Subnetting Made Easy

NetworkNewb

Ty for bumping that Deathmage! got my exam in a few days and still needed to work on my subnetting. I really like this explanation of it to!!

Definitely should be stickied

Deathmage

Been using this workbook the past 24 hours and I'm half way thru it and it's really helping me breeze through subnetting; I learned subnetting before I switched gears onto the VCP5-DCV track and lost touch with it but from my superb notes and this work book that I noted in my 5-star massive 6 subject notebook to do in-case I forgot (and see I knew I would) I'm picking it up again.. Right now I can do one in about 45 seconds; there is still 45 pages left of worksheets so maybe I'll get better at it.

Here is a link: Let me google that for you

Deathmage

Has anyone found a way to remember subnetting for class A?

See I can subnet Class B and C fine in binary but once I get to Class A remembering those large A** numbers is daunting to say the least. Curious if the CCENT/CCNA will have Class A's on it or if it's really just tons and tons of practice to just remember them?

The past 3 weeks I've been doing on average 20 or so subnetting practices a day (when I have down-time at work) between my workbook above and subnetting.net; and watch every subnetting video on youtube.

Jon_Cisco

I don't expect you will need anything past a /22.
If you do then you can revert to the math quickly on the whiteboard they give you. If you are familiar with binary up to like 4096 (/20) I don't think you will have any problems.

Deathmage

Jon_Cisco wrote: »

I don't expect you will need anything past a /22.
If you do then you can revert to the math quickly on the whiteboard they give you. If you are familiar with binary up to like 4096 (/20) I don't think you will have any problems.

Thanks Jon, I can do Class A but with a calculator. I've been just doing Class B and C's and really honing them. I mean If I had to do a Class A I could do it but with long math, just it's tedious to remember those numbers and I suck with multiplication. I haven't used math is like 6 years since college....

Deathmage

making sure I got this Class A address right:

number of needed hosts: 29
Network Address: 23.0.0.0

so.....

Address Class: A
Default Mask: 255.0.0.0
Custom Mask: 255.255.255.224
Total number of Subnets: 524,288
Total number of Host addresses: 32
Total usable addresses: 30
Number of bits borrowed: 19

Does that look right; I can remember up to 65,538 but need a calculator to go any higher. Really hope the exam doesn't have anything higher than 65,538.

I go with the

(Left to Right)
Subnets: X. 2 4 8 16 32 64 128 256 . 512 1024 2048 4096 8192 16384 32,768 65,536 . 131,072 262,144 524,288 1,048,576 (you all know the rest)

(Right to Left)
Hosts: X. (the big number for 8 blocks) .65,536 32,768 16,384 8,192 4096 2048 1025 512 . 256 128 64 32 16 8 4 2 1

now since I only need up to 32 hosts (32-2 would equal 30 and I need 29) and that's 3 bits into the 4th octet that would be 128+64+32=224 as the custom mask, so it would be 255.255.255.224.

been subnetting with this workbook day and night for weeks and I think it finally clicked. now I just reprinted it and going to do it again and again until I get really fast at it...

Maybe I'll try the power of 2 method but I got binary in my head right now, is binary good enough for the exam or is it still too long?

been finding that 5 to 10 minutes at work doing a few subnetting problems really helps with keep the skillz in my head; i feel like I'm dreaming binary anymore...


just to add to this here is a Class B:

number of needed subnets: 2000
number of needed hosts: 15
Network Address: 178.100.0.0

so.....

Address Class: B
Default Mask: 255.255.0.0
Custom Mask: 255.255.224.0
Total number of Subnets: 2048
Total number of Host addresses: 32
Total usable addresses: 30
Number of bits borrowed: 11

am I on the right track?

been using this: Let me google that for you

karthikheb

Hi guys,

I understood how to do the example originally posted and went through most of the examples posted by others , but i could not find answer to this type of question and this is the tricky ones that I am not able to get my head around
What is the broadcast address of the network 172.28.74.192 255.255.255.192?
Answer: 172.28.74.255

What i did was took the block size as 64

and my subnets were

172.28.0.0
172.28.64.0
172.28.128.0
so going by this logic the address specified in the question belongs to 172.28.64.0 and the broadcast address will be 172.28.127.255 .

