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For the subnet address scheme to work, every machine on the network must know which part
of the host address will be used as the subnet address. This is accomplished by assigning a
subnet
mask
to each machine. 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.
The network administrator creates a 32-bit subnet mask composed of 1s and 0s. The 1s in
the subnet mask represent the positions that refer to the network or subnet addresses.
Not all networks need subnets, meaning they use the default subnet mask. This is basically
the same as saying that a network doesn’t have a subnet address. Table 3.1 shows the default
subnet masks for Classes A, B, and C. These default masks cannot change. In other words, you
can’t make a Class B subnet mask read 255.0.0.0. If you try, the host will read that address
as invalid and usually won’t even let you type it in. For a Class A network, you can’t change
the first byte in a subnet mask; it must read 255.0.0.0 at a minimum. Similarly, you cannot
2
4
= 16
2
5
= 32
2
6
= 64
2
7
= 128
2
8
= 256
2
9
= 512
2
10
= 1,024
2
11
= 2,048
2
12
= 4,096
2
13
= 8,192
2
14
= 16,384
Before you get stressed out about knowing all these exponents, remember that it’s helpful to
know them, but it’s not absolutely necessary. Here’s a little trick since you’re working with 2s:
Each successive power of 2 is double the previous one.
For example, all you have to do to remember the value of 2
9
is to first know that 2
8
= 256. Why?
Because when you double 2 to the eighth power (256), you get 2
9
(or 512). To determine the
value of 2
10
, simply start at 2
8
= 256, and then double it twice.
You can go the other way as well. If you needed to know what 2
6
is, for example, you just cut
256 in half two times: once to reach 2
7
and then one more time to reach 2
6
.
120
Chapter 3
IP Subnetting, Variable Length Subnet Masks (VLSMs),
assign 255.255.255.255, as this is all 1s—a broadcast address. A Class B address must start
with 255.255.0.0, and a Class C has to start with 255.255.255.0.
Classless Inter-Domain Routing (CIDR)
Another term you need to familiarize yourself with is
Classless Inter-Domain Routing
(CIDR)
. It’s basically the method that ISPs (Internet service providers) use to allocate an
amount of addresses to a company, a home—a customer. They provide addresses in a certain
block size, something I’ll be going into in greater detail later in this chapter.
When you receive a block of addresses from an ISP, what you get will look something like
this: 192.168.10.32/28. This is telling you what your subnet mask is. The slash notation (/)
means how many bits are turned on (1s). Obviously, the maximum could only be /32 because
a byte is 8 bits and there are 4 bytes in an IP address: (4
×
8 = 32). But keep in mind that the
largest subnet mask available (regardless of the class of address) can only be a /30 because
you’ve got to keep at least 2 bits for host bits.
Take, for example, a Class A default subnet mask, which is 255.0.0.0. This means that the
first byte of the subnet mask is all ones (1s), or 11111111. When referring to a slash notation,
you need to count all the 1s bits to figure out your mask. The 255.0.0.0 is considered a /8
because it has 8 bits that are 1s—that is, 8 bits that are turned on.
A Class B default mask would be 255.255.0.0, which is a /16 because 16 bits are ones (1s):
11111111.11111111.00000000.00000000.
Table 3.2 has a listing of every available subnet mask and its equivalent CIDR slash notation.
TABLE 3 . 1
Default Subnet Mask
Class Format Default Subnet Mask
A
network.node.node.node
255.0.0.0
B
network.network.node.node
255.255.0.0
C
network.network.network.node
255.255.255.0
TABLE 3 . 2
CIDR Values
Subnet Mask CIDR Value
255.0.0.0 /8
255.128.0.0 /9
255.192.0.0 /10
Subnetting Basics
121
255.224.0.0 /11
255.240.0.0 /12
255.248.0.0 /13
255.252.0.0 /14
255.254.0.0 /15
255.255.0.0 /16
255.255.128.0 /17
255.255.192.0 /18
255.255.224.0 /19
255.255.240.0 /20
255.255.248.0 /21
255.255.252.0 /22
255.255.254.0 /23
255.255.255.0 /24
255.255.255.128 /25
255.255.255.192 /26
255.255.255.224 /27
255.255.255.240 /28
255.255.255.248 /29
255.255.255.252 /30
TABLE 3 . 2
CIDR Values
(continued)
Subnet Mask CIDR Value
122
Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
The /8 through /15 can only be used with Class A network addresses. /16 through /23 can
be used by Class A and B network addresses. /24 through /30 can be used by Class A, B, and
C network addresses. This is a big reason why most companies use Class A network addresses.
Since they can use all subnet masks, they get the maximum flexibility in network design.
No, you cannot configure a Cisco router using this slash format. But wouldn’t
that be nice? Nevertheless, it’s really important for you to know subnet masks
in the slash notation (CIDR).
