The first question concerns what constitutes a Class A, or a Class B, etc. network. Novices have trouble remembering where each class begins and ends. Table 3 shows a schema to help with this. First, let's discuss some basics of binary numbers.
A byte is a grouping of eight binary bits. Since a binary bit is either a 0 or a 1, a byte consists of eight 0s and/or 1s. No mystery here. So 10010101 is one byte and 11100000 is another. How do we convert these to decimal numbers? It turns out that the right-most bit has a weight of 1 (2<+>0<+>). The next bit to its left has a weight of 2 (2<+>1<+>), the next has a weight of 4 (2<+>2<+>), i.e., two raised to the second power and so on:
128 64 32 16 8 4 2 1 : decimal weightsThus, the binary number 10101001 has a decimal equivalent of
1x1 + 1x8 + 1x32 + 1x128 = 169If you assign contiguous 1s starting from the right, the above diagram can be used as a kind of calculator. Let's say you have 00001111 binary bits. To get the decimal equivalent, you could do the calculations the hard way, that is:
1x1 + 1x2 + 1x4 + 1x8 = 15or you could note the following (taking our number):
128 64 32 16 8 4 2 1 :decimal weightsIf you have all ones starting at the right side, you can simply take the weight of the first 0 bit (16 in this case), subtract 1, and you have 15—the decimal equivalent—without having to use a calculator. Thus, if all the bits on the right are 1s, you can determine the decimal value by using the above diagram as a kind of calculator. Note that the bits go up in powers of 2, so the ninth bit has a decimal weight of 256. So if you have a byte with all ones, i.e., 11111111, then it has a decimal value of 255 (256 -1). 255 appears many times in IP addressing.
0 0 0 0 1 1 1 1 :binary number
Table 1. Decimal Equivalents for Netmasking
|128 64 32 16 8 4 2 1||decimal|
|1 0 0 0 0 0 0 0||128|
|1 1 0 0 0 0 0 0||192|
|1 1 1 0 0 0 0 0||224|
|1 1 1 1 0 0 0 0||240|
|1 1 1 1 1 0 0 0||248|
|1 1 1 1 1 1 0 0||252|
|1 1 1 1 1 1 1 0||254|
|1 1 1 1 1 1 1 1||255|
Table 2. Shortened Netmask Table
|128 64 32 16 8 4 2 1||Binary|
|128 192 224 240 248 252 254 255||Decimal|
This table works fine, but is a bit unwieldy. Table 2 shows a short version. It says that if your byte is 11100000, then the decimal equivalent value is 224. If this bothers you, just use Table 1.
Table 3. Classes of IP Addresses
How did I get Table 3? I had to remember only a couple of pieces of information, then I constructed the rest. I know there are five classes of IP addresses, and the first byte of the IP address tells you to which class it belongs. I also know the schema for the binary starting value of the first byte, i.e., 0, 10, 110, etc. Because of the way it follows a schema, the second column is easy to construct. Now, using Table 2, it was easy to construct the third column.
Next, note that the fourth column (ending point) follows naturally by simply subtracting one from the beginning of the next class. A class C begins at 192, while a class D begins at 224. Hence, a class C must end at 223. Now you have no excuses about forgetting the beginning and ending points of each class; merely remember the binary schema, and take a minute to construct the table. On a side note, you don't have to worry about Classes D and E, except that the beginning of Class D tells you where Class C ends by subtracting 1.
Table 4. Bit-Wise Logical AND Truth Table
Bitwise ANDs work bit by bit. So, if you AND a 1 with a 1, you get a 1. If you AND two 0s, a 1 and a 0, or a 0 and a 1, however, you get a 0. Table 4 illustrates this operation.
Now let's take a whole byte and do a Logical AND with another byte. Suppose the first byte is 10110010 and the second byte is 01100111. Working from the right, note that the first byte has a decimal value of
0*1 + 1*2 + 0*4 + 0*8 + 1*16 + 1*32 + 0*64 + 1*128 = 178while the second byte has a decimal value of
1*1 + 1*2 + 1*4 + 0*8 + 0*16 + 1*32 + 1*64 + 0*128 = 103.Now, AND the two bytes:
1 0 1 1 0 0 1 0 178 decimal, ANDed withAs a second example, let's AND 178 with 255.
0 1 1 0 0 1 1 1 103 decimal
0 0 1 0 0 0 1 0 34 decimal
1 0 1 1 0 0 1 0 178 decimal, ANDed withWe know, then, that when you bit-wise AND any byte (number) with 255, you get the number dropping through, i.e., the result is merely the number again.
1 1 1 1 1 1 1 1 255 decimal
1 0 1 1 0 0 1 0 178 decimal
Table 5. Default Netmasks, etc.
So, what does this mean and what do we do with it? Let's work through Table 5. If we take the sample Class A address, 10.0.1.23 and bit-wise AND it with its default netmask, we obtain 10.0.0.0. What is 10.0.0.0? It's the network address—look at the last column.
Notice that the first byte gives the network address when ANDing a Class A network with its default netmask, while the first two bytes give the network address when ANDing a Class B IP address with the default Class B netmask. Hence, we say that the first byte of a Class A IP address gives the network address, and the three remaining bytes give the host addresses, i.e., a Class A address has the form N.H.H.H where N stands for Network and H stands for Host. Likewise, the first two bytes of a Class B IP address pertain to the network, and the last two bytes pertain to the host address, i.e., N.N.H.H. Finally, the first three bytes of a Class C IP address pertain to the network, while the last byte pertains to the host, i.e., N.N.N.H.