But the answer says its 172.28.74.255 where the 3rd octet remained unchanged. But if it is a Class B Address (starting with 172) then should the 3rd octet change ?

Phara0h

You are designing a subnet mask for the 172.16.0.0 network. You want 120 subnets with up to 300 hosts on each subnet. What subnet mask should you use?


Answer: 255.255.254.0

How you do work out the subnet mask in this iteration.

I know for 300 hosts you will be taking 9bits ( or on the 2*9 = 512 ).
I know that its a Class B address ( by default first two octets are network )

But cannot make the translation to equate- 254 in the third octet

aloch

There are 7 bits set in the third octet, adding them together results in 254.

Here's a **** sheet I made up: Subnet **** Sheet

Phara0h

aloch wrote: »

There are 7 bits set in the third octet, adding them together results in 254.

Here's a **** sheet I made up: Subnet **** Sheet

Thank you for your sheet but am not clear on how your table translates. I know that by default that a CLass B would be a /16. With getting 300 hosts this would now become a /25 ( Am thinkinh... not sure ) BUT how do you take /25 to make up the netmask- 255.255.254.0


Edit. No answers required any more. Worked it out, albeit using another method. Throwing into questions now to get it all nailed down.

Thanks

chanrong

hi , can someone calculate step by step this ip 10.0.0.0/8 . i need only 100000 host.

please help me

Thanks

Chev Chellios

Dude there is plenty of material here to help with that, give it a go and learn from the process. We have all had to do it the hardway!

nster

Phara0h wrote: »

Thank you for your sheet but am not clear on how your table translates. I know that by default that a CLass B would be a /16. With getting 300 hosts this would now become a /25 ( Am thinkinh... not sure ) BUT how do you take /25 to make up the netmask- 255.255.254.0


Edit. No answers required any more. Worked it out, albeit using another method. Throwing into questions now to get it all nailed down.

Thanks

For anyone reading this, where he went wrong is that you add subnets to the subnet mask or subtract host from /32, so 32 - 9 = /23 OR /16 + 7 bits of subnets (128 subnets) = /23

Deathmage wrote: »

making sure I got this Class A address right:

number of needed hosts: 29
Network Address: 23.0.0.0

so.....

Address Class: A
Default Mask: 255.0.0.0
Custom Mask: 255.255.255.224
Total number of Subnets: 524,288
Total number of Host addresses: 32
Total usable addresses: 30
Number of bits borrowed: 19

Does that look right; I can remember up to 65,538 but need a calculator to go any higher. Really hope the exam doesn't have anything higher than 65,538.

Again for anyone reading... I have a few power of two "anchor points" if you will, and think of them as MegaBytes for thinking.

So
2^3 = 8MB
2^6 = 64MB
2^10 = 1GB (1024)

For 2^ 10, 11, 12, 13, it's really easy, it's 1024, 1024*2, *4, or *8

For mental math, this is how I do the bigger numbers, it seems to work best for my mind, though I'll switch it up with different techniques at times. Useful for interviews I guess? I tend to do mental over calculator by habit

For 14 to 23, you do -10 to the exponent. Let's call it x. We want to multiply it by 1024 (2^10). x*1000 + 2x*10 + 2*2x

Let's take 4 examples, 14 16 19 and 23:

For 2^14:
2^4 = 16, so 16,000 + 320 + 64 = 16,384

for 2^16
2^6 = 64, so 64,000 + 1,280 + 256 = 65,536

for 2^19
2^9 = 512, so 512,000 + 10,240 + 2048 = 524,288

and for the hardest, 2^23
2^13 = 8,192 so 8,192,000 + 163,840 + 32,768

I usually calculate it backwards. So say for 2^15, my thought process is, 2^5 = 32, so 32,64,128. 128+640=768, +32000 = 32,768. For 2^21, 2^11 = 2048. 2048;4096;8192. 8,192+ 40,960 = 49,152 + 2,048,000 = 2,096,000 + 1,152 = 2,097,152

Blang008

Thanks for this! This has made my life so much easier. I do have a question though. Is there an easier way to figure out the type of questions below other than just counting up in your block size???