Subnetting Class C Addresses
There are many different ways to subnet a network. The right way is the way that works best
for you. In a Class C address, only 8 bits are available for defining the hosts. Remember that
subnet bits start at the left and go to the right, without skipping bits. This means that the only
Class C subnet masks can be the following:
Binary Decimal CIDR
---------------------------------------------------------
00000000 = 0 /24
10000000 = 128 /25
11000000 = 192 /26
11100000 = 224 /27
11110000 = 240 /28
11111000 = 248 /29
11111100 = 252 /30
We can’t use a /31 or /32 because we have to have at least 2 host bits for assigning IP addresses
to hosts. In the past, I never discussed the /25 in a Class C network. Cisco always had been concerned
with having at least 2 subnet bits, but now, because of Cisco recognizing the ip subnetzero
command in its curriculum and exam objectives, we can use just 1 subnet bit.
In the following sections, I’m going to teach you an alternate method of subnetting that
makes it easier to subnet larger numbers in no time. Trust me, you need to be able to subnet fast!
Subnetting a Class C Address: The Fast Way!
When you’ve chosen a possible subnet mask for your network and need to determine the number
of subnets, valid hosts, and broadcast addresses of each subnet that the mask provides, all you
need to do is answer five simple questions:
How many subnets does the chosen subnet mask produce?
How many valid hosts per subnet are available?
What are the valid subnets?
What’s the broadcast address of each subnet?
What are the valid hosts in each subnet?
Subnetting Basics 123
At this point, it’s important that you both understand and have memorized your powers of 2.
Please refer to the sidebar “Understanding the Powers of 2” earlier in this chapter if you need some
help. Here’s how you get the answers to those five big questions:
How many subnets? 2x = number of subnets. x is the number of masked bits, or the 1s.
For example, in 11000000, the number of 1s gives us 22 subnets. In this example, there
are 4 subnets.
How many hosts per subnet? 2y – 2 = number of hosts per subnet. y is the number of
unmasked bits, or the 0s. For example, in 11000000, the number of 0s gives us 26 – 2
hosts. In this example, there are 62 hosts per subnet. You need to subtract 2 for the subnet
address and the broadcast address, which are not valid hosts.
What are the valid subnets? 256 – subnet mask = block size, or increment number. An
example would be 256 – 192 = 64. The block size of a 192 mask is always 64. Start counting
at zero in blocks of 64 until you reach the subnet mask value and these are your subnets:
0, 64, 128, 192. Easy, huh?
What’s the broadcast address for each subnet? Now here’s the really easy part. Since we
counted our subnets in the last section as 0, 64, 128, and 192, the broadcast address is
always the number right before the next subnet. For example, the 0 subnet has a broadcast
address of 63 because the next subnet is 64. The 64 subnet has a broadcast address of 127
because the next subnet is 128. And so on. And remember, the broadcast of the last subnet
is always 255.
What are the valid hosts? Valid hosts are the numbers between the subnets, omitting the all
0s and all 1s. For example, if 64 is the subnet number and 127 is the broadcast address, then
65–126 is the valid host range—it’s always the numbers between the subnet address and the
broadcast address.
I know this can truly seem confusing. But it really isn’t as hard as it seems to be at first—
just hang in there! Why not try a few and see for yourself?
Subnetting Practice Examples: Class C Addresses
Here’s your opportunity to practice subnetting Class C addresses using the method I just
described. Exciting, isn’t it! We’re going to start with the first Class C subnet mask and work
through every subnet that we can using a Class C address. When we’re done, I’ll show you how
easy this is with Class A and B networks too!
Practice Example #1C: 255.255.255.128 (/25)
Since 128 is 10000000 in binary, there is only 1 bit for subnetting and 7 bits for hosts. We’re
going to subnet the Class C network address 192.168.10.0.
192.168.10.0 = Network address
255.255.255.128 = Subnet mask
Now, let’s answer the big five:
How many subnets? Since 128 is 1 bit on (10000000), the answer would be 21 = 2.
How many hosts per subnet? We have 7 host bits off (10000000), so the equation would
be 27 – 2 = 126 hosts.
124 Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
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 zero 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:
Before moving on to the next example, take a look at Figure 3.1. Okay, looking at a Class C
/25, it’s pretty clear there are two subnets. But so what—why is this significant? Well actually,
it’s not, but that’s not the right question. What you really want to know is what you would do
with this information!
FIGURE 3 . 1 Implementing a Class C /25 logical network
I know this isn’t exactly everyone’s favorite pastime, but it’s really important, so just hang in
there; we’re going to talk about subnetting—period. You need to know that the key to understanding
subnetting is to understand the very reason you need to do it. And I’m going to demonstrate
this by going through the process of building a physical network—and let’s add a router. (We now
have an internetwork, as I truly hope you already know!) Alright, because we added that router,
in order for the hosts on our internetwork to communicate, they must now have a logical network
addressing scheme. We could use IPX or IPv6, but IPv4 is still the most popular, and it also just
happens to be what we’re studying at the moment, so that’s what we’re going with. Okay—now
take a look back to Figure 3.1. There are two physical networks, so we’re going to implement a logical
addressing scheme that allows for two logical networks. As always, it’s a really good idea to
look ahead and consider likely growth scenarios—both short and long term, but for this example,
a /25 will do the trick.
Subnet 0 128
First host 1 129
Last host 126 254
Broadcast 127 255
.2 .3 .4 .130 .131 .132
Router#show ip route
[output cut]
C 192.168.10.0 is directly connected to Ethernet 0.
