Ch. 8 Flashcards
1
Q
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? A. 14 B. 15 C. 16 D. 30 E. 31 F. 62
A
- 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.
2
Q
2. You have a Class A host of 10.0.0.110/25. It needs to communicate to a host with an IP address of 10.0.0.210/25. Which of the following devices do you need to use in order for these hosts to communicate? A. A Layer 2 switch B. Router C. DNS server D. Hub
A
- B. Don’t freak because this is a Class A. What is your subnet mask? 255.255.255.128. Regardless of the class of address, this is a block size of 128 in the fourth octet. The subnets are 0 and 128. The 0 subnet host range is 1–126, with a broadcast address of 127. The 128 subnet host range is 129–254, with a broadcast address of 255. You need a router for these two hosts to communicate because they are in different subnets.
3
Q
3. 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
A
- 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, and so on). The host is in the 64 subnet.
4
Q
4. 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
A
- F. 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.
5
Q
5. You receive a call from a user who is complaining that they cannot get on the Internet. You have them verify their IP address, mask, and default gateway. The IP address is 10.0.37.144, with a subnet mask of 255.255.254.0. The default gateway is 10.0.38.1. What is the problem? A. Incorrect DNS server address B. Invalid subnet mask C. Incorrect gateway IP D. IP address and mask not compatible
A
- C. The host ID of 10.0.37.144 with a 255.255.254.0 mask is in the 10.0.36.0 subnet (yes, you need to be able to subnet in this exam!). Do not stress that this is a class A, what we care about is that the third octet has a block size of 2, so the next subnet is 10.0.38.0, which makes the broadcast address 10.0.37.255. The default gateway address of 10.0.38.1 is not in the same subnet as the host. Even though this is a Class A address, you still should easily be able to subnet this because you look more at the subnet mask and find your interesting octet, which is the third octet in this question. 256 – 254 = 2. Your block size is 2.
6
Q
6. 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
A
- 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, and so on. Address 14 is obviously in the 12 subnet.
7
Q
7. On a network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses? A. /27 B. /28 C. /29 D. /30 E. /31
A
- D. A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides two hosts per subnet.
8
Q
8. On which of the following devices are you most likely to be able to implement NAT? A. Hub B. Ethernet switch C. Router D. Bridge
A
- C. Devices with Layer 3 awareness, such as routers and firewalls, are the only ones that can manipulate the IP header in support of NAT.
9
Q
9. 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 router interface? A. 6 B. 8 C. 30 D. 62 E. 126
A
- 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.
10
Q
- When configuring the IP settings on a computer on one subnet to ensure that it can communicate with a computer on another subnet, which of the following is desirable?
A. Configure the computer with the same default gateway as the other computer.
B. Configure the computer with the same subnet mask as the other computer.
C. Configure the computer with a default gateway that matches the IP address of the interface of the router that is attached to the same subnet as the computer.
D. Configure the computer with a subnet mask that matches the IP address of the router’s interface that is attached to the same subnet as the computer.
A
- C. A computer should be configured with an IP address that is unique throughout the reachable internetwork. It should be configured with a subnet mask that matches that of all other devices on its local subnet, but not necessarily one that matches the mask used on any other subnet. It should also be configured with a default gateway that matches its local router’s interface IP address.
11
Q
11. 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. 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, and so on. 10 is in the 8 subnet. The next subnet is 16, so 15 is the broadcast address.
12
Q
12. What is the highest usable address on the 172.16.1.0/24 network? A. 172.16.1.255 B. 172.16.1.254 C. 172.16.1.253 D. 172.16.1.23
A
- B. A 24-bit mask, or prefix length, indicates that the entire fourth octet is used for host identification. In a special case, such as this, it is simpler to visualize the all-zeros value (172.16.1.0) and the all-ones value (172.16.1.255). The highest usable address, the last one before the all-ones value, is 172.16.1.254.
13
Q
- A network administrator is connecting two hosts directly through their Ethernet interfaces, as shown in the illustration. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts? (Choose two.)
g0801.eps
A. A crossover cable should be used in place of the straight-through cable.
B. A rollover cable should be used in place of the straight-though cable.
C. The subnet masks should be set to 255.255.255.192.
D. A default gateway needs to be set on each host.
E. The subnet masks should be set to 255.255.255.0.
A
- A, E. First, if you have two hosts directly connected, as shown in the graphic, then you need a crossover cable. A straight-through cable won’t work. Second, the hosts have different masks, which puts them in different subnets. The easy solution is just to set both masks to 255.255.255.0 (/24).
14
Q
14. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the subnet address of this host? 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. 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. Because 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 because 128 is the next subnet.
15
Q
- 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 0 subnet should not be considered valid for this question.
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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. 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). We start at 16 (remember, the question stated that we will not use subnet 0, 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.
