Networking - Subnetting - Discussion

Good, Thanks all.

16 hosts < 32, 32 = 2^5, so it takes 5 bits, and each subnet has 30 hosts.
32 " 5 = 27, it means the subnet-mask has /27 bits.

5 subnets < 8, 8 = 2^3,
therefore, 27 " 3 = 24 bits,
= 255.255.255.0.

8 subnets x 32 IPs " 2 = 254, the answer should be 255.255.255.0.

For example class c IP, 192.168.1.1/24 with subnet-mask 255.255.255.224.

16 hosts < 32, 32 = 2^5, so it takes 5 bits, and each subnet has 30 hosts.
32 " 5 = 27, it means the subnet-mask has /27 bits.

5 subnets < 8, 8 = 2^3,
therefore, 27 " 3 = 24 bits,
= 255.255.255.0.

8 subnets x 32 IPs " 2 = 254, the answer should be 255.255.255.0.

For example class c IP, 192.168.1.1/24 with subnet-mask 255.255.255.224.

4 subnets, each with at least 20 hosts. Which subnet mask? Please explain about it.

5 subnets 16 hosts

First, you have to meet the number of the host which is 16. By borrowing 4 bits we will meet 16 hosts but there will be no room for the network and broadcast address so we will use 5 bits that has 32 hosts. Using the 2^5 -2 will give us 30 which satisfy our required number of hosts.

Secondly, we will use the formula an actual number of network = 32 (by default) minus no. of borrowed bits so 32-5 equals 27.

Our new subnet mask is /27 which is equivalent to 255.255.255.224.

I need formula 2^16=32 value of network.

So I taken 128.64.32. value mention after another 0.0.0.0.0 calculate 128+64+32=224.
Ans 255.255.255.224.

225.255.255.192=/26 so 24+2(2N we need to borrow) formula 2^2=4 subnets.
for hosts 256-192=64 hosts.
255.255.255.224=/27 so 24+3(3n we need to borrow)formula 2^n=2^3=8 sub.
for hosts 256-224=32 hosts.
255.255.255.240=/28 so 24+4(4N we need to borrow) formula 2^N=16 subnets.
for hosts 256-240=16 hosts.

@Roopa

Why the best value for h is 5?

Isn't it would be 9 because h=18/9.

Thank you @Peeyush.

Prove if a host on a network has the address 172.16.45.14/30, the broadcast address of the subnet is 172.16.45.15.