C 192.168.10.128 is directly connected to Ethernet 1.
192.168.10.0 .1 .129 192.168.10.128
Subnetting Basics 125
Practice Example #2C: 255.255.255.192 (/26)
In this second example, we’re going to subnet the network address 192.168.10.0 using the
subnet mask 255.255.255.192.
192.168.10.0 = Network address
255.255.255.192 = Subnet mask
Now, let’s answer the big five:
How many subnets? Since 192 is 2 bits on (11000000), the answer would be 22 = 4 subnets.
How many hosts per subnet? We have 6 host bits off (11000000), so the equation would
be 26 – 2 = 62 hosts.
What are the valid subnets? 256 – 192 = 64. Remember, we start at zero and count in our
block size, so our subnets are 0, 64, 128, and 192.
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 zero subnet,
the next subnet is 64, so the broadcast address for the zero subnet is 63.
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,
64, 128, and 192 subnets, the valid host ranges of each, and the broadcast address of
each subnet:
Okay, again, before getting into the next example, you can see that we can now subnet a /26.
And what are you going to do with this fascinating information? Implement it! We’ll use Figure 3.2
to practice a /26 network implementation.
The /26 mask provides four subnetworks, and we need a subnet for each router interface.
With this mask, in this example, we actually have room to add another router interface.
Practice Example #3C: 255.255.255.224 (/27)
This time, we’ll subnet the network address 192.168.10.0 and subnet mask 255.255.255.224.
192.168.10.0 = Network address
255.255.255.224 = Subnet mask
How many subnets? 224 is 11100000, so our equation would be 23 = 8.
How many hosts? 25 – 2 = 30.
What are the valid subnets? 256 – 224 = 32. We just start at zero and count to the subnet
mask value in blocks (increments) of 32: 0, 32, 64, 96, 128, 160, 192, and 224.
The subnets (do this first) 0 64 128 192
Our first host (perform host addressing last) 1 65 129 193
Our last host 62 126 190 254
The broadcast address (do this second) 63 127 191 255
126 Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
FIGURE 3 . 2 Implementing a Class C /26 logical network
What’s the broadcast address for each subnet (always the number right before the
next subnet)?
What are the valid hosts (the numbers between the subnet number and the broadcast
address)?
To answer the last two questions, first just write out the subnets, then write out the broadcast
addresses—the number right before the next subnet. Last, fill in the host addresses. The
following table gives you all the subnets for the 255.255.255.224 Class C subnet mask:
Practice Example #4C: 255.255.255.240 (/28)
Let’s practice on another one:
192.168.10.0 = Network address
255.255.255.240 = Subnet mask
Subnets? 240 is 11110000 in binary. 24 = 16.
Hosts? 4 host bits, or 24 – 2 = 14.
Valid subnets? 256 – 240 = 16. Start at 0: 0 + 16 = 16. 16 + 16 = 32. 32 + 16 = 48. 48 + 16
= 64. 64 + 16 = 80. 80 + 16 = 96. 96 + 16 = 112. 112 + 16 = 128. 128 + 16 = 144. 144 +
16 = 160. 160 + 16 = 176. 176 + 16 = 192. 192 + 16 = 208. 208 + 16 = 224. 224 + 16 = 240.
The subnet address 0 32 64 96 128 160 192 224
The first valid host 1 33 65 97 129 161 193 225
The last valid host 30 62 94 126 158 190 222 254
The broadcast address 31 63 95 127 159 191 223 255
.66 .67 .68 .130 .131 .132
Router#show ip route
[output cut]
C 192.168.10.0 is directly connected to Ethernet 0
C 192.168.10.64 is directly connected to Ethernet 1
C 192.168.10.128 is directly connected to Ethernet 2
192.168.10.64
.2 .3 .4 .5
192.168.10.0
.65 .129 192.168.10.128
.1
Subnetting Basics 127
Broadcast address for each subnet?
Valid hosts?
To answer the last two questions, check out the following table. It gives you the subnets,
valid hosts, and broadcast addresses for each subnet. First, find the address of each subnet
using the block size (increment). Second, find the broadcast address of each subnet increment
(it’s always the number right before the next valid subnet), then just fill in the host addresses.
The following table shows the available subnets, hosts, and broadcast addresses provided
from a Class C 255.255.255.240 mask:
Cisco has figured out that most people cannot count in 16s and therefore have
a hard time finding valid subnets, hosts, and broadcast addresses with the
Class C 255.255.255.240 mask. You’d be wise to study this mask.
Practice Example #5C: 255.255.255.248 (/29)
Let’s keep practicing:
192.168.10.0 = Network address
255.255.255.248 = Subnet mask
Subnets? 248 in binary = 11111000. 25 = 32.
Hosts? 23 – 2 = 6.
Valid subnets? 256 – 248 = 0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120,
128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, and 248.
Broadcast address for each subnet?
Valid hosts?
