This document gives you basic information needed in order to
configure your router for routing IP, such as how addresses are broken
down and how subnetting works. You learn how to assign each interface on
the router an IP address with a unique subnet. There are many examples
to help tie everything together.
Cisco recommends that you have knowledge of these topics:
-
Basic understanding of binary and decimal numbers.
This document is not restricted to specific software and hardware versions.
If definitions are helpful to you, use these vocabulary terms to get you started:
-
Address—The unique number ID assigned to one host or interface in a network.
-
Subnet—A portion of a network sharing a particular subnet address.
-
Subnet mask—A 32-bit combination used to describe which portion of an address refers to the subnet and which part refers to the host.
-
Interface—A network connection.
If you have already received your legitimate address(es) from the
Internet Network Information Center (InterNIC), you are ready to begin.
If you do not plan to connect to the Internet, Cisco strongly suggests
that you use reserved addresses from
RFC 1918 
.
Refer to
Cisco Technical Tips Conventions for more information on document conventions.
An IP address is an address used in order to uniquely identify a
device on an IP network. The address is made up of 32 binary bits, which
can be divisible into a network portion and host portion with the help
of a subnet mask. The 32 binary bits are broken into four octets (1
octet = 8 bits). Each octet is converted to decimal and separated by a
period (dot). For this reason, an IP address is said to be expressed in
dotted decimal format (for example, 172.16.81.100). The value in each
octet ranges from 0 to 255 decimal, or 00000000 - 11111111 binary.
Here is how binary octets convert to decimal: The right most bit, or least significant bit, of an octet holds a value of 2
0. The bit just to the left of that holds a value of 2
1. This continues until the left-most bit, or most significant bit, which holds a value of 2
7. So if all binary bits are a one, the decimal equivalent would be 255 as shown here:
1 1 1 1 1 1 1 1
128 64 32 16 8 4 2 1 (128+64+32+16+8+4+2+1=255)
Here is a sample octet conversion when not all of the bits are set to 1.
0 1 0 0 0 0 0 1
0 64 0 0 0 0 0 1 (0+64+0+0+0+0+0+1=65)
And this is sample shows an IP address represented in both binary and decimal.
10. 1. 23. 19 (decimal)
00001010.00000001.00010111.00010011 (binary)
These octets are broken down to provide an addressing scheme that can
accommodate large and small networks. There are five different classes
of networks, A to E. This document focuses on addressing classes A to C,
since classes D and E are reserved and discussion of them is beyond the
scope of this document.
Note: Also note that the terms "Class A, Class B" and so on
are used in this document to help facilitate the understanding of IP
addressing and subnetting. These terms are rarely used in the industry
anymore because of the introduction of
classless interdomain routing (CIDR).
Given an IP address, its class can be determined from the three high-order bits.
Figure 1
shows the significance in the three high order bits and the range of
addresses that fall into each class. For informational purposes, Class D
and Class E addresses are also shown.
Figure 1
In a Class A address, the first octet is the network portion, so the Class A example in Figure 1
has a major network address of 1.0.0.0 - 127.255.255.255. Octets 2, 3,
and 4 (the next 24 bits) are for the network manager to divide into
subnets and hosts as he/she sees fit. Class A addresses are used for
networks that have more than 65,536 hosts (actually, up to 16777214
hosts!).
In a Class B address, the first two octets are the network portion, so the Class B example in
Figure 1
has a major network address of 128.0.0.0 - 191.255.255.255. Octets 3
and 4 (16 bits) are for local subnets and hosts. Class B addresses are
used for networks that have between 256 and 65534 hosts.
In a Class C address, the first three octets are the network portion. The Class C example in
Figure 1
has a major network address of 192.0.0.0 - 233.255.255.255. Octet 4 (8
bits) is for local subnets and hosts - perfect for networks with less
than 254 hosts.
A network mask helps you know which portion of the address identifies
the network and which portion of the address identifies the node. Class
A, B, and C networks have default masks, also known as natural masks,
as shown here:
Class A: 255.0.0.0
Class B: 255.255.0.0
Class C: 255.255.255.0
An IP address on a Class A network that has not been subnetted would
have an address/mask pair similar to: 8.20.15.1 255.0.0.0. To see how
the mask helps you identify the network and node parts of the address,
convert the address and mask to binary numbers.
