Struggling with this subnet question. Anyone know where i went wrong?

If an IP address is 172.16.112.1 (/25) what would be the subnet address of this host?

so far I have worked out that the total number of subnets is 512 and the total number of hosts is 126 per subnet
I then worked out that the subnet mask is 255.255.255.128

but then I need to work out what the increment is. In my sybex network plus book it tells me to take 256 and subtract the subnet ID from it. This means 256-128. Which means I can have networks of 0.0 and 128.0
But that would means theres only two subnetworks

can anyone please help.
I know the answer is 172.16.112.0 but have no idea how that came to be.

so far i have read sybex comptia net+ book on subnetting, also watched the cbt nuggets and train signal vids on subnetting but i seem to be failing when it comes to the more trickier looking questions. I fail with the questions that mix different class addresses with different class subnet masks.

Find more posts tagged with

Save $250 on 2025 certification boot camps from Infosec!

Book now with code EOY2025

Button

Comments

cyberguypr

I am also working on subnetting right now so I may not know what I'm talking about. Shouldn't the second be 0.128? Maybe first network would be 172.16.0.0, second one 172.16.0.128, third 172.16.1.0 and so on.

Let's see what the experts say.

Todd Burrell

If you are given an address of 172.16.112.1 (/25) then the network address of this network would be 172.16.112.0, the broadcast address would be 172.16.112.127 and the hosts for this network would be 172.16.112.1 - 172.16.112.126, so you would have 126 total hosts with one network address and one broadcast address.

The /25 indicates that there would be 128 addresses for the network and the subnet mask would be 255.255.255.128.

I'm not sure what this question was really asking, but I suspect they wanted the network address for this host with this subnet mask?

Hope this helps - and I think my math is right...

sandman748

MrXpert wrote: »

If an IP address is 172.16.112.1 (/25) what would be the subnet address of this host?

so far I have worked out that the total number of subnets is 512 and the total number of hosts is 126 per subnet
I then worked out that the subnet mask is 255.255.255.128

but then I need to work out what the increment is. In my sybex network plus book it tells me to take 256 and subtract the subnet ID from it. This means 256-128. Which means I can have networks of 0.0 and 128.0
But that would means theres only two subnetworks

can anyone please help.
I know the answer is 172.16.112.0 but have no idea how that came to be.

so far i have read sybex comptia net+ book on subnetting, also watched the cbt nuggets and train signal vids on subnetting but i seem to be failing when it comes to the more trickier looking questions. I fail with the questions that mix different class addresses with different class subnet masks.

You were along the right lines, just working with the wrong octet. Anything /24 and higher is subnetting the fourth octet. Therefore your two networks are 172.16.112.0 and 172.16.112.128

Keep in mind that the class B address you started with is 172.16.0.0 which is why there 512 subnets of 128 each (172.16.0.0....172.16.0.128....172.16.1.0....all they way up to 172.16.112.0 and further)

When given an address and asked to find the network, you can basically ignore the whole class thing. Just know that each /8 is a different octet and work with a smaller scale.

CodeBlox

That "incrementer" can be referred to as a "multiplier" as well. And the way you get the "multiplier" is by taking the place value of the least significant subnet bit.

11111111 11111111 11111111 10000000
Red bits are subnet bits and you have 25 of them. The "special octet" here is the 4th one. The place value of the least significant subnet bit is 128 here. That only gives you two subnets changes in the 4th octect however there is a possible 512 subnets with that mask.
172.16.0.0 <- subnet zero
172.16.0.128
172.16.1.0
172.16.1.128
...
...
...
172.16.254.0
172.16.254.128
172.16.255.0
172.16.255.128 <- broadcast subnet

The host 172.16.112.1 /25 falls in to the 172.16.112.0 subnet.
That subnets broadcast address would be 172.16.112.127

I hope I didn't make things more confusing but that's the way I learned this.

mella060

Basically, you just need to find the increment and which octet it is occurring in. This octet is commonly called the "interesting octet", since that is where all the action is happening.

In this example, the increment is occurring in the last or 4th octet so that is the interesting octet.

The way i find the increment is to work out the last bit position in the subnet mask where the value is a 1.

So your subnet mask is 255.255.255.128...in binary, that is

11111111.11111111.11111111.10000000

128 64 32 16 8 4 2 1

See where the last 1 is ?

So then your increment would be 128 and in the 4th octet. Remember that since the value in the 3rd octet is 255, it will just increment by 1 (256-255).

As mentioned above you would start at...

172.16.0.0
172.16.0.128
172.16.1.0
172.16.1.128
172.16.2.0
172.16.2.128
...................
...................
...................
...................
172.16.110.0
172.16.110.128
172.16.111.0
172.16.111.128
172.16.112.0
172.16.112.128

Subnetting takes a little while to master, but once you do you realize how easy it is...

MrXpert

hi,
thanks for all your help. I finally managed to work it out. I've now also done all 14 questions in the wendall odom icnd1 book on subnetting. Got them all right. Very pleased. Did a few others by Chris bryant and any random ones i could find online.Got those right too.
Can't believe i finally can do it properly.Thanks again

CodeBlox

It would probably be a good idea to do them daily until it's second nature. There are websites that push out endless subnetting questions testing your very knowledge of it asking for information like the subnet mask (given a prefix length), subnet, broadcast addr and so on...

I used these two every day:

Subnetting Quiz #1
Subnetting Quiz #2 (CIDR)

I can subnet in my head and in less than 15 seconds too

2URGSE

First of all, you need to be comfortable wit CIDR notation, you need to know it like the back of your hand.

172.16.112.1/25 --> 172.16.112.1 with a mask of 255.255.255.128

172 16 112 1
255 255 255 128

Rules of engagement:

If subnet # = 255, leave the # above it alone.

If subnet # = 0, the number above it will be zero

If subnet # is neither 255 or 0, this is your special octet.

So in this case, your special octet is the last one, since it's not 255 and it's not 0.

Take that #, subtract is from 256.

256-128 = 128, that's your magic #.

Back to the IP address, the # you were dealing with is 1.

You start counting from 0, than add 128 so...

0
128
256....etc.

But!, the value you need to get to it 1, so 128 is too big. 0 is the only option left.

So there's your answer, first subnet is 172.16.112.0.

Want to find the broadcast address? Add the magic # to the last octet, so it's 128, subtract 1, Broadcast is 172.16.112.127.

Valid range for this subnet is 172.16.112.1-172.16.112.126.