Take a look at the following table. It shows some of the subnets (first four and last four
only), valid hosts, and broadcast addresses for the Class C 255.255.255.248 mask:
Subnet 0 16 32 48 64 80 96 112 128 144 160 176 192 208 224 240
First host 1 17 33 49 65 81 97 113 129 145 161 177 193 209 225 241
Last host 14 30 46 62 78 94 110 126 142 158 174 190 206 222 238 254
Broadcast 15 31 47 63 79 95 111 127 143 159 175 191 207 223 239 255
Subnet 0 8 16 24 … 224 232 240 248
First host 1 9 17 25 … 225 233 241 249
Last host 6 14 22 30 … 230 238 246 254
Broadcast 7 15 23 31 … 231 239 247 255
128 Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
Practice Example #6C: 255.255.255.252 (/30)
Just one more:
192.168.10.0 = Network address
255.255.255.252 = Subnet mask
Subnets? 64.
Hosts? 2.
Valid subnets? 0, 4, 8, 12, etc., all the way to 252.
Broadcast address for each subnet (always the number right before the next subnet)?
Valid hosts (the numbers between the subnet number and the broadcast address)?
The following table shows you the subnet, valid host, and broadcast address of the first
four and last four subnets in the 255.255.255.252 Class C subnet:
Subnet 0 4 8 12 … 240 244 248 252
First host 1 5 9 13 … 241 245 249 253
Last host 2 6 10 14 … 242 246 250 254
Broadcast 3 7 11 15 … 243 247 251 255
Should We Really Use This Mask That Provides Only Two Hosts?
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 don’t know what to use
on your point-to-point WAN links. Is the 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 would 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. This is a really
important subject, one that we’ll address in a lot more detail in the section on VLSM network
design later in this chapter.
Subnetting Basics 129
Subnetting in Your Head: Class C Addresses
It really is possible to subnet in your head. Even if you don’t believe me, I’ll show you how.
And it’s not all that hard either—take the following example:
192.168.10.33 = Node address
255.255.255.224 = Subnet mask
First, determine the subnet and broadcast address of the above IP address. You can do this
by answering question 3 of the big five questions: 256 – 224 = 32. 0, 32, 64. The address of
33 falls between the two subnets of 32 and 64 and must be part of the 192.168.10.32 subnet.
The next subnet is 64, so the broadcast address of the 32 subnet is 63. (Remember that the
broadcast address of a subnet is always the number right before the next subnet.) The valid
host range is 33–62 (the numbers between the subnet and broadcast address). This is too easy!
Okay, let’s try another one. We’ll subnet another Class C address:
192.168.10.33 = Node address
255.255.255.240 = Subnet mask
What subnet and broadcast address is the above IP address a member of? 256 – 240 = 16.
0, 16, 32, 48. Bingo—the host address is between the 32 and 48 subnets. The subnet is
192.168.10.32, and the broadcast address is 47 (the next subnet is 48). The valid host range
is 33–46 (the numbers between the subnet number and the broadcast address).
Okay, we need to do more, just to make sure you have this down.
You have a node address of 192.168.10.174 with a mask of 255.255.255.240. What is the
valid host range?
The mask is 240, so we’d do 256 – 240 = 16. This is our block size. Just keep adding 16
until we pass the host address of 174, starting at zero, of course: 0, 16, 32, 48, 64, 80, 96, 112,
128, 144, 160, 176. The host address of 174 is between 160 and 176, so the subnet is 160. The
broadcast address is 175; the valid host range is 161–174. That was a tough one.
One more—just for fun. This is the easiest one of all Class C subnetting:
192.168.10.17 = Node address
255.255.255.252 = Subnet mask
What subnet and broadcast address is the above IP address a member of? 256 – 252 = 0
(always start at zero unless told otherwise), 4, 8, 12, 16, 20, etc. You’ve got it! The host address
is between the 16 and 20 subnets. The subnet is 192.168.10.16, and the broadcast address is 19.
The valid host range is 17–18.
Now that you’re all over Class C subnetting, let’s move on to Class B subnetting. But before
we do, let’s have a quick review.
What Do We Know?
Okay—here’s where you can really apply what you’ve learned so far, and begin committing
it all to memory. This is a very cool section that I’ve been using in my classes for years. It will
really help you nail down subnetting!
130 Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
When you see a subnet mask or slash notation (CIDR), you should know the following:
/25 What do we know about a /25?
128 mask
1 bits on and 7 bits off (10000000)
Block size of 128
2 subnets, each with 126 hosts
/26 What do we know about a /26?
192 mask
2 bits on and 6 bits off (11000000)
Block size of 64
4 subnets, each with 62 hosts
/27 What do we know about a /27?
224 mask
3 bits on and 5 bits off (11100000)
Block size of 32
8 subnets, each with 30 hosts
/28 What do we know about a /28?
240 mask
4 bits on and 4 bits off
Block size of 16
16 subnets, each with 14 hosts
/29 What do we know about a /29?
248 mask
5 bits on and 3 bits off
Block size of 8
32 subnets, each with 6 hosts
/30 What do we know about a /30?
252 mask
6 bits on and 2 bits off
Block size of 4
64 subnets, each with 2 hosts
Subnetting Basics 131
Regardless of whether you have a Class A, Class B, or Class C address, the /30 mask will
provide you with only two hosts, ever. This mask is suited almost exclusively—as well as suggested
by Cisco—for use on point-to-point links.