8.20.15.1 = 00001000.00010100.00001111.00000001
255.0.0.0 = 11111111.00000000.00000000.00000000
Once you have the address and the mask represented in binary, then
identifying the network and host ID is easier. Any address bits which
have corresponding mask bits set to 1 represent the network ID. Any
address bits that have corresponding mask bits set to 0 represent the
node ID.
8.20.15.1 = 00001000.00010100.00001111.00000001
255.0.0.0 = 11111111.00000000.00000000.00000000
-----------------------------------
net id | host id
netid = 00001000 = 8
hostid = 00010100.00001111.00000001 = 20.15.1
Subnetting allows you to create multiple logical networks that exist
within a single Class A, B, or C network. If you do not subnet, you are
only able to use one network from your Class A, B, or C network, which
is unrealistic.
Each data link on a network must have a unique network ID, with every
node on that link being a member of the same network. If you break a
major network (Class A, B, or C) into smaller subnetworks, it allows you
to create a network of interconnecting subnetworks. Each data link on
this network would then have a unique network/subnetwork ID. Any device,
or gateway, connecting
n networks/subnetworks has
n distinct IP addresses, one for each network / subnetwork that it interconnects.
In order to subnet a network, extend the natural mask using some of
the bits from the host ID portion of the address to create a subnetwork
ID. For example, given a Class C network of 204.17.5.0 which has a
natural mask of 255.255.255.0, you can create subnets in this manner:
204.17.5.0 - 11001100.00010001.00000101.00000000
255.255.255.224 - 11111111.11111111.11111111.11100000
--------------------------|sub|----
By extending the mask to be 255.255.255.224, you have taken three
bits (indicated by "sub") from the original host portion of the address
and used them to make subnets. With these three bits, it is possible to
create eight subnets. With the remaining five host ID bits, each subnet
can have up to 32 host addresses, 30 of which can actually be assigned
to a device
since host ids of all zeros or all ones are not allowed (it is very important to remember this). So, with this in mind, these subnets have been created.
204.17.5.0 255.255.255.224 host address range 1 to 30
204.17.5.32 255.255.255.224 host address range 33 to 62
204.17.5.64 255.255.255.224 host address range 65 to 94
204.17.5.96 255.255.255.224 host address range 97 to 126
204.17.5.128 255.255.255.224 host address range 129 to 158
204.17.5.160 255.255.255.224 host address range 161 to 190
204.17.5.192 255.255.255.224 host address range 193 to 222
204.17.5.224 255.255.255.224 host address range 225 to 254
Note: There are two ways to denote these masks. First, since
you are using three bits more than the "natural" Class C mask, you can
denote these addresses as having a 3-bit subnet mask. Or, secondly, the
mask of 255.255.255.224 can also be denoted as /27 as there are 27 bits
that are set in the mask. This second method is used with
CIDR.
With this method, one of these networks can be described with the
notation prefix/length. For example, 204.17.5.32/27 denotes the network
204.17.5.32 255.255.255.224. When appropriate the prefix/length notation
is used to denote the mask throughout the rest of this document.
The network subnetting scheme in this section allows for eight subnets, and the network might appear as:
Figure 2
Notice that each of the routers in Figure 2
is attached to four subnetworks, one subnetwork is common to both
routers. Also, each router has an IP address for each subnetwork to
which it is attached. Each subnetwork could potentially support up to 30
host addresses.
This brings up an interesting point. The more host bits you use for a
subnet mask, the more subnets you have available. However, the more
subnets available, the less host addresses available per subnet. For
example, a Class C network of 204.17.5.0 and a mask of 255.255.255.224
(/27) allows you to have eight subnets, each with 32 host addresses (30
of which could be assigned to devices). If you use a mask of
255.255.255.240 (/28), the break down is:
204.17.5.0 - 11001100.00010001.00000101.00000000
255.255.255.240 - 11111111.11111111.11111111.11110000
--------------------------|sub |---
Since you now have four bits to make subnets with, you only have four
bits left for host addresses. So in this case you can have up to 16
subnets, each of which can have up to 16 host addresses (14 of which can
be assigned to devices).
