2960 words - 12 pages

Unit 1 assignment

1. C

2. All of them are wrong the smallest measurement in the answers is a kilobyte and that is actually 1024 bytes not 106.

3. C

4. A,E

5. A

6. C

7. D

8. A

9. A,B,D

10. A

11. A

12. B,D

13. A,C

14. A,D

15. A

16. D

17. B

18. C

19. C,D

20. A,B

Lab 1.1

Exercise 1.1.1

103 > 1000 x 2 = 2000

102 > 100 x 9 = 900

101 > 10 x 3 = 30

100 > 1 x 1 = 1

2931

Exercise 1.1.2

22 > 4 x 1 = 4

21 > 2 x 1 = 2

20 > 1 x 0 = 0

4+2+0= 6

Exercise 1.1.3

21 > 2 x 1 =2

20 > 1 x 1= 1

2+1= 3

Exercise 1.1.4

24 > 16 x 1 = 16

23 > 8 x 0 = 0

22 > 4 x 0 = 0

21 > 2 x 1 ...view middle of the document...

1.10 see excel sheet

Exercise 1.1.11 see excel sheet

Exercise 1.1.12 see excel sheet

Lab 1.1 review

1. 127 to binary

127 > 27 > 128 x 0 = 0

127 > 26 > 64 x 1 = 64/ 127-64=63

63 > 25 > 32 x 1 = 32/ 63-32=31

31 > 24 > 16 x 1 = 16/ 31-16=15

15> 23 > 8 x 1 = 8/ 15-8=7

7 > 22 > 4 x 1 = 4/ 7-4=3

3> 21 > 2 x 1 = 2/ 3-2=1

1 > 20 > 1 x 1 = 1/ 1-1=0

011111112 = 127

The process of this is simply subtracting starting from the highest bit down to the lowest and, if the bit is higher than the decimal then you can subtract that so it ends up being a zero. But you just keep going until the decimal is now zero. There is also the divide by 2 method where you divide the decimal by 2 and every time you have a remainder then that is a 1 with no remainder it is a 0.

2. 102 and 00102 are the same because the first 2 digits of 00102 do not matter due to them being a 0 meaning there is no data there anyway.

3. In a base 5 numbering system using the first 4 digits of 5 you would only ever use the first 3 digits in binary being that 22 is 4 and anything higher would not work with it.

4. Creating a decimal to binary converstion in excel would be much harder because you would need to be able to get excel to realize it cannot subtract 128 from let’s say 70 or lower and also using the divide by 2 method would also be difficult seeing as it would have to realize the remainder and what that means getting it to display binary.

Lab 1.2

Exercise 1.2.1

1111

Exercise 1.2.2

1011

Exercise 1.2.3

1110

Exercise 1.2.4

111

Exercise 1.2.5

100

Exercise 1.2.6

0110

Exercise 1.2.7

1100

To repeatedly increment by a number of 2 you need to have 2 binary codes that have 1’s all in a row like 1111 +1111 then they would all increment

Exercise 1.2.8

1100

What I can conclude on any string of 1’s using AND, is that if both values are a 1 then it will be a 1 and if there is a 1 and a 0 then it will be a 0

Exercise 1.2.9

1111

What I can conclude about any string of 1’s with OR, is that if there is a 1 in the value then it will be a 1 and if there is no 1 then it will be a 0.

Lab 1.2 review

1. 11111110

1+0=1

0+1=1

0+1=1

1+0=1

0+1=1

0+1=1

0+1=1

0+0=0

2. 11001100 AND 11111100

1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |

1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 |

1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |

11001100

3. 11001100 OR 11111100

1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |

1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 |

1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 |

11111100

4. NOT (11001100 AND 11111100)

1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |

1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 |

1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |

NOT 11001100 = 00110011

Lab 1.3

Exercise 1.3.1

The decimal value of 1 byte is 255 and the decimal value of 2 bytes is 65535

Exercise 1.3.2

The decimal equivalent of this would be 8733, the bytes 1 and 2 as individual bytes would be. Byte 1: 25 standalone and 6670 as is, Byte 2: 233

Exercise 1.3.3

120MB

120x1024x1024

...

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