Subnetting Questions and Answers

Answer: C

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.

Answer: A

Correct answer: 6

A /29 leaves three host bits, so \[ 2^{3} - 2 = 8 - 2 = 6 \] usable addresses - the block of eight less the subnet address and the broadcast. The router interface is one of the six, so five hosts may join it. The 8 offered is the block size before those two are removed.

Answer: B

The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host 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.

Answer: 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.

Answer: D

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 host bits (6 hosts per subnet). This is the best answer.

Answer: C

The router's IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. This makes the third octet a block size of 2. The router's interface is in the 2.0 subnet, and 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.

Answer: B

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.

Answer: D

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.

Answer: D

Correct answer: /30

A point-to-point link needs exactly two usable addresses, and a /30 provides exactly two - four in the block, less the subnet and broadcast. Anything shorter wastes addresses on a link that can never use them, and there is nothing shorter that works, since a /31 has no room for a subnet and broadcast address under the classic rules.

Answer: A

Correct answer: 192.168.192.15

A /29 gives blocks of eight, so the subnets in the last octet run 0, 8, 16, 24 and so on. The address .10 falls in the block beginning at 8, whose broadcast is the last address in it: \[ 8 + 8 - 1 = 15. \] Working out the block size from the mask and finding which block the address falls in answers every question of this type.