A network administrator is troubleshooting an addressing issue on a branch office switch. … — Cisco CCNA

A network administrator is troubleshooting an addressing issue on a branch office switch. A host computer on VLAN 10 has been manually configured with the IP address $172.25.107.191$ and subnet mask $255.255.255.192$ . The switch SVI interface serving as the default gateway for VLAN 10 is assigned $172.25.107.129/26$ . The host is unable to ping its default gateway or communicate beyond its local segment. Which statement correctly identifies the cause of this connectivity failure?

The assigned address $172.25.107.191$ is the broadcast address for the subnet and cannot be utilized as a valid host IP address.Answer
B
The host address $172.25.107.191$ falls outside the RFC 1918 private IPv4 range, causing the private default gateway to drop non-routable public packets.
C
The default gateway $172.25.107.129/26$ provides a maximum of 62 usable addresses, meaning $172.25.107.191$ is the first host address of the next adjacent subnet.
D
The host IP address $172.25.107.191$ and gateway address $172.25.107.129$ belong to different subnets because a /26 prefix divides the third octet into 64-host boundaries.

Answer

The host address 172.25.107.191 is the broadcast address for the 172.25.107.128/26 subnet and is invalid for host assignment.

The correct answer accurately identifies that 172.25.107.191 is the directed broadcast address for the 172.25.107.128/26 network segment. For a /26 prefix, the block size is 64, giving a network ID of 172.25.107.128 and a broadcast ID of 172.25.107.191. Because broadcast addresses cannot be configured as valid unicast endpoints, the host network stack fails to process traffic correctly.

Step-by-Step Solution

Determine the subnet mask prefix length and block size.

A subnet mask of

255.255.255.192

corresponds to CIDR notation

/26

. The block size in the fourth octet is calculated as

256 - 192 = 64

Calculating the block size identifies subnet boundaries.

Calculate the subnet boundaries for the gateway IP

172.25.107.129

Subnet multiples of 64 in the 4th octet are 0, 64, 128, and 192. The gateway IP

172.25.107.129

falls into the network

172.25.107.128/26

Identifying the network ID establishes the IP range for VLAN 10.

Calculate the network address, usable host range, and broadcast address.

Network Address:

172.25.107.128

. First Usable Host:

172.25.107.129

. Last Usable Host:

172.25.107.190

. Broadcast Address:

172.25.107.191

Formula

2^n - 2

yields 62 usable host IPs (

2^6 - 2 = 62

Evaluate the validity of the host's manually assigned IP address

172.25.107.191

The address

172.25.107.191

matches the subnet broadcast address exactly.

Broadcast addresses are reserved for network-wide broadcasts and cannot be bound to network adapters as valid unicast host IPs.

Key Concept

Subnet Broadcast Address Identification and Usable Host Range Calculation

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