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Understanding IP Routing and Subnet Masks For reference, the following is a complete list of all possible masks:
Bits Mask Hex Bits Mask Hex
8 255.0.0.0 FF000000 20 255.255.240.0 FFFFF000
10 255.192.0.0 FFC00000 21 255.255.248.0 FFFFF800
11 255.224.0.0 FFE00000 22 255.255.252.0 FFFFFC00
12 255.240.0.0 FFF00000 23 255.255.254.0 FFFFFE00
13 255.248.0.0 FFF80000 24 255.255.255.0 FFFFFF00
14 255.252.0.0 FFFC0000 25 255.255.255.128 FFFFFF80
15 255.254.0.0 FFFE0000 26 255.255.255.192 FFFFFFC0
16 255.255.0.0 FFFF0000 27 255.255.255.224 FFFFFFE0
17 255.255.128.0 FFFF8000 28 255.255.255.240 FFFFFFF0
18 255.255.192.0 FFFFC000 29 255.255.255.248 FFFFFFF8
19 255.255.224.0 FFFFE000 30 255.255.255.252 FFFFFFFC
Certain combinations of subnet masks and remote addresses cannot be supported, due to Internet requirements. These combinations depend on the value of the first octet (first number in dotted IP notation) of the IP address. These restrictions are: Octet values Notes
0, 127, 224-255 These are illegal network numbers; do not use
1-126 Class A network, any of above masks will work
128-191 Class B network, 8-15 and 17 bits disallowed
192-223 Class C network, 8-23 and 25 bits disallowed
For each of the legal network classes and subnets, here are the total possible number of subnets and host addresses per subnet. This table gives information on how to use the host bits and subnet bits to allow the maximum number of hosts or subnets. Class A: Bits Subnets Hosts Bits Subnets Hosts
8 1 1,777,214 20 4094 4,094
10 2 4,194,302 21 8190 2,046
11 6 2,097,150 22 16382 1,022
12 14 1,048,574 23 32766 510
13 30 524,286 24 65534 254
14 62 262,142 25 131070 16
15 126 131,070 26 262142 62
16 254 65,534 27 524286 30
17 510 32,766 28 1048574 14
18 1,022 16,382 29 2097150 6
19 2,046 8,190 30 4194302 2
Class B: Bits Subnets Hosts Bits Subnets Hosts
16 1 65,534 24 254 254
18 2 16,382 25 510 126
19 6 8,190 26 1,022 62
20 14 4,094 27 2,046 30
21 30 2,046 28 4,094 14
22 62 1,022 29 8,190 6
23 126 510 30 16,382 2
Class C: Bits Subnets Hosts Bits Subnets Hosts
24 1 254 28 14 14
26 2 62 29 30 6
27 6 30 30 62 2
It is a trivial optimization exercise to show that the total number of available host addresses is maximized if subnetting is unused. In the presence of subnetting, the total number of available host addresses is maximized if the number of subnet bits is equal to the number of host bits. Thus, "optimum" mask for class A networks is 255.255.240.0, for class B networks it is 255.255.255.0, and for class C networks it is 255.255.255.240. Example of Class C Complex Subnet Masking Complex subnet masking is used to increase the number of usable subnets with less hosts per subnet or to increase the number of hosts per subnet with less actual subnets. In a Class C address the first 3 octets can not be changed, only the last octet can be changed. Anyone with a Class C network address has to use complex subnet masking in order to get any subnets. Complex masking may also be used in a Class A or a Class B address environment. The following is an example of Class C subnet masking. It tries to show, in actual bits and bytes, how to subnet a Class C address into 62 subnets, with 2 hosts per subnet. The example also shows why you lose some of your available addresses or hosts when you do complex masking. Network #: 197.129.59.0 Class: C Subnet Mask: 255.255.255.252
# of Subnets: 62
# of Host per Subnet: 2
Example: IP Address: 197.129.59.6
# Subnet Bits: 6
Subnet Address: 197.129.59.4
Subnet Mask: 255.255.255.252
Subnet Broadcast Address: 197.129.59.7
The IP Address broken down into binary is: IP Address Network Subnet Host
197.129.59.6 11000101.10000001.00111011 000001 10
The Subnet Mask can be obtained from the IP Address when you know how many bits you want to subnet on. We are subnetting on the first 6 bits of the last octet, and regulating 2 bits for the host field. For example: 11000101.10000001.00111011. | 000001 | 10 Where the last 2 bits are regulated for the host bits. We then change all the subnet bits to 1's and host bits to 0 to come up with the subnet mask of: 255.255.255.252 11111111.11111111.11111111. | 111111 | 00 To find the Subnet address, AND the Subnet bits of the IP address with the subnet mask. The first subnet address is: 197.129.59.6 11000101.10000001.00111011. | 000001 | 10
197.129.59.4 11000101.10000001.00111011. | 000001 | 00 Note: A logical AND returns a 1 when both values are one, otherwise it returns a 0. To find the Subnet Broadcast Address, AND the Subnet bits of the IP Address with 1 and change the host bits to 1. The subnet broadcast turns out to be: 197.129.59.7 11000101.10000001.00111011. | 000001 | 11 Therefore, the network breakdown for this subnet is: 197.129.59.4 reserved for Subnet Address Valid Host bits are: 197.129.59.5 through 197.129.59.6 197.159.29.7 reserved for subnet broadcast This same scenario is repeated for each subnet. The last subnet available is the 197.129.59.248 subnet address. In binary this address looks like: 11000101.10000001.00111011. | 111110 | 00 The last 2 bits, which are the host bits, are 0. This indicates a network address. Enough said on that. So where did the rest of the numbers go? There are a total of 255 addresses, but with this numbering scheme we only have 124 addresses. Where did the 130 other addresses go too? Every subnet must have a reserved address for a subnet address and a reserved address for a subnet broadcast. Class B addressing scheme, masking on the third octet with 8 bits, the subnet address usually looks like 134.141.35.0, with a subnet broadcast of 134.141.35.255. Class C address using a 6 bit mask, gives 62 different subnets. Each subnet must have 62 subnet addresses and 62 subnet broadcasts. Well, were down to 7 missing addresses. We can also take away the 197.129.59.252 address because it is also a reserved address. 11000101.10000001.00111011. | 111111 | 00 This is the all 1's subnet. Because of confusion with the subnet mask, it can not be used as a subnet address. That brings us down to 6 missing addresses. The addresses 197.129.59.1 thru 197.129.59.3 and 197.129.59.253 thru 197.129.59.255 seem like they should be valid addresses, but tyey are not. Here is why. The address of 197.129.59.1 in binary is: 11000101.10000001.00111011. | 000000 | 01 There are no 1's in the 4th octet part of the subnet field. All 0's in the subnet field do not make up a recognizable subnet. So we cannot use those 3 addresses. These addresses are referred to the 0's subnet. The address of 197.129.59.253 in binary is: 11000101.10000001.00111011. | 111111 | 01 If you look closely at this address you see that the subnet portion of the 4th octet is all 1's in the subnet field. This is exactly how the subnet mask was defined. So we cannot use the last three addresses. This brings us down to 0 lost addresses. |
