This appendix provides background information for readers unfamiliar with the considerations involved in assigning IP addresses to networks and hosts.
For additional information on IP addressing, see the following:
For each LS2020 node, you must provide from one to four IP addresses and the associated network masks, as follows:
Additional information on management addresses can be found in the LightStream 2020 Configuration Guide.
An IP address is a 32-bit identifier assigned to a host that uses the Internet Protocol. The IP address is represented by four octets (8-bit fields). In decimal form, an IP address consists of four fields separated by dots, where each field contains a value in the range 0 - 255. This is called dotted decimal notation.
Each host ID must be unique within a given network, and each network number must be unique within a given internet. Host IDs are assigned by the network administrator. The network number is assigned by the inter-network administrator. For a public network on the Internet, you must obtain a network number assigned by the Network Information Center (NIC).
An IP address consists of two parts. The first part of the address, called the network number, identifies a network on the internet; the remainder, called the host ID, identifies an individual host on that network. Historically, three classes of IP addresses have been defined:
An address mask determines which portion of an IP address identifies the network and which portion identifies the host. Like the IP address, the mask is represented by four octets. (An octet is an 8-bit binary number equivalent to a decimal number in the range 0 - 255). If a given bit of the mask is 1, the corresponding bit of the IP address is in the network portion of the address, and if a given bit of the mask is 0, the corresponding bit of the IP address is in the host portion.
The following table shows the mask 255.255.255.0 in both decimal and binary form, aligned with the class C address 192.15.28.16, also in both decimal and binary form:
Element | Network | Host | ||
|---|---|---|---|---|
| Mask | 255 | .255 | .255 | .0 |
| | 11111111 | 11111111 | 11111111 | 00000000 |
| Address | 192 | .15 | .28 | .16 |
| | 11000000 | 00001111 | 00011100 | 00010000 |
If a field of the network address is entirely used for the network number, the corresponding field of the mask has the decimal value 255 (binary 11111111), and if an address field is entirely used for the host ID, the corresponding field of the mask has the decimal value 0 (see the following table).
| Decimal Value in Field of Mask | Binary Value in Field of Mask | Function |
|---|---|---|
| 255 | 11111111 | Identify network number |
| 0 | 00000000 | Identify host ID |
Accordingly, the address masks for the three network classes described above are as follows:
| Address Class | Address Mask |
|---|---|
| A | 255.0.0.0 |
| B | 255.255.0.0 |
| C | 255.255.255.0 |
The boundary between the network portion and the host portion of an IP address can be shifted by replacing the first zero in the host portion with a different number. By this means, more than one physically distinct network can be managed under a single class A, class B, or class C network number.
When a modified address mask is used to partition a network into subnets, it is called a subnet mask. The simplest subnet masks for the three classes of networks are as follows, where n is a number other than zero:
| Address Class | Address Mask |
|---|---|
| A | 255.n.0.0 |
| B | 255.255.n.0 |
| C | 255.255.255.n |
(The other zeros in the mask for a class A or class B address can also be replaced by a different number, but, for simplicity, that possibility is ignored.)
The following is a simple example of subnetting partitions in a class B network.Set the third octet of the mask to 255, that is, use the address mask of a class C network. The following table shows the mask 255.255.255.0 in both decimal and binary form, aligned with the class B address 128.89.2.26, also in both decimal and binary.
Element | Network | Host | ||
|---|---|---|---|---|
| Mask | 255 | .255 | .255 | .0 |
| | 11111111 | 11111111 | 11111111 | 00000000 |
| Address | 128 | .89 | .2 | .26 |
| | 10000000 | 01011001 | 00000010 | 00011010 |
Outside the network, packets are routed to net 128.89. It is only when packets reach a router that is directly connected to the network that routing software takes account of the subnet mask. The mask partitions net 128.89 into 254 subnets (numbers 0 and 255 are reserved), each supporting up to 254 hosts.
However, values other than 255 may be used, and, obviously, 255 cannot be used in the host field of a class C network. The mask value n that is substituted for 0 must be one of the decimal values shown in column 1 of the following table.
| Column 1 | Column 2 | Column 3 | Column 4 |
|---|---|---|---|
| Decimal Value | Binary Value | Number of Subnets | Number of Host IDs |
| 192 | 11000000 | 2 | 62 |
| 224 | 11100000 | 6 | 30 |
| 240 | 11110000 | 14 | 14 |
| 248 | 11111000 | 30 | 6 |
| 252 | 11111100 | 62 | 2 |
Column 2 shows the values of n as binary numbers. These binary values are used in routing computations. If a given bit of the mask is 1, the corresponding bit of the IP address is part of the network ID. If a given bit of the mask is 0, the corresponding bit of the IP address is part of the host ID.
