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 = 2
20 > 1 x 0 = 0
16+0+0+2+0=18
Exercise 1.1.5
11100010
27 > 128 x 1 = 128
26 > 64 x 1 = 64
25 > 32 x 1 = 32
24 > 16 x 0 = 0
23 > 8 x 0 = 0
22 > 4 x 0 = 0
21 > 2 x 1 = 2
20 > 1 x 0 = 0
128+64+32+0+0+0+2+0=226 >
Exercise 1.1.6 156
156 > 27 > 128 x 1 = 128/ 156-128= 28
28 > 26 > 64 x 0 = 0
28 > 25 > 32 x 0 =0
28 > 24 > 16 x 1 = 16/ 28-16=12
12 > 23 > 8 x 1 = 8/ 12-8=4
4 > 22 > 4 x 1 = 4/ 4-4=0
0 > 21 > 2 x 0 = 0
0 > 20 > 1 x 0 =0
100111002 = 156
Exercise 1.1.7 255
255 > 27 > 128 x 1 = 128/ 255-128=127
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
111111112 = 255
Exercise 1.1.8 200
200 > 27 > 128 x 1 = 128/ 200-128=72
72 > 26 > 64 x 1 = 64/ 72-64=8
8> 25 > 32 x 0 = 0
8 > 24 > 16 x 0 = 0
8 > 23 > 8 x 1 = 8/ 8-8=0
0 > 22 > 4 x 0 = 0
0 > 21 > 2 x 0 = 0
0 > 20 > 1 x 0 = 0
110010002 = 200
Exercise 1.1.9 see excel sheet
Exercise 1.1.10 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
122880x1024
125829120 Bytes
Exercise 1.3.4
16GB and 32 bit word size
To find how many bytes per word 32/8=4
(16x1024x1024x1024)/4
(16,384x1024x1024)/4
(1,677,216x1024)/4
17,179,869,184/4
4,294,967,296 words
Exercise 1.3.5
1101102
0011,0110
27 > 128 x 0 = 0
26 > 64 x 0 = 0
25 > 32 x 1 = 32/16=2
24 > 16 x 1 = 16/16=1
2+1=3
23 > 8 x 0 = 0
22 > 4 x 1 = 4/1=4
21 > 2 x 1 = 2/1=2
20 > 1 x 0 = 0
4+2=6
3616
Exercise 1.3.6
F616 to decimal and binary
F = 15 6=6
15x16=240 + 6 = 246
240 > 27 > 128x 1 = 128/ 240-128=112
112 > 26 > 64 x 1 = 64/ 112-64=48
48 > 25 > 32 x 1 = 32/ 48-32=16
16 > 24 > 16 x 1 = 16/ 16-16=0
11112
6 > 23 > 8 x 0 = 0
6 > 22 > 4 x 1 = 4/ 6-4=2
2 > 21 > 2 x 1 = 2/ 2-2= 0
0 > 20 > 1 x 0 = 0
01102
111101102
Lab 1.3 review 1. The reason it is important to know how many system words will fit on a primary storage is because you need to know if you will be able to fit the operating system and other programs on the drive as well. 2. The reason you can fit more information when 2 bytes are joined together, is that when the 2 bytes are joined the individual bits multiply far past 128 3. Because each place of a hexadecimal number represents its own number where as decimal is one number in all. When you convert the hexadecimal to binary it takes 1 byte and cuts it in half so there are 4 bits to each placement. 4. 327 to binary then hexadecimal it will require joining 2 bytes together since 255 is the max for a single byte
327 > 28 > 256 x 1 = 256/ 327-256=71
71 > 27 > 128 x 0 = 0
71 > 26 > 64 x 1 = 64/ 71-64=7
7 > 25 > 32 x 0 = 0
7 > 24 > 16 x 0 = 0
7 > 23 > 8 x 0 = 0
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
0000,0001 0100,0111
28 > 256 x 1 = 256/256=1
1 = 1
27 > 128 x 0 = 0
26 > 64 x 1= 64/16=4
25 > 32 x 0 = 0
24 > 16 x 0 = 0
4
23 > 8 x 0 = 0
22 > 4 x 1 = 4/1=4
21 > 2 x 1 = 2/1=2
20 > 1 x 1 = 1/1=1
7
14716
Lab 1.4
Exercise 1.4.1
20, a blank space is important because it is needed to keep the code making sense
Exercise 1.4.2 ASCII | HEX | 0 | 30 | 1 | 31 | 2 | 32 | 3 | 33 | 4 | 34 | 5 | 35 | 6 | 36 | 7 | 37 | 8 | 38 | 9 | 39 | Exercise 1.4.3
Binary: 01001011
Decimal: 75 Exercise 1.4.3
ASCII is used as a coding for most languages using the standard alphabet and each character can be held in 8 bits, UCS allows for every written language and each character is potentially held in 32 bits but utilizes BMP currently which holds each character in 16 bits. Exercise 1.4.3
UTF- replaces ASCII in the same way as UCS with smaller character storage being 8 bits and it’s compatibility also made it a good choice to replace ASCII Lab 1.4 review 1. William Techow – 57 69 6C 6C 69 61 6D 20 54 65 63 68 6F 77
14 Bytes 2. 9 bits 3. N = 4E =01001110
E = 45 = 01000101
T = 54 = 01010100
W = 57 = 01010111
O = 4F = 01001111
R = 52 = 01010010
K = 4B = 01001011
Lab 1.5
Exercise 1.5.1
Local disk c: > users >will >documents> classes
Exercise 1.5.2
Local disk c: > users >will >documents> classes> networking
This is the only thing that I could find that changed from before
Exercise 1.5.3
Exercise 1.5.4
Disk fragmentation occurs because files are often broken up when stored on your hard drive. This process slows down your computers overall performance, making it take much longer to access the files making your CPU work harder to locate the target file. Changing the file structure will not change the fragmentation of your hard disk. It will slow down your PC all the same, as this is a common occurrence. Running a disk defragmentation will help fix this.
