More subnetting help please.

mikejj83

Hello,

I first posted here last week and I am still a beginner. I have been working on the subnetting from Todd Lammle's book for the CCENT. I've got a good grasp on the basic concept, but was hoping someone could show me how to work this problem:

Question: What valid host range is the IP address 172.24.73.50 255.255.252.0 a part of?

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I think the increment used in the ranges is 4, but I don't know how to quickly find the ranges without writing them all down from the beginning.

Any help would be greatly appreciated.

Thank you.

dynamik

Multiply. You know 4*20 is 80, so back off two steps until you get to 72.

kryolla

the third octet is the interesting octet so convert 73 into binary and 252 into binary then do the AND operation you'll come up with a network of 72.0/22 so now you make all host bits 1's and it will be 75.255 so the range is 172.24.72.1 to 172.24.75.254 I havent checked my work so hopefully Im right

phoeneous

kryolla wrote: »

the third octet is the interesting octet so convert 73 into binary and 252 into binary then do the AND operation you'll come up with a network of 72.0/22 so now you make all host bits 1's and it will be 75.255 so the range is 172.24.72.1 to 172.24.75.254 I havent checked my work so hopefully Im right

He said "quickly".

mikejj83, if you know your increment is 4, then count by 4's until you find your range. What range will .73 be in if you count by 4?

mikejj83

What about this one?

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Question: What valid host range is the IP address 172.24.254.92/22 a part of?
Answer: 172.24.252.1 through to 172.24.255.254
----

The increment should be 4, right? And then wouldn't I just start going up in increments of 4 in the 4th octet:
172.24.254.0
172.24.254.4
172.24.254.8
........
?

I'm having trouble doing any address other than a class C, because it's not clear to me how to carry numbers over into another octet.
Also in the problem above I don't see how the answer has that range with the 252 and 255 in the third octets.

dynamik

As kryolla mentioned, you just need to be concerned with the "interesting octet".

This is the second octet when your subnet bits are 9-16, the third for 17-24, and the fourth for 25-32.

As I mentioned earlier, just multiply. You know 50*4=200. You know 10*4=40. That gives you 240, so you just have a little ways to go. The next one is 244, then 248, then 252.

mikejj83

Thank you all for your patients. It is greatly appreciated.


Dynamik, I understand what you’re saying about the interesting octet. For the above scenario I’m going up increments of 4 in the “interesting octet” like this:
172.24.0.0
172.24.4.0
172.24.8.0
……
172.24.240.0
172.24.244.0
172.24.248.0
172.24.252.0 (and I stop here b/c I have reached 252 in the interesting octet, right?)
Then where do I go from here?
Do I start going up by 4 in the 4th octet now? 172.24.252.0, .4, .8??

dynamik

No. Once you know the starting point, the ending point is simply one less than the next subnet's starting address. In this case, you're at the last subnet (there is no 256), so the end of the range is 172.24.255.255.

Lets say the address was in the 172.24.148.0 subnet. You know the next subnet is 172.24.252.0, so you just need to back off by one. This will give you 172.24.251.255.

I know you're looking for the quick way to do this, but I really encourage you to spend awhile going through things the long way and writing everything out in binary (like kryolla suggested). After you do that for awhile, it will become much more apparent why things work the way they do. Skipping that step will really mess with your genuine understanding of the material.

mikejj83

My gosh I think I got it! Thank you so much for your help. I can't thank you enough!
I'm calling it a night for now, may have some broadcast address questions tomorrow, but I am much better off now. Thanks again!

Mike

kryolla

phoeneous wrote: »

He said "quickly".

mikejj83, if you know your increment is 4, then count by 4's until you find your range. What range will .73 be in if you count by 4?

you must be slow at things

miller811

here is a good site to test your skills
IP Subnet Practice

phoeneous

kryolla wrote: »

you must be slow at things

No, you just dont know how to comprehend what people type. Work smarter, not harder.

dynamik

Um, ok? Kryolla has consistently proven that he has a solid handle on Cisco technologies. I think you're knowledgeable and respect your opinions, but I'm not sure why you're getting into it with him, especially when what he said was correct...

NightShade1

Do some search in this site i posted how to do it

After reading that you wont be needing other explanaitions

tim100

mikejj83 wrote: »

Hello,

I first posted here last week and I am still a beginner. I have been working on the subnetting from Todd Lammle's book for the CCENT. I've got a good grasp on the basic concept, but was hoping someone could show me how to work this problem:

Question: What valid host range is the IP address 172.24.73.50 255.255.252.0 a part of?

----
I think the increment used in the ranges is 4, but I don't know how to quickly find the ranges without writing them all down from the beginning.

Any help would be greatly appreciated.

Thank you.

Since your subnet mask is 255.255.252.0 your network bits in the third octet are limited to:
128 64 32 16 8 4

A quick way would be to just take the closest bit(s) to your IP in question so take 64 and add 8 which makes:

172.24.72.0/22 (255.255.252.0)

If you add 4 to that the next network is:

172.24.76.0/22 (255.255.252.0) which is beyond the IP in question

So your network address and host range for that IP is:

Network address:
172.24.72.0/22 (255.255.252.0)

Host range:
172.24.72.1 - 172.24.75.254

172.24.75.255 is your broadcast