Your task is to troubleshoot a user’s PC with IP 192.168.100.1/28. Based on the given details choose the number of the network (N), subnet (S), and host (H) bits.

(A) Network Bits = 28 | Subnet Bits = 1 | Host Bits = 2
(B) Network Bits = 20 | Subnet Bits = 5 | Host Bits = 7
(C) Network Bits = 24 | Subnet Bits = 4 | Host Bits = 4
(D) Network Bits = 30 | Subnet Bits = 1 | Host Bits = 1

Network Bits = 24 | Subnet Bits = 4 | Host Bits = 4 is the correct answer.

The size of the network part is always either 8, 16, or 24 bits, based on whether it is Class A, B, or C, respectively.
In this case, we have a Class C address, so the network part is 24bits. N = 24
The prefix format is /28 so the mask is 255.255.255.240.
The number of subnet bits is the difference between the prefix length (28) and N (24).
S = prefix – N
S = 28 – 4 bits
S = 4
The number of the host bits is the total number of IP address bits(32) which is always 32bits minus the prefix length(28).
H = Address bits – prefix
H = 32 – 28
H = 4