Column 3 indicates the number of subnets into which the mask partitions the network. If the first n mask bits are 1s, there are 2n-2 subnets. (Two subnet numbers are reserved.)
Column 4 indicates the number of distinct host IDs that can be specified for each subnet using the remaining bits of this field. If the last n mask bits are 0s, there are 2n-2 hosts. (Two host IDs are reserved.)
Decimal value 254 (binary 11111110) is omitted from the table because it allows 0 hosts. Decimal value 0 (binary 00000000) and decimal value 128 (binary 10000000) are omitted because they allow 0 subnets.
The reserved network numbers and host IDs that were subtracted from 2n in columns 3 and 4 are as follows:
You know how to partition a network with a subnet mask, but you may not be able to determine which IP addresses fall in each of the subnets. The following example clarifies this.
Suppose you want to partition class C network 192.15.28 into five physical networks. Look in column 3 (Number of Subnets) of the last table in the Subnet Masks section. The closest approximation is to configure six subnets. To do this, enter 224 in the last field of the subnet mask. Thus, 255.255.255.224 is the subnet mask that you must specify to partition a class C network into six subnets.
The following table shows the mask 255.255.255.224 in both decimal and binary form, aligned with the class C address 192.15.28.16, also in both decimal and binary form.
Element | Network | Host | ||
|---|---|---|---|---|
| Mask | 255 | .255 | .255 .224 | |
| | 11111111 | 11111111 | 11111111 111 | 00000 |
| Address | 192 | .15 | .28 .16 | |
| | 11000000 | 00001111 | 00011100 000 | 10000 |
By putting 224 in the last field of the mask, you tell the software to use the leftmost three bits of the last octet (the high-order bits) to differentiate subnets. In an 8-bit field, in binary notation, the three high-order bits naturally partition the range of numbers (and the range of possible IP addresses) into eight equal parts, corresponding to the eight binary numbers 000 through 111.
If you are not familiar with manipulating binary numbers and the example above is not clear to you, refer to the following table.
| Decimal Range | Binary Range | High Bits |
|---|---|---|
| 0 - 31 32 - 63 64 - 95 96 - 127 128 - 159 160 - 191 192 - 223 224 - 255 | 00000000 - 00011111 00100000 - 00111111 01000000 - 01011111 01100000 - 01111111 10000000 - 10011111 10100000 - 10111111 11000000 - 11011111 11100000 - 11111111 | 000 001 010 011 100 101 110 111 |
The three high-order bits of the IP address field thus partition the full range of 256 possible 8-bit numbers that can appear in the last octet of an IP address into eight groups of 32 addresses (256/8 = 32), as shown in the following table.
| IP Address Ranges | Subnet | High Order Three Bits |
|---|---|---|
| 192.15.28.0 - 192.15.28.31 | 0 | 000 |
| 192.15.28.32 - 192.15.28.63 | 1 | 001 |
| 192.15.28.64 - 192.15.28.95 | 2 | 010 |
| 192.15.28.96 - 192.15.28.127 | 3 | 011 |
| 192.15.28.128 - 192.15.28.159 | 4 | 100 |
| 192.15.28.160 - 192.15.28.191 | 5 | 101 |
| 192.15.28.192 - 192.15.28.223 | 6 | 110 |
| 192.15.28.224 - 192.15.28.255 | 7 | 111 |
However, addresses reserved for broadcast and for "this net" must be excluded. These exclusions are as follows:
Suppose you want to subnet class B network 128.89 into six subnets. Refer to Column 3 (Number of Subnets) in the last table in the section entitled "Subnet Masks." You can see that the number to substitute for n is 224 if you want six subnets. The mask is 255.255.224.0 with six subnets. Recall that the mask is 255.255.0.0 with no subnetting. These two masks have the following binary representations:
255.255.0.0 = 11111111.11111111.00000000.00000000
255.255.224.0 = 11111111.11111111.11100000.00000000
With subnet mask 255.255.224.0, the first three bits of the third octet are included in the network identifier. You must divide 256 by the number of subnets to get the ranges of addresses for each subnet. The mask divides the range of numbers 0 - 255 that can appear in the third field of an IP address into eight ranges of 32 numbers: 0 - 31, 32 - 64, etc.
