Subnetting question...

dadaji

.... although I am learning it for 70-291 and hence it is in this section.

Which subnet does the host 10.155.254.236/20 belong to?
My answer: 10.155.128.0

The answer given: 10.155.240.0

Can someone explain? By the way, this question is from subnettingquestions.com

I am totally confused and trying to get my head around it.

Thank you.

Ahriakin

The quick and dirty way (as valuable as it is to understand the Binary, and you do need to in order to really 'get it', this is how most folks will work subnets mentally)...

Break out the octets you don't have to touch, so out of /20 that's the first 2 (/16)
So now we know it has to be 10.155.x.x . Since the mask does not touch the 4th octet for the sake of the question it can be ignored, you answer will hinge on the 3rd octet (allowing for trick answers using the subnet/broadcast address).

What's the subnet mask for 4 bits? .240 (128+64+32+16)
Subtract this from 256, = 16 . You now know that this network demarcs on the 3rd octet in periods of 16. So it's 10.155.0.0 - 10.155.15.255 , 10.155.16.0-10.155.31.255 etc. The quick way to get all the way up to the address in question is divide 16 into the 3rd octet, and pick the last number that it divides into evenly.
254/16 .. the closest evenly divided number is 240 (15 times 16). This is the 3rd octet of your subnet.
Finally following the basic/classical rules for subnetting:
Subnet number = 10.155.240.0
1st Usable address = 10.155.240.1
Last usable address = 10.155.255.254
Broadcast = 10.155.255.255

Your address falls within this range.

miller811

check out this link to help get you up to speed
http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

Devilsbane

The first thing I did was write down the subnet mask. /20 corresponds to 20 bits on, which I like to write like
255.255.11110000.0
Remember there are 4 octets, and 8 bits in each one. All 8 bits being a 1 corresponds to 255, which is why I simplified the first 2 octets.

Now you need to find the value of the last bit borrowed, or the position weight of the last one. In this case it is 16 because there are 4 bits in the third octet turned on. (they have the weights of 128, 64, 32 and 16. 16 is the last so it is the one we want)
This is your counter. Every network address will 16 higher than the previous, so lets start writing out our network addresses.
10.155.0.0
10.155.16.0
10.155.32.0
10.155.48.0
..... Keep increasing by 16...
10.155.208.0
10.155.224.0
10.155.240.0
10.155.256.0 (This one doesn't actually exist, but would be the next in line and can help make finding the broadcast address easier.)
Once you have built this list, I think it is pretty clear to tell that 10.155.254.236 is going to fall in between 10.155.240.0 and 10.155.256.0 which means that it will fall in the 10.155.240.0 subnet.

Hope this helps. I'm not a perfect subnetter, and am in a very similar boat as you. I can usually do the problems, but I do have to walk through them step by step.

Some other helpful hints.
The reason that we are adding 16 to the third octet, is becuase that is where the first host bit (the first 0) is found. If the subnet mast was a /28 or 255.255.255.140 then we would be adding 16 to the 4th octet. Remember to write it the easy way 255.255.255.11110000 now the network addresses would still increase by 16, but this time in the 4th octet.
Also remember that the value that is added is always found by determining the positional weight of the last 1.

phoeneous

Eventually you will figure out that the quickest way to answer most subnetting questions is to look at the subnet mask first. A /20 or a 255.255.240.0 will increment its subnets by 16. The quick math on that is 256-240. You know that subnetting will start on the third octet because it is the first octet that isnt 255.

So, with the original network ID of 10.155.0.0, count in increments of 16 in the third octet.

10.155.0.0
10.155.16.0
10.155.32.0
10.155.48.0
10.155.......
10.155.......

and so on until...

10.155.224.0
10.155.240.0

It is not possible to have a 10.155.256.0 so the host is in range 10.155.240.0 - 10.155.255.255.

dadaji

Thank you guys for explaining it. I am confused as to which method to use because I read like 3 different methods and tried to solve the problems. I did look at http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html, and tried to solve some problems but I am still having some difficulty understanding it. I may have some more questions for you guys.

Ahriakin

Nah, it's really just been reiterated 3 times . Each post has stated the same principle in slightly different terms.
You can work in binary which is the simplest but most time consuming or you can practice this 'trick' until it's 2nd nature, once you get the sequence straight you'll be subnetting with ease.

dadaji

Thanks Ahriakin.
I started out with binary from the MS Press book and I can do binary pretty well. But it takes a long time and in the exams, there is not much time to spend on this doing the binary way. SO I decided on the http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html method.
I am stumped by one more question.
Can someone explain me this in detail please? Its taken from the same thread as above.

What if they give me the subnet mask in dotted decimal?

If you're lucky and they give you a mask in dotted decimal format then you should have an even easier time. All you need again is your block size.

Let's say they have given a mask of 255.255.255.248 and you wish to know the block size. Here's the technique:

1. Starting from the left of the mask find which is the first octet to NOT have 255 in it.
2. Subtract the number in that octet from 256 to get your block size e.g. above it is 256 - 248 = block size of 8.
3. Count up from zero in your block size in the octet identified in step 1 as you have learned above (the example above would be in the last octet).

Another example is a mask of 255.255.192.0 - you would simply count up in 256 - 192 = 64 in the third octet.

One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the second octet.

Also how would you go about solving the following using the above method, it will help me a lot.
What valid host range is the IP address 172.23.146.221 255.255.254.0 a part of?

Thank you again.

Devilsbane

Speed really isn't that important of an issue, accuracy is much more important. Lets consider two people, person A has their CCNA, person B is you or me. Person A can do this question in 20 seconds, you can do it in 2 or 3 minutes.

You will probably have less than 5 subnetting questions on this test. So person A will do those 5 questions in 1:40, while it takes you a whopping 10-15 minutes. You have I think 2 hours to answer roughly 40 questions, which allows you 3 minutes per question. So you are still ok. Don't be in such a hurry that you make stupid mistakes.

phoeneous

dadaji wrote: »

Also how would you go about solving the following using the above method, it will help me a lot.
What valid host range is the IP address 172.23.146.221 255.255.254.0 a part of?

You already know the answer to this, I think you're just making the process complicated for yourself.

Ask yourself these questions:

What increment will my subnets count by?
Which is the first octet that isnt 255? (from left to right)

dadaji

I think, I am making it a bit complicated. Frankly, I don't know the answer to this question but I got the answer.
The valid host range is 172.23.146.1 - 172.23.147.254
Can someone confirm this?
Thank you.

phoeneous

dadaji wrote: »

I think, I am making it a bit complicated. Frankly, I don't know the answer to this question but I got the answer.
The valid host range is 172.23.146.1 - 172.23.147.254
Can someone confirm this?
Thank you.

You are correct. Now, how did you get to that answer?