Subnetting in Your Head: Class A Addresses
Again, I know this sounds hard, but as with Class C and Class B, the
numbers are the same; we just start in the second octet. What makes this
easy? You only need to worry about the octet that has the largest block
size, which is typically called the interesting octet, and one that is
something other than 0 or 255, such as, for example, 255.255.240.0 (/20)
with a Class A network. The second octet has a block size of 1, so any
number listed in that octet is a subnet. The third octet is a 240 mask,
which means we have a block size of 16 in the third octet. If your host ID
is 10.20.80.30, what is your subnet, broadcast address, and valid host
range?
The subnet in the second octet is 20 with a block size of 1, but the third
octet is in block sizes of 16, so we’ll just count them out: 0, 16, 32, 48, 64,
80, 96… voilà! By the way, you can count by 16s by now, right? Good!
This makes our subnet 10.20.80.0, with a broadcast address of
10.20.95.255 because the next subnet is 10.20.96.0. The valid host range
is 10.20.80.1 through 10.20.95.254. And yes, no lie! You really can do this
in your head if you just get your block sizes nailed!
Let’s practice on one more, just for fun!
Host IP: 10.1.3.65/23
First, you can’t answer this question if you don’t know what a /23 is. It’s
255.255.254.0. The interesting octet here is the third one: 256 – 254 = 2.
Our subnets in the third octet are 0, 2, 4, 6, etc. The host in this question
is in subnet 2.0, and the next subnet is 4.0, so that makes the broadcast
address 3.255. And any address between 10.1.2.1 and 10.1.3.254 is
considered a valid host.
Summary
Did you read Chapters 3 and 4 and understand everything on the first
pass? If so, that is fantastic—congratulations! However, you probably
really did get lost a couple of times. No worries because as I told you,
that’s what usually happens. Don’t waste time feeling bad if you have to
read each chapter more than once, or even 10 times, before you’re truly
good to go. If you do have to read the chapters more than once, you’ll be
seriously better off in the long run even if you were pretty comfortable the
first time through!
This chapter provided you with an important understanding of IP
subnetting—the painless way! And when you’ve got the key material
presented in this chapter really nailed down, you should be able to subnet
IP addresses in your head.
This chapter is extremely essential to your Cisco certification process, so
if you just skimmed it, please go back, read it thoroughly, and don’t forget
to do all the written labs too!
Exam Essentials
Identify the advantages of subnetting. Benefits of subnetting a
physical network include reduced network traffic, optimized network
performance, simplified management, and facilitated spanning of large
geographical distances.
Describe the effect of the
ip subnet-zero
command. This command
allows you to use the first and last subnet in your network design.
Identify the steps to subnet a classful network. Understand how
IP addressing and subnetting work. First, determine your block size by
using the 256-subnet mask math. Then count your subnets and
determine the broadcast address of each subnet—it is always the number
right before the next subnet. Your valid hosts are the numbers between
the subnet address and the broadcast address.
Determine possible block sizes. This is an important part of
understanding IP addressing and subnetting. The valid block sizes are
always 2, 4, 8, 16, 32, 64, 128, etc. You can determine your block size by
using the 256-subnet mask math.
Describe the role of a subnet mask in IP addressing. A subnet
mask is a 32-bit value that allows the recipient of IP packets to
distinguish the network ID portion of the IP address from the host ID
portion of the IP address.
Understand and apply the 2
x
– 2 formula. Use this formula to
determine the proper subnet mask for a particular size network given the
application of that subnet mask to a particular classful network.
Explain the impact of Classless Inter-Domain Routing (CIDR).
CIDR allows the creation of networks of a size other than those allowed
with the classful subnetting by allowing more than the three classful
subnet masks.
Written Labs
In this section, you’ll complete the following labs to make sure you’ve got
the information and concepts contained within them fully dialed in:
Lab 4.1: Written Subnet Practice #1
Lab 4.2: Written Subnet Practice #2
Lab 4.3: Written Subnet Practice #3
You can find the answers to these labs in Appendix A, “Answers to
Written Labs.”
Written Lab 4.1: Written Subnet Practice #1
Write the subnet, broadcast address, and a valid host range for question 1
through question 6. Then answer the remaining questions.
