How many IP addresses are in a "IP Address Block (/28)"?
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This is a follow-up question to .
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Answer:
The /28 tells you the size of the subnet mask. This is a mask value that you can apply in binary to find out which bits of the 32 bit address are used for network selection and which are used for host selection. 28 in this sense tells you that the mask is 11111111 11111111 11111111 11110000 (28 1's and 4 0's) For any given IP address, if you AND the mask and the address together, you get the network id, and if you NAND the mash and the address together you get the host id. Because you only have 4 bits left over to address the hosts that gives you a theoretical maximum of 2 to the power 4 (2*2*2*2) or 16 addresses. But since the address that is all 0's and the address that is all 1's is reserved, you only have 14 usable addresses in this block. If your network was 192.168.1. (4) then your broadcast for that network would be: 192.168.1.79 (which is 01001111 in decimal), and your network address would be 192.168.1.64 (01000000), everything between that range is in the same subnet.
Michael Brunton-Spall at Quora Visit the source
Other answers
4 bits of host address means 2^4 possible addresses (16). The all-ones host address is reserved for broadcast, and thus cannot be used for an actual host address, ergo you can have 15 hosts on a LAN with a /28 address block. The host zero address used to be a problem (it was used in old software as a broadcast address); see https://tools.ietf.org/html/rfc1122#section-3.3.6, section 3.3.6
Erik Fair
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