Address Space Concept - Help

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Overview
Address Space Notation

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Overview

In NetDB2, sets of about 250 addresses were often referred to as "subnets" or "networks". For example, 36.53 or 171.64.3 would be called subnets. As a new feature, NetDB3 supports classless IP addressing, also known as variable length subnet masking. These sets of addresses are called "address spaces". One or many address spaces may be grouped into a Network record which corresponds with an actual network. For example, the Ivory Tower building has one network which contains 2 address spaces--171.64.10.0/24 (the regular address space) and 172.24.10.0/24 (private address space).

Address Space Notation

Address spaces are generally specified with 2 parameters:

Network Number- first address in the address space
Example: 171.64.11.0, 128.12.75.0, 172.24.33.12

Subnet mask or Network Prefix- determines size of address space by specifying division between network ID and host ID

  • subnet mask
    32 bit number expressed as 4 decimal numbers separated by dots
    examples: 255.0.0.0, 255.255.0.0
  • network prefix
    front slash with decimal number specifying how many bits are in the network ID
    sometimes called "slash notation"
    examples: /24, /23

Addresses

Subnet Mask

Network Prefix

1024

255.255.252.0

/22

512

255.255.254.0

/23

256

255.255.255.0

/24

128

255.255.255.128

/25

64

255.255.255.192

/26

32

255.255.255.224

/27

16

255.255.255.240

/28

8

255.255.255.248

/29


Given subnet mask, figure out address space size:

  1. Convert subnet mask to binary
  2. Count number of zeros
  3. Raise 2 to this power, This number gives you the available addresses.
  4. The first address in an address space is the network address and cannot be used as a host address.
  5. The last address in an address space is the broadcast address and cannot be used as a host address.
  6. At Stanford, the standard is to use the first available host address as the router address.
  7. The total available host addresses is 3 less than the total available addresses.

Given network prefix, figure out address space size:

  1. Subtract network prefix from 32
  2. Raise 2 to this power
  3. The first address in an address space is the network address and cannot be used as a host address.
  4. The last address in an address space is the broadcast address and cannot be used as a host address.
  5. At Stanford, the standard is to use the first available host address as the router address.
  6. The total available host addresses is 3 less than the total available addresses.

Example 1- Network # = 36.53.0.0, Subnet Mask = 255.255.255.0

In binary, 255.255.255.0 = 11111111.11111111.11111111.00000000. Note that there are 24 ones followed by 8 zeros. The transition between 1's and 0's marks the division between the network ID and the host ID. Therefore, in IP address 36.53.0.2, 36.53.0 is the network ID and 2 is the host ID. The first address in this address space is 36.53.0.0 (because it's the network number). The following addresses are :

36.53.0.1

36.53.0.2

36.53.0.3

..

..

36.53.0.254

36.53.0.255 (as last address, reserved for broadcast address)

Note that the number of addresses in this address space is 256 or 2 to the 8th power. The first address (network address) and the last address (broadcast address) are reserved for special purposes which leaves 256-2 = 254 addresses for actual computers.

Using the formula above,

  1. 255.255.255.0 = 11111111.11111111.11111111.00000000
  2. 255.255.255.0 has 8 zeros
  3. 2 to the 8th power is 256 addresses which is the size of the address space
  4. 36.53.0.0 is the network address
  5. 36.53.255.255 is the broadcast address
  6. 36.53.0.1 is the router address
  7. The number of available host address is 256-3=253.

Example 2 - Address Space = 171.64.20.0/23

Using the formula above,

  1. 32-23=9
  2. 2 to the 9th power is 512 addresses
  3. 171.64.20.0 is the network address
  4. 171.64.21.255 is the broadcast address
  5. 171.64.20.1 is the router address
  6. The number of available host addresses is 512-3 = 509.



 
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Last modified October 30, 2000