Subnetting class b, network address 172.23.0.0/27

Subnetting is not my greatest strength

I need to subnet 172.23.0.0/27 the subnet mask is 255.255.255.224, I can not figure out what is the first and what is the last usable subnet.

The questions I have are: what is the last usable first subnet? What is the last usable second subnet? What is the last usable third, forth and fifth subnets? What are first usable subnets for the first, second, third, fourth and fifth?

can i get some help on this?

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Comments

royal

255.255.255.224 = 11111111.1111111.11111111.11100000

This means that, we'll use the 1st 3 bits when figuring out our IP range we can use.

172.23.0.xxx000000

Our starting subnet range would be:
172.23.0.00100000

Our ending subnet range would be:
172.23.0.11100000

So 172.23.0.32 would be the starting range while 172.23.0.224 would be the ending range.

The reason is because since we're working with only the 1st 3 octets, 001 would be a 1 which is the smallest # in binary, and all 1s would be the largest # we can have. To find out the 2nd subnet, we'd just do 010 since binary goes from 128 64 32 16 8 4 2 1. 3rd subnet would be 011, 4th would be 100, and so on.

Does that help?

jmrseadogg

Thanks for the help, I am starting to get it, but can you help me with the rest?

first usable subnet would be 001 32
second usable subnet would be 010 64
third usable subnet would be 011 96
fourth usable subnet would be 100 128
fifth usable subnet would be 110 192

would this be correct?

what woudl the last first subneet be then? Do you work backwords

jmrseadogg

So I was right on the rest of them? the fifth is 101 that would make it 160

So what is the last first usable Subnet, how do I reach that answer if the first usable first subnet is 32?

I am understanding this more, thanks for you time. I think I am understanding how your first the first usable, but not the last usable.

royal

I made a mistake, subnets can use all 0s and all 1s --- 2n. So first subnet would be 000. 2n would be 2x2x2 (2n and there's 3 bits) = 8 = 111

It's when doing client IPs you do 2n = 2.

So all your subnets would be:

000 = 1st subnet
001 = 2nd subnet
010 = 3rd subnet
011 = 4th subnet
100 = 5th subnet
101 = 6th subnet
110 = 7th subnet
111 = 8th subnet

Man it's been a long time since I've subnetted, you're making me think!

jmrseadogg

Wow,

So 8 subnets

So when I am asked,

last usable subnet of the first subnet, what does that mean?

royal

jmrseadogg wrote:

Wow,

So 8 subnets

So when I am asked,

last usable subnet of the first subnet, what does that mean?

That doesn't make sense. All those questions ask something such as, last Client IP Address of the 1st subnet, etc...

AlanJames

dont forget about the broadcast address

network 0

1st useable = .1
broadcast address = .31
last useable = .30

network 32

1st useable = .33
broadcast = .63
last useable .62

crstof

Be a little bit careful with this - if you are doing ICND/CCNA course you get the concept of "usable subnets" as well as "usable hosts".

So I'd agree with Royal that in practice you get the 8 subnets total - but in exam format the "usable subnets" is actually 8-2 = 6. This can be very mis-leading and I have been trying to find verification on what the official line in the CCNA is, but one of the practice exams I did definitely had the concept of "usable" subnets.

crstof

Check out the following

http://www.mcmcse.com/articles/subnetting.shtml

phantasm

Determining Subnets:
Us the equation 2^x = s where x is the number of borrowed bits (the 1's) and s is the number of subnets you will have.

Let's look at it like this: 172.23.0.11100000

Since there are three 1's in the last octect; x = 3.

So, 2^3 = 8 total subnets

Determining Hosts per Subnet:
To determine the number of hosts per subnet use the equation 2^n - 2 = y. N in this case is the number of unmasked bits (the 0's that follow the 1's in the last octect). So n in this case is 5.

So (2^5) - 2 = 30 usable hosts per subnet.

Example:

IP Address: 172.23.0.0 /24

Step 1: 2^3 = 8
Step 2: (2^5) - 2 = 30

So we will have 8 subnets which will have 30 usable IP addresses each.

The subnets are:
172.23.0.0
172.23.0.32
172.23.0.64
172.23.0.96
172.23.0.128
172.23.0.160
172.23.0.192
172.23.0.224

The IP's associated with your first network are:
Network Address: 172.23.0.0 /27
Usbale IP Range: 172.23.0.1 /27 to 172.23.0.30 /27
Broadcast IP: 172.23.0.31 /27

Frank_R

Hello to everybody.

I am studying for the CCNA esxam 640-802 and 640-816: I attended Cisco lessons but nobody told me how to subnet a class b network.
I have no problem to subnet a class C but I really am not able to subnet a class C.

I have this network

172.16.0.0

255.255.128.0/17

I know there are 2 subnets, and 32,766 host

I do not know how to calculate the valid host range for each subnet

Is there anybody who can help me?

thank you

Francesco Ruosi

Naples (Italy)