A device has an address of 192.168.144.21 and a mask of 255.255.255.240. What will be the broadcast address for the subnet to which this device is attached?
A.
192.168.144.23
B.
192.168.144.28
C.
192.168.144.31
D.
192.168.144.32
Explanation:
The broadcast address for the subnet to which this device is attached will be 192.168.144.31.
To determine the broadcast address of a network where a specific address resides, you must first determine the network ID of the subnetwork where the address
resides. The network ID can be obtained by determining the interval between subnet IDs. With a 28-bit mask, the decimal equivalent of the mask will be
255.255.255.240. The interval between subnets can be derived by subtracting the value of the last octet of the mask from 256. In this case, that operation would be
256 – 240. Therefore, the interval is 16.
The first network ID will always be the classful network you started with (in this case 192.168.144.0). Then each subnetwork ID in this network will fall at 16-bit
intervals as follows:
192.168.144.0
192.168.144.16
192.168.144.32
192.168.144.48
At 192.168.144.48 we can stop, because the address that we are given as a guide is in the network with a subnet ID of 192.168.144.16. Therefore, since the
broadcast address for this network will be 1 less than the next subnet ID (192.168.144.32), the broadcast address for the subnet to which this device is attached is
192.168.144.31.All the other options are incorrect because none of these will be the broadcast address for the subnet to which this device is attached.
Objective:
Network Fundamentals
Sub-Objective:
Applytroubleshooting methodologies to resolve problemshttps://www.cisco.com/c/en/us/support/docs/ip/routing-information-protocol-rip/13788-3.html#ustand_ip_add