"The Great Exception" --CBTNuggets

Kingfisher

I was going well with subnetting learning from CBTNuggets videos with the practice problems and I was getting all of them correct. However, now that I come to the "Great Exception" section, I find myself lost. They are basically stating:

"When subnetting based on the number of networks, SUBTRACT 1 from the number.
When subnetting based on the number of hosts per network, ADD 1 to the number."

I mean its a simple rule to just blindly follow, but the explanation is hard to digest and frankly, the accompanying document for this is not clear as well.

For example, in the document it states that "If I were to break a network into 8 subnets, you would assume it takes four bits since 8 in binary is (00001000). However you can achieve this requirement with only three bits since 0-7 is really 8 numbers (0,1,2,3,4,5,6,7). If you work out the problem by reserving only three bits, you will get exactly eight subnets."

The problem is that I solved this problem using 3 bits like they state and I do get 8 subnets. However, in previous problems, I did not follow this rule (subtract 1 for networks) and I got correct answers (according to the video). How is using 4 bits for 8 networks now wrong and now I need to use 3 bits. And if so, how come this wasn't taught at the very beginning of the section. Or, do I need to follow this if the exam asks for specific numbers.

If anyone can help me get this through my head I would appreciate it

dynamik

IMHO, these "short-cuts" are more confusing and convoluted than they're worth.

All I've done is memorize the values associated with 1-12 bits.

bits - networks (or hosts if you subtract 2)
1 - 2
2 - 4
3 - 8
4 - 16
5 - 32
6 - 64
7 - 128
8 - 256
9 - 512
10 - 1024
11 - 2048
12 - 4096

You don't even have to memorize all of them. The previous bit is always going to be half the value of the one you're on, and the next bit is going to be double the value. If you just remember that 5 is 32 and 10 is 1024, you can double or halve your way to any value you need. I don't see any reason why you need to mess around with any of those tricks. If you need 54 networks, well you'll obviously need six bits. The only change you have to make with hosts is to add two to whatever the required number of hosts is in order to accommodate the broadcast and network addresses. For example, if you were required to have 127 usable hosts, you'd have to add two and go with eight bits.