Curriculum for IP class:

IP Subnetting Class

2/27/01                                                                                                                                    By: Christophor Ronk

Section 1: Theory

Computers use the binary numbering system in everything that is done, even a task as simple as pressing a key on your keyboard. For Example: when you press the letter “A” on your keyboard, the keyboard controller sends a hex message with a series of 1’s and 0’s. The computer then looks those 1’s and 0’s up on a chart to determine which character was sent.

This same method is used throughout the entire computer and IP subnetting is based on this system. A binary digit (called a bit) can only be a 0 or a 1.  If humans counted in this manner, and one counted from 1 to 10, it would look like this: 1,10,11,100,101,110,111,1000,1001,1011. In computers everything is sent in bytes; 1 byte contains 8 bits of data. In our previous example, the letter A is sent through the computer using 8 bits of data: 01000001. All of this data combined is 1 byte of information. Any digits not used, are sent regardless. For instance, the binary number “4” is expressed as  00000100 so that all 8 bits within the byte are used. Let us digress a bit and talk about storage. When a hard drive is said to have a capacity of 20 Gigabytes, (20,000,000,000 bytes) in reality that hard drive stores 160 billion 1’s and 0’s; even CD rom drives use lasers to determine 1’s and 0’s. This is the essence of digital. Any digital signal contains a mark or a space, which is just another term in radio for 1 and 0.

Now with basic binary out of the way, let us look more closely at IP addressing:

You see:                      Computer Sees:

63.199.71.1                 00111111.11000111.01000111.00000001

255.255.255.0                          11111111.11111111.11111111.00000000

In the example above you see that the subnet mask lines up perfectly with each bit in the preceding IP address.

A subnet mask will always be consecutive 1’s followed by consecutive 0’s, never mixed. For example, let us look at the binary equivalent of a different subnet mask.

255.255.255.0             11111111.11111111.11111111.00000000

255.255.255.240                  11111111.11111111.11111111.11110000

As you can see, the 1’s and 0`s are segregated and never mix. A subnet is a group of IP addresses. Due to the fact that we will only deal with a class C address, the first 3 octets will ALWAYS be 1’s (on a class C). For simplicity we will drop the first 3 octets and only deal with the last.

Because there are only 8 bits per octet, and the 1’s and 0’s never mix, there are only 8 possible numbers for a subnet mask. They are as follows:

0

 

00000000

128

 

10000000

192

 

11000000

224

 

11100000

240

 

11110000

248

 

11111000

252

 

11111100

254

 

11111110

255

 

11111111

Picture a subnet like a cucumber. Every time you add a 1 to the subnet mask, you slice every subnet in half. And create two new subnets. For Example: The subnet mask 00000000 means that there are 256 possible binary numbers with those 8 available digits (because 11111111 = 255 in binary + 1 because 0 is also valid). This means there is 1 subnet containing 256 valid numbers. Lets assign one of those 0’s as a subnet ID using the subnet mask.  Instead of the subnet mask being 00000000 we are going to change it to 10000000.  That means one of the digits is being used to separate the class C into 2 groups, group 0 and group 1 (because it can only be a 0 or a 1). To figure out how many valid numbers per group, you see how many possible numbers exist. We have 7 bits to work with so we have 0000000 through 1111111.  1111111 in binary is 127 so there are 128 valid numbers per subnet.

Lets break it down even further.  Lets add another 1 to the subnet mask making it 11000000. This has taken our two groups, and broken them in half making 4 groups: group 00,01,10,11. Now we can figure out how many valid numbers per group there are. Out of 8 bits, two have been taken for subnetting, leaving us 6 additional bits to work with. With 6 bits, the highest that can be counted in binary is 63 + 1 (accounting for 0) meaning 64 valid numbers per subnet. As you can see, adding 1’s to the subnet mask simply further divides the groups of Ips into equal parts.

Section 2: Application

Now that we have studied the theory behind IP Subnetting, we can now learn how to apply that theory to IP sub-netting. Let us further examine the nature of the bits. Because of the nature of binary code, each bit can be assigned a value. For this we start with the first bit.  In binary the first bit is the bit farthest to the right. That bit has 2 possible values (0, and 1) and there are no bits to the right of it, so the value of that bit is 2^0. The second bit has 2 possible values and there is 1 bit to the right of it so its value is 2^1 and so on until one gets to the last bit which is 2^7.

0          0          0          0          0          0          0          0

2^7      2^6      2^5      2^4      2^3      2^2      2^1      2^0

If one does the math, the bit’s values are as follows:

0          0          0          0          0          0          0          0

128      64        32        16        8          4          2          1

With this knowledge, you can now count in binary as computers do.

Look at the binary number:       00000001

According to the chart, the 1st bit has a value of 1.Therefore, the decimal equivalent to 00000001 is 1.

