Frustration with subnetting

veritas_libertas

Other than subnetting I think can be ready for the CCENT in another month. What is driving me crazy is what I consider overlapping Octets and I'm sure I'm going to be criticized for using that term since it's less than accurate or scientific

It's the only way I can think of to describe it!!!

For instance lets say we have this from Subnettingquestions.com:

Question: What is the broadcast address of the network?

172.19.210.0 255.255.254.0

Answer: 172.19.211.255




Now if the question had been something like:



Question: What valid host range is the IP address 192.168.106.136 255.255.255.224 a part of?


or any standard Class mask it doesn't throw me off my game at all.


Any help is greatly appreciated.

impz

Question: What valid host range is the IP address 192.168.106.136 255.255.255.224 a part of?

256-224=32
32*5=160 but that wouldnt work cuz its over 136 but 32*4 is 128
so subnet number will be 192.168.106.128
broadcast is (128+32)-1 = 159

So, first valid host is 192.168.106.129, last valid host would be 192.168.106.158

veritas_libertas

impz wrote: »

Question: What valid host range is the IP address 192.168.106.136 255.255.255.224 a part of?

256-224=32
32*5=160 but that wouldnt work cuz its over 136 but 32*4 is 128
so subnet number will be 192.168.106.128
broadcast is (128+32)-1 = 159

So, first valid host is 192.168.106.129, last valid host would be 192.168.106.158

That is not really my struggle. What would throw me off is something like:

192.168.106.136 255.255.240.0

What valid host range is the above IP address a part of? Now I realize I could covert the third octets to binary and then just do the math, but that is time intensive. Is there any shortcut to this?

impz

veritas_libertas wrote: »

That is not really my struggle. What would throw me off is something like:

192.168.106.136 255.255.240.0

What valid host range is the above IP address a part of? Now I realize I could covert the third octets to binary and then just do the math, but that is time intensive. Is there any shortcut to this?

256-240=16
subnet no. would be 192.168.96.0
so first valid host address is 192.168.96.1

broadcast 192.168.[(96+16)-1].255 -->192.168.111.255
so last valid address would be 192.168.111.254

veritas_libertas

impz wrote: »

256-240=16
subnet no. would be 192.168.96.0
so first valid host address is 192.168.96.1

broadcast 192.168.[(96+16)-1].255 -->192.168.111.255
so last valid address would be 192.168.111.254

How did you get that? That is where I struggle, and I'm even more confused with how you got the last valid address.

miller811

veritas_libertas wrote: »

That is not really my struggle. What would throw me off is something like:

192.168.106.136 255.255.240.0

What valid host range is the above IP address a part of? Now I realize I could covert the third octets to binary and then just do the math, but that is time intensive. Is there any shortcut to this?

find the interesting octet - the non 255 one

which in this case is x.x.240.0
so you know the answer will be 192.168.x.x
256 - (the interesting octet) 240 = 16 in this case is the block size

so you have
192.168.0.0
192.168.16.0
192.168.32.0
192.168.48.0
192.168.64.0
192.168.80.0
192.168.96.0
192.168.112.0
192.168.128.0
192.168.144.0
etc....

so your address is in the 192.168.96.0 subnet

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

1. so 192.168.96.1 in the first available host
2. 192.168.111.255 would be the broadcast address
3. 192.168.111.254 would be the last available host address


you can apply this technique to any similar question.

impz

192.168.106.136 255.255.240.0

What valid host range is the above IP address a part of?

kk. time to explain a bit more in depth.

192.168.106.136
255.255.240.0

So any octet which corresponds with 255 will just be repeated back :

192.168.x.x

Now to find the subnet number it's simple.

240 would be the "interesting octet"
Determine the magic number (256-interesting octet); 256 - 240 = 16
16*6=96 so you will end up with

192.168.96.x

Since the last octet in the subnet mask is 0 then the last octet on the subnet number will also be zero.

Therefore, subnet number = 192.168.96.0

Now to find the broadcast address, you need to add the magic number to 96 and minus one.

(96+16)-1=111

so we end up with this : 192.168.111.x

Now the subnet mask is 255.255.240.0 and because the last octet in the subnet mask is 0 then for the last octet of the broadcast address it becomes 255.

Therefore, broadcast address would be 192.168.111.255

The first valid host would be 192.168.96.1 (the first ip address after the subnet number) and the last would be 192.168.111.254 (1 ip address before the broadcast address)


Similarly, 192.170.106.136 255.252.0.0

First off, 192.x.x.x

Interesting Octet: 252
Magic number: 256-252=4
4*42=168

Then 192.168.x.x

The rest of the subnet mask is 0 so just fill in the rest of the ip address with 0 to get the subnet address or whatever you wanna call it.

