Network+: ch 8 Written Labs and Review Questions Flashcards
Practice Example #1C: 255.255.255.128 (/25)
192.168.10.0 = Network address
255.255.255.128 = Subnet mask
Answer the big 5 from network + and the four from subnetting .com
Because 128 is 10000000 in binary, there is only 1 bit for subnetting, and there are 7 bits or hosts. We’re going to subnet the Class C network address 192.168.10.0.
- 168.10.0 = Network address
- 255.255.128 = Subnet mask
Now, let’s answer the big five: How many subnets? Because 128 is 1 bit on (10000000), the answer is 2^1 = 2. Only looking at the fourth octet in class C, but last 2 in Class b, last 3 a???
How many hosts per subnet? We have 7 host bits off (10000000), so the equation is 2^7 – 2 = 126 hosts.
What are the valid subnets? 256 – 128 = 128. Remember, we’ll start at zero and count in our block size, so our subnets are 0, 128.
What’s the broadcast address for each subnet? The number right before the value of the next subnet is all host bits turned on and equals the broadcast address. For the 0 subnet, the next subnet is 128, so the broadcast of the 0 subnet is 127.
What are the valid hosts? These are the numbers between the subnet and broadcast address. The easiest way to find the hosts is to write out the subnet address and the broadcast address. This way, the valid hosts are obvious. The following table shows the 0 and 128 subnets, the valid host ranges of each, and the broadcast address of both subnets:
Subnet 0 128
First host 1 129
Last host 126 254
Broadcast 127 255
Should We Really Use This Mask That Provides Only Two Hosts?
Imagine you are the network administrator for Acme Corporation in San Francisco, with dozens of WAN links connecting to your corporate office. Right now your network is a classful network, which means that the same subnet mask is on each host and router
interface. You’ve read about classless routing where you can have different size masks, but you don’t know what to use on your point-to-point WAN links. Is 255.255.255.252 (/30) a helpful mask in this situation?
Yes, this is a very helpful mask in wide area networks.
If you use the 255.255.255.0 mask, then each network will have 254 hosts, but you only use 2 addresses with a WAN link! That is a waste of 252 hosts per subnet. If you use the 255.255.255.252 mask, then each subnet has only 2 hosts, and you don’t waste precious
addresses.
Write the subnet, broadcast address, and valid host range 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.
Write the subnet, broadcast address, and valid host range 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.
Write the subnet, broadcast address, and valid host range 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.
Write the subnet, broadcast address, and valid host range 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.
Write the subnet, broadcast address, and valid host range 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.
Write the subnet, broadcast address, and valid host range 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.
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.
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.
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 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.
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 host bits would never change.
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
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 differenet subnets.
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, and so on. The host is in the 64 subnet.
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
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.
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?
A. Invalid IP address
B. Invalid subnet mask
C. gateway IP is incorrect
D. IP address and mask are not compatible.
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!). 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.
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, and so on. Address 14 is obviously
in the 12 subnet.
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
D. A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides
two hosts per subnet.
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
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.
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 /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.
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. 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
router’s interface that is attached to the same subnet as the computer.
D. Configure the computer with a subnet mask that matches the IP
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.
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, and so on. 10 is in the 8 subnet. The next subnet is 16, so 15 is
the broadcast address.
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
B. 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.
