets and 30 hosts

- How is it 2048 subnets? The subnet mask is in the 4th octet so shouldn't it be only 8 subnets? I understand that it's taking the 8 bits from the 3rd octet and 3 bits from the 4th octet making it 2^11 = 2048 but why is it doing that considering the way the subnet mask is setup? I'm obviously missing a huge step here.

Question: What valid host range is the IP address 172.28.248.229/28 a part of?
Answer: 172.28.248.225 through to 172.28.248.238

- I'm wondering how to do questions like this in my head. I understand that the increment is 16 given the /28 CIDR. Now how do I quickly figure out that .225 is the start of the network(first host)? I can't sit there and write out all the network ranges during the test. Sometimes for questions like this I'll do some multiplication in my head but doing multiples of 16 takes too much time. Is there a better way?

I hope I made sense, let me know if you need me to clarify what I'm asking. I appreciate the help here so much.

Question 1:
For your first question, you are correct in realizing that you're missing a huge step. I think that you're forgetting that the first octet of the IP address is a "172", which means it is a Class B IP address. This gives you a default Class B Subnet Mask of 255.255.0.0. Contrast that with the /27 that they give you and you come out with the following:



Question 2:
Now for your second question, I can't help you do the problem in your head because I am not skilled enough for that, but I can help break down the steps that I take to achieve this answer.

Step 1: Write down the given IP address and Subnet mask in Binary where applicable. For example, I only convert the last byte of (229), in the IP address, to binary since the rest have 255 in the subnet mask. Same thing for converting the Subnet Mask, there is no need to write out all 1s for 255 since you already know that.

Step 2: Find the Network Subnet Number. To do this, look at the octet where it is not 255 in the Subnet mask, in this case that is the last byte of 229 in the IP address. Now, you will compare the IP address byte of 229 to the Subnet address byte of 240. To find the network number, change all the bits in the IP byte to "0" that do not have a 1 in the Subnet byte. So, in our case it would look like 11100000. When transferring that back to decimal it would convert to 224.

Step 3: Now that you've found the network number you need to find the broadcast number. Compare the IP address byte of 229 to the subnet byte of 240. Everytime you see a "0" in the subnet byte you will change it to a "1" in the IP byte. In this case we would have 11101111 as the IP byte. You need to convert that binary number back to decimal. You can either add all the 1s up, or you can subtract the 0 bits from 255 (which is how I do it to save time). So, 255- the one bit that isn't turned on (16) leaves us with 239. So 239 is the broadcast IP address.

Step 4: This is the easy part. Now just add one more than the Network number for the First IP address in the range, giving us ".255". Now, for the last IP address in the range just subtract one from the broadcast IP, which gives us ".238".



I really hope that this explanation helps you out some as I know that IP Subnetting can be quite confusing at first. It took me a few days of doing these problems for hours straight in order to get to a decent speed. I still can't do them in my head, but I normally use this route and can have them done in under 20 seconds.

Moki99

Thank you all so much for your help. I appreciate it.

Okay my first question is resolved. I get that now. I'm still having a hard time doing problems like my second one quickly even when using your method. I can do the "How many subnets and hosts per subnet can you get" kind of questions in 10 seconds. I guess I need to keep practicing. Thanks again.

ciscoman2012

Moki99 wrote: »

Thank you all so much for your help. I appreciate it.

Okay my first question is resolved. I get that now. I'm still having a hard time doing problems like my second one quickly even when using your method. I can do the "How many subnets and hosts per subnet can you get" kind of questions in 10 seconds. I guess I need to keep practicing. Thanks again.

Yup just keep practicing and it will get easier and easier.

Just remember, say you have a subnet of /18, which is 255.255.192.0.

Keep in mind that when you use the method I showed you, the last octet would simply be a "0" for the network number, "1" for the First IP address, "255" for the Broadcast #, and then "254" for the Last IP address in the range.

jabvasquez

How many subnets and hosts per subnet can you get from the network 10.0.0.0/20?
Answer: 4096 subnets and 4094 hosts


I am a beginner in Subnetting...could anybody help me how it arrived on that answers. I understood how it got the number of Hosts but not the number of subnets....Your help will be highly appreciated.

simonmoon

jabvasquez wrote: »

How many subnets and hosts per subnet can you get from the network 10.0.0.0/20?
Answer: 4096 subnets and 4094 hosts

10.0.0.0 is normally a Class A address which would be /8. That means you have 8 bits of network, 12 bits of subnet, 12 bits of hosts. To find the number of subnets is 2^subnet-bits. So it's 2^12=4096.

jabvasquez

simonmoon wrote: »

10.0.0.0 is normally a Class A address which would be /8. That means you have 8 bits of network, 12 bits of subnet, 12 bits of hosts. To find the number of subnets is 2^subnet-bits. So it's 2^12=4096.

Thank you so much for your informational explanation...I understood it very well... God Bless you Sir Simonmoon!

hzhalite1

I solved this question.
Question: How many subnets and hosts per subnet can you get from the network 172.30.0.0/27?
Answer: 2048 subnets and 30 hosts



While I did solve it, I have no idea how I would start doing 2048 subnet combinations with 30 hosts inbetween. Could someone help me please?

Excellent1

hzhalite1 wrote: »

I solved this question.
Question: How many subnets and hosts per subnet can you get from the network 172.30.0.0/27?
Answer: 2048 subnets and 30 hosts



While I did solve it, I have no idea how I would start doing 2048 subnet combinations with 30 hosts inbetween. Could someone help me please?

