Networking - Networking Basics - Discussion

Can anybody explain me this question in more elaborate way from the scratch?

First of all you have to find the base number for that the formula:-.

256-mask = base number will be used.
Now that will be 256-192 = 64.

Now to find the number of valid subnet's do the following.

Subnet 1: 64.
Subnet 2: 64+64 = 128.
Subnet 3: 128+64 = 192 (not valid as it comes upto subnet 's mask address).

So 2 valid subnets.

Next find the broad cast address that would be 127 and 191.

So first and last valid address of the subnet would be.

65, 129 (first valid address).
And
126, 190 (last valid address).

So the range would be 129 to 190.
Hence the answer. :).

@Richa.

Where did you get that 191?

As you said n/w add starts from 128 and end to 191 how?

To find valid host range, we need subnet mask.

Here subnet mask 255.255.255.192 (11111111.11111111.11111111.1100000).

So for 192 (128+64+0+0+0+0+0+0). Here magic number is 64.

0-63 (192.168.168.0 (n/w address) & 192.168.168.63 (broadcast address). So we shouldn't use as IP address in between we can use these IP addresses).

64-127.
128-191.
192-255.

Our question where 192.168.168.188 address lies in these range?

Answer is 192.168.168.129-192.168.168.190.

I hope you understood well.

Thank you.

Ravi sankar.

I really don't understand these answer correctly. Please explain it in brief.

Please explain other very simple method.

@Ravi Sankar.

Why you call 64 as magic number and why not 192? Please help me.

64 - 2 = 62 then b is the answer.

Hi, @All.

Subtract 192 from 256 = 64,

Means, the subnetting has been done by dividing into four parts, 64 * 4 = 256.
Now see the option.
64 * 2 = 128,
And 64 * 3 = 192.

So IP add should be in between.

192.168.168.129 to 192.168.168.191.

@Shyamnarayan.

Your answer is perfect with the simple solution.