If you can memorize this “What Do We Know?” section, you’ll be much better off in your
day-to-day job and in your studies. Try saying it out loud, which helps you memorize things—
yes, your significant other and/or coworkers will think you’ve lost it, but they probably already
do if you are in the networking field. And if you’re not yet in the networking field but are studying
all this to break into it, you might as well have people start thinking you’re an odd bird now
since they will eventually anyway.
It’s also helpful to write these on some type of flashcards and have people test your skill.
You’d be amazed at how fast you can get subnetting down if you memorize block sizes as well
as this “What Do We Know?” section.
Subnetting Class B Addresses
Before we dive into this, let’s look at all the possible Class B subnet masks first. Notice that
we have a lot more possible subnet masks than we do with a Class C network address:
255.255.0.0 (/16)
255.255.128.0 (/17) 255.255.255.0 (/24)
255.255.192.0 (/18) 255.255.255.128 (/25)
255.255.224.0 (/19) 255.255.255.192 (/26)
255.255.240.0 (/20) 255.255.255.224 (/27)
255.255.248.0 (/21) 255.255.255.240 (/28)
255.255.252.0 (/22) 255.255.255.248 (/29)
255.255.254.0 (/23) 255.255.255.252 (/30)
We know the Class B network address has 16 bits available for host addressing. This means
we can use up to 14 bits for subnetting (because we have to leave at least 2 bits for host addressing).
Using a /16 means you are not subnetting with Class B, but it is a mask you can use.
By the way, do you notice anything interesting about that list of subnet values—
a pattern, maybe? Ah ha! That’s exactly why I had you memorize the binary-todecimal
numbers at the beginning of this section. Since subnet mask bits start on
the left and move to the right and bits can’t be skipped, the numbers are always
the same regardless of the class of address. Memorize this pattern.
The process of subnetting a Class B network is pretty much the same as it is for a Class C,
except that you just have more host bits and you start in the third octet.
Use the same subnet numbers for the third octet with Class B that you used for the fourth
octet with Class C, but add a zero to the network portion and a 255 to the broadcast section
132 Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
in the fourth octet. The following table shows you an example host range of two subnets used
in a Class B 240 (/20) subnet mask:
Just add the valid hosts between the numbers and you’re set!
The preceding example is true only until you get up to /24. After that, it’s
numerically exactly like Class C.
Subnetting Practice Examples: Class B Addresses
This section will give you an opportunity to practice subnetting Class B addresses. Again, I
have to mention that this is the same as subnetting with Class C, except we start in the third
octet—with the exact same numbers!
Practice Example #1B: 255.255.128.0 (/17)
172.16.0.0 = Network address
255.255.128.0 = Subnet mask
Subnets? 21 = 2 (same as Class C).
Hosts? 215 – 2 = 32,766 (7 bits in the third octet, and 8 in the fourth).
Valid subnets? 256 – 128 = 128. 0, 128. Remember that subnetting is performed in the
third octet, so the subnet numbers are really 0.0 and 128.0, as shown in the next table.
These are the exact numbers we used with Class C; we use them in the third octet and add
a 0 in the fourth octet for the network address.
Broadcast address for each subnet?
Valid hosts?
The following table shows the two subnets available, the valid host range, and the broadcast
address of each:
Okay, notice that we just added the fourth octet’s lowest and highest values and came up
with the answers. And again, it’s done exactly the same way as for a Class C subnet. We just
use the same numbers in the third octet and add 0 and 255 in the fourth octet—pretty simple
First subnet 16.0 32.0
Second subnet 31.255 47.255
Subnet 0.0 128.0
First host 0.1 128.1
Last host 127.254 255.254
Broadcast 127.255 255.255
Subnetting Basics 133
huh! I really can’t say this enough: It’s just not hard; the numbers never change; we just use
them in different octets!
Practice Example #2B: 255.255.192.0 (/18)
172.16.0.0 = Network address
255.255.192.0 = Subnet mask
Subnets? 22 = 4.
Hosts? 214 – 2 = 16,382 (6 bits in the third octet, and 8 in the fourth).
Valid subnets? 256 – 192 = 64. 0, 64, 128, 192. Remember that the subnetting is performed
in the third octet, so the subnet numbers are really 0.0, 64.0, 128.0, and 192.0,
as shown in the next table.
Broadcast address for each subnet?
Valid hosts?
The following table shows the four subnets available, the valid host range, and the broadcast
address of each:
Again, it’s pretty much the same as it is for a Class C subnet—we just added 0 and 255 in
the fourth octet for each subnet in the third octet.
Practice Example #3B: 255.255.240.0 (/20)
172.16.0.0 = Network address
255.255.240.0 = Subnet mask
Subnets? 24 = 16.
Hosts? 212 – 2 = 4094.
Valid subnets? 256 – 240 = 0, 16, 32, 48, etc., up to 240. Notice that these are the same
numbers as a Class C 240 mask—we just put them in the third octet and add a 0 and 255
in the fourth octet.
Broadcast address for each subnet?
Valid hosts?
