Is Subnetting hard?

[r1ckyr4y EBK]

So, I'm currently working on my A.S. and then eventually bachelors. I'm only on my 3rd semester and I got stuck in a more advanced class because none of the ones I should have started with were available for summer. We are doing Subnetting and I have no problem finding resources online to use, I was just wondering if anybody here knows if its going to be super difficult for me or not. My professor is a bit of a douche with a "you should've had IT experience before taking my class, you should know this" attitude. Well, I'm going to school to get the damn experience. Anyways, here is a part of the assignment.. I don't want the answer outright.. but if anybody has a link to a resource that I can use to learn it would be very much appreciated.

To help develop your subnet-calculation skills, you are now challenged with a practice subnetting exercise.
Using a separate sheet of paper, solve the following scenario:
Based on your network design requirements, you determine that you should use a 26-bit subnet mask applied to your 192.168.0.0/24 network. You now need to calculate each of the created subnets. Additionally, you want to know the broadcast address and the range of usable addresses for each of the created subnets.

[r1ckyr4y EBK]

here is the rest, just in case it doesn't make sense from what I gave you:

Advanced Subnetting Practice: Exercise 1 Solution
As described earlier, you can go through the following four-step process to determine the subnet address:
Step 1. The subnet mask (in binary) is 11111111.11111111.11111111.11000000.
The interesting octet is the fourth octet because the fourth octet contains the last 1 in the subnet mask.
Step 2. The decimal value of the fourth octet in the subnet mask is 192 (11000000 in decimal). Therefore, the block size is 64 (256 – 192 = 64).
Step 3. The first subnet is 192.168.0.0/26 (the value of the original 192.168.0.0 network with the borrowed bits [the first 2 bits in the last octet] set to 0).
Step 4. Counting by 64 (the block size) in the interesting octet (the fourth octet) allows you to calculate the remaining subnets, resulting in the following subnets:
192.168.0.0
192.168.0.64
192.168.0.128
192.168.0.192
The directed broadcast addresses for each of the preceding subnets can be calculated by adding 63 (that is, one less than the block size) to the interesting octet for each subnet address. Excluding the subnet addresses and directed broadcast addresses, a range of usable addresses can be calculated, the results of which are seen in Table 1 .
Complete the table by replaing the X with a decimal value
Table 1 Usable IP Address Ranges for the 192.168.0.0/26 Subnets

[Omega]

Subnetting is actually very easy once you figure it out. Most IT courses give only a few lessons in subnetting and then you are expected to know how it works.

You can learn it by watching a few videos online.

You should be able to learn it in 15-30 minutes, the hardest part is keeping this memorized.

[BloodShed]

Omega lähetti viestin:

.... the hardest part is keeping this memorized.

I remeber the time I had to memorize the OSI model as part of a Cisco course.

Always make notes to make it easier to remember.

[Washell]

It's not hard. Just practice converting binary to decimal and back for an octet and learn the basic steps.

Honestly, if your logic and math skills have trouble with subnetting, you should reconsider your career choice.

[r1ckyr4y EBK]

Washell lähetti viestin:

It's not hard. Just practice converting binary to decimal and back for an octet and learn the basic steps.

Honestly, if your logic and math skills have trouble with subnetting, you should reconsider your career choice.

I never said I had trouble with it, this is my first time dealing with it and I just wanted opinions on the difficulty. No need for the sassy comment.

[Washell]

It's implied by coming here and asking if it's hard.

[Aliquis Freedom & Ethnopluralism]

The description seem unnecessarily complex. Then again I don't know how natural this comes for you.

I guess you are aware that 0-255 is done with 8 bits and as such 8+8+8+8 bits = 32 bits. 24 bits is the first three numbers. As the subnet mask is 26 that will cover two more binary digits and with that you can do four numbers, or if you so will each subnet will have 256/4 = 64 adresses.

Your subnets will be:
192.168.0.0 with a broadcast address of 192.168.0.63 and I guess the usable range is 192.168.0.1-196.168.0.62.
192.168.0.64 with a broadcast address of 192.168.0.127 with a usable range of 196.168.0.65-192.168.0.126.
192.168.0.128 with a broadcast address of 192.168.0.191 with a ...
192.168.0.192

Unless I guess/remember way too poorly.

[Aliquis Freedom & Ethnopluralism]

... if you wanted to do it binary then:
192: 11000000
168: 10101000
0: 00000000

So the first 24 bits are:
11000000.10101000.00000000

As you'll be using 26 bits the subnet addresses will start with:
11000000.10101000.00000000.00
11000000.10101000.00000000.01
11000000.10101000.00000000.10
11000000.10101000.00000000.11

If 000000 for each of those are reserved for the network (is it?) and 111111 is a broadcast address then those are already used for that leaving everything inbetween.

So first network address: 11000000.10101000.00000000.00000000
First usable IP: 11000000.10101000.00000000.00000001
Last usable IP: 11000000.10101000.00000000.001111110
And broadcast: 11000000.10101000.00000000.001111111

Second network address: 11000000.10101000.00000000.01000000
First usable IP: 11000000.10101000.00000000.01000001
Last usable IP: 11000000.10101000.00000000.011111110
Broadcast: 11000000.10101000.00000000.01111111

And so on.

It's weird how it's 17-21 years or something since I studied such things, addresses was limited back then and IPv6 was a thing already and yet IPv4 is still used ...