# Subnetting - Impending aneurysm..

nhen
Posts:

**8**Member ■□□□□□□□□□
Afternoon,

Not sure if this is in the incorrect place, I was only able to post in a few places so excuse me if it is.

I am approaching completion of the CBT Nuggets ICND1 (new) course material and hoping to take my 100-101 exam next month. Through that material I have picked up some subnetting and I am able to create appropriate subnets given network/host sizes required, complete VLSM questions and do

However, I keep running into issues with certain questions in practice tests I am taking, as well as some questions on here - subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online

I have googled and googled, but I am still unable to find the missing piece that is causing me to get correct answers sometimes when reverse engineering subnet problems but incorrect answers other times. My head is spinning with the many different ways to approach this and I feel like there is something so simple I am missing that once understood will result in everything falling into place.

For instace this one:

The way I was taught to do these is to work out the increment, start at 0 in the octet that subnet mask relates to and work your way up using the increment.

Doing that in this instance with working..

/28 = 255.255.255.240.

4 bits in the last octet due to the first three being 8 + 8 + 8 = 24 (28 - 24 = 4). 4 bits = 240 as (128 + 64 + 32 + 16 = 240)

So far so good.. I think.

Now to find the increment, can be done easily 32 - 28 = 4 (2^4 = 16)

So we count by 16 from 0 in the 4th octet, resulting in:

172.17.217.0

.16

.32

.48

.64

.96

.112

.128

.144

.160

.176

.192

.208

.224

.240

.256

Doing this and we obviously discover that .144 is a broadcast address which cannot be correct. Making the method of working this out incorrect.

If I hadn't hit a brick wall the next step would have been to take the subnet below or above to work out the range. Eg.

172.17.217.128 - 143 (usable addresses .127-.142)

or

172.17.217.144 - 159 (usable addresses .145 - .15

So next I think well this is a Class B address so perhaps I have to add my increments of 16 from the 3rd octet not the 4th?

But to do this and reach the last valid host of the 172.17.217.144/28 would take forever! Obviously not a viable way of doing this.

What am I missing? How in the world does one accomplish this?

The way I am doing this works sometimes but not in others.. I think it has to do with the Class of IP address but I am so confused now, my brain is mush..

Any help would be greatly appreciated!!

Not sure if this is in the incorrect place, I was only able to post in a few places so excuse me if it is.

I am approaching completion of the CBT Nuggets ICND1 (new) course material and hoping to take my 100-101 exam next month. Through that material I have picked up some subnetting and I am able to create appropriate subnets given network/host sizes required, complete VLSM questions and do

__reverse engineering of subnets when given an IP and subnet mask.__*some*However, I keep running into issues with certain questions in practice tests I am taking, as well as some questions on here - subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online

I have googled and googled, but I am still unable to find the missing piece that is causing me to get correct answers sometimes when reverse engineering subnet problems but incorrect answers other times. My head is spinning with the many different ways to approach this and I feel like there is something so simple I am missing that once understood will result in everything falling into place.

For instace this one:

**Question:***What is the last valid host on the subnetwork 172.17.217.144/28?***Answer:**172.17.217.158The way I was taught to do these is to work out the increment, start at 0 in the octet that subnet mask relates to and work your way up using the increment.

Doing that in this instance with working..

/28 = 255.255.255.240.

4 bits in the last octet due to the first three being 8 + 8 + 8 = 24 (28 - 24 = 4). 4 bits = 240 as (128 + 64 + 32 + 16 = 240)

So far so good.. I think.

Now to find the increment, can be done easily 32 - 28 = 4 (2^4 = 16)

So we count by 16 from 0 in the 4th octet, resulting in:

172.17.217.0

.16

.32

.48

.64

.96

.112

.128

.144

.160

.176

.192

.208

.224

.240

.256

Doing this and we obviously discover that .144 is a broadcast address which cannot be correct. Making the method of working this out incorrect.

If I hadn't hit a brick wall the next step would have been to take the subnet below or above to work out the range. Eg.

172.17.217.128 - 143 (usable addresses .127-.142)

or

172.17.217.144 - 159 (usable addresses .145 - .15

So next I think well this is a Class B address so perhaps I have to add my increments of 16 from the 3rd octet not the 4th?

But to do this and reach the last valid host of the 172.17.217.144/28 would take forever! Obviously not a viable way of doing this.

