DevNet Associate (DEVASC) v1.0Network FundamentalsHard
A network device is configured with the IP address 192.168.1.10 and a subnet mask of 255.255.255.240. Which of the following IP addresses would be considered part of the same network segment as this device?
- A192.168.1.14
- B192.168.1.33
- C192.168.1.17
- D192.168.1.1
Show answer & explanationAnswer & explanation
Correct answer: A. 192.168.1.14
The subnet mask 255.255.255.240 is a /28. This means the last octet has 4 network bits (11110000) and 4 host bits. The block size for a /28 in the last octet is 16. The network address for 192.168.1.10 is 192.168.1.0 (10 AND 240 in binary). The next network address would be 192.168.1.16. Therefore, devices from 192.168.1.1 to 192.168.1.14 are in the 192.168.1.0/28 network.
Why the other options are wrong
- B. 192.168.1.33 is in a much later network segment, 192.168.1.32/28.
- C. 192.168.1.17 is in the next network segment, 192.168.1.16/28.
- D. 192.168.1.1 is in the range 192.168.1.0/28 (0-15).
Network Address Calculation
To find the network address, perform a bitwise AND operation between the IP address and the subnet mask.
- Network address is the first address in a subnet, with all host bits set to 0.
- Broadcast address is the last address in a subnet, with all host bits set to 1.
- Usable host addresses are between the network and broadcast addresses.
Memory trick: Subnetting: 'AND' to find Network, 'OR' to find Broadcast, 'Hosts' in between.