Now let us look at the binary number:   10000000

According to the chart, the 8th bit has a value of 128. Therefore, the decimal equivalent to 10000000 is 128.

Finally, lets look at the binary number 10010101

According to the chart, the 8th bit has a value of 128 so we add 128. The 5th bit has a value of 16 so we add 16. The 3rd bit has a value of 4 so we add that. The first bit has a value of 1 so we add that as well. Add all of it up and you get 128+16+4+1 = 149.  Therefore, the binary equivalent to 10010101 is 149.

Lets take this the other way. Lets say you are given 149 and asked the binary equivalent. In order to do this one must start with the bit that holds the greatest value. This is the eighth bit. It is worth 128. Is 128 < 149? Yes it is, so assign a 1 to the eighth bit and subtract it from the total.   Ok, we now know that the eighth bit is a 1 and the remainder is 21. Now lets move to the 7th bit. The value of this bit is 64. Is 64 < 21? No it is not so this bit is a 0. We now know that the 8th and 7th bits are 1 and 0. Lets continue to the sixth bit. Is 32 < 21? No it is not, the 6th bit is a 0. Is 16 < 21? Yes it is, so we make the 5th bit a 1 and subtract 16 from the remainder leaving 5. We have now constructed the last 4 digits of this binary number (1001) and if you continue you will find that the end result is indeed 10010101. (note that this will only work for numbers up to 255. In order to count higher, a ninth bit would be required).

Section 3: To top it off

For each subnet two valid numbers are used in overhead. The first IP address is used for subnet ID and the last IP is used for broadcast. Lets look at the class C   63.199.71.x .  If the subnet mask is 00000000 then that means all 256 valid numbers (0-255) are in the same group. The first number is 0 and is used for the subnet ID. The last number is 255 and it is used for the broadcast leaving 1-254 as valid IP addresses.

If we break that group into 2 subnets by changing the subnet mask to 10000000 that breaks the 256 numbers into two groups, 0-127 and 128-255. In each subnet, the first and last numbers are used. This means the two subnets have usable ip addresses 1-126 and 129-254. Lets break it down by IP address.

63.199.71.0                                         Subnet ID for first subnet

63.199.71.1-- 63.199.71.126              Usable IP address first subnet

63.199.71.127                                     Broadcast IP address for first subnet

63.199.71.128                                                                          Subnet ID for second subnet

63.199.71.129—63.199.71.254          Usable IP address second subnet

63.199.71.255                                                                          Broadcast IP address for second subnet

The further you subnet, the more IP addresses are used for subnet ID and broadcast. It is a number that doubles every time you further subnet the class C.  The following is a chart in relation to the number of IP addresses lost to overhead:

Subnet Mask              IP addresses lost

00000000                    2

10000000                    4

11000000                    8

11100000                    16

11110000                    32

11111000                    64

11111100                    128

11111110                    256

As you can tell, the last subnet mask is never used and 11111100 is rarely used due to the gross waste of IP addresses.

Now that we have learned the core of IP subnetting, we can express the subnet mask in its decimal counterpart to save space. (Refer to page 2).

Lets look at the ever important bit chart (originally on page 3).

128      64        32        16        8          4          2          1

This chart is the base of subnetting. If you have a class C available and a customer wants 4 consecutive IP addresses we want to see if this is possible. If you start on the first bit and count over to 4, you move two spaces to the left. This means the first two bits in the mask are 0’s and the rest are 1’s.  The mask is now 1111100 or 252.  That means four IP addresses? No, that means 4 valid binary numbers. Remember,  the first and last numbers are used for subnet id/broadcast. This leaves 2 usable IP addresses. We have to go to the next subnet mask, which is 248. 248 is 11111000.  The first 3 bits are zero so start at 1 and count over 3 spaces. You land on 8.  This means there are 8 valid numbers. Subtract 2 to get the total valid IP addresses per subnet. The total is 6 usable. In this case, the customer would have to pay for 6 ip addresses as apposed to the 4.

Now lets see the subnets in relationship to each other.  If you took the 63.199.71.x class C and subnetted it with a 224 subnet mask, how many subnets would there be? The binary equivalent of 224 is 11100000. Using the table we count over 4 spaces and that takes us to 32.  That means that the subnet ID for each subnet is separated by 32 ip addresses.  The subnets would be as follows:

63.199.71.0

63.199.71.32

63.199.71.64

63.199.71.96

63.199.71.128

63.199.71.192

There are a total of six subnets with 30(32-2) usable IP addresses per subnet.

Useful Tables:

Subnet Mask                Binary Equivalent

0

 

00000000

128

 

10000000

192

 

11000000

224

 

11100000

240

 

11110000

248

 

11111000

252

 

11111100

Bit Chart:

128      64        32        16        8          4          2          1