Subnetting Questions and Answers

Q: To test the IP stack on your local host, which IP address would you ping?

Correct answer: 127.0.0.1127.0.0.1 is the loopback address, and pinging it exercises the local TCP/IP stack without touching any network hardware - a reply proves the software is installed and running. 127.0.0.0 is the network address of the loopback block and 127.0.0.255 a broadcast within it; neither is the host address to test.

Q: If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?

Correct answer: 172.16.45.12A /30 gives blocks of four, so the boundaries in the last octet fall at 0, 4, 8, 12, 16 and so on. The address .14 lies between 12 and 15, so its subnet address is 172.16.45.12, its broadcast is .15, and the two usable addresses are .13 and .14.

Q: What is the subnetwork address for a host with the IP address 200.10.5.68/28?

Correct answer: 200.10.5.64A /28 gives blocks of sixteen, so the boundaries are 0, 16, 32, 48, 64, 80 and so on. The address .68 falls in the block beginning at 64, which is therefore the subnet address; the broadcast is .79 and the usable range .65 to .78. The 56 offered is not a multiple of sixteen and could not be a /28 boundary at all.

31.

To test the IP stack on your local host, which IP address would you ping?

A.127.0.0.0
B.1.0.0.127
C.127.0.0.1
D.127.0.0.255

Answer: C

Correct answer: 127.0.0.1

127.0.0.1 is the loopback address, and pinging it exercises the local TCP/IP stack without touching any network hardware - a reply proves the software is installed and running. 127.0.0.0 is the network address of the loopback block and 127.0.0.255 a broadcast within it; neither is the host address to test.

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32.

If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?

A.172.16.45.0
B.172.16.45.4
C.172.16.45.8
D.172.16.45.12

Answer: D

Correct answer: 172.16.45.12

A /30 gives blocks of four, so the boundaries in the last octet fall at 0, 4, 8, 12, 16 and so on. The address .14 lies between 12 and 15, so its subnet address is 172.16.45.12, its broadcast is .15, and the two usable addresses are .13 and .14.

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33.

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.192.168.10.142
B.192.168.10.66
C.192.168.100.254
D.192.168.10.143

Answer: A

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.

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34.

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?

A.Router(config)#ip classless
B.Router(config)#no ip classful
C.Router(config)#ip unnumbered
D.Router(config)#ip subnet-zero

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.

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35.

Using the illustration from the previous question, what would be the IP address of S0 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 be considered valid for this question.

A.192.168.10.24
B.192.168.10.62
C.192.168.10.30
D.192.168.10.127

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.

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36.

What is the subnetwork address for a host with the IP address 200.10.5.68/28?

A.200.10.5.56
B.200.10.5.32
C.200.10.5.64
D.200.10.5.0

Answer: C

Correct answer: 200.10.5.64

A /28 gives blocks of sixteen, so the boundaries are 0, 16, 32, 48, 64, 80 and so on. The address .68 falls in the block beginning at 64, which is therefore the subnet address; the broadcast is .79 and the usable range .65 to .78. The 56 offered is not a multiple of sixteen and could not be a /28 boundary at all.

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