Subnetting Questions and Answers

Q: 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?

Correct answer: 192.168.192.15A /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.

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.

21.

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?

A.192.168.192.15
B.192.168.192.31
C.192.168.192.63
D.192.168.192.127

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.

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22.

What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?

A.14
B.15
C.16
D.30

Answer: D

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 number of host bits would never change.

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23.

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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24.

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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