**Order instructions**

This exercise will focus on your ability to accomplish the following tasks:

- Work with base 10 exponents (powers of 10) and understand how position defines value
- Work with base 2 exponents (powers of 2) and understand how position defines value
- Manually convert simple binary numbers and decimal numbers
- Manually convert 32-bit binary IP addresses and dotted decimal IP addresses
- Describe the differences between binary and decimal numbering systems
__Step 1 – Decimal Numbers__

Decimal Number Conversion Example

The following chart shows how the decimal number system represents the number 352,481. This will help in understanding the binary numbering system.

Exponent | 10^{6} |
10^{5} |
10^{4} |
10^{3} |
10^{2} |
10^{1} |
10^{0} |

Position | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

Value | 1000000 | 100000 | 10000 | 1000 | 100 | 10 | 1 |

Number | 0 | 3 | 5 | 2 | 4 | 8 | 1 |

0 x 1,000,000 | 3 x 100,000 | 5 x 10,000 | 2 x 1,000 | 4 x 100 | 8 x 10 | 1 x 1 |

The number 352,481 if read from right to left would be (1 x 1) + (8 x 10) + (4 x 100) + (2 x 1,000) + (5 x 10,000) + (3 x 100,000) for a total of 352,481 (a six-digit number).

Here is another way to look at it that makes it easier to add up the decimal number values:

Position of digit (from right) | Value of bit position (10^X or ten to the power of) | Number value from 0 to 9 | Calculation | Decimal Value |

1st Decimal Digit | 10^ 0 or 1 | 1 | 1 x 1 | 1 |

2nd Decimal Digit | 10^ 1 or 10 | 8 | 8 x 10 | 80 |

3rd Decimal Digit | 10^ 2 or 100 | 4 | 4 x 100 | 400 |

4th Decimal Digit | 10^ 3 or 1000 | 2 | 2 x 1,000 | 2,000 |

5th Decimal Digit | 10^ 4 or 10000 | 5 | 5 x 10,000 | 52,000 |

6th Decimal Digit | 10^ 5 or 100000 | 3 | 3 x 100,000 | 300,000 |

Decimal Value (Total of 6 digits) | 352,481 |

__Step 2 – Binary Numbers__

Binary Number Conversion Example

The following table shows the detail calculations (starting from the right side) to convert the binary number __10011100__ into a decimal number.

Position of digit (from right) | Value of bit position (two to the power of) | Is bit a One (on) or a Zero (Off) | Calculation | Decimal Value |

1st Decimal Digit | 2^ 0 or 1 | 0 | 0 x 1 | 0 |

2nd Decimal Digit | 2^ 1 or 2 | 0 | 0 x 2 | 0 |

3rd Decimal Digit | 2^ 2 or 4 | 1 | 1 x 4 | 4 |

4th Decimal Digit | 2^ 3 or 8 | 1 | 1 x 8 | 8 |

5th Decimal Digit | 2^ 4 or 16 | 1 | 1 x 16 | 16 |

6th Decimal Digit | 2^ 5 or 32 | 0 | 0 x 32 | 0 |

7th Decimal Digit | 2^ 6 or 64 | 0 | 0 x 64 | 0 |

8th Decimal Digit | 2^ 7 or 128 | 1 | 1 x 128 | 128 |

Decimal Value (Total of 8 digits) | 156 |

__Step 3 – Binary to Decimal Practice Exercises__

Task: Practice converting the 4 binary octets of an IP address to the dotted decimal equivalent.

Explanation: Look at the Binary number bit status. If there is a ONE in a position add the value shown. If there is a ZERO in a position then do not add it. Note that 8 bits cannot represent a decimal number greater than 255 (If all 8 positions are ones then 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255).

10011100 . 11100011 . 01110000 . 11011010

__Now, Solve for the 1st,2nd, 3rd and 4th octet Decimal value and Enter the value in the Underlined__

Exponent | 2^{7} |
2^{6} |
2^{5} |
2^{4} |
2^{3} |
2^{2} |
2^{1} |
2^{0} |

Bit Position | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

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

Binary Number Bit Status | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 |

1st Octet Decimal Value:____________________

Exponent | 2^{7} |
2^{6} |
2^{5} |
2^{4} |
2^{3} |
2^{2} |
2^{1} |
2^{0} |

Bit Position | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

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

Binary Number Bit Status | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |

2nd Octet Decimal Value:____________________

Exponent | 2^{7} |
2^{6} |
2^{5} |
2^{4} |
2^{3} |
2^{2} |
2^{1} |
2^{0} |

Bit Position | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

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

Binary Number Bit Status | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |

3rd Octet Decimal Value:____________________

Exponent | 2^{7} |
2^{6} |
2^{5} |
2^{4} |
2^{3} |
2^{2} |
2^{1} |
2^{0} |

Bit Position | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

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

Binary Number Bit Status | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 |

4th Octet Decimal Value:____________________

__Step 4 – Decimal to Binary Practice Exercises.__

**Task**: Practice converting the following decimal values __90, 99, 200, 254__to the binary octet equivalent.

Explanation: Look at the Decimal value and then subtract binary values starting from 128 (the highest value binary bit). If the number is larger than 128 then put a one in the first position binary number bit status. Subtract 128 from the number and then see if there is a 64 left. If there is put a one there otherwise put a zero and see if there is a 32. Continue until all 8 bits are defined as either a zero or a one.

__Now, Solve the 1st,2nd, 3rd and 4th octet Decimal value to binary bit number and Enter the binary values in the Underlines.__

Exponent | 2^{7} |
2^{6} |
2^{5} |
2^{4} |
2^{3} |
2^{2} |
2^{1} |
2^{0} |

Bit Position | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

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

Binary Number Bit Status |

90 -à1st Octet Binary Value:____________________

Exponent | 2^{7} |
2^{6} |
2^{5} |
2^{4} |
2^{3} |
2^{2} |
2^{1} |
2^{0} |

Bit Position | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

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

Binary Number Bit Status |

99-à2nd Octet Binary Value:____________________

Exponent | 2^{7} |
2^{6} |
2^{5} |
2^{4} |
2^{3} |
2^{2} |
2^{1} |
2^{0} |

Bit Position | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

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

Binary Number Bit Status |

200 à3rd Octet Binary Value:____________________

Exponent | 2^{7} |
2^{6} |
2^{5} |
2^{4} |
2^{3} |
2^{2} |
2^{1} |
2^{0} |

Bit Position | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |

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

Binary Number Bit Status |

254-à4th Octet Binary Value:____________________

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