Subnetting Question

So I've been battling subnetting for the last couple days as I'm learning the material for the Net+ exam. I read the quick guide here, I've looked at professor messers and trainsignals videos. I have a couple charts and for the life of me, I still can't grasp the rules for figuring out host address ranges based on specific subnets. Here's an example.

Question: What is the first valid host on the subnetwork
that the node 192.168.65.232 255.255.255.240 belongs to?

Answer: 192.168.65.225

How does one come up with this? I'm stumped...

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Comments

earweed

First off your block size will be 16 (256 - 240 from the subnet address). Your valid subnets are 192.168.65. 0, 192.168.65.16, 192.168.65.32, (skip a few for brevity)...192.168.65.224, 192.168.65.240 (which is too big) and your subnet will be 192.168.65.224 which has address range of 192.168.65.224 - 192.168.65.239. You can't use the subnet address (or the last address of your subnet as that's the broadcast address) so the first valid host is 192.168.65.225

djfunz

Thank you, thank you! I think I'm gonna try a few more to be sure but I am starting to get it. I was not grasping the fact that the number of networks on a specific network was actually the "block size" that one is allowed to use in the range of a subnet. Since they give the node information in the question, then I know that I can't use the range from 192.168.65.80 - 192.168.65.96.

So if the question was, what is the first valid host on the subnetwork
that the node 192.168.65.92 255.255.255.240 belongs to?

Then the answer would be 192.168.65.81. Right?

I think I understand better now. Thanks earweed.

Darril

Earweed's method is certainly valid. Another method I sometimes use is to bit it out longhand.

For this, you'd need to know the network ID of 192.168.65.232 255.255.255.240.

240 is 1 1 1 1 0 0 0 0
so the first four bits of this octal are used in the subnet mask.

The decimal values of these bits are 128 64 32 16 (we're not concerned with the last four at the moment since we're only looking for the network ID).

232 is 1 1 1 0 1 0 0 0 but the subnet mask is 240, so the focus is only on the first four bits.

1 1 1 0 0 0 0 0 = 224 This gives you the last decimal in the network ID.

The network ID of 192.168.65.232 255.255.255.240 is 192.168.65.224.
The first IP address in the range is the network ID + 1 or 192.168.65.225.

You can bit out all the network IDs with
0 0 0 0 Decimal 0, mask is 192.168.65.0, first IP 192.168.65.1
0 0 0 1 Decimal 16 mask is 92.168.65.16, first IP 192.168.65.17
Note: This also tells you the last IP in the previous subnet. It's the network ID - 2
192.168.65.16 - 2 = 192.168.65.14
In other words the range of the first subnet is 192.168.65.1 through 192.168.65.14

0 0 1 0 Decimal 32, mask is 192.168.65.32, first IP 192.168.65.33
. . .
1 1 1 0 Decimal 224, mask is 192.168.65.224, first IP 192.168.65.225
1 1 1 1 Decimal 240, mask is 192.168.65.240, first IP 192.168.65.241

There are many shortcuts, but I find that people often remember the shortcuts better after they've learned the longhand methods.

HTH,

Darril Gibson
Blog

Darril

djfunz,

I think you and I were writing our posts at the same time. Glad you got it, and yes, for 192.168.65.92 255.255.255.240 it is 192.168.65.81.

Darril

djfunz

I hear ya Darril. I appreciate the shortcuts and charts. They are helpful and seem to work but my problem is there are so many methods out there and everyone seems to explain it a different way. I can't of course use these calculators and charts on the test so it's down to understanding and memorization I guess.

Darril

Is that true that the calculator is not available on the exam? It's been a long time since I took the Network+ exam but I recently took the SY0-301 exam and when it showed a subnetting question, it provided access to a calculator.

Darril Gibson

KrisA

djfunz wrote: »

They are helpful and seem to work but my problem is there are so many methods out there and everyone seems to explain it a different way.

I hear that, after reading 200+ pages(from different sources) on the topic is can become all one big garbled mess.

drkat

Question: What is the first valid host on the subnetwork
that the node 192.168.65.232 255.255.255.240 belongs to?

192.168.65.232 since our subnet mask is 255.255.255.240 we're only concerned with the last octet

232 = 11101000 in binary so we use the left most significant bits to determine our network value

240 = 11110000
==================
11100000 = 224

Sorry forgot to actually put the math in there

DigitalZeroOne

Here is a good way to remember this:

128
64
32
16
8
4
2
1

128
192
224
240
248
252
254
255

The bottom line are subnets, the top line is your increment.

You want the valid host on 192.168.65.232 - 255.255.255.240

You can't do anything with the first 3 octets, so we are only working with the last one which is 240. If you look, 240 increments by 16

So it would go something like this 192.168.65.0 - 192.168.65.15
192.168.65.16 - 192.168.65.31
192.168.65.32 - 192.168.65.47

In this example, remember, you need a subnet that can't be used, and a broadcast that can't be used.
192.168.65.0 is the Subnet, 192.168.65.15 is the broadcast
192.168.65.16 is the Subnet, 192.168.65.31 is the broadcast...and so on.
So you get the ranges of 192.168.65.1 - 192.168.55.14, 192.168.65.17-192.168.65.30 and so on.

If we kept going we would get to:

192.168.65.224 - 192.168.65.239
192.168.65.240 - 192.168.65.255

Remember, you can't use the first or the last, so you get 192.168.65.225 - 192.168.65.238 as your ranges and 192.168.65.241 - 192.168.65.254

So for the machine that has 192.168.65.232 with a subnet mask of 255.255.255.240 it falls in between 192.168.65.224 - 192.168.65.239, you can't use the very first subnet and the very last broadcast so just look at the range.

I hope this looks good once I hit reply.