Examine the partial output of the show ip interface command below.

What is the subnet broadcast address of the LAN connected to the router from which the command was
executed?

A.
192.168.93.15

B.
192.168.93.255

C.
1.1.1.255

D.
1.1.1.127

Explanation:
In the output we can see there are two interfaces, a serial interface (which goes to another router) and a
GigabitEthernet interface (the LAN interface). The LAN interface has an address of 192.168.93.1/28, which is a
mask of 255.255.255.240. When this mask is used against the 192.168.93.0 classful network, it yields the
following subnets:
192.168.93.0
192.168.93.16
192.168.93.32
192.168.93.48
and so on, incrementing in intervals of 16 in the last octet.
Since the LAN interface has an address of 192.168.93.1, the interface is in the 192.168.93.0/28 network. That
networks broadcast address is the last address before the next subnet address of 192.168.93.16. Therefore,
the broadcast address of the LAN connected to the router from which the command was executed is
192.168.93.15.
The address 192.168.93.255 is not the broadcast address. If a standard 24-bit mask were used instead of
the /28, this would be the broadcast address.
The address 1.1.1.255 is the broadcast address of the network in which the Serial interface resides. The
question asked for the LAN interface.
The address 1.1.1.127 would be the broadcast address of the network in which the Serial interface resides if
the mask used on the interface were 255.255.255.128. However, that is not the mask, and the question asked
for the LAN interface.
Objective:
Network Fundamentals
Sub-Objective:
Configure, verify, and troubleshoot IPv4 addressing and subnetting