# Subnetting question

Member Posts: 142
Hello Everyone,

Can someone help clarify this question for me ? I am reading the 70-293 Syngress book and there is a question about subnetting that isn't making sense.

Question-

You are given a task to create eight subnets on your LAN, and you have been assigned the address space 172.16.128.0 /23. How many hosts will you have and what is the CIDR notation for the new subnet's address space ?

A. 2032 hosts on 172.16.128.0 /24
B. 240 hosts on 172.16.128.0 /27
C. 496 hosts on 172.16.128.0 /26
D. 48 hosts on 172.16.128.0 /29

Answer: C

Here are my thoughts on this question..

If they are asking me to create eight subnets and giving me 172.16.128.0 why would they give me the subnet mask already? To me it defeats the purpose of asking the question if give me the subnet mask already.

Also 172.16.128.0 /23 gives me 128 subnets and 510 hosts per subnet which meets their requirements. Secondly 172.16.128.0 /26 is 1024 subnets and 62 hosts per subnet and they have it listed as 496 hosts 0_o

Can someone please tell me what I am missing here ?

Thanks!

## Comments

• Banned Posts: 12,312 ■■■■■■■■■□
That is the starting network you have to work with, so you have nine bits to meet those requirements. There is always a subnet mask, even with the classful networks.

Therefore, you need three bits to satisfy the eight-subnet requirement, which will make your resulting networks /26. The remaining six bits will give you 62 (usable) hosts per network, or 496 total.
• Member Posts: 142
dynamik wrote: »
That is the starting network you have to work with, so you have nine bits to meet those requirements. There is always a subnet mask, even with the classful networks.

Therefore, you need three bits to satisfy the eight-subnet requirement, which will make your resulting networks /26. The remaining six bits will give you 62 (usable) hosts per network, or 496 total.

I understand what you are saying but I guess I am still not understanding this question completely. Using what you said if I have 9 bits to work with using 1 bit to make /24 gives me 8 subnets(or more) and 2032 hosts and therefor answer A would also work ... right ?
• Banned Posts: 12,312 ■■■■■■■■■□
No, because you need to split that into eight subnets, which requires three bits.

You have a /23 to work with. There are 32 bits total, so you have nine to work with. You need three to meet the subnet requirement, so that leaves you with six for the hosts.

Here's a scenario: Suppose your company starts with a typical Class-B address. They subnet that to /23 and give one of those networks to each office location. You're a network administrator at one of those offices, and you need to subnet that address because your location requires eight different networks.
• Member Posts: 278
Zaits wrote: »
Hello Everyone,

Can someone help clarify this question for me ? I am reading the 70-293 Syngress book and there is a question about subnetting that isn't making sense.

Question-

You are given a task to create eight subnets on your LAN, and you have been assigned the address space 172.16.128.0 /23. How many hosts will you have and what is the CIDR notation for the new subnet's address space ?

A. 2032 hosts on 172.16.128.0 /24
B. 240 hosts on 172.16.128.0 /27
C. 496 hosts on 172.16.128.0 /26
D. 48 hosts on 172.16.128.0 /29

Answer: C

Here are my thoughts on this question..

If they are asking me to create eight subnets and giving me 172.16.128.0 why would they give me the subnet mask already? To me it defeats the purpose of asking the question if give me the subnet mask already.

Also 172.16.128.0 /23 gives me 128 subnets and 510 hosts per subnet which meets their requirements. Secondly 172.16.128.0 /26 is 1024 subnets and 62 hosts per subnet and they have it listed as 496 hosts 0_o

Can someone please tell me what I am missing here ?

Thanks!

They give you /23 as a starting point, which means you have to start your calculations from there, leaving you 9 bits to make your calculations, so just double.... start with 2, then 4, then 8, leaving you at /26 since 8 is how many subnets it says it wants created, if it wanted 16 subnets it would be /27, and so on and so forth...

