Networking - Networking Basics - Discussion

Thank you @Lavanya S S.

Why other options are in correct, anyone can explain it briefly?

Here are the answers using 255.255.255.192:

How many subnet bits are used in this mask? Answer: 2 2^2-2=2 subnets.

How many host bits are available per subnet? Answer: 6 2^6-2=62 hosts per subnet.

What are the subnet addresses?Answer: 256-192=64 (the first subnet)64+64=128 (the second subnet)64+128=192. However, although 192 is the subnet mask value, it's not a valid subnet. The valid subnets are 64 and 128.

What is the broadcast address of each subnet?Answer: 64 is the first subnet and 128 is the second subnet. The broadcast address is always the number before the next subnet. The broadcast address of the 64 subnet is 127. The broadcast address of the 128 subnet is 191.

What is the valid host range of each subnet? Answer: The valid hosts are the numbers between the subnet number and the mask. For the 64 subnet, the valid host range is 64-126. For the 128 subnet, the valid host range is 129-190.

I can't understand. Please anyone help me.

I think answer should be B. Because 2nd subnet should be 128-192. I. E 128 is network address and 192 is broadcast address. Hence valid host range should be 129-191. So answer is B.

Subnet should be C 192.168.168.128-190.

Thank you @Jitendra.

Where did 156-192 come from?

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.

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.