Which two of the addresses below are available for host addresses on the subnet 192.168.15.19/28?

A. 192.168.15.17
B. 192.168.15.14
C. 192.168.15.29
D. 192.168.15.16
E. 192.168.15.31
F. None of the above

Answer: A, C


Explanation:
The network uses a 28bit subnet (255.255.255.240). This means that 4 bits are used for the networks and 4 bits for the hosts. This allows for 14 networks and 14 hosts (2n-2).
The last bit used to make 240 is the 4th bit (16) therefore the first network will be 192.168.15.16. The network will have 16 addresses (but remember that the first address is the network address and the last address is the broadcast address). In other words, the networks will be in increments of 16 beginning at 192.168.15.16/28. The IP address we are given is 192.168.15.19. Therefore the other host addresses must also be on this
network. Valid IP addresses for hosts on this network are:

192.168.15.17-192.168.15.30.


Incorrect Answers:
B. This is not a valid address for this particular 28 bit subnet mask. The first network address should be 192.168.15.16.
D. This is the network address.
E. This is the broadcast address for this particular subnet.