Subnet 0.0 64.0 128.0 192.0
First host 0.1 64.1 128.1 192.1
Last host 63.254 127.254 191.254 255.254
Broadcast 63.255 127.255 191.255 255.255
134 Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
The following table shows the first four subnets, valid hosts, and broadcast addresses in a
Class B 255.255.240.0 mask:
Practice Example #4B: 255.255.254.0 (/23)
172.16.0.0 = Network address
255.255.254.0 = Subnet mask
Subnets? 27 = 128.
Hosts? 29 – 2 = 510.
Valid subnets? 256 – 254 = 0, 2, 4, 6, 8, etc., up to 254.
Broadcast address for each subnet?
Valid hosts?
The following table shows the first five subnets, valid hosts, and broadcast addresses in a
Class B 255.255.254.0 mask:
Practice Example #5B: 255.255.255.0 (/24)
Contrary to popular belief, 255.255.255.0 used with a Class B network address is not called a
Class B network with a Class C subnet mask. It’s amazing how many people see this mask used
in a Class B network and think it’s a Class C subnet mask. This is a Class B subnet mask with
8 bits of subnetting—it’s considerably different from a Class C mask. Subnetting this address is
fairly simple:
172.16.0.0 = Network address
255.255.255.0 = Subnet mask
Subnets? 28 = 256.
Hosts? 28 – 2 = 254.
Valid subnets? 256 – 255 = 1. 0, 1, 2, 3, etc., all the way to 255.
Broadcast address for each subnet?
Valid hosts?
Subnet 0.0 16.0 32.0 48.0
First host 0.1 16.1 32.1 48.1
Last host 15.254 31.254 47.254 63.254
Broadcast 15.255 31.255 47.255 63.255
Subnet 0.0 2.0 4.0 6.0 8.0
First host 0.1 2.1 4.1 6.1 8.1
Last host 1.254 3.254 5.254 7.254 9.254
Broadcast 1.255 3.255 5.255 7.255 9.255
Subnetting Basics 135
The following table shows the first four and last two subnets, the valid hosts, and the
broadcast addresses in a Class B 255.255.255.0 mask:
Practice Example #6B: 255.255.255.128 (/25)
This is one of the hardest subnet masks you can play with. And worse, it actually is a really
good subnet to use in production because it creates over 500 subnets with 126 hosts for each
subnet—a nice mixture. So, don’t skip over it!
172.16.0.0 = Network address
255.255.255.128 = Subnet mask
Subnets? 29 = 512.
Hosts? 27 – 2 = 126.
Valid subnets? Okay, now for the tricky part. 256 – 255 = 1. 0, 1, 2, 3, etc. for the third
octet. But you can’t forget the one subnet bit used in the fourth octet. Remember when I
showed you how to figure one subnet bit with a Class C mask? You figure this the same
way. (Now you know why I showed you the 1-bit subnet mask in the Class C section—
to make this part easier.) You actually get 2 subnets for each third octet value, hence the
512 subnets. For example, if the third octet is showing subnet 3, the 2 subnets would actually
be 3.0 and 3.128.
Broadcast address for each subnet?
Valid hosts?
The following table shows how you can create subnets, valid hosts, and broadcast
addresses using the Class B 255.255.255.128 subnet mask (the first eight subnets are shown,
and then the last two subnets):
Practice Example #7B: 255.255.255.192 (/26)
Now, this is where Class B subnetting gets easy. Since the third octet has a 255 in the mask section,
whatever number is listed in the third octet is a subnet number. However, now that we
Subnet 0.0 1.0 2.0 3.0 ... 254.0 255.0
First host 0.1 1.1 2.1 3.1 ... 254.1 255.1
Last host 0.254 1.254 2.254 3.254 ... 254.254 255.254
Broadcast 0.255 1.255 2.255 3.255 ... 254.255 255.255
Subnet 0.0 0.128 1.0 1.128 2.0 2.128 3.0 3.128 ... 255.0 255.128
First host 0.1 0.129 1.1 1.129 2.1 2.129 3.1 3.129 ... 255.1 255.129
Last host 0.126 0.254 1.126 1.254 2.126 2.254 3.126 3.254 ... 255.126 255.254
Broadcast 0.127 0.255 1.127 1.255 2.127 2.255 3.127 3.255 ... 255.127 255.255
136 Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
have a subnet number in the fourth octet, we can subnet this octet just as we did with Class
C subnetting. Let’s try it out:
172.16.0.0 = Network address
255.255.255.192 = Subnet mask
Subnets? 210 = 1024.
Hosts? 26 – 2 = 62.
Valid subnets? 256 – 192 = 64. The subnets are shown in the following table. Do these
numbers look familiar?
Broadcast address for each subnet?
Valid hosts?
The following table shows the first eight subnet ranges, valid hosts, and broadcast
addresses:
Notice that for each subnet value in the third octet, you get subnets 0, 64, 128, and 192 in
the fourth octet.
Practice Example #8B: 255.255.255.224 (/27)
This is done the same way as the preceding subnet mask, except that we just have more subnets
and fewer hosts per subnet available.
172.16.0.0 = Network address
255.255.255.224 = Subnet mask
Subnets? 211 = 2048.
Hosts? 25 – 2 = 30.
Valid subnets? 256 – 224 = 32. 0, 32, 64, 96, 128, 160, 192, 224.
Broadcast address for each subnet?
Valid hosts?
