Subnetting Help

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This is the method I have been using for subnetting.

192.168.10.0/25
192.168.10.0 Network ID
/25 CIDR = 255.255.255.128

Because 192 falls between 192.0.1.1 to 223.255.254.254 it is defined as a class C address.
(Reference: IP Reference)

Given that it's a class C then it follows this convention Network.Network.Network.Host
First Question: How many Subnets?
128 in binary = 1000 0000
So 2^1 = 2 subnets (only 1 bit turned on)

Next, How many Hosts?
2^7 - 2 = 126 (the minus 2 is to take away the 0 and 255 -- network and broadcast)

Ok, now what are the valid subnets?
256-128 = 128 so 0 and 128

Now, the broadcast for each subnet and the valid hosts

Subnet 0 128
First host 1 129
Last host 126 254
Broadcast 127 255

OK, I'm fine with that.... However what's with this classless subnetting?

Also, when I go to practice problems at:
Subnetting Practice Generator
What's going on?
138.36.41.10/28
Shouldn't this be defined as a class B address? But when I check my answers, why does it have the form network.network.network.host? (shouldn't it be network.network.host.host?), also why doesn't this calculator show the result for each subnet instead of just the 0?


Thanks in advance for the help!

CIDR was designed to replace classes. There will be addresses in CIDR block 25 that correspond to all 5 of the old classes but none of the classes have 25 bit subnets (well D and E could, but you know what I mean).

Thanks, so when you I see CIDR notation (/x) think "classless"?

Also, if I am just given a IP address and subnet, how can I tell if it's classful or classless?

Could you please do 138.36.41.10/28 and show your steps?

Thanks again.

If they give you a subnet mask or you see CIDR notation then think classless. Pretty much always think classless for subnetting questions (classes are just special cases of classless subnetting: /8, /16, and /24 for A, B, and C).

So 138.36.41.10/28 is a sub-netted class C address (it has a 24-bit network part (making it class C), a 3-bit subnetwork part, and a 5-bit host part).

3 bits gives you 8 subnets (2^3) and 5 bits gives you 30 hosts per subnet (2^5 - 2).

Valid subnets are 0 (0b000000), 128 (0b10000000), 64 (0b01000000), 192 (0b11000000), 32 (0b00100000), 160 (0b10100000), 96 (0b01100000), and 224 (0b11100000).

I'm just going to do the broadcast and hosts for one subnet, 64:

Subnet: 64 (0b01000000)
First Host: 65 (0b01000001)
Last Host: 94 (0b01011110)
Broadcast: 95 (0b01011111)

"a 3-bit subnetwork part, and a 5-bit host part)."

Can you please tell me how you determined the subnetwork portion and the host portion?

Thanks again.

The /28 indicates that the network part is 28 bits. This takes up the first three bytes and three of the eight bits from the last byte. This leaves the last 5 bits for the host part.