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Discovering computer networks: hands on in the Open Networking Lab
Discovering computer networks: hands on in the Open Networking Lab

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7.1 Base arithmetic

In this part there is a slight side track from the focus on communications networks: instead you’ll take a short time to look at some arithmetic that will help you to understand the representation of network addresses. This is very fundamental arithmetic about how we count and how we represent numbers, and you may already be familiar with this. If this is the case, then you can treat this part as revision.

There are three videos in this part – each video is followed by an activity.

Watch the video below, which is about 3 minutes long.

Box _unit5.1.1 Base arithmetic

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Example _unit5.1.1 Activity 1 Test yourself

5 minutes

ITQ _unit5.1.1

  • In the base 10 (decimal) number 242, the symbol 2 represents different values depending on its position. What are these values?

  • The symbol ‘2’ on the left is in third place so a weighting of 100 is applied to it, giving it a value of 200. The symbol ‘2’ on the right is in first place so a weighting of 1 is applied to it giving it a value of 2.

ITQ _unit5.1.2

  • Working in the base 10 (decimal) number system, in a six-symbol number such as 495,362, what weighting is applied to the leftmost symbol? What does the symbol ‘4’ represent in the number 495,362?

  • The leftmost symbol in our example is in sixth place so a weighting of 100,000 is applied to this symbol. So in the decimal number 495,362, the ‘4’ represents 400,000.

Watch the video below, which is about 2 minutes long.

Box _unit5.1.2 Exponential notation

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Example _unit5.1.2 Activity 2 Test yourself

5 minutes

Give your answers to the following:

ITQ _unit5.1.3

  • 8 × 103 =

  • 8 × 103 = 8000

ITQ _unit5.1.4

  • 4 × 105 =

  • 4 × 105 = 400,000

ITQ _unit5.1.5

  • 7 × 107 =

  • 7 × 107 = 70,000,000

Watch the video below, which is about 3 minutes long.

Box _unit5.1.3 Base 2 arithmetic

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Example _unit5.1.3 Activity 3 Test yourself

5 minutes

Give the denary equivalents of the following binary numbers:

ITQ _unit5.1.6

  • 1110

  • 1110 = (1 × 23) + (1 × 22) + (1 × 21) + (0 × 20) = 8 + 4 + 2 + 0 = 14

ITQ _unit5.1.7

  • 1001

  • 1001 = (1 × 23) + (0 × 22) + (0 × 21) + (1 × 20) = 8 + 0 + 0 + 1 = 9

ITQ _unit5.1.8

  • 101101

  • This is more difficult as you need to apply some higher binary weightings of 25 (16) and 26 (32).

    So 101101 = (1 × 25) + (0 × 24) + (1 × 23) + (1 × 22) + (0 × 21) + (1 × 20) = 32 + 0 + 8 + 4 + 0 + 1 = 45.