Take a look at how a Class B network might be subnetted. If you have
network 172.16.0.0 ,then you know that its natural mask is 255.255.0.0
or 172.16.0.0/16. Extending the mask to anything beyond 255.255.0.0
means you are subnetting. You can quickly see that you have the ability
to create a lot more subnets than with the Class C network. If you use a
mask of 255.255.248.0 (/21), how many subnets and hosts per subnet does
this allow for?
172.16.0.0 - 10101100.00010000.00000000.00000000
255.255.248.0 - 11111111.11111111.11111000.00000000
-----------------| sub |-----------
You are using five bits from the original host bits for subnets. This allows you to have 32 subnets (2
5).
After using the five bits for subnetting, you are left with 11 bits for
host addresses. This allows each subnet so have 2048 host addresses (2
11), 2046 of which could be assigned to devices.
Note: In the past, there were limitations to the use of a
subnet 0 (all subnet bits are set to zero) and all ones subnet (all
subnet bits set to one). Some devices would not allow the use of these
subnets. Cisco Systems devices allow the use of these subnets when the
ip subnet zero command is configured.
Now that you have an understanding of subnetting, put this knowledge
to use. In this example, you are given two address / mask combinations,
written with the prefix/length notation, which have been assigned to two
devices. Your task is to determine if these devices are on the same
subnet or different subnets. You can do this by using the address and
mask of each device to determine to which subnet each address belongs.
DeviceA: 172.16.17.30/20
DeviceB: 172.16.28.15/20
Determining the Subnet for DeviceA:
172.16.17.30 - 10101100.00010000.00010001.00011110
255.255.240.0 - 11111111.11111111.11110000.00000000
-----------------| sub|------------
subnet = 10101100.00010000.00010000.00000000 = 172.16.16.0
Looking at the address bits that have a corresponding mask bit set to
one, and setting all the other address bits to zero (this is equivalent
to performing a logical "AND" between the mask and address), shows you
to which subnet this address belongs. In this case, DeviceA belongs to
subnet 172.16.16.0.
Determining the Subnet for DeviceB:
172.16.28.15 - 10101100.00010000.00011100.00001111
255.255.240.0 - 11111111.11111111.11110000.00000000
-----------------| sub|------------
subnet = 10101100.00010000.00010000.00000000 = 172.16.16.0
From these determinations, DeviceA and DeviceB have addresses that are part of the same subnet.
Given the Class C network of 204.15.5.0/24, subnet the network in order to create the network in
Figure 3 with the host requirements shown.
Figure 3
Looking at the network shown in Figure 3,
you can see that you are required to create five subnets. The largest
subnet must support 28 host addresses. Is this possible with a Class C
network? and if so, then how?
You can start by looking at the subnet requirement. In order to
create the five needed subnets you would need to use three bits from the
Class C host bits. Two bits would only allow you four subnets (2
2).
Since you need three subnet bits, that leaves you with five bits for
the host portion of the address. How many hosts does this support? 2
5 = 32 (30 usable). This meets the requirement.
Therefore you have determined that it is possible to create this
network with a Class C network. An example of how you might assign the
subnetworks is:
netA: 204.15.5.0/27 host address range 1 to 30
netB: 204.15.5.32/27 host address range 33 to 62
netC: 204.15.5.64/27 host address range 65 to 94
netD: 204.15.5.96/27 host address range 97 to 126
netE: 204.15.5.128/27 host address range 129 to 158
In all of the previous examples of subnetting, notice that the same
subnet mask was applied for all the subnets. This means that each subnet
has the same number of available host addresses. You can need this in
some cases, but, in most cases, having the same subnet mask for all
subnets ends up wasting address space. For example, in the
Sample Exercise 2
section, a class C network was split into eight equal-size subnets;
however, each subnet did not utilize all available host addresses, which
results in wasted address space.
Figure 4 illustrates this wasted address space.
Figure 4
Figure 4
illustrates that of the subnets that are being used, NetA, NetC, and
NetD have a lot of unused host address space. It is possible that this
was a deliberate design accounting for future growth, but in many cases
this is just wasted address space due to the fact that the same subnet
mask is being used for all the subnets.
Variable Length Subnet Masks (VLSM) allows you to use different masks for each subnet, thereby using address space efficiently.
