Data2chap3

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.) 172.16.0.5 172.16.1.100 172.16.1.198 172.16.2.255 172.16.3.0 172.16.3.255

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.

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? 172.16.112.0 172.16.0.0 172.16.96.0 172.16.255.0 172.16.128.0

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 128 is the next subnet.

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? 192.168.192.15 192.168.192.31 192.168.192.63 192.168.192.127 192.168.192.255

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.

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? 6 8 30 62 126

A. A /29 (255.255.255.248), regardless of the class of address, has only 3 host bits. Six hosts 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.) 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. 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. 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 need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use? 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248

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 address for a host with the IP address 200.10.5.68/28? 200.10.5.56 200.10.5.32 200.10.5.64 200.10.5.0

C. 200.10.5.64 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.

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

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.

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? 192.168.19.0 255.255.255.0 192.168.19.33 255.255.255.240 192.168.19.26 255.255.255.248 192.168.19.31 255.255.255.248 192.168.19.34 255.255.255.240

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.

To test the IP stack on your local host, which IP address would you ping? 127.0.0.0 1.0.0.127 127.0.0.1 127.0.0.255 255.255.255.255

C. To test the local stack on your host, ping the loopback interface of 127.0.0.1.

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? 14 15 16 30 31 62

D. 30 /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.

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? 2 3 4 5 6 7

D. 5 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.

If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to? 172.16.45.0 172.16.45.4 172.16.45.8 172.16.45.12 172.16.45.16

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.

On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses? /27 /28 /29 /30 /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 subnet of 172.16.17.0/22. Which is the valid host address? 172.16.17.1 255.255.255.252 172.16.0.1 255.255.240.0 172.16.20.1 255.255.254.0 172.16.16.1 255.255.255.240 172.16.18.255 255.255.252.0 172.16.0.1 255.255.255.0

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

The network address of 172.16.0.0/19 provides how many subnets and hosts? 7 subnets, 30 hosts each 7 subnets, 2,046 hosts each 7 subnets, 8,190 hosts each 8 subnets, 30 hosts each 8 subnets, 2,046 hosts each 8 subnets, 8,190 hosts each

F. 8 subnets, 8,190 hosts each 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.