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11. 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?
Solution:
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.
12. You have an interface on a router with the IP address of 192.168.192.10/29. What is the broadcast address the hosts will use on this LAN?
Solution:
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.
13. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
Solution:
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.
14. On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
Solution:
A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.
15. To test the IP stack on your local host, which IP address would you ping?
Solution:
To test the local stack on your host, ping the loopback interface of 127.0.0.1.
16. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
Solution:
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.
17. 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
Solution:
A /28 is a 255.255.255.240 mask. Let's count to the ninth subnet (we need to find the broadcast address of the eighth subnet, so we need to count to the ninth subnet). Starting at 16 (remember, the question stated that we will not use subnet zero, so we start at 16, not 0), 16, 32, 48, 64, 80, 96, 112, 128, 144. The eighth subnet is 128 and the next subnet is 144, so our broadcast address of the 128 subnet is 143. This makes the host range 129-142. 142 is the last valid host.
18. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224?
Solution:
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets, each with 30 hosts. However, if the command ip subnet-zero is not used, then only 6 subnets would be available for use.
19. Using the illustration from the previous question, what would be the IP address of SO if you were using the first subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. Again, the zero subnet should not b
Solution:
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.
20. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
Step 1:
The given IP address is 200.10.5.68 with a subnet mask of /28.
Step 2:
Convert the subnet mask from its decimal representation to binary.
The subnet mask /28 corresponds to 255.255.255.240 in decimal, which is 11111111.11111111.11111111.11110000 in binary.
Step 3:
Perform a bitwise 'AND' operation between the IP address and the subnet mask.
Step 4:
The result of the 'AND' operation is the subnetwork address.
Therefore, the subnetwork address for the host with the IP address 200.10.5.68/28 is:
C. 200.10.5.64
This output corresponds to the subnetwork address obtained by applying the subnet mask to the given IP address.
Solution:
Explanation:Step 1:
The given IP address is 200.10.5.68 with a subnet mask of /28.
Step 2:
Convert the subnet mask from its decimal representation to binary.
The subnet mask /28 corresponds to 255.255.255.240 in decimal, which is 11111111.11111111.11111111.11110000 in binary.
Step 3:
Perform a bitwise 'AND' operation between the IP address and the subnet mask.
IP address: 200.10.5.68 -> 11001000.00001010.00000101.01000100Subnet mask: 255.255.255.240 -> 11111111.11111111.11111111.11110000Subnet (AND): 200.10.5.64 -> 11001000.00001010.00000101.01000000Step 4:
The result of the 'AND' operation is the subnetwork address.
Therefore, the subnetwork address for the host with the IP address 200.10.5.68/28 is:
C. 200.10.5.64
This output corresponds to the subnetwork address obtained by applying the subnet mask to the given IP address.
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