The following table shows the first eight subnets:
Subnet 0.0 0.64 0.128 0.192 1.0 1.64 1.128 1.192
First host 0.1 0.65 0.129 0.193 1.1 1.65 1.129 1.193
Last host 0.62 0.126 0.190 0.254 1.62 1.126 1.190 1.254
Broadcast 0.63 0.127 0.191 0.255 1.63 1.127 1.191 1.255
Subnet 0.0 0.32 0.64 0.96 0.128 0.160 0.192 0.224
First host 0.1 0.33 0.65 0.97 0.129 0.161 0.193 0.225
Last host 0.30 0.62 0.94 0.126 0.158 0.190 0.222 0.254
Broadcast 0.31 0.63 0.95 0.127 0.159 0.191 0.223 0.255
Subnetting Basics 137
This next table shows the last eight subnets:
Subnetting in Your Head: Class B Addresses
Are you nuts? Subnet Class B addresses in our heads? It’s actually easier than writing it out—
I’m not kidding! Let me show you how:
Question: What subnet and broadcast address is the IP address 172.16.10.33
255.255.255.224 (/27) a member of?
Answer: The interesting octet is the fourth octet. 256 – 224 = 32. 32 + 32 = 64. Bingo: 33
is between 32 and 64. However, remember that the third octet is considered part of the
subnet, so the answer would be the 10.32 subnet. The broadcast is 10.63, since 10.64 is
the next subnet. That was a pretty easy one.
Question: What subnet and broadcast address is the IP address 172.16.66.10
255.255.192.0 (/18) a member of?
Answer: The interesting octet is the third octet instead of the fourth octet. 256 – 192 = 64.
0, 64, 128. The subnet is 172.16.64.0. The broadcast must be 172.16.127.255 since
128.0 is the next subnet.
Question: What subnet and broadcast address is the IP address 172.16.50.10
255.255.224.0 (/19) a member of?
Answer: 256 – 224 = 0, 32, 64 (remember, we always start counting at zero). The subnet
is 172.16.32.0, and the broadcast must be 172.16.63.25 since 64.0 is the next subnet.
Question: What subnet and broadcast address is the IP address 172.16.46.255
255.255.240.0 (/20) a member of?
Answer: 256 – 240 = 16. The third octet is interesting to us. 0, 16, 32, 48. This subnet
address must be in the 172.16.32.0 subnet, and the broadcast must be 172.16.47.255
since 48.0 is the next subnet. So, yes, 172.16.46.255 is a valid host.
Question: What subnet and broadcast address is the IP address 172.16.45.14
255.255.255.252 (/30) a member of?
Answer: Where is the interesting octet? 256 – 252 = 0, 4, 8, 12, 16 (in the fourth octet).
The subnet is 172.16.45.12, with a broadcast of 172.16.45.15 because the next subnet is
172.16.45.16.
Question: What is the subnet and broadcast address of the host 172.16.88.255/20?
Answer: What is a /20? If you can’t answer this, you can’t answer this question, can you?
A /20 is 255.255.240.0, which gives us a block size of 16 in the third octet, and since no
Subnet 255.0 255.32 255.64 255.96 255.128 255.160 255.192 255.224
First host 255.1 255.33 255.65 255.97 255.129 255.161 255.193 255.225
Last host 255.30 255.62 255.94 255.126 255.158 255.190 255.222 255.254
Broadcast 255.31 255.63 255.95 255.127 255.159 255.191 255.223 255.255
138 Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
subnet bits are on in the fourth octet, the answer is always 0 and 255 in the fourth octet.
0, 16, 32, 48, 64, 80, 96…bingo. 88 is between 80 and 96, so the subnet is 80.0 and the
broadcast address is 95.255.
Question: A router receives a packet on an interface with a destination address of
172.16.46.191/26. What will the router do with this packet?
Answer: Discard it. Do you know why? 172.16.46.191/26 is a 255.255.255.192 mask,
which gives us a block size of 64. Our subnets are then 0, 64, 128, 192. 191 is the broadcast
address of the 128 subnet, so a router, by default, will discard any broadcast packets.
Subnetting Class A Addresses
Class A subnetting is not performed any differently than Classes B and C, but there are 24 bits
to play with instead of the 16 in a Class B address and the 8 in a Class C address.
Let’s start by listing all the Class A masks:
255.0.0.0 (/8)
255.128.0.0 (/9) 255.255.240.0 (/20)
255.192.0.0 (/10) 255.255.248.0 (/21)
255.224.0.0 (/11) 255.255.252.0 (/22)
255.240.0.0 (/12) 255.255.254.0 (/23)
255.248.0.0 (/13) 255.255.255.0 (/24)
255.252.0.0 (/14) 255.255.255.128 (/25)
255.254.0.0 (/15) 255.255.255.192 (/26)
255.255.0.0 (/16) 255.255.255.224 (/27)
255.255.128.0 (/17) 255.255.255.240 (/28)
255.255.192.0 (/18) 255.255.255.248 (/29)
255.255.224.0 (/19) 255.255.255.252 (/30)
That’s it. You must leave at least 2 bits for defining hosts. And I hope you can see the pattern
by now. Remember, we’re going to do this the same way as a Class B or C subnet. It’s just
that, again, we simply have more host bits and we just use the same subnet numbers we used
with Class B and C, but we start using these numbers in the second octet.
