233 lines
6.9 KiB
Markdown
233 lines
6.9 KiB
Markdown
| Roll Number | 00677611 |
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| ----------- | -------- |
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| ID | CHC119 |
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# 1 - Linux Bash
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### 1.1 - If Statements
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### 1.2 - Processing Variable Number of Arguments with While, Using if
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# 2 - Linux System Administration
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#### Setup:
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- Create **sysadmin** user:
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- Add **sysadmin** to the **root** group
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- Create global_aliases.sh and set permissions
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- Create aliases:
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- Add source for existing users:
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- Create users:
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- Create group for Hugh and Ryan
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- Add Hugh and Ryan to the group
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- Create shared directory
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- Set permissions to directory to allow root and sharedaccess to read, write and execute, others can read
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- Set permissions of all folders in /home/ to rwx for exclusively the user that owns it. Other users may not read, write or execute
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#### Testing:
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##### Aliases
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- Test **ll** alias:
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- Test **la** alias:
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- Test other uses can use alias:
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##### Shared Directory:
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- Both Ryan and Hugh can read and write inside the shared directory.
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- Rob can read from the directory:
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- Rob cannot write to the directory:
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##### Home Directories:
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- Users cannot access each other's home directories.
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- However, the owner may access the home directory.
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#### bash_history
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# CSI Task 1
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### a)
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#### i)
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```
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X = (A ∧ B ∧ C) ∨ ( ¬A ∧ B ∧ ¬C) ∨ ( ¬A ∧ ¬B ∧ C)
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```
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#### ii)
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```csvtable
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source: 1a.csv
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```
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### b)
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Truth Table:
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```csvtable
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source: 1b.csv
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```
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Karnaugh Graph:
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| | | AB | | | |
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| --- | --- | --- | --- | --- | --- |
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| | | 00 | 01 | 11 | 10 |
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| C | 0 | 1 | 0 | 1 | 0 |
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| | 1 | 1 | 0 | 1 | 0 |
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As we can see from the Karnaugh Graph, the output of C is completely irrelevant.
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Boolean Expression:
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```
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OUT = (IN1 ∧ IN2) ∨ ( ¬IN1 ∧ ¬IN2)
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```
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Logic Circuit:
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# CSI Task 2
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### a)
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| Name | George |
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| ------- | ---------------------------------------------------------------------------------------- |
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| Decimal | 71,101,111,114,103,101 |
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| Binary | 1000111,1100101,1101111,1110010,1100111,1100101 |
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| 8N2 | 0(Idle) 1(Start) 01000111 00(Stop) 1(Start) 01100101 00(Stop) 1(Start) 01101111 00(Stop) |
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| | 1(Start) 01110010 00(Stop) 1(Start) 01100111 00(Stop) 1(Start) 01100101 00(Stop) 0(Idle) |
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- 
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### b)
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| Sent | UTF-8 | Binary | UTF-8 | Received |
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| -------- | ----- | ------------ | ----- | -------- |
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| G | 71 | 0100 0111 | 71 | G |
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| e | 101 | 0110 0101 | 101 | e |
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| o | 111 | 0110 1111 | 111 | o |
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| r | 114 | 0111 0010 | 114 | r |
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| g | 103 | 0110 0111 | 103 | g |
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| e | 101 | 0110 0101 | 101 | e |
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| Sum | 601 | 10 0101 1001 | 601 | |
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| Checksum | 89 | | 89 | |
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601 $mod$ 256 = 89
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or,
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10 0101 1001 AND
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00 1111 1111
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00 0101 1001 = 89
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Example:
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| Sent | UTF-8 | Binary | UTF-8 | Received |
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| -------- | ----- | ------------- | ----- | -------- |
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| G | 71 | 0100 0111 | 71 | G |
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| e | 101 | 011**1** 0101 | 117 | u |
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| o | 111 | 0110 11**0**1 | 109 | m |
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| r | 114 | 0111 0010 | 114 | r |
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| g | 103 | 01**0**0 0111 | 71 | G |
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| e | 101 | 0110 0101 | 101 | e |
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| Sum | 601 | 10 0100 0111 | 583 | |
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| Checksum | 89 | | 71 | |
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Here we can detect an error in transmission, since the checksum differs on each side of the message. A checksum however cannot detect *where* an error occurs, just the fact that it has.
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### c)
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$Baud Rate = \frac{Bitrate}{Bits Per Symbol}$
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$Bit Rate = 2MBps$
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$Bits Per Symbol = 8 data + 1 start + 2 stop = 11 bits$
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$Baud Rate = \frac{2MBps}{11b} = 0.1818MHz = 181.8kHz$
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$5F = 5 * \frac{Baud Rate}{2}$
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$5F = 5 * 90.9kHz = 454.5kHz$
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Required Bandwidth: 454.5kHz
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# CSI Task 3
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### a)
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#### Page Accesses
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1. 1 2 3 4 5
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2. 2 3 4 1 5
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3. 3 4 1 2 5
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Using the clock algorithm, there are 7 page faults total with this sequence of page accesses.
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### b)
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Since A has a runtime of 4 seconds, it is impossible to allocate 4-7 (4, 5, 6, 7 -> 5 processes) second runtime with no conflicts, so I will allocate 4 in a process of my choice.
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$Response Ratio = \frac{Wait Time + Run Time}{Run Time}$
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| Process | Runtime | Start Time | Response Ratio |
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| ------- | ------- | ---------- | -------------- |
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| A | 4 | 0 | 1 |
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| B | 5 | 1 | 1.2 |
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| C | 6 | 2 | 1.33 |
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| D | 7 | 3 | 1.43 |
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| E | 4 | 4 | 2 |
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Process E has the highest Response Ratio, and will be scheduled first.
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Then Process D, Process C, B and finally A.
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| t (Time) | Complete Process | Start Process |
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| -------- | ---------------- | ------------- |
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| 0 | | A |
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| 4 | A | E |
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| 8 | E | D |
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| 15 | D | C |
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| 21 | C | B |
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| 26 | B | |
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Since the only available process at t=0 is A, it must still be performed first, regardless of where it sits in the queue.
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E is then available at t=4, when A is completed, allowing the remaining queue to run in descending order of response ratio.
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