Let me know if you want further clarification.
• Member Posts: 142
Alright I completely understand your logic and I am on board with how you are getting /26
3 bits = (2x2x2) = 8 subnets 3 + 23 = /26 etc..

If I use 1 bit ( /24) that is only 2 subnets and therefor not meeting my requirements of 8 subnets.

The way I was looking at it was 172.16.128.0 / 24 gets broken down like so..

172.16.128.0 - 172.16.128.255
172.16.129.0 - 172.16.129.255
172.16.130.0 - 172.16.130.255
172.16.131.0 - 172.16.131.255
172.16.132.0 - 172.16.132.255
172.16.133.0 - 172.16.133.255
172.16.134.0 - 172.16.134.255
172.16.135.0 - 172.16.135.255

This gives me 8 subnets and 254 usable hosts = 2032 total
172.16.128.0 /26 breaks down to

172.16.128.0 - 172.16.128.63
172.16.128.64 - 172.16.128.127
172.16.128.128 - 172.16.128.191
172.16.128.192 - 172.16.128.255
172.16.129.0 - 172.16.129.63
172.16.129.64 - 172.16.129.127
172.16.129.128 - 172.16.129.191
172.16.129.192 - 172.16.129.255

This gives me 8 subnets and 62 usable hosts. = 496 total

The way I break it down it looks like both meet the requirements of eight subnets..
• Member Posts: 142
Zaits wrote: »
Alright I completely understand your logic and I am on board with how you are getting /26
3 bits = (2x2x2) = 8 subnets 3 + 23 = /26 etc..

If I use 1 bit ( /24) that is only 2 subnets and therefor not meeting my requirements of 8 subnets.

The way I was looking at it was 172.16.128.0 / 24 gets broken down like so..

172.16.128.0 - 172.16.128.255
172.16.129.0 - 172.16.129.255
172.16.130.0 - 172.16.130.255
172.16.131.0 - 172.16.131.255
172.16.132.0 - 172.16.132.255
172.16.133.0 - 172.16.133.255
172.16.134.0 - 172.16.134.255
172.16.135.0 - 172.16.135.255

This gives me 8 subnets and 254 usable hosts = 2032 total
172.16.128.0 /26 breaks down to

172.16.128.0 - 172.16.128.63
172.16.128.64 - 172.16.128.127
172.16.128.128 - 172.16.128.191
172.16.128.192 - 172.16.128.255
172.16.129.0 - 172.16.129.63
172.16.129.64 - 172.16.129.127
172.16.129.128 - 172.16.129.191
172.16.129.192 - 172.16.129.255

This gives me 8 subnets and 62 usable hosts. = 496 total

The way I break it down it looks like both meet the requirements of eight subnets..

Well I guess after taking a step back and looking at this question again 172.16.128.0 /23 that subnet would be 172.16.128.0 - 172.16.129.255 which is exactly what I have listed above as my 8 subnets...
• Banned Posts: 12,312 ■■■■■■■■■□
172.16.128.0/23 split into /24 will only give you the following two networks:

172.16.128.0-172.168.128.255
172.16.129.0-172.168.129.255

172.16.10000000.00000000-172.16.10000000.11111111
172.16.10000001.00000000-172.16.10000001.11111111

The next one would require that you have a /22 and ten bits total:

172.168.130.0-172.168.130.255
172.16.10000010.00000000-172.16.10000010.11111111

• Member Posts: 142
dynamik wrote: »
172.16.128.0/23 split into /24 will only give you the following two networks:

172.16.128.0-172.168.128.255
172.16.129.0-172.168.129.255

172.16.10000000.00000000-172.16.10000000.11111111
172.16.10000001.00000000-172.16.10000001.11111111

The next one would require that you have a /22 and ten bits total:

172.168.130.0-172.168.130.255
172.16.10000010.00000000-172.16.10000010.11111111

Ya I think I got it now just took me a bit to figure my logic out on my end. This is the reason why I keep visiting this website because of its great community and people willing to help.

Thanks!!
Sign In or Register to comment.