5 Written questions
5 Matching questions
- Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host ID's on the LAN interface attached to the router?
- You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?
- On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
- Unable to Ping Local Host
- Unable to Ping Remote Destination
- a The routers IP address on the E0 interface is 172.16.2.1/23, which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet, 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.
- b A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.
- c Problem with the NIC; Replace the NIC
- d A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.
- e Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
5 Multiple choice questions
- A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0, 8, 16, 24, etc. 10 is in the 8 subnet. The next subnet is 16, so 15 is the broadcast address.
- each network segment can use a different subnet mask
- 2y - 2 where y is the number of unmasked bits (or 0's)
- This is a pretty simple question. A /28 is 255.255.255.240, which means that our block size is 16 in the fourth octet. 0, 16, 32, 48, 64, 80, etc. The host is in the 64 subnet.
5 True/False questions
The network address of 172.16.0.0/19 provides how many subnets and hosts? → A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits but provides 13 host bits, or 8 subnets, each with 8,190 hosts.
To test the IP stack on your local host, which IP address would you ping? → A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.
/25 → 11000000 = 192
Why subnet? → Facilitated spanning of large geographical distances
/28 → 11110000 = 240