Networking - Networking Basics - Discussion

Please explain other very simple method.

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

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.

@Richa.

Where did you get that 191?

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

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. :).

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

I don't know that how we can find the IP address so please explain in details?

192.168.168.188 255.255.255.192
mask 26.

And now have to find first address,
Applying AND operation with first address.

192.168.168.188
8 8 8 2

188 = 10111100.
And last 2 bit of mask,
11.

Hence first address,
192.168.168.128.

And last address,

Adding complement of mask into first address.

00000000 00000000 00000000 0011111111.
(63).

Since 1 address 192.168.168.128.

+ 0 0 0 63.

192.168.168.191.

sbnet mask=255.255.255.{192} equiv to [11111111.11111111.11111111.11]00 0000 in binary. Now the binary values in [] represents the n/w portion i.e. first 26 bits is allocated for n/w portion and remaining 6 bits for the host portion.

Now take a look into the IP address.. 192.168.168.188 where the 1st 26 bits represent d N/W portion. So the binary representation of IP address is [11000000.10101000.10101000.10]111100,the bits in [] again representing the N/W portion.

There are four imp. things dat can be concluded from an IP address given with subnet mask,namely,
-> N/W ADDRESS
Represent d IP address in binary form and convert d HOST portion into ZEROS,and leave d N/W portion as it is. for eg. 11000000.10101000.10101000.10"111100" (192.168.168.188) is changed to 11000000.10101000.10101000.10"000000"(192.168.168.128;represents d n/w address). NOTICE THE CONVERSIONS REPRESENTED BY ".."

->1ST USABLE ADDRESS
The next IP address immediately after the N/W address is d 1st usable address. For eg. 192.168.168.129 is the 1st usable address in this case.

->LAST USABLE ADDRESS
Note that d host portion is of 6 bits i.e.2^6(=64) hosts can be connected. So d host portion can be divided in 64 blocks each i.e 0-63,64-127,128-191..and so on. Here in this case,since the n/w add. started from .128 so it must end on .191. So the LAST USABLE IP ADDRESS IS THE IP ADDRESS WHICH COMES IMMEDIATELY BEFORE THE BROADCAST ADDRESS. So the last usable IP address is 192.168.168.190.

->BROADCAST ADDRESS
It refers to d last IP address of the block. In this eg, the broadcast address is 192.168.168.191 since d block is 128-191 as seen in the previous step.

I hope u must have understood clearly.

Subnet masks 255.255.255.192 means that it has 6 bits to represent hosts. which means in the last octet: first two bits represent network bits and other 6 host bits.

Now you are looking to fit 192.168.168.188.

Out of two bits you can have : either 10 or 11( 11 is not possible it will go beyond 192) so network bit is 10 , which makes it

128.( 255.255.255.128)

Now , possible values for hosts (for 6 bits) will start with: all zeroes to all 11s( but zero and all ones are meant for this network and broadcast)

Hence the possible values are: from 1 to 62

Which will be: 129 to 190