Hey ..
When they say gained, do they mean the difference between the /23 and the /27 mask?
Not the total, the difference. Or the amount gained from the original.
What do you think?
Main Topics
Browse All TopicsHi,
1)This is related to the CCNA Exam.
2) In one of the prep test:
Question: How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many usable host addresses will there be per subnet?
The Given answer: 16 subnets, 30 hosts
3) My question: i) Why does the answer are 16 subnets?, ii) Per my understanding, it is 6 subnets and 30 hosts, am i right?, iii) Please confirm or provide the right answers with the explanation.
4) Thank you
tjie
This Question has been solved and asker verified All Experts Exchange premium technology solutions are available to subscription members.
Experts Exchange has been collecting answers to technology questions since 1996…3 million and counting! If you have a question, chances are we already have your answer.
If you can't find the exact answer you're looking for, ask our exclusive community of 50,000 experts. You’ll get a personalized answer from a trusted professional.
Thousands of free tech tips, tricks, how-to’s and tutorials are available in our peer reviewed articles section. See for yourself how smart our experts are, no login required.
Access the answers to your technology questions today.
30-day free trial. Register in 60 seconds.
Members of the expert community talk about why the experience at Experts Exchange is different than what you will find anywhere else.

Try it out and discover for yourself.
30-day free trial. Register in 60 seconds.
Join the community of experts here and help other tech pros by answering question in your area of expertise. You can earn FREE access to all Experts Exchange's premium features and resources.
And ..
Here's the answer:
http://www.allinterview.co
Check it out.
Since I'm not a fan of clicking on links to get answers (it's not that I don't trust you, but...). Here is the deal:
172.17.32.0/23 is the address space you start with and represents 2^(32-23) ip addresses - or 2^9=512 addresses (including network and broadcast addresses). Or you can say you have ONE subnet with 510 hosts
If you split this address space in two using a /24 mask, you get two subnets, which equates to one **extra** subnet (2-1=1), therefore you **gain** one subnet. Each subnet will have 254 hosts (2^8-2)
If you split these two subnets again, you end up with 4 subnets with a /25 mask, giving you 4 subnets, so a **gain** of 3 subnets. Each subnet has 126 (2^7-2) hosts.
If you split these two subnets again, you end up with 8 subnets with a /26 mask, giving you 8 subnets, so a **gain** of 7 subnets. Each subnet has 62 (2^6-2) hosts.
If you split these two subnets again, you end up with 16 subnets with a /27 mask, giving you 16 subnets, so a **gain** of 15 subnets. Each subnet has 30 (2^5-2) hosts.
So the **CORRECT** answer is that you **gain** 15 subnets (remember you had 1 to start with, and you haven't lost it) and each subnet has 30 hosts.
If the CORRECT answer is not available for a given question, you have to guess what the examiner thinks is the "most correct" - in this case the idiot who set the exam probably couldn't count the original subnet you started with, so 16 subnets, each with 30 usable addresses is probably the best answer.
The multiples 2, 4, 8 , 16, 32, 64, 128 (and 256, 512, 1024 ...) come up time and time again in subnetting, so it is worth learning your 2, 4, 8 , 16, 32, 64, and 128 times tables - as far as 256 anyway. Pretty soon you can work out that:
subnets that have a mask of /30 ALL start with a multiple of 4 (0, 4, 8, 12, 16, ...)
subnets that have a mask of /29 ALL start with a multiple of 8 (0, 8, 16, 24, 32, ...)
subnets that have a mask of /28 ALL start with a multiple of 16 (0, 16, 32, 48, 64, ...)
subnets that have a mask of /27 ALL start with a multiple of 32 (0, 32, 64, 96, 128, ...)
subnets that have a mask of /26 ALL start with a multiple of 64 (0, 64, 128, 192)
subnets that have a mask of /25 ALL start with a multiple of 128 (0, 128)
subnets that have a mask of /24 ALL start with a multiple of 1 (0, 1, 2, 3, 4, ... - in the 3rd octet)
subnets that have a mask of /23 ALL start with a multiple of 2 (0, 2, 4, 6, 8, ... - in the 3rd octet)
subnets that have a mask of /22 ALL start with a multiple of 4 (0, 4, 8, 12, 16, ... - in the 3rd octet)
and so on. If you look at the mask in decimal, there is a formula to work out the multiple I've used above. It is 256-(the last non-zero octet of the mask). So if the mask is /19 aka 255.255.224.0, the multiple for your subnets is 256-224=32 - in the 3rd octet. So you your subnets will be x.x.0.0, x.x.32.0, x.x.64.0, x.x.96.0 etc
OOOPS - correction: Subtle difference, but then I am pedantic. Corrections in bold italics
Here is the deal:
172.17.32.0/23 is the address space you start with and represents 2^(32-23) ip addresses - or 2^9=512 addresses (including network and broadcast addresses). Or you can say you have ONE subnet with 510 hosts
If you split this address space in two using a /24 mask, you get two subnets, which equates to one **extra** subnet (2-1=1), therefore you **gain** one subnet. Each subnet will have 254 hosts (2^8-2)
If you split these two subnets again, you end up with 4 subnets with a /25 mask, giving you 4 subnets, so a **gain** of 3 subnets. Each subnet has 126 (2^7-2) hosts.
If you split these four subnets again, you end up with 8 subnets with a /26 mask, giving you 8 subnets, so a **gain** of 7 subnets. Each subnet has 62 (2^6-2) hosts.
If you split these eight subnets again, you end up with 16 subnets with a /27 mask, giving you 16 subnets, so a **gain** of 15 subnets. Each subnet has 30 (2^5-2) hosts.
So the **CORRECT** answer is that you **gain** 15 subnets (remember you had 1 to start with, and you haven't lost it) and each subnet has 30 hosts.
Business Accounts
Answer for Membership
by: uucknaaaPosted on 2009-06-24 at 22:18:54ID: 24708360
Hi
or.com/
This is the second time today I've used this. See at this link:
http://www.subnet-calculat
It'll give you all the answers.