5 Written questions
5 Matching questions
- Which two statements describe the IP address 10.16.3.65/23?
- If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
- Unable to Ping Remote Destination
- Why subnet?
- a Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
- b 11100000 = 224
- c Facilitated spanning of large geographical distances
- d A /30, regardless of the class of address, has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0, 4, 8, 12, 16, etc. Address 14 is obviously in the 12 subnet.
- e The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254.
5 Multiple choice questions
- A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets, each with 30 hosts. Does it matter if this mask is used with a Class A, B or C network address? Not at all. The amount of hosts bits would never change.
- A 240 mask is 4 subnet bits and provides 16 subnets, each with 14 hosts. We need more subnets, so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).
- /28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0, 16, 32, 48. The host is in the 32 subnet, with a broadcast address of 47. Valid hosts 33-46
- 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 True/False questions
What is the subnet and broadcast address of the host 172.16.88.255/20? → 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.
/28 → 10000000 = 128
Why subnet? → 2^x where x is the number of masked bits (or 1's)
Classful Routing → all nodes in the network use the same subnet mask
You have a Class B network and need 29 subnets. What is your mask? → Use different size masks on each router interface