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Understanding IP Addressing: Everything You Ever Wanted To Know

VLSM Example

Given
An organization has been assigned the network number 140.25.0.0/16 and it plans to deploy VLSM. Figure 20 provides a graphic display of the VLSM design for the organization.

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Figure 20: Address Strategy for VLSM Example

The first step of the subnetting process divides the base network address into 16 equal-sized address blocks. Then Subnet #1 is divided it into 32 equal-sized address blocks and Subnet #14 is divided into 16 equal-sized address blocks. Finally, Subnet #14-14 is divided into 8 equal-sized address blocks.

Define the 16 Subnets of 140.25.0.0/16
The first step in the subnetting process divides the base network address into 16 equal-size address blocks. This is illustrated in Figure 21.

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Figure 21: Define the 16 Subnets for 140.25.0.0/16

Since 16 = 2 4 , four bits are required to uniquely identify each of the 16 subnets. This means that the organization needs four more bits, or a /20, in the extended-network-prefix to define the 16 subnets of 140.25.0.0/16. Each of these subnets represents a contiguous block of 2 12 (or 4,096) network addresses.

The 16 subnets of the 140.25.0.0/16 address block are given below. The subnets are numbered 0 through 15. The italicized portion of each address identifies the extended-network- prefix, while the bold digits identify the 4-bits representing the subnet-number field:

Base Network: 10001100.00011001 .00000000.00000000 = 140.25.0.0/16
Subnet #0: 10001100.00011001.0000 0000.00000000 = 140.25.0.0/20
Subnet #1: 10001100.00011001.0001 0000.00000000 = 140.25.16.0/20
Subnet #2: 10001100.00011001.0010 0000.00000000 = 140.25.32.0/20
Subnet #3: 10001100.00011001.0011 0000.00000000 = 140.25.48.0/20
Subnet #4: 10001100.00011001.0100 0000.00000000 = 140.25.64.0/20
:
:
Subnet #13: 10001100.00011001.1101 0000.00000000 = 140.25.208.0/20
Subnet #14: 10001100.00011001.1110 0000.00000000 = 140.25.224.0/20
Subnet #15: 10001100.00011001.1111 0000.00000000 = 140.25.240.0/20

Define the Host Addresses for Subnet #3 (140.25.48.0/20)
Let’s examine the host addresses that can be assigned to Subnet #3 (140.25.48.0/20). This is illustrated in Figure 22.

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Figure 22: Define the Host Addresses for Subnet #3 (140.25.48.0/20)

Since the host-number field of Subnet #3 contains 12 bits, there are 4,094 valid host addresses (2 12 -2) in the address block. The hosts are numbered 1 through 4,094.
The valid host addresses for Subnet #3 are given below. The italicized portion of each address identifies the extended-network-prefix, while the bold digits identify the 12-bit host-number field:

Subnet #3: 10001100.00011001.0011 0000.00000000 = 140.25.48.0/20
Host #1: 10001100.00011001.0011 0000.00000001 = 140.25.48.1/20
Host #2: 10001100.00011001.0011 0000.00000010 = 140.25.48.2/20
Host #3: 10001100.00011001.0011 0000.00000011 = 140.25.48.3/20
:
:
Host #4093: 10001100.00011001.0011 1111.11111101 = 140.25.63.253/20
Host #4094: 10001100.00011001.0011 1111.11111110 = 140.25.63.254/20

The broadcast address for Subnet #3 is the all 1′s host address or:

10001100.00011001.0011 1111.11111111 = 140.25.63.255

The broadcast address for Subnet #3 is exactly one less than the base address for Subnet #4 (140.25.64.0).

Define the Sub-Subnets for Subnet #14 (140.25.224.0/20)
After the base network address is divided into sixteen subnets, Subnet #14 is further subdivided into 16 equal-size address blocks. This is illustrated in Figure 23.

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Figure 23: Define the Sub-Subnets for Subnet #14 (140.25.224.0/20)

Since 16 = 2 4 , four more bits are required to identify each of the 16 subnets. This means that the organization will need to use a /24 as the extended-network-prefix length.
The 16 subnets of the 140.25.224.0/20 address block are given below. The subnets are numbered 0 through 15. The italicized portion of each sub-subnet address identifies the extended-network-prefix, while the bold digits identify the 4-bits representing the sub-subnet-number field:

Subnet #14: 10001100.00011001.1110 0000.00000000 = 140.25.224.0/20
Subnet #14-0: 10001100.00011001.1110 0000 .00000000 = 140.25.224.0/24
Subnet #14-1: 10001100.00011001.1110 0001 .00000000 = 140.25.225.0/24
Subnet #14-2: 10001100.00011001.1110 0010 .00000000 = 140.25.226.0/24
Subnet #14-3: 10001100.00011001.1110 0011 .00000000 = 140.25.227.0/24
Subnet #14-4: 10001100.00011001.1110 0100 .00000000 = 140.25.228.0/24
:
:
Subnet #14-14: 10001100.00011001.1110 1110 .00000000 = 140.25.238.0/24
Subnet #14-15: 10001100.00011001.1110 1111 .00000000 = 140.25.239.0/24

Define Host Addresses for Subnet #14-3 (140.25.227.0/24)
Let’s examine the host addresses that can be assigned to Subnet #14-3 (140.25.227.0/24). This is illustrated in Figure 24.

