Hardest 4 questions from CCNA 1-2. Need Good explainer!!

remmy7593

The above ways of inverting your bits to get the broadcast will work fine.

For myself, the quickest way I have found is the following..I'll use this paste from a subnet calculator to explain.

The subnet is this example is based on the fourth octet...(The octet where the range is anything other than 255).

In this case it's 240. So, with that in mind, subtract 240 from 255 which gives you 15.

Add 15 to 32 (192.168.15.32)

This gives you your broascast of 192.168.15.47

To summarize, just subtract the last number in your mask from 255, and add it to the NETWORK number.

Address: 192.168.15.32
Netmask: 255.255.255.240 = 28
Wildcard: 0.0.0.15
Network: 192.168.15.32/28
Broadcast: 192.168.15.47

remmy7593

Another easy method for me in dealing with network increments, is very simple.

Let's say you have an address of 192.168.1.17 with a mask of 255.255.255.224

the mask in binary is 11111111.11111111.11111111.11100000

128 64 32 16 8 4 2 1

1 1 1 0 0 0 0 0

Your network number increments are always based on the position of the last "1" in the mask.

So with that in mind, 224 means that the last "1" is under the 32 postion. This means your network increments in 32.

If it was 240, your network would increment by 16. It's just that simple.

markzab

remmy7593 wrote:

Another easy method for me in dealing with network increments, is very simple.

Let's say you have an address of 192.168.1.17 with a mask of 255.255.255.224

the mask in binary is 11111111.11111111.11111111.11100000

128 64 32 16 8 4 2 1

1 1 1 0 0 0 0 0

Your network number increments are always based on the position of the last "1" in the mask.

So with that in mind, 224 means that the last "1" is under the 32 postion. This means your network increments in 32.

If it was 240, your network would increment by 16. It's just that simple.

I like that. Good call.

ishaq0925

the ip is classB and the defalt mask for it is 16 and the question show 192 mean /18 the networks can be 0 64 128 and 192 so how 151.100.255.1 is possible?

ishaq0925

the ip is classB and the defalt mask for it is 16 and the question show 192 mean /18 the networks can be 0 64 128 and 192 so how 151.100.255.1 is possible?

albinorhino187

151.100.255.1 is just one of the many addresses in the 151.100.192.0/18 network. That network includes all IPs from 151.100.192.0 - 151.100.255.255. Even though the network number starts at 192, that doesn't mean it's only 192. The mask is 255.255.192.0. Meaning for each subnet, first, the 255s don't change, so it'll always be 151.100.x.x. The 192 means each subnet's third octet covers (256 - 192 = 64) 64 numbers. And then the 0 in the fourth octet is all 256.

So the first IP in the 151.100.192.0/18 network is 151.100.192.1, then .2, .3, all the way to 151.100.192.255. And 151.100.192.255 isn't the broadcast IP, it's just another usable IP in the subnet. And the very next usable IP, which is a valid host IP, and still in the same 151.100.192.0/18 network, is 151.100.193.0. You could give a computer 151.100.193.0/18 and ping it. And then .1, .2 etc, and then you're at 151.100.193.255. The next ip...151.100.194.0. And you'll keep doing that, incrementing the third octet after the 4th octet rolls over, for the same 151.100.192.0/18 subnet until finally you're in the 151.100.255.X range. At this point, you'll go through the .1, .2, etc, and then at 151.100.255.254 is the last usable IP in the 151.100.192.0/18 network, and 151.100.255.255 is the broadcast.