Subnet question

Posts: 6Registered Users ■□□□□□□□□□
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!

• Posts: 55Member ■■□□□□□□□□
bundleboy wrote: »
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!

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
• Posts: 326Member
This is the way I do it, hope it helps.
A /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 [ ]
• Posts: 326Member
@capitanuionut
Pretty 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 [ ]
• Posts: 33Banned ■■□□□□□□□□
I followed along pretty well, except I don't understand where or why you got 2^5 in both examples.

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?
• Posts: 55Member ■■□□□□□□□□
SneakDogg wrote: »
I followed along pretty well, except I don't understand where or why you got 2^5 in both examples.

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?

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.
• Posts: 897Member
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• Posts: 6Registered Users ■□□□□□□□□□
thanks guys.. still confused but with practice i hope i get it..