1. 192.168.100.25/30
2. 192.168.100.37/28
3. 192.168.100.66/27
4. 192.168.100.17/29
5. 192.168.100.99/26
6. 192.168.100.99/25
7. You have a Class B network and need 29 subnets. What is your mask?
8. What is the broadcast address of 192.168.192.10/29?
9. How many hosts are available with a Class C /29 mask?
10. What is the subnet for host ID 10.16.3.65/23?
Written Lab 4.2: Written Subnet Practice #2
Given a Class B network and the net bits identified (CIDR), complete the
following table to identify the subnet mask and the number of host
addresses possible for each mask.
Classful
Address
Subnet
Mask
Number of Hosts per Subnet (2
x
–
2)
/16
/17
/18
/19
/20
/21
/22
/23
/24
/25
/26
/27
/28
/29
/30
Written Lab 4.3: Written Subnet Practice #3
Complete the following based on the decimal IP address.
Decimal IP
Address
Address
Class
Number of
Subnet and
Host Bits
Number
of
Subnets
(2
x
)
Number
of Hosts
(2
x
– 2)
10.25.66.154/23
172.31.254.12/24
192.168.20.123/28
63.24.89.21/18
128.1.1.254/20
208.100.54.209/30
Review Questions
The following questions are designed to test your
understanding of this chapter’s material. For more information on
how to get additional questions, please see
www.lammle.com/ccna
.
You can find the answers to these questions in Appendix B, “Answers to
Review Questions.”
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
2. 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
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
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
5. 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.
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
7. 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
8. 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
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 the router interface?
A. 6
B. 8
C. 30
D. 62
E. 126
10. 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
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
12. 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
13. 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.
14. 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
15. 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
16. 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
17. 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
18. 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
19. 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
20. 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
Chapter 5
VLSMs, Summarization, and Troubleshooting
TCP/IP
THE FOLLOWING ICND1 EXAM TOPICS ARE
COVERED IN THIS CHAPTER:
Network Fundamentals
1.7 Apply troubleshooting methodologies to resolve problems
1.7.a Perform fault isolation and document
1.7.b Resolve or escalate
1.7.c Verify and monitor resolution
1.8 Configure, verify, and troubleshoot IPv4 addressing and
subnetting
Now that IP addressing and subnetting have
been thoroughly covered in the last two chapters, you’re fully prepared
and ready to learn all about variable length subnet masks (VLSMs). I’ll
also show you how to design and implement a network using VLSM in
this chapter. After ensuring you’ve mastered VLSM design and
implementation, I’ll demonstrate how to summarize classful boundaries.
We’ll wrap up the chapter by going over IP address troubleshooting,
focusing on the steps Cisco recommends to follow when troubleshooting
an IP network.
So get psyched because this chapter will give you powerful tools to hone
your knowledge of IP addressing and networking and seriously refine the
important skills you’ve gained so far. So stay with me—I guarantee that
your hard work will pay off! Ready? Let’s go!
To find up-to-the minute updates for this chapter, please see
www.lammle.com/ccna
or the book’s web page at
www.sybex.com/go/ccna
.
Variable Length Subnet Masks (VLSMs)
Teaching you a simple way to create many networks from a large single
network using subnet masks of different lengths in various kinds of
network designs is what my primary focus will be in this chapter. Doing
this is called VLSM networking, and it brings up another important
subject I mentioned in Chapter 4, “Easy Subnetting,” classful and
classless networking.
Older routing protocols like Routing Information Protocol version 1
(RIPv1) do not have a field for subnet information, so the subnet
information gets dropped. This means that if a router running RIP has a
subnet mask of a certain value, it assumes that all interfaces within the
classful address space have the same subnet mask. This is called classful
routing, and RIP is considered a classful routing protocol. We’ll cover RIP
and the difference between classful and classless networks later on in
Chapter 9, “IP Routing,” but for now, just remember that if you try to mix
and match subnet mask lengths in a network that’s running an old
routing protocol, such as RIP, it just won’t work!
However, classless routing protocols do support the advertisement of
subnet information, which means you can use VLSM with routing
protocols such as RIPv2, Enhanced Interior Gateway Protocol (EIGRP),
and Open Shortest Path First (OSPF). The benefit of this type of network
is that it saves a bunch of IP address space.
As the name suggests, VLSMs can use subnet masks with different
lengths for different router interfaces. Check out
Figure 5.1
to see an
example of why classful network designs are inefficient.