Since there are 11 bits in all for the host ID (eight in the last field, plus three in the third field), the mask 255.255.128.0 provides for 2046 (211 -2) hosts in each of the eight subnets. The range of addresses for each subnet is shown in the following table.
| IP Address Ranges | Subnet | High Order Three Bits |
|---|---|---|
| 128.89.0.1 - 128.89.31.254 | 0 | 000 |
| 128.89.32.1 - 128.89.63.254 | 1 | 001 |
| 128.89.64.1 - 128.89.95.254 | 2 | 010 |
| 128.89.96.1 - 128.89.127.254 | 3 | 011 |
| 128.89.128.1 - 128.89.159.254 | 4 | 100 |
| 128.89.160.1 - 128.89.191.254 | 5 | 101 |
| 128.89.192.1 - 128.89.223.254 | 6 | 110 |
| 128.89.224.1 - 128.89.255.254 | 7 | 111 |
Subnet 0 is reserved because the addresses 128.89.0.1 - 128.89.31.254 have 000 as the high-order bits. Subnet 7 is reserved because the addresses 128.89.224.1 - 128.89.255.254 have 111 as the high-order bits.
The first address in each subnet address range is reserved for reference to the subnet as a whole because the last five bits of the address field are zeros. The last address in each subnet address range is reserved as a broadcast address for the subnet because the last five bits of the address field are ones. For example, in subnet 3, 128.89.96.0 is the subnetwork address and 128.89.127.255 is the broadcast address.
Suppose you have an ADSL modem with a four port router (e.g., a D-Link
DSL504 ADSL Modem/Router). You've bought a second router (e.g., a
Belkin 54Mbps Wireless 802.11g) and want to plug this into the
network to add in and to share more local machines, and to share the
Internet connection. Let's refer to the first ADSL router as router
A and the second as router B.
Configure router A to issue DHCP addresses in some range that
does not include one IP address that we will use for router B
For example, router A might only issue IP's in the range
starting at 192.168.0.2 and ending at 192.168.0.33 and we'll configure
router B with 192.168.0.40. This is all the setup that is
required for router A, which otherwise has DHCP enabled and
its usual WAN setup for your ISP.
Disable DHCP for router B, and configure its WAN
(Wide Area Network) to any STATIC IP. Specify a gateway IP of 0.0.0.0
(or perhaps 192.168.111.1, if your router will not allow
0.0.0.0). This will stop it sending traffic to its WAN (we won't be
using this router's WAN connection). Further configure the WAN Type
to be Static with a WAN IP of 192.168.111.2 perhaps (should be
different to the A network), and a Subnet Mask of
255.255.255.0.
The LAN (local area network) configuration for router
B should be set to STATIC with an IP address within the
subnet range of router A but outside its DHCP range. We might
set the LAN IP to 192.168.0.40 with a Subnet Mask of 255.255.255.255
(or 255.255.255.254 perhaps if that doesn't work) and with DHCP
Disabled. In fact, router A will serve as the DHCP server for
anything connected to router B.
Make sure that nothing is plugged into router B's WAN.
Connect a LAN ethernet port of router B to a LAN ethernet
port of router A to have them talking to each other, using
the usual ethernet cable that you would use to plug your computer into
the router.
Router A Router B
WAN: --> ISP modem WAN: Empty ethernet
Configured for ISP Static IP with Gateway 0.0.0.0
LAN: LAN:
IP=192.168.0.1 IP=192.168.0.233
Subnet 255.255.255.0 Subnet 255.255.255.255
DHCP: Enabled DHCP: Disabled
LAN Ethernet Cabling:
(1) <======================> (1) -->
(2) --> PC1 (2) --> PC4
(3) --> PC2 (3) --> VoIP
(4) --> PC3 (4) -->
That's it! (But check out Section 67.3.5 for
details on protecting your wireless connection from random access.)
Now, computers serviced by router B (directly connected by
Ethernet cable or else connected wirelessly) will be assigned DHCP by
router A, within the 192.168.0.* network, together with DNS
assignments. Router B is just another IP node on
A's network. Any LAN computer can access and configure
router B by accessing it as 192.168.0.233. All computers
will be on the same network subnet and so they will have access to
each other for file and printer sharing.
