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Difficulty: EasyIPv4 Addressing and Subnetting

A network administrator assigns an IP address of 192.168.50.77/27192.168.50.77/27 to a server interface. What is the broadcast address for the subnet to which this server belongs?

  1. A
    192.168.50.63192.168.50.63
  2. 192.168.50.95192.168.50.95Answer
  3. C
    192.168.50.96192.168.50.96
  4. D
    192.168.50.127192.168.50.127

Answer

The broadcast address for the subnet containing host 192.168.50.77/27192.168.50.77/27 is 192.168.50.95192.168.50.95.
A /27/27 subnet prefix leaves 55 host bits, yielding a subnet block size of 25=322^5 = 32 addresses. Subnets increment by 3232 in the fourth octet (192.168.50.0192.168.50.0, 192.168.50.32192.168.50.32, 192.168.50.64192.168.50.64, 192.168.50.96192.168.50.96). The host address 192.168.50.77192.168.50.77 lies between 192.168.50.64192.168.50.64 and 192.168.50.95192.168.50.95. The final address in this range, 192.168.50.95192.168.50.95, is the broadcast address.

Step-by-Step Solution

1
Determine the subnet block size from the prefix length.
A /27/27 prefix leaves 32−27=532 - 27 = 5 host bits. The subnet block size (increment) is 25=322^5 = 32.
Calculating the block size identifies the boundary increments in the fourth octet.
2
Identify the subnet network boundaries in the fourth octet.
Subnet network addresses increment by 3232: .0,.32,.64,.96,.128.0, .32, .64, .96, .128, and so forth.
Listing the multiples of 3232 establishes exact subnet boundary ranges.
3
Locate the specific subnet range containing 192.168.50.77192.168.50.77.
Since 7777 falls between 6464 and 9595, the network address is 192.168.50.64192.168.50.64 and the broadcast address is 192.168.50.95192.168.50.95.
The last address in a subnet block range is reserved as the broadcast address.

Key Concept

IPv4 Subnet Calculation and Broadcast Address Identification
Estimated Time:45s
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Topics

IPv4 Addressing and Subnetting

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