What am I missing? How in the world does one accomplish this?

The way I am doing this works sometimes but not in others.. I think it has to do with the Class of IP address but I am so confused now, my brain is mush..

Any help would be greatly appreciated!!

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## Comments

6,833Mod ModGo check this thread out: http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html. You can thanks us later.

4Registered Users ■■■□□□□□□□339MemberAs a suggestion, I would simply memorize your netmasks/powers of 2/increment by 16 if you are going for your CCENT/CCNA, that will make your life easier as the test is hard and time is critical. The faster you can do these, the better!

As for brain mush, been there done that, especially with subnetting. Take a break, regroup, tackle it again! Good luck!

Online IP Subnet Calculator is useful for checking your work but obviously you need to know if you are doing classless or not, they have a calc for that on that page too....

8Member ■□□□□□□□□□clowreed1985 Ah.. I get it! That was in the wording.. Well my misreading it. I was interpreting that as a valid host IP I was being given and then needing to work out the next valid host. I did mean to write subnet ID for .144 not broadcast address.

I have another which is confusing me though..

The IP address of a host is 192.168.152.177 /28 and I need to work out what subnet that address is part of and find another host.

So the increment is 16 and the subnet mask is 255.255.255.240.

But if I count in 16s from 192.168.152.0 my working tells me that the range .177 is in in:

192.168.152.160-177 (usable hosts being 161-176)

Meaning that 177 is the broadcast address? That can't be right.

Am I wrong in counting from 192.168.152.0 in this instance? If so, where do I start adding my increments from?

Thanks again!

339MemberStart your counting from the increments of 16 you listed above, each one listed is the start of a new subnetwork.

8Member ■□□□□□□□□□OK let me write the working for that:

192.168.152.0

.16

.32

.48

.64

.80

.96

.112

.128

.144

.160

.178

.194

So that tells me that .160-.177 is one subnetwork. So .176 is the last address of that subnetwork, not the start of the next one?

Then .178-.193 the next.

That's here I'm confused? I did this on the calculator and it gave me what you have written above, but I can't see how it works with my way of working.

Should I not be counting from 192.168.152.0 ?

8Member ■□□□□□□□□□1,462Member ■■■■■■■■□□This x 1000. I failed the CCNA twice because I couldn't subnet fast enough. This thread literally saved my CCNA hopes last year. I also refer co-workers to this thread when the subject comes up.

339MemberRemember to check your increments. 160-178 is 18 by my count, not 16, that's likely where you are going wrong.

It should be 160+16=176

So 175 is the broadcast address

and 174 is the last useable host.

And 176 will be the start of the new subnetwork, with 177-190 as valid host addresses and a broadcast address of 191.

Your process for working out does seem solid but check your math and take your time. If something doesn't seem right, then take a moment to check over your calculations. You seem to be doing very well and it's good you are putting so much effort in to this, it will pay off!

8Member ■□□□□□□□□□Ah! Haha yes the break was needed, I had been at practice questions and then subnetting practice for 8 hours straight without a break. No wonder my math went to buggery. Thank you for pointing that out and thank you for the encouragement, very much appreciated and motivating!

And thank you YFXblu I will continue on the thread tomorrow once I go over everything from today.

OK calling it for today, back to it tomorrow.

44Member ■■□□□□□□□□1. spot the magic number, which is 16(on the 5th leftmost bit)

2. determine the last subnet within, which is 172.17.217.16X9=172.17.217.144

3. host part is the last 4 bits, so it should be 1110(in binary) = 14

4. so the last host = 172.17.217.144+14 = 172.17.217.158

241MemberSorry for the movie references...

44Member ■■□□□□□□□□yes, yours the quickest way.

196MemberIt takes time. It took my around three weeks of doing exercises every day on paper until one day it just 'clicked'. Make sure you really understand the powers of 2. It is very important and helps make life easier. After working with subnetting for a while and the binary math, you should be able to look at a question and pretty much straight away know what the answer is.

I used the subnetting chapter in Todd Lammles CCNA study guide to learn subnetting. It is excellent and he makes it easy to understand and there are quite a few exercises in there to do as well. I did the exercises again and again until i really understood what was going on. Rinse. Repeat

But when first learning make sure you write everything down on paper. It really helps.

