Networking - Networking Basics - Discussion

How do know the subnet mask is /26? Please explain the concept in clear.

S1
192.168.168.188.
255.255.255.192.

Net id 192.168.168.128.
1011 0010.
1100 0000 AND LOGIC.
100000. = 128.

S2 255-192 = 63.
BROAD CAST 192.168.168.191.
.129 TO .190 is my valid host range.

188 = 10111100.
192 = 11000000.
255.255.255.192 means that network id will be of 26 bits and host id will be of 6 bits
we know that 2^6=64 IP addresses can be generated.

10"111100" is the binary form of 188

Put all zeroes in the double quotes, we get the first IP address which is the network address i.e 192.168.168.128.

So the next IP will be 192.168.168.129
put all ones in the double quotes, we get the last IP address, which is the broadcast address i.e 10"111111", 192.168.168.191.

So the valid IP range will be excluding the network id and broadcast id.
Therefore, 192.168.168.128-192.168.168.190 is the required solution.

The answer is correct.

Subnet Mask = 255.255.255.192 where. CIDR /26 means, 11111111.11111111.11111111.11000000.
The default subnet mask is 255.255.255.0, So, We can only change the last byte.

Subnet = 2^2 = 4 Total Subnets.
Valid IP Addresses = 2^6 = 64 Valid IP's.
Valid Host Ranges = 2^6 - 2 = 62 Host addresses.

Subnet Number ----- First network adrs ----- Valid Host Range ----- Broadcast adrs
1. 192.168.168.0 ----- 192.168.168.1 ----- 192.168.168.62 ----- 192.168.168.63
2. 192.168.168.64 ----- 192.168.168.65 - ----- 192.168.168.126 ----- 192.168.168.127
3. 92.168.168.128 ----- 192.168.168.129 - ----- 192.168.168.190 ----- 192.168.168.191( 192.168.168.188 is here)
4.192.168.168.192 ----- 192.168.168.191 ----- 192.168.168.254 ----- 192.168.168.255.

Hence, Ans is correct.

Answer (A) is right.

Because valid host means usable ip.
The CIDR is;

Subnet=192.168.168.128
First Usable IP= 192.168.168.129
Last usable IP is= 192.168.168.190
Broadcast IP is= 192.168.168.191.
so, answer is correct.

The subnet mask is 26, but the actual number of bits is 32 so 32-26 =6 bits for the one subnet. So 2^6 is 64 one subnet will have 64 addresses.

So, 0-63,64-127, 128-191 and 192-255 is the 4 subnets, in that 128 lies in 128-191 subnet.

This is the trick for this one.

The given is 192.168.188.188 with a subnet of 255.255.255.192. This subnet is /26. You should know the value of the increment of each subnet mask. In this case, it's 64.

Since it's in the fourth octet, we can now find the CLOSEST number which can be divided by the increment value. Only below the range of the given IP.

In this case, 128 is the CLOSEST number to the given 188 that CAN BE DIVIDED by the INCREMENT VALUE which is 64.

So we know that the IP will be 192.168.188.188 /26.

Since the increment value is 64 which is also the number of IP addresses, we will just ADD 64 to 128. Which will give us the last IP 192.168.188.192.

So, what do we have now?

First IP: 192.168.188.128 which is the NETWORK ADDRESS.
Last IP: 192.168.188.192 which is the BROADCAST ADDRESS.

These 2 IPs cannot be used.

So the usable IP range will be 192.168.188.129 to 192.168.188.191

Always remember that the first and last IP of the IP range is the Network and Broadcast addresses, respectively. And they are cannot be used.

I hope this will be a clear solution.

IP address is 192.168.168.188 .188 in binary is (1 0 1 1 1 1 0 0) =.188.
Subnet mask is 255.255.255.192 .192 in binary is (1 1 0 0 0 0 0 0) =.192.
as .255 is all ones, last octet of first network ID will be (1 0 0 0 0 0 0 0 )=.128.
with 256 - 192 = 64, N=(64) which is our magic number. N-2 = 62 hosts/network.
first N/W ID 192.168.168.128,

1st Host ID 192.168.168.129.
last Host ID 192.168.168.190.

Broadcast 192.168.168.191.
with (128+64=192), 2nd N/W ID will be 192.168.168. 192.

Where did 156-192 come from?

Thank you @Jitendra.