Question: How many subnets and hosts per subnet can you get from the network 172.29.0.0/21?


Answer: 32 subnets and 2046 hosts

madafakafaka

Hello, can someone done this task for me?
An ISP is granted a block of addresses starting with 158.78.0.0/16. The ISP wants to distribute
these blocks to 3200 customers as follows.
a. The first group has 400 medium-size businesses; each needs 64 addresses.
b. The second group has 600 small businesses; each needs 16 addresses.
c. The third group has 80 households; each needs 256 addresses.
Design the subblocks and give the slash notation for each subblock. Find out how many
addresses are still available after these allocations

toasterboy1

Oh my! This made things so easy. Been away from networking for a while and this made it seem like a walk in the park. Thanks!

raruku

Hi all, can you please check if below answers are correct, as I have been given below for next coming interview questions.
Thanks in advance for the assistance.

1. Enter the last valid host on the network 192.168.92.0 255.255.255.224?
Increment 32, so the last valid host is 192.168.92.30

2. What is the maximum number of valid subnets one will have from the network 10.29.252.144/13? Assume this is a class A address.
2^5 = 32 maximum number of valid subnets, since borrowing 5 bits

3. What is the most efficent subnet if you need 110686 usable hosts on a subnet? Present your answer as a subnet mask?
2^17 - 2 = 131070. So the subnet mask will be 255.254.0.0, as we need to have 17 zero bits for the host.

4. What is the broadcast address of the network 10.121.108.13/14?
Network ID: 10.120.0.0, 10.124.0.0
So, the broadcast is 10.123.255.255

5.What is the maximum number of valid hosts one will have from the network 192.168.124.0 255.255.255.224?
5 bits zero for the host, so the answer is 30

6. What is the number of network IDs in a Class C network? 3, because of 255.255.255.0

dforeman12

Wow, so glad I came across this post from 2008. Brilliant.

koolaid002

Lordflasheart, I don't know if you came up with this on your own? But it's amazing. I'm not as advanced as some (I'm working on my net+ cert) but this is very helpful and getting the basics down. Thanks, I really appreciate it!

MAC_Addy

Can we vote to make this a sticky?

Josh77

Question: What is the first valid host on the subnetwork that the node 172.18.186.50/23 belongs to?

MAC_Addy

Josh77 wrote: »

Question: What is the first valid host on the subnetwork that the node 172.18.186.50/23 belongs to?

I'm confused a little - you've asked the question that is clearly copied and from a website, on a forum that is about technology. AND on a subnetting thread that TEACHES you how to subnet. This one is easy to answer. How about you have a little stab at it. If you get it wrong, we can help you.

maketwinkiesgreat

Can't thank you enough for this. It's cleared up so much confusion. Even the cisco netacad class I'm in at the community college didn't teach it this well.

mani123

kindly explain how to subnet if 1000 hosts are required. ip adress is 150.150.0.0/16. please explain. im doing it for your method but its not done

mani123

kindly explain how to subnet if 1000 hosts are required. ip adress is 150.150.0.0/16. please explain. im doing it for your method but its not done

Brada130

can someone help with this:


1) You’ve been given the Subnet ID 192.168.1.0/24, and are asked to divide this into 5 subnets of equal size. What are the network addresses, broadcast addresses, and ranges for each of these subnets?

2) You are given a network of 10.50.24.0/21, which contains 2,048 addresses. What subnet mask should you use to divide this into four subnets of 512 addresses each?

3) You are given the network 172.16.14.0/23, and asked to create subnets to meet the following requirements:

a. 1 LAN that can accommodate 30 hosts
b. 1 LAN that can accommodate 50 hosts
c. 2 LANs that can accommodate 100 hosts each
d. 2 Point-to-point networks

albinorhino187

For mani123:

So you need 1000 hosts, and you're starting with 150.150.0.0/16. This is a pretty common type of question where one will say you need X # of networks, or Y # of hosts. Since this one is asking for hosts, we'll approach it that way.

