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Using VLSM to Extend the IP Addresses Scheme

As IP subnets have grown, administrators have looked for ways to use their address space more efficiently and flexibly. One of the resulting techniques is called variable-length subnet masking (VLSM). VLSM is a method of designating a different subnet mask for the same network number on different subnets. Thus, a network administrator can use a long mask on networks with few hosts and a shorter mask on subnets with many hosts. If VLSM is not supported on your routers, your organization is stuck with the prospect of using a single subnet mask for an entire growing Class A, B, or C network. VLSM facilitates a hierarchical design that is more representative of an organization's logical network topology. Cisco supports VLSM with RIP version2, Open Shortest Path First (OSPF), Integrated Intermediate System-to-Intermediate System (IS-IS), Enhanced Interior Gateway Routing Protocol (EIGRP), and Border Gateway Protocol version 4 (BGP4), as well as with static routing. RIP version 1 and IGRP do not allow for VLSM because they are classful protocols, which do not transport subnet mask information in their updates. In Figure 3.2, RouterA has three subnets with two masks (/24 and /30).

Obviously, you want to use the available IP address space as efficiently as possible. A VLSM scheme is a much better method of allocating existing address space. In other words, you want to use the fewest number of subnet bits as necessary and the fewest number of host bits as necessary. A simple example of VLSM can be demonstrated through a connection between two Cisco routers over a wide area network (WAN) serial link. If you used a network address such as 172.16.2.0 with an 8-bit subnetting mask (the entire third octet), you would be wasting 252 addresses for a WAN point-to-point link that needs only two addresses. Because you need only two host addresses, you could use the following formula:

2n – 2 = 2 evaluates to:
2n = 4 which results in
n = 2

Figure 3.2Figure 3.2 A router with three subnets using different masks.


You need only 2 bits to represent two host addresses. Therefore, the subnet mask used for only two hosts is 255.255.255.252, or /30. If your network was originally using subnet 172.16.2.0 /24, you could reassign the following to your first two WAN interfaces using VLSM, as shown in Table 3.3 and Figure 3.3.



Table 3.3 Using VLSM to Apply a 30-bit Mask to Subnet 172.16.2.0 for More Efficient WAN Link Addressing

Description

Decimal

Subnetwork Binary

Subnetwork address

172.16.2.0 /30

.00000010.00000000

First WAN interface

172.16.2.1 /30

.00000010.00000001

Second WAN interface

172.16.2.2 /30

.00000010.00000010

Broadcast address

172.16.2.3 /30

.00000010.00000011



Figure 3.3 Figure 3.3 Using VLSM for WAN point-to-point serial links for more efficient address allocation.

You could then use the remaining subnetwork addresses for additional WAN point-to-point serial links as necessary. Here's an example:

172.16.2.4 /30
172.16.2.8 /30
172.16.2.12 /30
...
172.16.2.252 /30

Addresses need to be planned and assigned carefully because VLSM can add a higher level of complexity to your network administration. To really understand this process of subnetting a previously subnetted address, let's look at a practical scenario. Assume that you have been tasked by your CIO to redesign your existing Class B subnet of 172.16.1.0 /25. This 172.16.1.0 /25 scheme already gives your organization more than 500 subnets of 126 hosts each. However, you have been mandated to divide this one subnetwork into three departmental broadcast domains of 30 hosts each. In addition, you need to use this same address space for several WAN point-to-point serial links. Currently, your Class B subnet address of 172.16.1.0 with a subnet mask of 255.255.255.128 leaves you 7 bits for host addresses, or 126 hosts (27 – 2 = 126). Table 3.4 shows your present addressing scheme for subnet 172.16.1.0 /25.

Table 3.4 The Addressing Scheme for Subnet 172.16.1.0 /25

Description

Decimal

Subnetwork Binary

Subnetwork address

172.16.1.0

.00000001.00000000

First available host

172.16.1.1

.00000001.00000001

Second available host

172.16.1.2

.00000001.00000010

Last available host

172.16.1.126

.00000001.01111110

Broadcast address

172.16.1.127

.00000001.01111111


To subnet 172.16.1.0 /25 down to a VLSM to gain four subnets, you need to first write down the original subnetwork in its binary representation, like so:

10101100.00010000.00000001.00000000

Next, identify the bits that make up the original subnet mask with italics:

10101100.00010000.00000001.00000000

Because you need to borrow some bits from the remaining 7 bits for a VLSM mask, calculate the number you need:

2n = 4 results in
n = 2

So, you need two more bits for your VLSM mask, like so:

1 0 1 0 1 1 0 0.0 0 0 1 0 0 0 0.0 0 0 0 0 0 0 1.0 0 0 0 0 0 0 0
_____________network___________|______subnet_____|VSM|__hosts__|

Now that you have 5 bits remaining for the hosts, you can determine the number of available host addresses:

n is the remaining subnet bits, therefore:
n = 5
2n – 2 = 30 hosts per subnetwork

Therefore, you need the last 5 of the 7 original host bits to generate these 30 hosts per subnet. This combination works well for your needs.

Your new VLSM mask is 255.255.255.224, or /27. This is now the subnet mask for all the hosts in your three departments. As shown in Table 3.5, of the original 7 hosts bits, the first 2 bits are used to give you four new subnetworks and the remaining 5 bits are used for the 30 hosts on each subnet.

