Hello Everyone

Ravenclaw

I am not sure if I am in the correct area but I am working on my CCENT and had a sample question and answer that
I cant figure how to solve it, any help would be greatly appreciated..:)

What valid host range is the IP address 172.30.98.55/23 a part of?
Answer: 172.30.98.1 through to 172.30.99.254

Futura

A /24 is 254 usable hosts so a /23 is 510 usable hosts.

So think about it again now.

Clues:
172.30.98.0 is the network, lowest value in the range

172.30.99.255 is the broadcast. highest value in the range


Learn all the /cidr notations and subnetting will be easy as pie.

Plantwiz

@ravenclaw,

What sort of study material are you using and how was the above explained to you in the source(s) you are using?

Taking a single question can be difficult to understand the reasoning behind the solution. If you can provide your sources, others who have the same source, or have used something similar may have a better way to assist you.

Ravenclaw

@futura
Ok thanks I may have to do that, I was thinking the range was 172.30.98.1 - 172.30.98.254

@plantwiz
I found a site packetlife.net it gives you random subnetting questions and the answers, just not how to solve them

Xyro

Ravenclaw wrote: »

What valid host range is the IP address 172.30.98.55/23 a part of?
Answer: 172.30.98.1 through to 172.30.99.254

Ravenclaw wrote: »

Ok thanks I may have to do that, I was thinking the range was 172.30.98.1 - 172.30.98.254

Did you alter your answer or was that a typo?

Ravenclaw

No I thought the answer would be a range of 172.30.98.1 - 172.30.98.254

Xyro

Ahh ok, well that's, of course, incorrect. You're somewhat "close" though. You just seem to be missing a portion of the idea.

"What valid host range is the IP address 172.30.98.55/23 a part of?"
"Answer: 172.30.98.1 through to 172.30.99.254"

172.30.98.55 is a B class address; therefore, the 172.30 (A.B) doesn't change. You have gotten this far since the beginning of the range is, in fact, 172.30.98.0 which will make the 1st usable address 172.30.98.1 as you have.

You wrote:

...I thought the answer would be a range of 172.30.98.1 - 172.30.98.254

The next item you need to do is look at that 98.

What I do is:

1. Convert the prefix to a subnet mask (in this case /23 = 255.255.254.0).
2. Then look at the 254 since it is the octet of the 98 (3rd in both the IP address & the subnet mask you see).
3. Minus the 254 from 256 (which = 2)
4. This means that each 3rd octet will "jump" by 2 (0,2,4,8, etc.)
5. You can count by 2s until you get to 98 lol, but it's much easier to simply look at 98 & realize it's an even # so therefore it will begin a range, not end it... since when you count by 2s the even # always begins a range & the odd # ends it.
6. You then know that it will be going from 98 to 99 (since the next beginning range will be 100).
7. Simply fill in the remaining octet(s) (in this case 1) to 255 (all binary 1s).
8. Then it's quite obvious you know how to find the usable addresses in a range, so I will omit this final item.

Hope this, at least, somewhat assists.

pamccabe

Welcome and enjoy the journey.

d6bmg

Understand the subnetting concepts first. Read Odom's official guides and practice the subnetting problems.

Ravenclaw

ok got it, thanks for the detailed explanation. I needed that broken down

Regards