Lab 1.5 review 1. It is important so that you can keep files organized and easy to find and use. If you make it too detailed you can forget what went where and that can cause you to take a long time to find the files that you want. 2. The way that it differs is that instead of going into the documents folder under users > will you go to the desktop folder. 3. When you defragment the disk it takes all of the scattered files and tries to put them all in the same area of the disk so it doesn’t have to search through the entire disk to find what it is looking for then it updates the memory references to reflect where it went. Doing this does not change how you find all of your files.
Lab 1.6
Exercise 1.6.1
117 GB total on C: drive 23 GB free
465 GB total on E: drive 256 GB free
931 GB total on F: drive 267 GB free
Exercise 1.6.2
I do not have a Linux or MAC OS computer.
Exercise 1.6.3
16GB DDR3 RAM 1600 GHz
Exercise 1.6.4
I do not have a Linux or MAC OS computer.
Exercise 1.6.5
AMD FX 6300 six-core processor 3.5 GHz
Exercise 1.6.6
I do not have a Linux or MAC OS computer.
Exercise 1.6.7
It shows all of the programs that are running and how much of the CPU, RAM, Disk and network it is taking up.

Exercise 1.6.8
Well the program that is using the CPU the most while I am typing is word. But while I am just sitting and looking at it, it will change between Norton, Chrome and task manager.
Exercise 1.6.9
I do not have a Linux or MAC OS computer.
Lab review 1.6 1. It is important to know this information to know what your computer is capable of, so that when you are loading either a new OS or program to your computer you know if you will be able to actually run them on that computer. 2. On a windows computer you can go into the task manager and left click then hit the end process button or you can hit the right click and then press the end process in the dropdown. In MAC you can choose force quit from the apple menu then you can click the app and hit force quit. The risk to doing this is you could stop a process that is vital for the computer to run. 3. To find updates to firmware you would usually go to the manufacturers website, or the website of the OS you have installed like Microsoft for example.
Lab 2.5
Exercise 2.4.1
I hit the windows key + r then typed cmd and hit enter
Then I typed ipconfig and hit enter then got this.
Microsoft Windows [Version 10.0.10240]
(c) 2015 Microsoft Corporation. All rights reserved.

C:\Users\will>ipconfig
Windows IP Configuration
Ethernet adapter Ethernet: Connection-specific DNS Suffix . : IPv6 Address. . . . . . . . . . . : 2601:406:4004:7370::3 IPv6 Address. . . . . . . . . . . : 2601:406:4004:7370:957a:da5d:8494:417e Temporary IPv6 Address. . . . . . : 2601:406:4004:7370:c9e1:92ab:9ef0:3920 Link-local IPv6 Address . . . . . : fe80::957a:da5d:8494:417e%3 IPv4 Address. . . . . . . . . . . : 10.0.0.15 Subnet Mask . . . . . . . . . . . : 255.255.255.0 Default Gateway . . . . . . . . . : fe80::e288:5dff:febb:bded%3 10.0.0.1
Tunnel adapter isatap.{481BAAE1-E197-43DA-A0F0-65E1BF21E144}: Media State . . . . . . . . . . . : Media disconnected Connection-specific DNS Suffix . :
Tunnel adapter Teredo Tunneling Pseudo-Interface: Connection-specific DNS Suffix . : IPv6 Address. . . . . . . . . . . : 2001:0:5ef5:79fb:32:3c86:bbda:13a1 Link-local IPv6 Address . . . . . : fe80::32:3c86:bbda:13a1%6 Default Gateway . . . . . . . . . :
C:\Users\will>
Lab 2.4 review 1. I only have windows so I can’t really compare the 2 operating systems 2. The other benefits are being able to check your disk for errors and cleaning the disk’s empty space and overwriting that information so that when you get rid of that drive no one can see what is or used to be on the disk. 3. The reason they are more graphical instead of command line based is that, it is more user friendly this makes it so that people will feel less confused while using the OS and will be more likely to want to buy it.
Lab 3.5
Exercise 3.5.1
I don’t have windows XP don’t really know of anyone who does anymore but I will see how much of this is the same in windows 10
Well there was very little that was the same as XP for windows 10, but the reason it is important to keep a record of this information is because, you want to be able to remember passwords and the setup that you have so that when there is a problem you can see if anything has been changed since it last worked.
Lab 3.5 review 1. Yes there is a risk of enableing file sharing on your PC it will make it easier to infect your PC with viruses and or get sensitive information from your computer. And the place that you can usually access the folders you share is in public under your username. 2. Yes there is a risk due to the fact that printers have a hard drive of their own to store documents sent to the printer, and someone can hack into it and view them.