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Computer Networking Basics

A host has IP address 192.168.10.77/27. What is the broadcast address of its subnet?

  1. A192.168.10.95
  2. B192.168.10.127
  3. C192.168.10.255
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Answer: A. 192.168.10.95

/27 leaves 32 − 27 = 5 host bits, so each subnet holds 2^5 = 32 addresses and the fourth octet starts at boundaries of 0, 32, 64, 96 and so on. 77 falls in the range 64–95, so the network address is 192.168.10.64 and the broadcast address, with the host part all 1s, is 64 + 32 − 1 = 95, giving 192.168.10.95. The assignable host addresses are the 30 addresses .65–.94. The subnet mask for /27 is 255.255.255.224.

  • A. Correct. It is the last address of the 64–95 subnet.
  • B. This is the broadcast address of the 64–127 subnet, from mistaking the prefix for /26 (blocks of 64).
  • C. This is the broadcast address if the prefix is mistaken for /24.
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