CH 4 REVIEW

You have a Class B network and need 29 subnets. What is your mask?

A default Class B is 255.255.0.0. A Class B 255.255.255.0 mask is 256 subnets, each with 254 hosts. We need fewer subnets. If we used 255.255.240.0, this provides 16 subnets. Let's add one more subnet bit. 255.255.248.0. This is 5 bits of subnetting, which provides 32 subnets. This is our best answer, a /21.

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 interface? A. 172.16.112.0 B. 172.16.0.0 C. 172.16.96.0 D. 172.16.255.0 E. 172.16.128.0

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 112.128 is the next subnet.

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 E. 192.168.10.126

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), we have 16, 32, 48, 64, 80, 96, 112, 128, 144, etc. 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.

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 E. 192.168.192.255

A. 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.

Given an IP address 172.16.28.252 with a subnet mask of 255.255.240.0, what is the correct network address? A. 172.16.16.0 B. 172.16.0.0 C. 172.16.24.0 D. 172.16.28.0

A. For this example, the network range is 172.16.16.1 to 172.16.31.254, the network address is 172.16.16.0, and the broadcast IP address is 172.16.31.255.

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

C. This is a pretty simple question. A /28 is 255.255.255.240, which means that our block size is 16 in the fourth octet. 0, 16, 32, 48, 64, 80, etc. The host is in the 64 subnet.

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? A. 6 B. 8 C. 30 D. 62 E. 126

A. A /29 (255.255.255.248), regardless of the class of address, has only 3 host bits. Six is the maximum number of hosts on this LAN, including the router interface.

Which two statements describe the IP address 10.16.3.65/23? (Choose two.) A. The subnet address is 10.16.3.0 255.255.254.0. B. The lowest host address in the subnet is 10.16.2.1 255.255.254.0. C. The last valid host address in the subnet is 10.16.2.254 255.255.254.0. D. The broadcast address of the subnet is 10.16.3.255 255.255.254.0. E. The network is not subnetted.

B, D. 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.

You have a network in your data center that needs 310 hosts. Which mask should you use so you waste the least amount of addresses? A. 255.255.255.0 B. 255.255.254.0 C. 255.255.252.0 D. 255.255.248.0

B. We need 9 host bits to answer this question, which is a /23.

You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use? A. 255.255.255.192 B. 255.255.255.224 C. 255.255.255.240 D. 255.255.255.248

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.

What is the subnetwork number of a host with an IP address of 172.16.66.0/21? A. 172.16.36.0 B. 172.16.48.0 C. 172.16.64.0 D. 172.16.0.0

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.

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

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.

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? A. 192.168.19.0 255.255.255.0 B. 192.168.19.33 255.255.255.240 C. 192.168.19.26 255.255.255.248 D. 192.168.19.31 255.255.255.248 E. 192.168.19.34 255.255.255.240

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.

The network address of 172.16.0.0/19 provides how many subnets and hosts? A. 7 subnets, 30 hosts each B. 7 subnets, 2,046 hosts each C. 7 subnets, 8,190 hosts each D. 8 subnets, 30 hosts each E. 8 subnets, 2,046 hosts each F. 8 subnets, 8,190 hosts each

C. A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.

You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive the following error: Bad mask /26 for address 192.168.10.62 A. Why did you receive this error? B. You typed this mask on a WAN link and that is not allowed. C. This is not a valid host and subnet mask combination. D. ip subnet-zero is not enabled on the router. E. The router does not support IP.

C. First, you cannot answer this question if you can't subnet. The 192.168.10.62 with a mask of 255.255.255.192 is a block size of 64 in the fourth octet. The host 192.168.10.62 is in the zero subnet, and the error occurred because ip subnet-zero is not enabled on the router.

Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host IDs on the LAN interface attached to the router? (Choose two.) A. 172.16.0.5 B. 172.16.1.100 C. 172.16.1.198 D. 172.16.2.255 E. 172.16.3.0 F. 172.16.3.255

D, E. 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.

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 E. 31 F. 62

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 subnet bits would never change.

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 E. 172.16.45.16

D. 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.

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? A. 2 B. 3 C. 4 D. 5 E. 6 F. 7

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.

Which mask should you use on point-to-point links in order to reduce the waste of IP addresses? A. /27 B. /28 C. /29 D. /30 E. /31

D. A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.

You have a network with a host address of 172.16.17.0/22. From the following options, which is another valid host address in the same subnet? A. 172.16.17.1 255.255.255.252 B. 172.16.0.1 255.255.240.0 C. 172.16.20.1 255.255.254.0 D. 172.16.16.1 255.255.255.240 E. 172.16.18.255 255.255.252.0 F. 172.16.0.1 255.255.255.0

E. 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 has the correct subnet mask listed, and 172.16.18.255 is a valid host.