Subnetting Practice Examples: Class A Addresses
When you look at an IP address and a subnet mask, you must be able to distinguish the bits
used for subnets from the bits used for determining hosts. This is imperative. If you’re still
struggling with this concept, please reread the section “IP Addressing” in Chapter 2. It shows
you how to determine the difference between the subnet and host bits and should help clear
things up.
Subnetting Basics 139
Practice Example #1A: 255.255.0.0 (/16)
Class A addresses use a default mask of 255.0.0.0, which leaves 22 bits for subnetting since
you must leave 2 bits for host addressing. The 255.255.0.0 mask with a Class A address is
using 8 subnet bits.
Subnets? 28 = 256.
Hosts? 216 – 2 = 65,534.
Valid subnets? What is the interesting octet? 256 – 255 = 1. 0, 1, 2, 3, etc. (all in the second
octet). The subnets would be 10.0.0.0, 10.1.0.0, 10.2.0.0, 10.3.0.0, etc., up to 10.255.0.0.
Broadcast address for each subnet?
Valid hosts?
The following table shows the first two and last two subnets, valid host range, and broadcast
addresses for the private Class A 10.0.0.0 network:
Practice Example #2A: 255.255.240.0 (/20)
255.255.240.0 gives us 12 bits of subnetting and leaves us 12 bits for host addressing.
Subnets? 212 = 4096.
Hosts? 212 – 2 = 4094.
Valid subnets? What is your interesting octet? 256 – 240 = 16. The subnets in the second
octet are a block size of 1 and the subnets in the third octet are 0, 16, 32, etc.
Broadcast address for each subnet?
Valid hosts?
The following table shows some examples of the host ranges—the first three and the
last subnets:
Subnet 10.0.0.0 10.1.0.0 … 10.254.0.0 10.255.0.0
First host 10.0.0.1 10.1.0.1 … 10.254.0.1 10.255.0.1
Last host 10.0.255.254 10.1.255.254 … 10.254.255.254 10.255.255.254
Broadcast 10.0.255.255 10.1.255.255 … 10.254.255.255 10.255.255.255
Subnet 10.0.0.0 10.0.16.0 10.0.32.0 … 10.255.240.0
First host 10.0.0.1 10.0.16.1 10.0.32.1 … 10.255.240.1
Last host 10.0.15.254 10.0.31.254 10.0.47.254 … 10.255.255.254
Broadcast 10.0.15.255 10.0.31.255 10.0.47.255 … 10.255.255.255
140 Chapter 3 IP Subnetting, Variable Length Subnet Masks (VLSMs),
Practice Example #3A: 255.255.254.0 (/23)
255.255.254.0 gives us 15 bits of subnetting and leaves us 9 bits for host addressing.
Subnets? 215 = 32,766.
Hosts? 29 – 2 = 510.
Valid subnets? What is your interesting octet? 256 – 254 = 2. The subnets in the second
octet are a block size of 1 and the subnets in the third octet are a block size of 2. Starting
at 0, the subnets are 0, 2, 4, 6, etc.
Broadcast address for each subnet?
Valid hosts?
The following table shows some examples of the host ranges—the first three and the
last subnets:
Practice Example #4A: 255.255.255.192 (/26)
Let’s do one more example using the second, third, and fourth octets for subnetting.
Subnets? 218 = 262,144.
Hosts? 26 – 2 = 62.
Valid subnets? In the second and third octet, the block size is 1, and in the fourth octet,
the block size is 64.
Broadcast address for each subnet?
Valid hosts?
The following table shows the first four subnets and their valid hosts and broadcast
addresses in the Class A 255.255.255.192 mask:
The following table shows the last four subnets and their valid hosts and broadcast addresses:
Subnet 10.0.0.0 10.0.2.0 10.0.4.0 … 10.255.254.0
First host 10.0.0.1 10.0.2.1 10.0.4.1 … 10.255.254.1
Last host 10.0.1.254 10.0.3.254 10.0.5.254 … 10.255.255.254
Broadcast 10.0.1.255 10.0.3.255 10.0.5.255 … 10.255.255.255
Subnet 10.0.0.0 10.0.0.64 10.0.0.128 10.0.0.192
First host 10.0.0.1 10.0.0.65 10.0.0.129 10.0.0.193
Last host 10.0.0.62 10.0.0.126 10.0.0.190 10.0.0.254
Broadcast 10.0.0.63 10.0.0.127 10.0.0.191 10.0.0.255
Subnet 10.255.255.0 10.255.255.64 10.255.255.128 10.255.255.192
First host 10.255.255.1 10.255.255.65 10.255.255.129 10.255.255.193
Variable Length Subnet Masks (VLSMs) 141
Subnetting in Your Head: Class A Addresses
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 (typically called the interesting octet; one that is something other than 0 or
255)—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?) This makes our subnet 10.20.80.0, with a broadcast 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!
Okay, let’s practice on one more, just for fun (please study this one!).
Host IP: 10.16.3.65/23
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 10.16.2.1 and 10.16.3.254 is considered
a valid host.
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