Your first subnet would be:
172.30.0.0: Hosts 172.30.0.1 through 172.30.0.30 (172.30.0.0 is the subnet address, 172.30.0.31 is the broadcast address)
2nd subnet would be:
172.30.0.32: Hosts 172.30.0.33 through 172.30.0.62 (172.30.0.32 is the subnet address, 172.30.0.63 is the broadcast address)

This would continue all the way through the last subnet:
172.30.255.224: Hosts would be 172.30.255.225 through 172.30.255.254 (172.30.255.224 is the subnet address, 172.30.255.255 is the broadcast address)

Todd Burrell

The quick way to get the number of subnets and hosts from 172.30.0.0/27 is this:

172.30 is a class B address, so the default subnet mask is /16. Given /27, that gives 11 bits for the subnet mask (27-16), and 2^11=2048 subnets.

The number of hosts is simply 2 raised to the power of 32-27=5. So 2^5=32 - then take away 2 addresses for the network and the broadcast address and you get 30.

So 172.30.0.0/27 gives 2048 subnets with 30 host addresses per subnet...

I'd recommend working as many examples as possible for this type of subnet problem - and if you have the Odom books they really explain it well as does the Lammle book.

hzhalite1

Thank you both for your answers, you've been very helpful.

whoeverBH

Could someone please help me get my head around this!!

I've got an IP of 192.10.10.0

I need 14 usable subnets
and 14 usable hosts persubnetwork


1/ This is Class C isnt it?

2/ The default subnet mask is 255.255.255.0?

3/ How many bits do you borrow for the subnets?

4/ What is the new subnet mask for this host in dotted decimal format

5/ What is the new subnet mask for this host in IP Prefix format

6/ How many subnets were created

7/ How many hosts per subnet?

8/ How many usable hosts per subnet?

I would be so Appreciative as i am so confused

Trifidw

whoeverBH wrote: »

Could someone please help me get my head around this!!

I've got an IP of 192.10.10.0

I need 14 usable subnets
and 14 usable hosts persubnetwork


1/ This is Class C isnt it? Yes.

2/ The default subnet mask is 255.255.255.0? Yes.

3/ How many bits do you borrow for the subnets? 256/14 = 18.blah so to split it up into 14 subnets you will need ranges of 16 addresses. So you need to borrow 4 bits (1 bit is 128, 2 bits is 64, 3 bits is 32...)

4/ What is the new subnet mask for this host in dotted decimal format 255.255.255.240
128+64+32+16=240

5/ What is the new subnet mask for this host in IP Prefix format /28
Class C default is 24, +4=28

6/ How many subnets were created? 16
256/16=16

7/ How many hosts per subnet? 16
working out in q3

8/ How many usable hosts per subnet? 14
Total minus network and broadcast address. Remember if you want to be able to route out of this you will need at least 1 IP address for the routed interface/vlan.

I would be so Appreciative as i am so confused

I done these in my head and haven't looked at subnetting without a calculator for 4 years so someone may want to double check them...

Edit:added some explanations for learning purposes.

drkat

192.10.10.0/24

1) this is a class C
2) yes /24
3) We need to see how many hosts we need - so we need 14 so we use a 240 mask
4) 255.255.255.240
5)/28
6) 16
7) 16
14

THE WORK:

192.10.10.0
255.255.255.0

256-14 = 242 so we use 255.255.255.240

256 - 240 = 16 hosts -2 for usable = 14

binary table:
128 64 32 16 8 4 2 1
1 1 1 1 0 0 0 0

this shows us borrowing 4 bits 2^4 = 16
shows 4 zeros which is our hosts 2^4 = 16

MAC_Addy

Can someone help me out on a way to figure this out easily?

I got this from subnetting questions - What is the first valid host on the subnetwork that the node 172.19.231.239/22 belongs to? I know the answer is 172.19.228.1 255.255.252.0

But the thing is, I know how to work it out, just not very fast. I had to write down on my pad starting from 4, 8, 16 and so on. Is there a quicker way of finding the answer? Subnetting has clicked in my head, i just can't do it very fast.

ciscoman2012

MAC_Addy wrote: »

Can someone help me out on a way to figure this out easily?

I got this from subnetting questions - What is the first valid host on the subnetwork that the node 172.19.231.239/22 belongs to? I know the answer is 172.19.228.1 255.255.252.0

But the thing is, I know how to work it out, just not very fast. I had to write down on my pad starting from 4, 8, 16 and so on. Is there a quicker way of finding the answer? Subnetting has clicked in my head, i just can't do it very fast.

I attached a picture of how I worked it out. Sorry I didn't really explain much in the picture but it was how I worked out the problem. For this question, you need to convert the IP address given to you as well as the subnet mask to binary. I've done that for you in the first two lines that say IP for IP address and Subnet for the subnet mask.

As you can see, the interesting octet that you want to focus on is the one where the subnet isn't all 1s, or octet number 3. Next, to find out the Network that the IP address is located in you will want to change the bits in the IP address octet 3 to "0" that appear when the subnet mask is not a "1".

If you do this, you will realize that the third octet looks like 11100100. That will be your network number along with the rest of your IP address. So 172.19.228.0.

After that it's easy, just go up one number in the fourth octet to find your first valid IP address.

MAC_Addy

Finally, I get it. Where'd you learn that from?

ciscoman2012

MAC_Addy wrote: »

Finally, I get it. Where'd you like that from?

What do you mean?

MAC_Addy

Dang it. I meant, learn, not like. Silly me.

ciscoman2012

MAC_Addy wrote: »

Dang it. I meant, learn, not like. Silly me.

Honestly, I'm not sure. I watched some CBT Nugget videos along with Train Signal / Chris Bryant and then a few people on Youtube and figured that is the way I like to do it best. Even though I work out each problem it seems to work pretty well. I don't trust myself doing math in my head haha.

drowningfish84

After using this method, how do I find the "number of bits borrowed", "number of hosts per network"?