A server administrator has been given the following information from the network team in order to configure IP
addresses on a new server:
Network Address: 199.90.10.32/28
The network team also instructed the server administrator to use the LAST usable address on the given
network. Which of the following IP address/subnet mask combinations should the server administrator use?
A.
199.90.10.49, 255.255.255.240
B.
199.90.10.47, 255.255.255.224
C.
199.90.10.47, 255.255.255.240
D.
199.90.10.48, 255.255.255.240
Answer C is not correct. The calculation is as follows -> CIDR /28 has 4 bits to the max of 32. Thats 2^4 which is 16. If you take the first number for network address and last number for broadcast address, you are left with 14 possbile host addresses. Then we can conclude that the address: 199.90.10.32/28 is the network address, 199.90.10.33 – 199.90.10.46 are host addresses, and the 199.90.10.47 is the last address used as a broadcast address. Thus the conclusion is that a server administrator would be able to use that LAST usable address on the network which is 199.90.10.46 and NOT 199.90.10.47 because that is the broadcast address of the /28 subnet.
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I agree, the last usable IP is .46 because the multiplier is 16.
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