Binary Numbering

Order instructions

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

  1. Work with base 10 exponents (powers of 10) and understand how position defines value
  2. Work with base 2 exponents (powers of 2) and understand how position defines value
  3. Manually convert simple binary numbers and decimal numbers
  4. Manually convert 32-bit binary IP addresses and dotted decimal IP addresses
  5. Describe the differences between binary and decimal numbering systems
  6. 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 106 105 104 103 102 101 100
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
  1. 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
  1. 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 27 26 25 24 23 22 21 20
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 27 26 25 24 23 22 21 20
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 27 26 25 24 23 22 21 20
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 27 26 25 24 23 22 21 20
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:____________________

  1. Step 4 – Decimal to Binary Practice Exercises.

Task: Practice converting the following decimal values 90, 99, 200, 254to 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 27 26 25 24 23 22 21 20
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 27 26 25 24 23 22 21 20
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 27 26 25 24 23 22 21 20
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 27 26 25 24 23 22 21 20
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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