FIGURE 5.1
Typical classful network
Looking at
Figure 5.1
, you can see that there are two routers, each with
two LANs and connected together with a WAN serial link. In a typical
classful network design that’s running RIP, you could subnet a network
like this:
192.168.10.0 = Network
255.255.255.240 (/28) = Mask
Our subnets would be—you know this part, right?— 0, 16, 32, 48, 64, 80,
etc., which allows us to assign 16 subnets to our internetwork. But how
many hosts would be available on each network? Well, as you know by
now, each subnet provides only 14 hosts, so each LAN has only 14 valid
hosts available (don’t forget that the router interface needs an address too
and is included in the amount of needed valid hosts). This means that one
LAN doesn’t even have enough addresses needed for all the hosts, and
this network as it is shown would not work as addressed in the figure!
Since the point-to-point WAN link also has 14 valid hosts, it would be
great to be able to nick a few valid hosts from that WAN link to give to
our LANs!
All hosts and router interfaces have the same subnet mask—again, known
as classful routing—and if we want this network to be efficient, we would
definitely need to add different masks to each router interface.
But that’s not our only problem—the link between the two routers will
never use more than two valid hosts! This wastes valuable IP address
space, and it’s the big reason you need to learn about VLSM network
design.
VLSM Design
Let’s take
Figure 5.1
and use a classless design instead, which will become
the new network shown in
Figure 5.2
. In the previous example, we wasted
address space—one LAN didn’t have enough addresses because every
router interface and host used the same subnet mask. Not so good. A
better solution would be to provide for only the needed number of hosts
on each router interface, and we’re going to use VLSMs to achieve that
goal.
FIGURE 5.2
Classless network design
Now remember that we can use different size masks on each router
interface. If we use a /30 on our WAN links and a /27, /28, and /29 on
our LANs, we’ll get 2 hosts per WAN interface and 30, 14, and 6 hosts per
LAN interface—nice (remember to count your router interface as a host)!
This makes a huge difference—not only can we get just the right amount
of hosts on each LAN, we still have room to add more WANs and LANs
using this same network!
To implement a VLSM design on your network, you need to
have a routing protocol that sends subnet mask information with the
route updates. The protocols that do that are RIPv2, EIGRP, and
OSPF. Remember, RIPv1 will not work in classless networks, so it’s
considered a classful routing protocol.
Implementing VLSM Networks
To create VLSMs quickly and efficiently, you need to understand how
block sizes and charts work together to create the VLSM masks.
Table 5.1
shows you the block sizes used when creating VLSMs with Class C
networks. For example, if you need 25 hosts, then you’ll need a block size
of 32. If you need 11 hosts, you’ll use a block size of 16. Need 40 hosts?
Then you’ll need a block of 64. You cannot just make up block sizes—
they’ve got to be the block sizes shown in
Table 5.1
. So memorize the
block sizes in this table—it’s easy. They’re the same numbers we used
with subnetting!
Table 5.1
Block sizes
Prefix Mask Hosts Block Size
/25
128
126
128
/26
192
62
64
/27
224
30
32
/28
240
14
16
/29
248
6
8
/30
252
2
4
The next step is to create a VLSM table.
Figure 5.3
shows you the table
used in creating a VLSM network. The reason we use this table is so we
don’t accidentally overlap networks.
You’ll find the sheet shown in
Figure 5.3
very valuable because it lists
every block size you can use for a network address. Notice that the block
sizes start at 4 and advance all the way up to a block size of 128. If you
have two networks with block sizes of 128, you can have only 2 networks.
With a block size of 64, you can have only 4, and so on, all the way to 64
networks using a block size of 4. Of course, this is assuming you’re using
the
ip subnet-zero
command in your network design.
So now all you need to do is fill in the chart in the lower-left corner, then
add the subnets to the worksheet and you’re good to go!
Based on what you’ve learned so far about block sizes and the VLSM
table, let’s create a VLSM network using a Class C network address
192.168.10.0 for the network in
Figure 5.4
, then fill out the VLSM table,
as shown in
Figure 5.5
.
In
Figure 5.4
, we have four WAN links and four LANs connected
together, so we need to create a VLSM network that will save address
space. Looks like we have two block sizes of 32, a block size of 16, and a
block size of 8, and our WANs each have a block size of 4. Take a look and
see how I filled out our VLSM chart in
Figure 5.5
.
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