Subnet number = 192.168.0.0

Now in the case of a broadcast address, the second octet would be [168 + magic number(4) ]- 1 =171

So we get 192.171.x.x

Now because the subnet mast have the last 2 octets as 0, the last 2 octets of the broadcast address would be 255. (which basically is the opposite of what we did for the subnet number)

Broadcast address = 192.171.255.255

First valid host : First ip address after the subnet number : 192.168.0.1
Last valid host : One ip address before the broadcast address : 192.171.255.254

I had the same problem when I first did subnetting but after rereading Odom's chapter on subnetting the the "subnetting made easy thread" I got the gist of it.

Technology1

The MASK OCTET is octet 3, which is 240. This question is already giving you the subnet mask as 255.255.240.0

Taking 240 and subtracting it from 256 gives you the Network Increment number of 16. The Networks are in increments of 16, in the 3rd octet. To be able to determine the host range, you first need to know what the Network ID and the Broadcast ID of this network is. Anything in between the Network ID and the Broadcast ID is the IP host range.

192.168.96.0 Network ID
192.168.96.1 First Valid Host
192.168.111.254 Last Valid Host
192.168.111.255 Broadcast ID (always 1 less than the next subnet)

192.168.112.0 (Next Subnet)
192.168.128.0 (Next Subnet)
192.168.144.0 (Next Subnet)
192.168.160.0 (Next Subnet)
192.168.176.0 (Next Subnet)
192.168.192.0 (Next Subnet)

The networks are incrementing by 16, in the 3rd octet.

veritas_libertas wrote: »

That is not really my struggle. What would throw me off is something like:

192.168.106.136 255.255.240.0

What valid host range is the above IP address a part of? Now I realize I could covert the third octets to binary and then just do the math, but that is time intensive. Is there any shortcut to this?

veritas_libertas

Argh, back to subnetting...

My newborn daughter has started sleeping more, and I have started drinking more coffee. It seems that the combination has allowed me to get back to my studies

wbosher

Check out CBT nuggets, that is by far the easiest way of doing subnetting I've come across. After a bit of practice, you'll learn some shortcuts without going into all the binary stuff.

It's pretty daunting at first, but when the penny drops it becomes second nature. Don't worry, you'll get there.

j-man

^^^^^

What wbosher said

Keep plugging away. It took about three weeks for me until the "eureka" moment came.

Odom's ICND1 book teaches both binary and decimal ways of tackling subnetting. When I finished that chapter, I was sure that binary was the way to go and I'll never look at decimal again. I was wrong. After learning the binary, decimal subnetting and all the shortcuts made sense.

Check out Odom's CCNA video mentor available from the cisco learning network. His subnetting really connected the dots for me.

veritas_libertas

impz wrote: »

192.168.106.136 255.255.240.0

What valid host range is the above IP address a part of?

kk. time to explain a bit more in depth.

192.168.106.136
255.255.240.0

So any octet which corresponds with 255 will just be repeated back :

192.168.x.x

Now to find the subnet number it's simple.

240 would be the "interesting octet"
Determine the magic number (256-interesting octet); 256 - 240 = 16
16*6=96 so you will end up with

192.168.96.x

Since the last octet in the subnet mask is 0 then the last octet on the subnet number will also be zero.

Therefore, subnet number = 192.168.96.0

Now to find the broadcast address, you need to add the magic number to 96 and minus one.

(96+16)-1=111

so we end up with this : 192.168.111.x

Now the subnet mask is 255.255.240.0 and because the last octet in the subnet mask is 0 then for the last octet of the broadcast address it becomes 255.

Therefore, broadcast address would be 192.168.111.255

The first valid host would be 192.168.96.1 (the first ip address after the subnet number) and the last would be 192.168.111.254 (1 ip address before the broadcast address)


Similarly, 192.170.106.136 255.252.0.0

First off, 192.x.x.x

Interesting Octet: 252
Magic number: 256-252=4
4*42=168

Then 192.168.x.x

The rest of the subnet mask is 0 so just fill in the rest of the ip address with 0 to get the subnet address or whatever you wanna call it.

Subnet number = 192.168.0.0

Now in the case of a broadcast address, the second octet would be [168 + magic number(4) ]- 1 =171

So we get 192.171.x.x

Now because the subnet mast have the last 2 octets as 0, the last 2 octets of the broadcast address would be 255. (which basically is the opposite of what we did for the subnet number)

Broadcast address = 192.171.255.255

First valid host : First ip address after the subnet number : 192.168.0.1
Last valid host : One ip address before the broadcast address : 192.171.255.254

I had the same problem when I first did subnetting but after rereading Odom's chapter on subnetting the the "subnetting made easy thread" I got the gist of it.

