Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host IDs on the LAN interface attached to the router?
172.16.1.100
172.16.1.198
172.16.2.255
172.16.3.0
Options
A. 1 only
B. 2 and 3 only
C. 3 and 4 only
D. None of the above
Correct Answer
3 and 4 only
Explanation
The router's IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. This makes the third octet a block size of 2. The router's interface is in the 2.0 subnet, and the broadcast address is 3.255 because the next subnet is 4.0. The valid host range is 2.1 through 3.254. The router is using the first valid host address in the range.
More questions
1. How many collision domains are created when you segment a network with a 12-port switch?
You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts-this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts. This is the best answer.
5. Which command is required for connectivity in a Frame Relay network if Inverse ARP is not operational?
If you have a router in your Frame Relay network that does not support IARP, you must create Frame Relay maps on your router, which provide known DLCI-to-IP address mappings.
6. What VTP mode allows you to change VLAN information on the switch?
A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.
8. Which command will allow you to see real-time translations on your router?
Any secondary route to a remote network is considered a feasible successor, and those routes are only found in the topology table and used as backup routes in case of primary route failure. You can see the topology table with the
show ip eigrp topology command.