Allocating IP Addresses in a Subnetted Environment
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9 hours 49 minutes
Now our next section is all about allocating these I p addresses in a submitted environment,
which means maybe you've created 10 subnets.
What are the I? P Range is valid in each sub net. The first thing we want to find is something called the Increment, which means in our network are significant. Octet will be implemented to a certain number.
Whatever that increments is is the number two, which we will add. I hope that makes sense, but let's take a look. Let's say I have the 10 network and it needs to be submitted so that I have 14 new subnets. I'll need a 255.24 oh sub net mask notice. I put a slash 12,
and that indicates 12 bits set to one in the sub net mask.
You can figure that out.
Eight bits and 255 4 bits and 24.
I'm going to start using that C. I. D. R. Notation more frequently because you want to get used to it.
So the sub net mask 255.240 point 0.0 with a significant octet to four. Oh,
there's a trick defining the increment. Here.
Take the number 256 a magic number. Just trust Kelly
and subtract the significant octet.
So 256 minus the significant octet is 16. As an instrument
for our first network idea, we start with a zero sum it. It's just 10.0 point zero.
Then the second network ID is 10.16. The 3rd, 10.32 then 10.48.
What's important? Remember each network increments in 16 in this case. So when you're looking to find the valid I P ranges everything from 10.0 point 0.1 through 10.15 point 255.254 will all be on the same network.
The next network ID at 10.16 point 0.0 has valid I P addresses of 10.16 point 0.1 through the highest address before moving to the next subject.
Why not 255.255? Because that is the broadcast address.
This is very typical of questions you'll see on the exam.
Here are a handful of subnets. What are the sub net masks? You'd use. And what are the increments
going back to the increments? I said 256 minus 240 Because 256 is magical.
It is to to the eighth power
to to the eighth power minus. How many bids we steal gives us the value we increments.
You can also look at this in binary if you want.
If you look at the address ranges, you'll see that up to the point of increment. The 16th character. Everything has to match. That may sound more complex, but I do want you to know you can prove this in binary.
What about more practice questions? What fun is a class without any test questions? Once again, I'd like you to pause the video, read and complete the questions. Then we'll go over them in a minute.
These questions take everything we've learned a step further. The first step is to figure out what sub net mask we should use going back to the chart. If you're unsure and remember it of stealing one bit and how many networks you get, be sure to have that down because without getting the sub net mask, right? Nothing will be right.
The sub net mask I need is 255.2 48 with 2 48 being significant. Opted.
Take the number 256 minus 248 and get the increment of eight.
My first network is 10.0 point 0.0 network, with the first valid I P address of 10.0 point 0.1 all the way up to the highest valid I p address 10.7 point 255.154
The next network is 10.8 then 10.8 point 01 through 10.15 point 255.254 I hope you got all these questions, right? I'll leave it here is you can pause and check your answers.
Look specifically at three and four, we have 192.168 point 1.0 network
by default. That is a class seat.
What that means is the first three octaves are already set and I'm stealing from the last opted.
I need support for seven subnets. I can get that if I steal three bits. If I steal three bits, my significant Oct it is 224.
The increment is 256 minus 224 32
Each network will increment in units of 32.
So for the I P addresses, the first network is 192.168 point 1.0, and the second I d will be 192.168 point 1.32.
So what's the I P address range? Between those,
the first is 192.168 point 1.1. The temptation is to say, 192.168 point 31 but 31 is the broadcast address
on a Class C network. Because we don't have any additional optics to steal from, we have to account for the network ID and the broadcast address. So the second range is 192.168 point 1.33. As the first valid address
all the way through 192.168 point 1, 10.62
Don't forget 0.1 point 63 is the broadcast address.
If we assume the same network as we saw in three increments of 32 the question asks, Are these two hosts on the same network?
I don't know. So let's take a look.
192.160 point 1.29 is on the 120 going back.
192.168 point 1.29 is on the 192.168 point 1.0 network, which is the first network ID containing valid addresses to 30.
The second I P address is on the 192.160 point 1.32 network.
When you're being asked if two hosts are on the same network, it helps to write these things out and figure out the network ID. Then you can figure out the 1st and 2nd range. In some cases, you may have to go up to the fourth range on the test.
Use these questions as practice and make sure you're solid.