Networking - Subnetting - Discussion

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.

As we require 16 Host in each subnet. We will first check Binary Table.

128 64 32 16 8 4 2 1.

If we take 16 we will not get 16 usable IP we will only get 14 usable IP. 2 ID will be taken for Subnet id and Broadcast ID.

So will take 32.

Now if we check mask it will be created like below:
11111111.11111111.11111111.11100000

By converting this into decimal it will then become: 255.255.255.224

This is the right answer.

Class c default subnet mask \24 like 255.255.255.0

Example:

Converting bit 1, n=2^1=2.

Converting bit 2, n=2^2=4.

Converting bit 3, n=2^3=8.

So first converting bit 1 is two subnets, second one is 4 subnets and third one is 8
Subnets support. We want 5 subnets so by using 3 converting bits.

Default network bits+converting bits, N+n = 24+3 = 27.

By using subnetmask 255.255.255.224

255.255.255.224 in binary would be written as:

11111111.11111111.11111111.11100000.

To find number of host:

Starting from the octet consisting both 1 and 0.

Count number of zeros to the end of the subnet mask.

The last octet consist of 5 zeros therefore.

2^h-2 (h = number of zeros) = 2^5-2 = 32 -2 = 30.

To find number of sub nets:

2^n (n = number of ones) 2^3 = 8.

Suppose you are given the following class C IP address: 192.100.1.1. You need to perform Subnetting in such a way that you get 3 (three) subnets such that each subnet contains 62 (sixty two) valid hosts.

You need to specify the network and broadcast address of each subnet and the valid host range in each of the subnet. Please solve it as early as possible.

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.

Suppose you are given the following class C IP address: 192.100.1.1. You need to perform Subnetting in such a way that you get 3(three) subnets such that each subnet contains 62 (sixty two) valid hosts. You need to specify the network and broadcast address of each subnet and the valid host range in each of the subnet.
solve it and give answer

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.

Please tell me if I'm wrong, but another way to think of it is:

128 64 32 16 8 4 2 1.

1 1 1 0 0 0 0 0.

32 is the smallest bit used so.

32 - 1 (domain) - 1 (server) = 30.

Honestly, I think I remember seeing it this way in a tutorial, so if I'm wrong please correct and/or email me.