# Subnet question

bundleboy
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hi guys could you be so kind as to help me with this question with a step by step explanation. I would very much appreciate it.

"How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask and how many usable host addreses will there be per subnet?"

Thank you!

"How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask and how many usable host addreses will there be per subnet?"

Thank you!

0

## Comments

55Member ■■□□□□□□□□First you need to get the network address.

- 172.17.32.0/23 is a subnet of the 172.17.0.0/16 class B network

This means you borrowed 7 bits from the host portion ->

a total of 128 subnets (2^7 )with 510(2^9-2) hosts each.

- If you want to subnet the 172.17.32.0/23 in a /27 you need to borrow 4 more bits.

This means that you will get 2^4 subnets - 16

and 2^5-2 hosts per subnet - 30

The subnets will be :

172.17.32.0/27

172.17.32.32/27

........................

172.17.32.224/ 27 --- 8 subnets

+

172.17.33.0/27

........................

172.17.33.224/27 --- another 8 subnets

326MemberA /27 bit mask means 27 bits are used for the network, leaving 5 for hosts 2^5=32 so there are 32 addresses per subnet, minus 1 for the network address and 1 for the broadcast address, so 30 useable host addresses.

Your original /27 mask gave you 512 (2^9) addresses. There are 2 ways now to find the number of subnets.

Either divide 512 by your calculated number of hosts (512/32)=16

Or use the difference in bits from your original /27 and your subnetted /23, leaving you 4 network bits for subnetting. 2^4=16....

So, you have 16 subnets each with 30 useable hosts.

1st subnet. 172.17.32.0 to 172.17.32.31

2nd subnet. 172.17.32.32 to 172.17.32.63

And so on

Last subnet. 172.17.33.224 to 172.17.33.255

hope this helps, I apologise if this doesn't make sense, I usually do it quickly in my head so conveying it in words is hard for me!!!

Going all out for Voice. Don't worry Data; I'll never forget you:study: CVoice [X] CIPT 1 [ ] CIPT 2 [ ] CAPPS [ ] TVOICE [ ]

326MemberPretty much the same method, by the time I'd typed all that on the phone, you'd already posted it!!

Going all out for Voice. Don't worry Data; I'll never forget you:study: CVoice [X] CIPT 1 [ ] CIPT 2 [ ] CAPPS [ ] TVOICE [ ]

33Banned ■■□□□□□□□□I understand you would use 2^5 if you know that you need 30 subnets, but how did you work your way backward to get it looking for how many subnets?

55Member ■■□□□□□□□□You need to look for the type of ip address you have.. 172.17.0.0/23 comes from a class B ip address where you have 16 bits for network and 16 bits for hosts.

172.17.0.0/23 means that you borrowed 7 bits (23-16) from the host portion...so you will have a total of 2^7 = 128 subnets

The number of hosts is 2^ (32-23) 9 = 512 -2 = 510

172.17.32.0/27 means you need 4 more bits...so for this network you will have 2^4 subnets = 16 and 2^(32-27)5 = 32-2 = 30 hosts.

Hope this is clearer.

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