Also, in regards to your question about which subnet, host 192.168.152.177/28 belongs to, after working with subnetting and increments for a while.

A quick way to find the increment/block size is to find the value in the subnet mask where the last bit position is a 1. For example, with 192.168.152.177/28

The subnet mask is /28 = 255.255.255.240 = 11111111.11111111.11111111.11110000

Remember that each octet has the values 128 64 32 16 8 4 2 1

So in the last octet where it is 11110000 you can look at it like...

128 64 32 16 8 4 2 1

11110000

The last bit position where the value is 1 is 16 which is the increment.

Also when working with higher numbers such as in your case .177, instead of starting at 192.168.152.0 a quicker way to find the answer is to go up by larger increments until you get close to your number.

So for example you can go up by 128 to get to 192.168.152.128

Then go up by 32 to get to 192.168.152.160

Then go up by 16 to get to 192.168.152.176 which is your answer. So then the next subnet would start at 192.168.252.192 and your usable host range would be...192.168.252.177-190

Again you really need to know your powers of 2...2 4 8 16 32 64 128 256 etc to make life a lot easier

8Member ■□□□□□□□□□Cheers steveyeung and bbarrick, ha to the movie references.. Jeremy loves those too.

Thanks mella060, I do write everything up just as you've done there. I find having two ways to find my increment means I can check I did it correctly if unsure. The first way was using the last bit in the working octet, then I implemented the other way from the thread everyone suggested as my checker.

Awesome with jumping to 128 etc, rather than going by 16 all the way.. So obvious too, all powers of 2.. Just didn't see that was possible.

Using bbarrick's way is quick for instances where you just need one subnet in a range, but if I need to find multiple surrounding subnets then what you suggested works a treat! The question I was doing yesterday with that 192 address required just that, so I finished it this morning in no time by jumping to 128 and going in 16s from there. Muchhh quicker.

I'll keep writing everything out until it starts to click and I can see it in my head. I think I will always want my 128 64 32 16 8 4 2 1 at the top of my page as it makes things so easy, and I'm a visual person.

Very easy to find the powers of 2 as well, as can just count from 2.. 2^5 (count 5 from 2) = 32. Quick and simple.

Thanks again everyone, I'll keep you posted on my progress!

8Member ■□□□□□□□□□Passed my ICND1 exam yesterday with flying colours! Thanks again for all of the helpful comments to my posts, once it was clear what my mistakes were I went over my study material again and again; practised with PT for my labs and did an 1h+ of subnettingquestions.com a day. Writing everything out so I could see any 'trying to go too quick mistakes'.

Nailed the IP addressing and services portion of the test thanks to that.

I did all of my ICND1 study through CBT Nuggets (Jeremy is a legend), and had planned to leave the ICND2 until I got back to Sydney at the start of December. However the place I did my test yesterday also teach CCNA, CCNP, CCIE etc. For ICND2 they do it over 4-5 days (full days), 1-1 with an instructor and get to use their proper labs. They have mock tests at the end and don't 'let' you take the real exam until you are achieving 90% on the mock exams.

If you don't pass the real exam they will retrain you for free but you pay for both exams regardless; training fee does not include exam.

They charge £359 for the ICND2 training, exam costs just over £100.

It's through these guys - Synergy : CCIE Lab Training | CCDE Lab Training and IT Solutions by CCIE Certified Consultants and Instructors

Has anyone heard of them? Any advice?

I'm not sure what do to, I do like the idea of having someone in person to bounce questions off and some real lab equipment to learn on. Plus that 5 day turnaround on the material is much quicker than I could get through it on my own.. Hmm.

44Member ■■□□□□□□□□339MemberSorry no experience with those guys!

I guess it would depend on if you really need to get the exam done sooner rather than later and how familiar you already are with the material!

I personally prefer to take things at my own pace and learn and practice things a few times before jumping in to an exam but that's just me and everyone is different!

Congrats again on your pass! Keep it up!

8Member ■□□□□□□□□□Thanks BGraves I think I'm with you on the last point; thought about it this morning and decided I will relax a bit now, get GNS3 running instead of PT and then spend a bit of time breaking and fixing things while slowly (comparatively speaking) going through the CBT Nuggets ICND2 material.

I'd rather have the CCNA feeling nice and confident, rather than squishing everything in sausage factory style!

Very much appreciated thank you for your help!