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Figure 24: Define the Host Addresses for Subnet #14-3 (140.25.227.0/24)

Each of the subnets of Subnet #14-3 has 8 bits in the host-number field. This means that each subnet represents a block of 254 valid host addresses (2 8 -2). The hosts are numbered 1 through 254.

The valid host addresses for Subnet #14-3 are given below. The italicized portion of each address identifies the extended-network-prefix, while the bold digits identify the 8- bit host-number field:

Subnet #14-3: 10001100.00011001.11100011 .00000000 = 140.25.227.0/24
Host #1 10001100.00011001.11100011 .00000001 = 140.25.227.1/24
Host #2 10001100.00011001.11100011 .00000010 = 140.25.227.2/24
Host #3 10001100.00011001.11100011 .00000011 = 140.25.227.3/24
Host #4 10001100.00011001.11100011 .00000100 = 140.25.227.4/24
Host #5 10001100.00011001.11100011 .00000101 = 140.25.227.5/24
.
.
Host #253 10001100.00011001.11100011 .11111101 = 140.25.227.253/24
Host #254 10001100.00011001.11100011 .11111110 = 140.25.227.254/24

The broadcast address for Subnet #14-3 is the all 1′s host address or:

10001100.00011001.11100011. 11111111 = 140.25.227.255

The broadcast address for Subnet #14-3 is exactly one less than the base address for Subnet #14-4 (140.25.228.0).

Define the Sub 2 -Subnets for Subnet #14-14 (140.25.238.0/24)
After Subnet #14 was divided into sixteen subnets, Subnet #14-14 is further subdivided into 8 equal-size address blocks. This is illustrated in Figure 25.

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Figure 25: Define the Sub 2 -Subnets for Subnet #14-14 (140.25.238.0/24)

Since 8 = 2 3 , three more bits are required to identify each of the 8 subnets. This means that the organization will need to use a /27 as the extended-network-prefix length.

The 8 subnets of the 140.25.238.0/24 address block are given below. The subnets are numbered 0 through 7. The italicized portion of each sub-subnet address identifies the extended-network-prefix, while the bold digits identify the 3-bits representing the subnet 2 -number field:

Subnet #14-14: 10001100.00011001.11101110 .00000000 = 140.25.238.0/24
Subnet#14-14-0: 10001100.00011001.11101110.000 00000 = 140.25.238.0/27 Subnet#14-14-1: 10001100.00011001.11101110.001 00000 = 140.25.238.32/27 Subnet#14-14-2: 10001100.00011001.11101110.010 00000 = 140.25.238.64/27 Subnet#14-14-3: 10001100.00011001.11101110.011 00000 = 140.25.238.96/27 Subnet#14-14-4: 10001100.00011001.11101110.100 00000 = 140.25.238.128/27 Subnet#14-14-5: 10001100.00011001.11101110.101 00000 = 140.25.238.160/27 Subnet#14-14-6: 10001100.00011001.11101110.110 00000 = 140.25.238.192/27 Subnet#14-14-7: 10001100.00011001.11101110.111 00000 = 140.25.238.224/27

Define Host Addresses for Subnet #14-14-2 (140.25.238.64/27)
Let’s examine the host addresses that can be assigned to Subnet #14-14-2 (140.25.238.64/27). This is illustrated in Figure 26.

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Figure 26: Define the Host Addresses for Subnet #14-14-2 (140.25.238.64/27)

Each of the subnets of Subnet #14-14 has 5 bits in the host-number field. This means that each subnet represents a block of 30 valid host addresses (2 5 -2). The hosts will be numbered 1 through 30.

The valid host addresses for Subnet #14-14-2 are given below. The italicized portion of each address identifies the extended-network-prefix, while the bold digits identify the 5-bit host-number field:

Subnet#14-14-2: 10001100.00011001.11101110.010 00000 = 140.25.238.64/27
Host #1 10001100.00011001.11101110.010 00001 = 140.25.238.65/27
Host #2 10001100.00011001.11101110.010 00010 = 140.25.238.66/27
Host #3 10001100.00011001.11101110.010 00011 = 140.25.238.67/27
Host #4 10001100.00011001.11101110.010 00100 = 140.25.238.68/27
Host #5 10001100.00011001.11101110.010 00101 = 140.25.238.69/27
.
.
Host #29 10001100.00011001.11101110.010 11101 = 140.25.238.93/27
Host #30 10001100.00011001.11101110.010 11110 = 140.25.238.94/27

The broadcast address for Subnet #14-14-2 is the all 1′s host address or:

10001100.00011001.11011100.010 11111 = 140.25.238.95

The broadcast address for Subnet #6-14-2 is exactly one less than the base address for Subnet #14-14-3 (140.25.238.96).

Additional Practice with VLSM
Please turn to Appendix D for practice exerciss to reinforce your understanding of VLSM.

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By Crimson – November 19, 2006

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