Strange IP address

[zed255]

Excuse me if I'm just a little dense. Where is the 2^5=32 coming from?
IIRC the binary reads as: 87654321
where:
place marker 1 =1
place marker 2 =2
place marker 3 =4
place marker 4 =8
place marker 5 =16
place marker 6 =32
place marker 7 =64
place marker 8 =128
and the sum of i.e. 11111111 binary = 255 decimal

so 00100000 binary = 32 decimal, right? So what's 2^5=32?
I do notice the pattern, but don't understand your math. =^..^=

The last five bits of the IP address are available to specify a specific host. The math is 2 to the power of 5, or 2x2x2x2x2=32, two valid states per bit and five bits. Don't forget that logically speaking values of zero count as a valid state, so even though if we added up the values of the bits 16+8+4+2+1=31 we still have 32 possible values since zero is valid. The left bit is always the MSB and the right is the LSB, as you have laid out above. So in the example we have been using:

Subnet mask is 255.255.255.224 in dotted decimal and in binary is 11111111 11111111 11111111 11100000

The ones represent where the network values in the IP address would be and the zeros where the host values would be in the IP address.

IP address of 172.24.16.171 in dotted decimal in binary would be 10101100 00011000 00010000 10101011

Network 10101100 00011000 00010000 101 and host 01011

The first 27 bits, as specified by the ones in the subnet mask represent the network portion of the IP address and will be the same for all hosts on that subnet. The last five bits specify the host, and must be unique to each host.

[blackcat4866]

For reference have a look at the following subnet mask values. Note that for subnetting a class C address space these are the only valid subnet mask values.

255.255.255.000 - total 256 - available 254 - one subnet
255.255.255.128 - total 128 - available 126 - two subnets
255.255.255.192 - total 64 - available 62 - four subnets
255.255.255.224 - total 32 - available 30 - eight subnets
255.255.255.240 - total 16 - available 14 - sixteen subnets
255.255.255.248 - total 8 - available 6 - thirty-two subnets
255.255.255.252 - total 4 - available 2 - sixty-four subnets

This might illustrate the relationship better, hope it helps rather than confuses.

So it's 256 - 224 - 2 = 30 That's math that I can understand.
I was looking for something more mysterious.

Thanks for explaining. =^..^=

[zed255]

Hope I didn't cook too many neurons out there. I just wanted to try to explain subnetting in a little more detail to hopefully give some extra understanding of where the numbers come from.

That method works, I was originally taught the 'hard' binary method. So a quick reference formula that come out of this discussion would be:

256 - Subnet Octet - 2 = Available Addresses

Note this is only valid for a class C address scheme.

[blackcat4866]

Hope I didn't cook too many neurons out there ...

There aren't that many neurons left.
Please try to excuse my ignorance. I have collected a few odd facts along the way, but nothing organized.
I've just finished pasting these posts into a ReadMe file called: Class C IP addresses.

I appreciate your taking the time to pound it into my head. =^..^=

[zed255]

Ignorance is nothing more than a lack of knowledge or understanding, and for folks who are motivated it's easy to fix by providing the information and explanations required. I have all the patience in the world to assist someone who just needs a little information or explanation and really wants to understand. It's only when someone wants to have me think for them and spoon feed them every step of the way because of being lazy or just outright incapable of wrapping their head around the issue that it chaps my ass.

I enjoyed the more involved interaction on this thread, I thank you for sticking with it. Hopefully others benefit from it before it's buried too deep.

[JR2ALTA]

Next semester, Variable Length Subnet Masking

[blackcat4866]

Go ahead junior, but slowly. My remaining neurons are still smarting from the last bit. =^..^=