You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
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.
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You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet. How many bits must you borrow from the host field to provide the correct subnet mask?
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.
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If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?
Answer: A
A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on-which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet.
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You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts can have IP addresses on the LAN attached to the router interface?
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.
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What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
Answer: C
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.
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The network address of 172.16.0.0/19 provides how many subnets and hosts?
Answer: B
Correct answer: 8 subnets, 8,190 hosts each
172.16.0.0 is a class B network with a natural /16, so a /19 borrows three bits and leaves thirteen: \[ 2^{3} = 8\ \mathrm{subnets}, \qquad 2^{13}-2 = 8190\ \mathrm{hosts}. \] All eight subnets are usable on any modern IOS, and the answers offering seven subnets come from the obsolete rule that excluded the all-zeros subnet.
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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?
Answer: C
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.
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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?
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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You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
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.
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On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
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.
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