Why do you multiply it by six?

jdfriesen

veritas_libertas wrote: »

Why do you multiply it by six?

You're trying to find the multiple of 16 that is less than or equal to 106 without going over. It's very useful to just be able to count by multiples of 2,4,8,16,32,64. Once you can do that, it becomes pretty obvious what the starting address will be.

veritas_libertas

jdfriesen wrote: »

You're trying to find the multiple of 16 that is less than or equal to 106 without going over. It's very useful to just be able to count by multiples of 2,4,8,16,32,64. Once you can do that, it becomes pretty obvious what the starting address will be.

ahhhhhhh, now I understand the reasoning behind that. Thanks

okplaya

veritas_libertas wrote: »

Why do you multiply it by six?

Well the original question asks to find the valid host range for 192.168.106.136 255.255.240.0.

Since the "interesting octet" is in the third octet, we know that octet of the IP address is the number we are interested in. That number is 106 (x.x.106.x) right? So that means we need to figure out which network that would be in based of the magic number of 16 - which gives us our block size for our subnets.

Now you could count in your head 0,16,32,48,64, etc. But that takes too long. So if you can multiply fairly well, you sort of just guesstimate as close as you can. Multiplying by 6 is not a requirement by any means, it is purely his/her way of getting the answer. Some people add, multiply, divide, heck some people have every multiple of 16, 32, 64 just memorized. It really is personal preference.

16 x 1 = 16 (no that's way too low go higher)
16 x 4 = 64 (hmmm I'm getting closer but need to go a little higher)
16 x 6 = 96 (192.168.96.0) <--sounds good what would be next?
16 x 7 = 112 (192.168.112.0) <--- Ah 112 is higher than 106, which is the number in my interesting octet so that's not right. Ok, I have my answer.

Note: The above can be done mentally in seconds once you're confident and have the methods down-packed. That was my lame attempt at re-enacting the thought process some people may do.

From there you can pretty much tell that 192.168.106.136 will have to be in the 192.168.96.0 network.

Jas21

wbosher wrote: »

Check out CBT nuggets, that is by far the easiest way of doing subnetting I've come across. After a bit of practice, you'll learn some shortcuts without going into all the binary stuff.

It's pretty daunting at first, but when the penny drops it becomes second nature. Don't worry, you'll get there.

Amen - it WILL click, and when it does you'll jump from your seat and shout "Eureka" - well, I did anyway!!

Keep at it

AvidNetworker

Well with the first question, my head would say this.

255.255.254.0 is 256*2 as far as finding hosts. So for each 256 hosts, I have to add to the 3rd octet. So since there is 256*2, I know that I need to increment by 2 in the 3rd octet. So I am covered with the subnet .0 and .1. Since I know the broadcast is the last in the subnet, then I know it is going to be .1.255. I honestly just kinda do it all in my head at this point. There wasn't a single time on the CCNA exam that I needed the provided pen and paper. I think the key to subnetting is thinking visually instead of solely concentrating on math and numbers. Visualize that address and subnet mask in your head.

veritas_libertas

Now that I've started studying Subnetting again I'm hitting a wall. I'm using Todd Lammle's book and he goes in your head. Unfortunately I'm not convinced that he goes into it deeply enough. Let's say you are given the following:

How many hosts do you have available with an IP of 192.168.0.0 with a subnet of 255.252.0.0. Now I know by memorizing this list that this means we have eighteen bits turned off:

Binary Decimal

---

00000000 = 0
10000000 = 128
11000000 = 192
11100000 = 224
11110000 = 240
11111000 = 248
11111100 = 252
11111110 = 254
11111111 = 255

That means the answer can be achieved by 2^18 = 262144. As I understand it we don't get access to the Windows calculator. That kind of makes me wonder how on earth I'm suppose to do that conversion in my head

ChickenNuggetz

veritas_libertas wrote: »

Now that I've started studying Subnetting again I'm hitting a wall. I'm using Todd Lammle's book and he goes in your head. Unfortunately I'm not convinced that he goes into it deeply enough. Let's say you are given the following:

How many hosts do you have available with an IP of 192.168.0.0 with a subnet of 255.252.0.0. Now I know by memorizing this list that this means we have eighteen bits turned off:

Binary Decimal

---

00000000 = 0
10000000 = 128
11000000 = 192
11100000 = 224
11110000 = 240
11111000 = 248
11111100 = 252
11111110 = 254
11111111 = 255

That means the answer can be achieved by 2^18 = 262144. As I understand it we don't get access to the Windows calculator. That kind of makes me wonder how on earth I'm suppose to do that conversion in my head

Heh, I was actually feeling the same way not too long ago. I was terrifried that I would get some subnetting question with some ridiculous number that I would have to "convert" in my head. The good news is, the subnetting you'll do on the exam, will most likely be using manageable numbers. Its still good to know your powers of 2 beyond 128, but I dont think you'll need as high as 262144!