The first thing I'd do is start counting in powers of two until you reach the requirement. That will tell you how many bits you need for your hosts. Be careful, though, because when you're counting for hosts, you must subtract two for your broadcast and network IDs. So 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024. That's 10 bits. 1024 - 2 is 1022, which is still enough to support our 1000 needed.

So we now know we need 10 bits to support our host requirement. We're working with 150.150.0.0/16, which means we have only 16 bits to play with. We need 10 of those 16 for our hosts, so that leaves 6 bits for our network.

Add those 6 bits to the /16, and you get a /22. So that's what our subnets will use to support 1000 hosts each.

albinorhino187

Brada130:

Question 1 seems a little tricky, but remember you're only sizing your networks based on powers of two. So you start with your original, 192.168.1.0/24. That's one network. Start counting your powers of 2 for bits to determine how many networks you can make with the 8 remaining bits which also satisfy the number of subnets required. You need 5 subnets, but counting our bits gives us 2, 4, 8.

So the only way to divide your original network into 5 equal parts is really to divide it into 8 equal parts and only use 5 of them. It took us 3 bits to get to that 8, so add that to your /24 for a /27, or 255.255.255.224 for the mask. That leaves 32 numbers for each network, so your networks would be 192.168.1.0 - 31, 192.168.1.32 - 63, 192.168.1.64 - 95, 192.168.1.96 - 127, and 192.168.1.128 - 159. There's still 3 other networks left in the original /24, but we're not using those because we only needed 5.

Question 2 is pretty simple. You have a /21, which means you have 11 bits left to work with. Just start counting powers of 2 until you hit your requirement. 2, 4, 8, 16, 32, 64, 128, 256, 512. So that's 9 bits we need for our hosts. We had 11 to work with, we need 9 for hosts, so we have 2 left for the subnet. Add that 2 to your original /21, so you'll use a /23. /23 is 255.255.254.0.

Question 3, you should always start out accommodating your largest networks first. So first you need networks for 100 hosts. Count out your bits, remembering for hosts you need to subtract 2. 2, 4, 8, 16, 32, 64, 128. 7 bits. You start with a /23, that leaves 9 bits to play with. Our first two networks will need 7 bits for the hosts, leaving 2 bits for the network. Add those 2 to the /23 for a /25, which is a mask of 225.255.255.128.

So, we take 172.16.14.0/25 as our first subnet, and 172.16.14.128 /25 as the second. That satisfies part c.

Now, we need a 50 host network. Count the bits. 2, 4, 8, 16, 32, 64. 6 bits. Now we're working with the /25 we left off from with part c, not the original /23. We only have 7 bits to play with, we need 6 for our hosts, so we'll add that remaining one to the /25 for a /26. We're also starting where the last network ended, which was 172.16.14.255.

So, this 50 host network will begin 172.16.15.0/26. That satisfies part b.

Now for the 30 host network. Count the bits. 2, 4, 8, 16, 32. 5 bits is just enough, because 32-2 is 30. We're starting with the previous /26, which leaves 6 bits total, we need 5 for our hosts, so add the remaining one to the /26 for the network and that gives /27. 255.255.255.224.

Starting where we left off would be 172.16.15.64/27. That leaves 172.16.15.96 /27 as the start of our first point to point. We need 2 hosts on our point-to-point subnets. 2, 4. 2 bits, and 4-2 is 2, so that's exactly enough. We had 5 bits to play with, we need 2 bits for our hosts, so that leaves 3 remaining network bits, which is a /30 (mask 255.255.255.252) when we add those three to our /27.

So, our first point to point network will be 172.16.15.96 /30, and that would make the second one 172.16.15.100 /30.

Technically, this is a little sloppy, because you'll end up with a lot of leftover IP space, and ideally you'd chunk it out better with future needs in mind, but this still gets the job done within the given network.

Harry Roles

Check out this free video course :

https://www.udemy.com/ipv4-and-ipv6-addressing-and-subnetting/

Slinkyslinkyslinky

Signed up just to say thank you for the advice LordFlasHeart. Never been able to subnet, find host ranges etc before and after about an hour of using your technique I can practically do it in my head.

Thanks again