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Difficulty: EasyIPv4 Addressing and Subnetting

A network engineer is configuring an interface on a Cisco router servicing an internal LAN segment. The interface is assigned the IPv4 address 192.168.1.37192.168.1.37 with a subnet mask of 255.255.255.240255.255.255.240 (/28/28). Which IPv4 address represents the broadcast address for this subnetwork?

  1. A
    192.168.1.32192.168.1.32
  2. 192.168.1.47192.168.1.47Answer
  3. C
    192.168.1.48192.168.1.48
  4. D
    192.168.1.63192.168.1.63

Answer

192.168.1.47192.168.1.47
The subnet mask 255.255.255.240255.255.255.240 (/28/28) provides a block size of 16 addresses per subnet. Multiples of 16 in the 4th octet yield network addresses of 192.168.1.0192.168.1.0, 192.168.1.16192.168.1.16, 192.168.1.32192.168.1.32, and 192.168.1.48192.168.1.48. The host 192.168.1.37192.168.1.37 resides in the 192.168.1.32/28192.168.1.32/28 subnet, making 192.168.1.47192.168.1.47 the broadcast address.

Step-by-Step Solution

1
Determine the block size from the subnet mask.
A subnet mask of 255.255.255.240255.255.255.240 corresponds to a /28/28 prefix, leaving 32−28=432 - 28 = 4 host bits. The block size is 24=162^4 = 16.
Calculating the block size defines the increment between subnetwork boundaries.
2
Identify the network boundary for IP address 192.168.1.37192.168.1.37.
Subnet multiples of 16 in the 4th octet are 0, 16, 32, 48. Since 37 falls between 32 and 47, the subnetwork ID is 192.168.1.32192.168.1.32.
The network address is the lowest boundary address for the subnet.
3
Calculate the broadcast address for the subnet.
The broadcast address is one less than the next network address (192.168.1.48−1=192.168.1.47192.168.1.48 - 1 = 192.168.1.47).
The broadcast address is the highest address in the block with all host bits set to 1.

Key Concept

IPv4 Subnet Masking and Broadcast Address Calculation
Estimated Time:1m 0s
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Topics

IPv4 Addressing and Subnetting

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