Question

Given the ip address 172.16.12.54 with the subnet mask of 255.255.255.240 , which of the following are valid host addresses on the same subnet ?

A: 172.16.12.64
B: 172.16.12.57
C: 172.16.12.49
D: 172.16.12.48
E: 172.16.12.63
F: 172.16.12.45

Answer 1

OK, do you know how to tell how many hosts are in each block based on the netmask?

Answer 2

For Class C and smaller networks, you are looking at the last octet. That means of the 255.255.255.x thing, you need to look at the x spot.

You start with 256 addresses from 0 to 255 (0 is the first, which is why it is 256 total). That is 255.255.255.0, or just .0 for making it short.

The next step is based off of taking half that. So, we have half of 256. That is 128. 0+128 is the new netmask and our blocks are now 128 addresses. So from 0 to 127 and 128 to 255.

Again, you take half that. Half of 128 is 64. 128+64 is 192, so the new netmask is .192. The blocks are now 64, so 0 to 63, 64 to 127, 128 to 191, and 192 to 255.

See the progression? The netmask is half the change in addresses added to the last netmask. So we do this again, and we have 32. 192+32 is what? 32 addresses make the blocks into what?

The best thing you can do is make a quick chart. Work on writing it out quickly so that if you need to do a test, you just work out the chart at the start and use it through the whole test. You may not even need to fill in all the start and stop blocks. Just get some size where it is easy to find the rest, and from the top and bottom of each of those blocks you can find the others very rapidly. At blocks of 16 or 32, this begins to happen. No matter what method you use, it should work out.

Answer 3

By progrssibly cutting each box in half you can make a chart. Upper left becomes the network address, the lower right the broadcast, and ebverything between those is a usable hoast address.