5 Written questions
5 Matching questions
- Given 172.16.0.0/18, What are the valid hosts?
- Additional Windows Troubleshooting
- You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?
- What is the subnet for host ID 10.16.3.65/23?
- Class C Subnet Masks
- a .0.1 - .63.254, .64.1 - .127.254, .128.1 - .191.254, .192.1 - .255.254
- b /25 (with ip subnet-zero), /26, /27, /28, /29, /30
- c /23 is 255.255.254.0. The third octet is a block size of 2. 0, 2, 4. The subnet is in the 16.2.0 subnet, the broadcast address is 16.3.255
- d A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask listed, and 172.16.18.255 is a valid host.
- e Traceroute, Arp -a, Ipconfig /all
5 Multiple choice questions
- RIPv2, EIGRP, or OSPF
- 2y - 2 where y is the number of unmasked bits (or 0's)
- A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on—which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.
- 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.
- The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
5 True/False questions
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? → 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.
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? → 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.
Classful Routing → all nodes in the network use the same subnet mask
Given 172.16.0.0/18, What's the broadcast address for each subnet? → 63.255, 127.255, 191.255, 255.255
Given 192.168.10.0/28, How many hosts per subnet? → 126