This is what I've always remembered:

8192 4096 2048 1024 512 256 128 64 32 16 8 4 2 1


I've never had to do any subnetting that gave numbers above 8192. The Cisco exams are pretty good about giving you "real world" scenarios where you might be dealing with hundreds of hosts, not really hundreds of thousands!

joshmadakor

miller811 wrote: »

find the interesting octet - the non 255 one

which in this case is x.x.240.0
so you know the answer will be 192.168.x.x
256 - (the interesting octet) 240 = 16 in this case is the block size

so you have
192.168.0.0
192.168.16.0
192.168.32.0
192.168.48.0
192.168.64.0
192.168.80.0
192.168.96.0
192.168.112.0
192.168.128.0
192.168.144.0
etc....

so your address is in the 192.168.96.0 subnet

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

1. so 192.168.96.1 in the first available host
2. 192.168.111.255 would be the broadcast address
3. 192.168.111.254 would be the last available host address


you can apply this technique to any similar question.

This just made things click for me. Thank you so much sir. Rep++

veritas_libertas

Glad my own frustrations helped someone else out

Thanks again to everyone who contributed to this thread.

veritas_libertas

Okay, so here is one that has me baffled:



I know that we have multiples of 16 because 256 - 240 = 16. So I would think the last subnet would be 10.188.240.254.

I would think it would be:

16
32
48
64
80
96
112
128
144
160
176
192
208
224
240

I wouldn't think it would go any higher because it would then be 256, which is too high.

xbuzz

veritas_libertas wrote: »

I wouldn't think it would go any higher because it would then be 256, which is too high.

It actually will go another subnet high in increment of 16. At the 16th it will reset to the next subnet.

10.188.128.0 255.255.240.0

You're right it is increments of 16 in the 3rd octet.. (256-240=16)

10.188.128.0 is the start of that subnet range.

To get the start of the NEXT subnet range, you add 16. So the START of the next subnet is:

10.188.144.0

Therefore the range of the first network range would be

10.188.128.0 - 10.188.143.255. (Which is 1 ip address below 10.188.144.0)

Since the last ip in the subet range is always a broadcast, that would mean the broadcast address would be 10.188.143.255 which would make the last HOST address as 10.188.143.254.

Hope this helps, only started this subnetting business yesterday so dunno if I explained it that well.

veritas_libertas

Okay. I was reading the question wrong. *palm to forehead* Now it seems so obvious

Thanks

veritas_libertas

I'm hoping you guys will be nice enough to help me with another question that has stumped me. I'm sure it's something simple I'm missing:

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

The hosts make complete sense to me, but the subnets through me for a loop. I'm looking at third octet and seeing:

11110000.00000000

In my mind that means 2^4=16. Where on earth does 4,096 come from?

MAC_Addy

It's a Class A address with a class B subnet, so 12 bits extra bits, 8 from the 2nd octet and 4 from the 3rd.

Turgon

If you guys really want to learn your stuff, check out this thread by my great friend, the brilliant Scott Morris. He sent me a Xyplex Terminal server across the pond in 2002 when I really couldn't afford a Cisco term server.. and it still works. Thanks Scott!

Binary Math, Part II

veritas_libertas

I don't understand how the class could change things up. I would think that /20 i.e. 240 would still stay in the third octet.

Just when I think I have it down I learn something new...

MAC_Addy

veritas_libertas wrote: »

I don't understand how the class could change things up. I would think that /20 i.e. 240 would still stay in the third octet.

Just when I think I have it down I learn something new...

I know your frustration! I'm right there with you. I've been through it, thought I knew it and then bam! There's something new. It's the same with Class A, B and C addresses. Here's a PDF that I created a while ago - it's actually copied from another website that I found it on. It may help a little.

veritas_libertas

@Turgon: Thanks, the more I read or watch something from INE, the more I'm impressed. They really do have quality material.

@MAC_Addy: Thanks, you are very right. I had no idea how deep Subnetting could get. I'm not sure I'm looking forward to studying VLSM :P

MAC_Addy

veritas_libertas wrote: »

I'm not sure I'm looking forward to studying VLSM :P

I thought the same, too. But really, I love it! I see you're reading the same book as me - Todd does a brilliant job of explaining this! I found it extremely interesting!