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Nt1210 Lab

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Fredrick Donaldson
January 1, 2015
Introduction to Networking
Chapter 1 Lab 1:1:1 Reading Binary

1) The mapping of 2931 should contain the values 2, 9, 3, and 1 in the lower-digits places and 2000, 900, 30, and 1 in the bottom boxes. The place values are the same regardless, so the only change is the addition of 103 as the leftmost digit value. 2 | 9 | 3 | 1 | 2,000 | 900 | 30 | 1 |

Exercise 1.1.2

The mapping of 1102 will have 1, 1, and 0 in the lower bit values and 4, 2, and 0 in the bottom bits, for a sum of 6. 1 | 1 | 0 | 4 | 2 | 0 |

Exercise1.1.3

The mapping of 112 will have a leading 0 to fit the mapping, which does not affect the results. The lower bits will have leading 0s and then 1 and 1 in the rightmost bits. This results in all 0s at the bottom, with 2 and 1 in the rightmost bits, for a sum of 3. 0 | 0 | 1 | 1 | 0 | 0 | 2 | 1 |

Exercise 1.1.4
The placeholder 24 = 16 is necessary for this mapping because it is five digits. The result is 16 + 2 = 18 16 | 8 | 4 | 2 | 1 | 1 | 0 | 0 | 1 | 0 |

Exercise 1.1.5

This value represents a full byte of information, with placeholder values 27 = 128, 26 = 64, 25 = 32, 24 = 16, 23 = 8, 22 = 4, 21 = 2, and 20 = 1. The bits should align left to right as 1, 1, 1, 0, and 0,0,1,0 leading to the bottom-row bits 128,64,32,0,0,0,2,0 = 226. 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 128 | 64 | 32 | 0 | 0 | 0 | 1 | 0 |

Exercise 1.1.6

For the value 156, the largest bit is 128 (or 27), leaving 28. For 28, the next smallest bit is 16 (or 24), leaving 12. For 12, the next bit is 8, leaving 4. 4 is a bit value with a 0 result, so the bit result is 100111002.

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