TCP/IP Address Classes



TCP/IP Address Classes

|Class |IP address |Network ID |Host ID |Valid Ranges |High-Order Bits |

| | | | | | |

|A |w.x.y.z |w. |x.y.z |001 – 127 |0 |

|B |w.x.y.z |w.x. |y.z |128 – 191 |10 |

|C |w.x.y.z |w.x.y. |z |192 – 223 |110 |

Octet High-Order Bit(s) Binary Representation

Items in boldface are always the value shown

|Class |Octet 1 |Octet 2 |Octet 3 |Octet 4 |

| | | | | |

| |wwwww |xxxxxxxx. |yyyyyyyy. |zzzzzzzz |

| | | | | |

|A |00000000. |00000000. |00000000. |00000000 |

|B |10000000. |00000000. |00000000. |00000000 |

|C |11000000. |00000000. |00000000. |00000000 |

|D |11100000. |00000000. |00000000. |00000000 |

|E |11110000. |00000000. |00000000. |00000000 |

Default Subnet Masks

|Class | |IP Address |Network ID |Host ID |

| | | | | |

|A | |w.x.y.z |w. |x.y.z |

| |Default Mask |255.0.0.0 |255. |0.0.0 |

| | | | | |

|B | |w.x.y.z |w.x. |y.z. |

| |Default Mask |255.255.0.0 |255.255. |0.0 |

| | | | | |

|C | |w.x.y.z |w.x.y. |z |

| |Default Mask |255.255.255.0 |255.255.255. |0 |

Address Summary

|Class |Number of Networks |Number of Hosts |Valid Range |Network Ids |

| | | |(First Octet |Only) |

| | | | | |

| | | |Decimal |Binary |

| | | | | |

|A |126 |16,777,214 |001. |00000001. |

| | | |126. |01111110. |

| | | | | |

|B |16,384 |65,534 |128. |10000000. |

| | | |191. |10111111. |

| | | | | |

|C |2,097,152 |254 |192. |11000000. |

| | | |223. |11011111. |

| | | | | |

Calculating Binary / Decimal Conversions

The following chart can be applied to any single octet:

|Bits |8 |7 |6 |5 |4 |3 |2 |1 |

| | | | | | | | | |

|Decimal Value |128 |64 |32 |16 |8 |4 |2 |1 |

| | | | | | | | | |

Example:

For a decimal IP address of 131.107.002.200

Octet 1: 131 = 10000011 or 128 + 2 + 1

Octet 2: 107 = 01101011 or 64 + 32 + 8 + 2 + 1

Octet 3: 002 = 00000010 or 2

Octet 4: 200 = 11001000 or 128 + 64 + 8

When converting decimal to binary or binary to decimal it is important to remember:

1. The decimal values shown in the table above are absolute and therefore never change

2. Binary numbers deal in only two discreet states; 1’s and 0’s or more commonly referenced as On or Off with 1 being On and 0 being Off

3. Always begin your conversion with the high-order bit, the left most one or bit 8 and continue working to the right

4. When converting decimal to binary you are given the decimal value of each octet. With that in mind for an example as shown above we can obtain the binary value of the 1st octet, 131., by applying the following formula:

a) 1st compare the decimal value (131) to the absolute decimal value of the high-order bit 8 which is 128. If it is higher then that bit must be On or equal to a 1, if it is lower then the bit must be Off or a 0. The binary value at this point has a “1” in the left most bit.

b) Next take the absolute decimal value of the following bit; remember you are working left to right. In the chart bit 7 is equal to 64. Add this value to our prior value which was on or equal to a 1, 128 + 64 = 192. Since 192 is greater than 131 bit 7 must be Off. At this point we have a binary value of “10” in the two left most bits.

c) We now move one more bit to the left or bit 6. As we did above we apply the absolute decimal value for that bit, which is 32, to the 128 we already have. So the calculation is 128 + 32 and that equals 160 which is still higher than 131. The result is bit 6 must also be Off. We now have a binary value of “100” in the three left most bits.

d) Continuing on this line, bit 5 equals a decimal value of 16. So 128 + 16 = 144 which is still higher than 131 so that bit must be Off also. Our binary value now has “1000” in the four left most bits.

e) Moving on to bit 4, which has a decimal value of 8, we add 128 + 8 and the total equals 136. It is still higher than 131 so bit 5 must be Off and the binary value now has “10000” in the five left most bits.

f) Turning now to bit 3, which has a decimal value of 4, we add 128 + 4 and the total equals 131. Close but it’s still higher than 131 so bit 4 must be Off and the binary value has “100000” in the six left most bits.

g) On to bit 2, which has a decimal value of 2, we add 128 + 2 and the total is 130. Since it’s lower than our 131 the 2nd bit must be on. Our binary value now has “1000001” in the seven left most bits.

h) Finally the last bit or bit 1, which has a decimal value of 1, we add 128 + 2 + 1 and the total is 131. I know where did the + 2 come from, remember we continually add all bits that are On. Our binary value is now complete and has a value of “10000011” which is equal to a decimal value of 131. Success!

5. When converting binary to decimal, this is the easy one; you are given the binary value of each octet. With that in mind for an example as shown above we can obtain the decimal value of the 2nd octet, 01101011, by applying the following formula:

a) Take the entire binary value and place it in the chart on page 2.

|Bits |8 |7 |6 |5 |4 |3 |2 |1 |

| | | | | | | | | |

|Decimal Value |128 |64 |32 |16 |8 |4 |2 |1 |

| | | | | | | | | |

|Binary Value |0 |1 |1 |0 |1 |0 |1 |1 |

b) All we need do is add the decimal values for all bits that are On and presto we have the decimal conversion done. So it looks like this: 64 + 32 + 8 + 2 + 1 = 107.

Now that you know how to do it, try doing it for the 3rd and 4th octets.

Try the formulas on these, if you dare!:

192.168.92.254

209.158.10.68

92.41.254.238

251.252.253.255

11111001.00100111.11111111.10101010

01010101.11001100.00110011.00001111

11111111.11111111.11111111.11111111

00000001.11111111.00000000.11111111

During class, I will give you some more to do and maybe even show you how to use a calculator to do it even faster.

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download