5 Written questions
5 Matching questions
- How many valid hosts per subnet?
- Given 172.16.0.0/18, How many hosts per subnet?
- You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts can have IP addresses on the LAN attached to router interface?
- How many hosts are available with a Class C /29 mask?
- a 2y - 2 where y is the number of unmasked bits (or 0's)
- b /25 is 255.255.255.128. The fourth octet is a block size of 128. 0, 128. The host is in the 0 subnet, broadcast of 127. Valid host 1-126
- c /29 is 255.255.255.248, which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
- d 16,382
- e A /29 (255.255.255.248), regardless of the class of address, has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN, including the router interface.
5 Multiple choice questions
- 32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address
- /25 (with ip subnet-zero), /26, /27, /28, /29, /30
- 0 & 128
- .0.1 - .63.254, .64.1 - .127.254, .128.1 - .191.254, .192.1 - .255.254
- 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.
5 True/False questions
Classless Routing Protocols → all interfaces within the classful address space have the same subnet mask
Using the following illustration, what would be the IP address of E0 if you were using the eighth subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. The zero subnet should not be considered valid for this question → A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32, so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.
/29 → 11111000 = 248
Given 192.168.10.0/28, What's the broadcast address for each subnet? → The number just before the next subnet; the broadcast of the last subnet is always 255
Able to ping but still unable to communicate → Possible DNS problem