Table 3.5 The Binary Representation of Your Four New VLSM Subnets

Network

Subnet

VLSM

Host

First subnet 172.16.1.0/27

10101100.00010000

.00000001.0

00

00000

Second subnet 172.16.1.32/27

10101100.00010000

.00000001.0

01

00000

Third subnet 172.16.1.64/27

10101100.00010000

.00000001.0

10

00000

Fourth subnet 172.16.1.96/27

10101100.00010000

.00000001.0

11

00000


To reiterate, your four new VLSM subnets are

  • 172.16.1.0 /27 (the all-zeros subnet)

  • 172.16.1.32 /27

  • 172.16.1.64 /27

  • 172.16.1.96 /27 (this subnet can be used for WAN links)

Table 3.6 shows your new addressing scheme for the first subnet (subnet zero) of 172.16.1.0 /27.

Table 3.6 The Addressing Scheme for the First of Your Four New VLSM Subnets(Subnet Zero)

Description

Decimal

Subnetwork Binary

Subnetwork address

172.16.1.0

.00000001.00000000

First available host

172.16.1.1

.00000001.00000001

Second available host

172.16.1.2

.00000001.00000010

Last available host

172.16.1.30

.00000001.00011110

Broadcast address

172.16.1.31

.00000001.00011111


Table 3.7 shows the second subnet of 172.16.1.32 /27.

Table 3.7 The Addressing Scheme for the Second of Your Four New VLSM Subnets

Description

Decimal

Subnetwork Binary

Subnetwork address

172.16.1.32

.00000001.00100000

First available host

172.16.1.33

.00000001.00100001

Second available host

172.16.1.34

.00000001.00100010

Last available host

172.16.1.62

.00000001.00111110

Broadcast address

172.16.1.63

.00000001.00111111


Table 3.8 shows the third subnet of 172.16.1.64 /27.

Table 3.8 The Addressing Scheme for the Third of Your Four New VLSM Subnets

Description

Decimal

Subnetwork Binary

Subnetwork address

172.16.1.64

.00000001.01000000

First available host

172.16.1.65

.00000001.01000001

Second available host

172.16.1.66

.00000001.01000010

Last available host

172.16.1.94

.00000001.01011110

Broadcast address

172.16.1.95

.00000001.01011111


Table 3.9 shows the fourth subnet (all ones) of 172.16.1.96 /27, which is used for further VLSM subnetting WAN serial links.

Table 3.9 The Addressing Scheme for the Last of Your Four New VLSM Subnets

Description

Decimal

Subnetwork Binary

Subnetwork address

172.16.1.96

.00000001.01100000

First available host

172.16.1.97

.00000001.01100001

Second available host

172.16.1.98

.00000001.01100010

Last available host

172.16.1.126

.00000001.01111110

Broadcast address

172.16.1.127

.00000001.01111111


With that task accomplished, you now need to accommodate for your WAN serial links. However, it is critical that you understand that you can only further subnet a subnetwork that is presently unused. Therefore, you must use the fourth VLSM subnet, 172.16.1.96 /27, to create your WAN addressing scheme. As mentioned, these subnetworks need only two hosts each. Let's use the formula with two hosts:

2n –2 = 2 evaluates to:
2n = 4 which results in:
n = 2

Table 3.10 shows the efficient WAN link address assignments you can use now because of the flexibility of variable-length subnet masking.

Table 3.10 An Addressing Scheme for the WAN Serial Connections Using VLSMSubnets

Description

Decimal

Subnetwork Binary

WAN subnet 1

172.16.1.96 /30

.00000001.01100000

First WAN link on subnet 1

172.16.1.97 /30

.00000001.01100001

Second WAN link on subnet 1

172.16.1.98 /30

.00000001.01100010

Broadcast address on subnet 1

172.16.1.99 /30

.00000001.01100011

WAN subnet 2

172.16.1.100 /30

.00000001.01100100

First WAN link on subnet 2

172.16.1.101 /30

.00000001.01100101

Second WAN link on subnet 2

172.16.1.102 /30

.00000001.01100110

Broadcast address on subnet 2

172.16.1.103 /30

.00000001.01100111


Because you needed only 2 bits to represent two host addresses, the subnet mask for your WAN links is 255.255.255.252, or /30. Your network was originally using subnet 172.16.1.0 /25 for a single subnetwork of 126 hosts. You have now used VLSM to more efficiently reallocate your address pool with several contiguous (23 = 8) IP addresses left over for WAN connections. The flexibility of VLSM should be obvious by now, and you will be leveraging these techniques throughout the remaining chapters of this book. You can now start reassigning the new IP addresses and masks to your router interfaces, as shown in Listing 3.1. Remember that you must be running a routing protocol that supports VLSM and classless addresses, such as EIGRP or OSPF to name a few.

Listing 3.1 A Sample Router Configuration Using VLSM

RouterA(config)# interface ethernet 0/0
RouterA(config-if)# ip address 172.16.1.1 255.255.255.224
RouterA(config)# interface ethernet 0/1
RouterA(config-if)# ip address 172.16.1.33 255.255.255.224
RouterA(config)# interface ethernet 0/2
RouterA(config-if)# ip address 172.16.1.65 255.255.255.224
RouterA(config)# interface serial 0/0
RouterA(config-if)# ip address 172.16.1.97 255.255.255.252
RouterA(config)# interface serial 0/1
RouterA(config-if)# ip address 172.16.1.101 255.255.255.252
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