Given the same network and requirements as in
Sample Exercise 2 develop a subnetting scheme with the use of VLSM, given:
netA: must support 14 hosts
netB: must support 28 hosts
netC: must support 2 hosts
netD: must support 7 hosts
netE: must support 28 host
Determine what mask allows the required number of hosts.
netA: requires a /28 (255.255.255.240) mask to support 14 hosts
netB: requires a /27 (255.255.255.224) mask to support 28 hosts
netC: requires a /30 (255.255.255.252) mask to support 2 hosts
netD*: requires a /28 (255.255.255.240) mask to support 7 hosts
netE: requires a /27 (255.255.255.224) mask to support 28 hosts
* a /29 (255.255.255.248) would only allow 6 usable host addresses
therefore netD requires a /28 mask.
The easiest way to assign the subnets is to assign the largest first. For example, you can assign in this manner:
netB: 204.15.5.0/27 host address range 1 to 30
netE: 204.15.5.32/27 host address range 33 to 62
netA: 204.15.5.64/28 host address range 65 to 78
netD: 204.15.5.80/28 host address range 81 to 94
netC: 204.15.5.96/30 host address range 97 to 98
This can be graphically represented as shown in Figure 5:
Figure 5
Figure 5 illustrates how using VLSM helped save more than half of the address space.
Classless Interdomain Routing (CIDR) was introduced to improve both
address space utilization and routing scalability in the Internet. It
was needed because of the rapid growth of the Internet and growth of the
IP routing tables held in the Internet routers.
CIDR moves way from the traditional IP classes (Class A, Class B,
Class C, and so on). In CIDR , an IP network is represented by a prefix,
which is an IP address and some indication of the length of the mask.
Length means the number of left-most contiguous mask bits that are set
to one. So network 172.16.0.0 255.255.0.0 can be represented as
172.16.0.0/16. CIDR also depicts a more hierarchical Internet
architecture, where each domain takes its IP addresses from a higher
level. This allows for the summarization of the domains to be done at
the higher level. For example, if an ISP owns network 172.16.0.0/16,
then the ISP can offer 172.16.1.0/24, 172.16.2.0/24, and so on to
customers. Yet, when advertising to other providers, the ISP only needs
to advertise 172.16.0.0/16.
For more information on CIDR, see
RFC 1518 and
RFC 1519 .
Routers A and B are connected via serial interface.
hostname routera
!
ip routing
!
int e 0
ip address 172.16.50.1 255.255.255.0
!(subnet 50)
int e 1 ip address 172.16.55.1 255.255.255.0
!(subnet 55)
int t 0 ip address 172.16.60.1 255.255.255.0
!(subnet 60) int s 0
ip address 172.16.65.1 255.255.255.0 (subnet 65)
!S 0 connects to router B
router rip
network 172.16.0.0
hostname routerb
!
ip routing
!
int e 0
ip address 192.1.10.200 255.255.255.240
!(subnet 192)
int e 1
ip address 192.1.10.66 255.255.255.240
!(subnet 64)
int s 0
ip address 172.16.65.2 (same subnet as router A's s 0)
!Int s 0 connects to router A
router rip
network 192.1.10.0
network 172.16.0.0
Class B Effective Effective
# bits Mask Subnets Hosts
------- --------------- --------- ---------
1 255.255.128.0 2 32766
2 255.255.192.0 4 16382
3 255.255.224.0 8 8190
4 255.255.240.0 16 4094
5 255.255.248.0 32 2046
6 255.255.252.0 64 1022
7 255.255.254.0 128 510
8 255.255.255.0 256 254
9 255.255.255.128 512 126
10 255.255.255.192 1024 62
11 255.255.255.224 2048 30
12 255.255.255.240 4096 14
13 255.255.255.248 8192 6
14 255.255.255.252 16384 2
Class C Effective Effective
# bits Mask Subnets Hosts
------- --------------- --------- ---------
1 255.255.255.128 2 126
2 255.255.255.192 4 62
3 255.255.255.224 8 30
4 255.255.255.240 16 14
5 255.255.255.248 32 6
6 255.255.255.252 64 2
*Subnet all zeroes and all ones included. These
might not be supported on some legacy systems.
*Host all zeroes and all ones excluded.
