Network+ Ch 8 Questions

192.168.100.17/29

192.168.100.17/29. A /29 is 255.255.255.248. The fourth octet is a block size of 8. 0, 8, 16, 24. The host is in the 16 subnet, broadcast of 23. Valid hosts are 17-22.

192.168.100.25/30

192.168.100.25/30. A /30 is 255.255.255.252. The valid subnet is 192.168.100.24, broadcast is 192.168.100.27, and valid hosts are 192.168.100.25 and 26.

On which of the following devices are you most likely to be able to implement NAT?

Devices with Layer 3 awareness, such as routers and firewalls, are the only ones that can manipulate the IP header in support of NAT.

You have one IP address provided from your ISP with a /30 mask. However, you have 300 users that need to access the Internet. What technology will you use to implement a solution?

Network Address Translation can allow up to 65,000 hosts to get onto the Internet with one IP address by using Port Address Translation (PAT).

192.168.100.37/28

192.168.100.37/28. A /28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0, 16, 32, 48. The host is in the 32 subnet, with a broadcast address of 47. Valid hosts are 33-46.

192.168.100.66/27

192.168.100.66/27. A /27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0, 32, 64. The host is in the 32 subnet, broadcast address of 63. The valid host range is 33-62.

192.168.100.99/25

192.168.100.99/25. A /25 is 255.255.255.128. The fourth octet is a block size of 128. 0, 128. The host is in the 0 subnet, broadcast of 127. Valid hosts are 1-126.

192.168.100.99/26

192.168.100.99/26. A /26 is 255.255.255.192. The fourth octet has a block size of 64. 0, 64, 128. The host is in the 64 subnet, broadcast of 127. Valid hosts are 65-126.

What is the subnet for host ID 172.16.3.65/23?

A /23 is 255.255.254.0. The third octet is a block size of 2. 0, 2, 4. The subnet is in the 16.2.0 subnet; the broadcast address is 16.3.255.

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 /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.

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 /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.

Using the illustration in question 16, 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.

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 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.

Using the following illustration, what would be the IP address of E0 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 /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet 0), 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 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 /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.

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 /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.

How many hosts are available with a Class C /29 mask?

A /29 is 255.255.255.248, which is 5 subnet bits and 3 host bits. This is only 6 hosts per subnet.

What is the broadcast address of 192.168.192.10/29?

A /29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0, 8, 16. The host is in the 8 subnet, and broadcast is 15.

If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?

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.

What is the highest usable address on the 172.16.1.0/24 network?

A 24-bit mask, or prefix length, indicates 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.

The network address of 172.16.0.0/19 provides how many subnets and hosts?

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 have a network with a subnet of 172.16.17.0/22. Which is the valid host addresses?

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.

When configuring the IP settings on a computer on one subnet to ensure it can communicate with a computer on another subnet, which of the following is desirable?

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.

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 use 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.

On a network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?

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

A network administrator is connecting hosts A and B 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.)

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).

If you are forced to replace a router that has failed to the point that you are unable to access its current configuration to aid in setting up interface addresses on the new router, which of the following can you reference for assistance?

The best method here is to check the configuration of devices that were using the old router as a gateway to the rest of the internetwork. Routers do not periodically cache their configurations to servers of any sort. You might have copied the old router's configuration to a TFTP server or the like; but failing that, you will have to rebuild the configuration from scratch, which might well be much more than interface addresses. Therefore, keeping a copy of the router's current configuration somewhere other than on the router is a wise choice. Routers don't auto-configure themselves; we wouldn't want them to.

You receive a call from a user that 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?

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!). The third octet has a block size of two, so the next subnet is 10.0.28.0, 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. Class A subnetting is covered in Appendix A.

What is the subnetwork address for a host with the IP address 200.10.5.68/28?

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.

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?

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 differenet subnets. 3

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.)

The router's IP address on the E0 interface is 172.16.2.1/23, which is a 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.