I am current taking the CCENT Course and I am viewing the video "Understand and apply subnetting Part 1 and 2"

and I am having a very difficult time following a particular subject in the video.

One of the examples is how do you make 8 networks with the provided IP below:

192.168.10.0

I understand how you start it, I understand converting it to binary. The part where I get lost is when they say we need 12 bits to make 8 networks. (How are they coming up with 12 bits)

Then in the video he goes with 3 bits? that does not make any sense to me?

00001100 (The one's being the 8 and 4 place holder value) wouldn't that be 4 bits(The last four digits being used

I don't understand that part, can someone assist

]]>I am current taking the CCENT Course and I am viewing the video "Understand and apply subnetting Part 1 and 2"

and I am having a very difficult time following a particular subject in the video.

One of the examples is how do you make 8 networks with the provided IP below:

192.168.10.0

I understand how you start it, I understand converting it to binary. The part where I get lost is when they say we need 12 bits to make 8 networks. (How are they coming up with 12 bits)

Then in the video he goes with 3 bits? that does not make any sense to me?

00001100 (The one's being the 8 and 4 place holder value) wouldn't that be 4 bits(The last four digits being used

I don't understand that part, can someone assist

]]>My assumption is that I'm a poor communicator with a poor example. I probably thinking about another problem as I said something like that. It wasn't intentional.

So, let me try again, with the following:

To create the 8 eight networks, you must borrow 3 bits from the host bits portion, when you do, you first determine the new subnet mask.

I'm assuming I gave a subnet mask with the example. If we start with a 255.255.255.0 (default subnet mask for a Class C). I would only need to borrow 3 host bits. Which would mean 255.255.255.224 as the subnet mask. From that point. I would change the 4th octet to binary. Then within the 1st 3bits I borrowed, I would come up with every possible combination:

192.168.10.**000**00000 = 192.168.10.0

192.168.10.**001**00000 = 192.168.10.32

192.168.10.**010**00000 = 192.168.10.64

192.168.10.**011**00000 = 192.168.10.96

192.168.10.**100**00000 = 192.168.10.128

192.168.10.**101**00000 = 192.168.10.160

192.168.10.**110**00000 = 192.168.10.192

192.168.10.**111**00000 = 192.168.10.224

From there you should be able to determine the following for each network

- Network ID (above)
- range of usable IP addresses within each range
- Broadcast ID
- subnet masks

You did a good job explaining, I just still don't understand where you are coming up with the 3 bits. Where is that number coming from? How do you know you need to borrow 3 bits from the host portion?

]]>To create 8 **additional** networks from the current given single network. You can only do so if you subdivide that network. In binary, we do that by taking bits from the Host bits. So with a 255.255.255.0= 1 network, but the last 8 bits are the host portion.

If I need an additional 8 networks, I must "steal" or "borrow" enough bits to create the 8 networks. This requires me ask how bit must I have to do so? 2^3 = 8 networks. So we borrow 3 bit and add that to the subnet mask= `255.255.255.11100000=255.255.255.224`

.

It also means that we must begin with the original given networks and now come up with every possible combination- seen the previous post. It is then we can create the 8 networks as seen.

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