If you are granted a full Class B suite of addresses with a network address of 22.214.171.124, what do you do with them? Remember, a Class B network has the form of N.N.H.H, i.e., the last two bytes can be used for assigning host IP addresses. This yields a network with 2<+>16<+> - 2 host addresses. The -2 comes from the fact that 126.96.36.199 is the network address, so it can't be assigned to a host; the last address on the network, 188.8.131.52, is used for broadcasts, so it also can't be assigned to a host.
This would be a very big network (65,534 host addresses), far too big to be practical. A very simple approach is to “borrow” one byte's worth of host addresses and assign them as network addresses. That would yield 2<+>8<+> = 256 networks with 254 hosts on each. Even here, these are large networks. This process of borrowing host addresses and using them for networks is called subnetting. We accomplish this by using a sub-netmask (SNM). In this case, we would use a sub-netmask of 255.255.255.0, which is the default Class C netmask. Hence, we have taken one Class B network and turned it into 256 Class C networks.
If we AND 184.108.40.206 with 255.255.255.0, we get a network address of 220.127.116.11 with the first available host address of 18.104.22.168 and the last of 22.214.171.124, since 126.96.36.199 is reserved for broadcasts. Another way of doing this is to start with the network address (188.8.131.52 in this case), turn all host bits into 1s, and obtain the broadcast address. Here, the last byte is used for host addresses, so turning them to ones gives 184.108.40.206. This type of broadcast is called a directed broadcast, meaning that it jumps routers while a local broadcast (which doesn't jump routers) has the form 255.255.255.255 no matter which class of network is involved.
If you're not too stunned at this point, you may wonder if you can subnet only on byte boundaries or if you can subnet a Class C network. The answers are “no” and “yes”, respectively; i.e., you can work in the middle of a byte.
Let's do some mathematics. If we have 4 bits for hosts, will it be enough? 2<+>4<+>-2 = 14 and is not enough. So, let's use 5 bits for hosts: 2<+>5<+>-2 = 30 which will work. However, we have 8 bits in the last byte for hosts, so let's borrow three bits for subnetworks; then we still have the requisite 5 bits for hosts. Great, but how many subnets do we have? How about 2<+>3<+> = 8? We have, then, eight subnetworks with 30 host addresses on each. If you are doing the math, you are probably saying, “but 8x30 is only 240 addresses; what happened to the others?” Valid question! Oops, don't get sore, but it's time to construct another table. Note that each address will have the form of 210.168.94.last byte, and the SNM (sub-netmask) will have the form 255.255.255.last byte. Let's just work with the last byte.
Table 6. Subnetworks for Class C Network (shows the last byte)
From Table 2 (or Table 1), we see the SNM will be 255.255.255.224. The 224 comes from the last byte being 11100000. So what are the subnets? Table 6 shows them (last byte only).
Let's detail a few. First, take the smallest. The full subnetwork address of the smallest is 220.127.116.11. The next one up is 18.104.22.168, and so on. Remember that with three bits to work with, we get 2<+>3<+> = 8 subnets, and looking at Table 6, you see them.
Table 7. Analysis of 256 Values of Last Byte
|0||invalid||first subnet address|
|1-30||valid||hosts on first subnet|
|31||invalid||broadcast address of first subnet|
|32||invalid||second subnet address|
|33-62||valid||hosts for second subnet|
|63||invalid||broadcast address of second subnet|
|64||invalid||third subnet address|
|65-94||valid||hosts for third subnet|
|95||invalid||broadcast address of third subnet|
|96||invalid||fourth subnet address|
|97-126||valid||hosts for fourth subnet|
|127||invalid||broadcast address of fourth subnet|
|128||invalid||fifth subnet address|
|129-158||valid||hosts for fifth subnet|
|159||invalid||broadcast address of fifth subnet|
|160||invalid||sixth subnet address|
|161-190||valid||hosts for sixth subnet|
|191||invalid||broadcast address of sixth subnet|
|192||invalid||seventh subnet address|
|193-222||valid||hosts for seventh subnet|
|223||invalid||broadcast address of seventh subnet|
|224||invalid||eighth subnet address|
|225-254||valid||hosts for eighth subnet|
|255||invalid||broadcast for eighth subnet|
Back to the question of why we get only 240 host addresses. “(Gasp)—another table!” Looking at the last byte, we get Table 7.
Now let's answer the question of what happened to the other addresses. To do this, tally all the “invalid addresses”, i.e., those that can't be used for host addresses.
First, we have eight subnets, each with a subnetwork address and a broadcast address. So we lose 8*2 = 16 addresses here. Now if we subtract these 16 from 256, we get 240 available host addresses.
Doing it the other way is much easier. We have eight subnetworks, each with 30 valid IP addresses; this gives us 8*30=240 valid IP addresses total, the magic number.
For fun, let's do one more thing: analyze the sixth subnetwork in a little more detail. The last byte is 10100000 binary or 160 decimal. The full subnet address is 22.214.171.124 decimal, and we use an SNM of 255.255.255.224. Remember, I said to take the subnet address, set all the host bits to 1s and add them to get the broadcast address. If we do this correctly, it should give the same result as Table 7.
We use five bits for host addresses, so the decimal value of the sixth bit is 32. Subtracting 1 gives 31. Thus, setting the five host bits to 1s, i.e., 00011111, gives a value of 31 decimal. Adding this to the last byte of the subnet address (160) gives 191 for the broadcast address, agreeing with Table 7. Here is the “whole Megillah”:
126.96.36.199 The Sub-Network address188.8.131.52-190 Valid host addresses184.108.40.206 Directed Broadcast address
One final point. Some authors use the term “sub-netmask” even when referring to the default netmasks—they are being just a tad loose with their terms. Happy IP addressing, and remember, Linux is inevitable.