G4G0-1/Semester 2/Computer Systems Internals & Linux/Trimester 2 Assignment/Trimester 2 Assignment.md

Roll Number	@00677611
ID	CHC119
Name	George Wilkinson

1 - Linux Bash

1.1 - If Statements

1.2 - Processing Variable Number of Arguments with While, Using if

2 - Linux System Administration

Setup:

• Create sysadmin user:
• Add sysadmin to the root group
• Create global_aliases.sh and set permissions
• Create aliases:
• Add source for existing users:
• Create users:
• Create group for Hugh and Ryan
• Add Hugh and Ryan to the group
• Create shared directory
• Set permissions to directory to allow root and sharedaccess to read, write and execute, others can read
• Set permissions of all folders in /home/ to rwx for exclusively the user that owns it. Other users may not read, write or execute

Testing:

Aliases

• Test ll alias:
• Test la alias:
• Test other uses can use alias:

Shared Directory:

• Both Ryan and Hugh can read and write inside the shared directory.

• Rob can read from the directory:

• Rob cannot write to the directory:

Home Directories:

• Users cannot access each other's home directories.

• However, the owner may access the home directory.

CSI Task 1

a)

i)

Since Boolean Algebra ( ., +, Line over ) is not possible to type in my text editor, I will use Boolean Expression Logic symbols ( ^, V, ¬) throughout the CSI task sections. I apologise for any inconvenience.

X = (A ^ B ^ C) v ( ¬A ^ B ^ ¬C) v ( ¬A ^ ¬B ^ C)

ii)

source: 1a.csv

b)

Truth Table:

source: 1b.csv

Karnaugh Graph:

		AB
		00	01	11	10
C	0	1	0	1	0
	1	1	0	1	0

As we can see from the Karnaugh Graph, the output of C is completely irrelevant.

Boolean Expression:

OUT = (IN1 ^ IN2) ∨ ( ¬IN1 ^ ¬IN2)

Logic Circuit:

CSI Task 2

a)

Name	George
Decimal	71,101,111,114,103,101
Binary	1000111,1100101,1101111,1110010,1100111,1100101
8N2	0(Idle) 1(Start) 01000111 00(Stop) 1(Start) 01100101 00(Stop) 1(Start) 01101111 00(Stop)
	1(Start) 01110010 00(Stop) 1(Start) 01100111 00(Stop) 1(Start) 01100101 00(Stop) 0(Idle)

• 

b)

Sent	UTF-8	Binary	UTF-8	Received
G	71	0100 0111	71	G
e	101	0110 0101	101	e
o	111	0110 1111	111	o
r	114	0111 0010	114	r
g	103	0110 0111	103	g
e	101	0110 0101	101	e
Sum	601	10 0101 1001	601
Checksum	89		89

601 mod 256 = 89 or, 10 0101 1001 AND 00 1111 1111 00 0101 1001 = 89

Example:

Sent	UTF-8	Binary	UTF-8	Received
G	71	0100 0111	71	G
e	101	0111 0101	117	u
o	111	0110 1101	109	m
r	114	0111 0010	114	r
g	103	0100 0111	71	G
e	101	0110 0101	101	e
Sum	601	10 0100 0111	583
Checksum	89		71

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.

c)

( Equations and formulae written with LaTeX )

Baud Rate = \frac{Bitrate}{Bits Per Symbol}

Bit Rate = 2MBps Bits Per Symbol = 8 data + 1 start + 2 stop = 11 bits

Baud Rate = \frac{16000000bps}{11b} = 1,454,545Baud = 1.454545MHz = 1454.545kHz

5F = 5 * \frac{Baud Rate}{2} 5F = 5 * 727.2725kHz = 3636.3625kHz = 3.636MHz

Required Bandwidth: 3.636MHz

CSI Task 3

a)

Page Accesses

1. 1 2 3 4 5
2. 2 3 4 1 5
3. 3 4 1 2 5

Using the clock algorithm, there are 7 page faults total with this sequence of page accesses.

b)

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.

Response Ratio = \frac{Wait Time + Run Time}{Run Time}

Process	Runtime	Start Time	Response Ratio
A	4	0	1
B	5	1	1.2
C	6	2	1.33
D	7	3	1.43
E	4	4	2

Process E has the highest Response Ratio, and will be scheduled first. Then Process D, Process C, B and finally A.

t (Time)	Complete Process	Start Process
0		A
4	A	E
8	E	D
15	D	C
21	C	B
26	B

Since the only available process at t=0 is A, it must still be performed first, regardless of where it sits in the queue. E is then available at t=4, when A is completed, allowing the remaining queue to run in descending order of response ratio.