Networking - Networking Basics - Discussion

128 is called a network address and 191 is called broadcast address.

Thanks it is easy to understand.

Not getting the solution, please explain me in detail.

The correct answer will be 192.168.168.128 - 191.

The answer is right they said "usable" ip address, so the network is 192.168.168.128 to .191, 128 is network address and 191 is broadcast address.

256-192 = 64 so increment by 64.

Networ - Broadcast
64 - 127
128 - 191 > so the usable host address range is 127 - 190 (answer is A).
192 - 255.

Everybody is confused because desimal binary conversation, here is the way to better understand this: subnet is all 1s but the last 6 digits determines hosts. You cant do anything for the fist 24 bits because they are fixed (255.255.255) you have play with the first 2 bits of the last octet, they can be:

00xxxxxx
01xxxxxx
10xxxxxx
11xxxxxx

In other word, you can create 4 different networks out of these bits, now convert them to decimal:
The first one's max value will be 63 (if all x were 1) which is not even close to 188.

Second posibility or network can be max 127 not close to 188.

Third one is good, it can be max 191 (don't forget you can't change the 0 on the second digit from left, it is belong to network portion not host, so max=1011111y. Here you cannot make y=1 because when it is 1 with other bits, it becomes network address.

11xxxxxx's min address is 192 so it will never contain 188 although other bits were 0.

It's simple guys.

Given IP address : 192.168.168.188.
Given subnet mask address:- 255.255.255.192.

To find the N/W ID
step1: Convert both addresses into binary form.
step2: Perform AND operation b/w them.
step3: Result will be your network ID i.e(192.168.168.128).
step4: So range of the N/W starts from 192.168.168.129 (starting host ID).
step5: The last id of the N/w is 192.168.168.191 (which is broadcast domain).
step6: So the last id of the N/W will be 192.168.168.190.

Thank you.

How you get 192.168.168.191 as broadcast id?

Hello, all.

The answer is obtained as follows:

IP: 192.168.168.188
Subnet: 255.255.255.192

192 in binary :
1 1 0 0 0 0 0 0
188 in binary:
1 0 1 0 1 1 0 0

192 AND 188:
1 1 0 0 0 0 0 0
1 0 1 0 1 1 0 0
Ans: 1 0 0 0 0 0 0 0

In 192 first 2 digits are 1 therefore in Ans 1 0 will be part of subnet id.

While remaining 6 will give host address;

First Host addr: 1 0 0 0 0 0 0 1 = 129.
Last host addr: 1 0 1 1 1 1 1 0 = 190.

And Broad cast addr will be : 1 0 1 1 1 1 1 1.

Hi, I am not getting this, Please explain it in detail.