ICT201: Assessment 2 XXXXXXXXXXCompiled by: Ali Noori XXXXXXXXXXT1 2020 T1 2020: ICT201 Computer Organisation and Architecture Assessment 2 Assessment type: Report (2,000 words) – individual...

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ICT201: Assessment 2 Compiled by: Ali Noori T1 2020 T1 2020: ICT201 Computer Organisation and Architecture Assessment 2 Assessment type: Report (2,000 words) – individual assignment Purpose: This assessment will allow students to demonstrate that they can identify and understand synchronisation and deadlocks. This assessment contributes to Learning Outcomes b and c. Value: 30% (Report 20%; Presentation 10%) Due Date: Report Submission via Moodle (Week 9 Sunday 23:55); Presentation (Week 10 – 11) Submission: Upload the completed report via Moodle. Task Details: The report will require analysis of an operating system scenario and a report on the systems and logical issues involved, as well as options for resolving the problem and subsequent implications. Referencing: Please remember that all sources used in assessment tasks must be suitably referenced. Harvard referencing is the required method. A Referencing Guide, “Harvard Referencing”, and a Referencing Tutorial can be found on the right-hand menu strip in Moodle on all subject pages. Assessment Topic: Analysis of an Operating System scenario, Computer Organisation and Architecture Report Assignment Details: 1- Given the following arrival times and CPU cycle times. Job Arrival Time CPU cycles required Priority A 0 3 3 (Silver) B 3 8 2 (Gold) C 4 3 1 (Diamond) D 7 14 2 (Gold) E 12 2 2 (Gold) a) Draw a timeline (Gantt-Chart) for each of the following scheduling algorithms. b) What is the Waiting Time and Turnaround Time of each process for each of the scheduling algorithms? (Details of the calculation is essential). i) First-Come-First-Served (FCFS) ii) Round Robin (RR) (use time quantum of 3) iii) Highest Response Ratio Next (HRRN) iv) Shortest Remaining Time (SRT) v) Shortest Process Next (SPN) ICT201: Assessment 2 Compiled by: Ali Noori T1 2020 2- Using either internet resources or books, understand the concept of Memory Management and Virtual Memory. Define those terms (Memory Management and Virtual Memory) in your own words. You must provide references and cite the sources that you consulted for this task. (Harvard referencing is the required method.) 3- Consider the directed resource graph shown below and answer the following questions: a. Is this system deadlocked? b. Which, if any, processes are blocked? c. What is the resulting graph after reduction? R1 R2 R3 R4 P1 P2 P3 Assessment 2 Marking Rubrics - Report: Criteria Unsatisfactory Satisfactory Effective Excellent Exceptional 30 % Content Fail (0 – 49%) Pass (50 – 64%) Credit (65 – 74%) Distinction (75 – 84%) High Distinction (85 – 100%) Depth of Content The explanation is sufficiently inaccurate, incomplete, or confusing that the reader gains little information from the report. It appears that little attempt has been made to help the reader understand the material. Some difficulty in reading, structure lacking in some parts Well written, structure not totally clear Well written and structured. An accurate and complete explanation of key concepts and theories is made, drawing on relevant literature. Enough detail is presented to allow the reader to understand the content and make judgments about it. In addition, applications of theory are included to illuminate issues. Readers gain insights. 6 Accuracy of Content Does not directly answer question Directly answers most of the question Directly answers the question Directly answers the question giving additional insights Information (names, facts, etc.) included in the report is consistently accurate. 6 ICT201: Assessment 2 Compiled by: Ali Noori T1 2020 Clarity of Purpose No support for the project’s objectives taken in the answer. Provides some support for the project’s objectives taken. The project’s objectives are presented. The motivation for pursuing the project and its relevance are addressed. The discussion is reasonably clear but not compelling. Provides excellent level of support for project’s objectives taken. The project’s objectives are clearly stated. The motivation for pursuing the project and its relevance are clearly and persuasively established by relating the project to current engineering problems. 6 Depth of Analysis No analysis presented Analysis based on some argument Analysis based on sound argument Analysis based on an extensive argument Results are carefully and objectively analysed. Interpretations are made using appropriate equations, models, or theories. 6 Use of Language: Word Choice, Grammar, and Sentence Structure Very little knowledge of subject matter demonstrated. Adequate knowledge of subject matter demonstrated. Sound knowledge of subject matter demonstrated. High level of knowledge of subject matter demonstrated. High level of knowledge of subject matter and synthesis with theory demonstrated 6
Answered Same DayMay 21, 2021

Answer To: ICT201: Assessment 2 XXXXXXXXXXCompiled by: Ali Noori XXXXXXXXXXT1 2020 T1 2020: ICT201 Computer...

Kshitij answered on May 23 2021
149 Votes
Computer organization and architecture
Student name
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Part 1
1. The concept relating to first come first serve: algorithm which is none premature
Here are the series stated for the concept of first come first serve
    Procedures
    A
    B
    C
    D
    E
    Time of arrival
    0
    3
    4
    7
    12
    Time of serving
    3
    8
    3
    14
    2
    Time for turnaround
    3
    8
    10
    21
    18
    Time
for waiting
    0
    0
    7
    7
    16
a. Giant chart
    A
    0-3
    B
    3-11
    C
    11-14
    D
    14-28
    E
    28-30
b. Time taken for turnaround
Time taken for turnaround= time for completion-time of arrival
    A
    B
    C
    D
    E
    3-0=3
    11-3=8
    14-4=10
    28-7=21
    30-12=18
Time for waiting = time taken in turnaround- time of burst
    
    A
    B
    C
    D
    E
    3-3=0
    8-8=0
    10-3=7
    21-14=7
    18-2=6
2. Round robin: premature algorithm with turnaround time =3
    Procedures
    A
    B
    C
    D
    E
    Time of arrival
    0
    3
    4
    7
    12
    Time of serving
    3
    8
    3
    14
    2
    Time for turnaround
    3
    8
    10
    21
    18
    Time for waiting
    0
    0
    7
    7
    16
a. Gantt chart
    A
    0-3
    B
    3-6
    C
    6-9
    D
    9-12
    E
    12-14
    B
    14-17
    D
    17-20
    B
    20-22
    D
    22-30
b. Time taken for turnaround
Time taken for turnaround= time for completion-time of arrival
    A
    B
    C
    D
    E
    3-0=0
    22-3=19
    9-4=5
    30-7=23
    14-12=2
Time for waiting = time taken in turnaround- time of burst
    A
    B
    C
    D
    E
    3-3=0
    19-8=11
    15-3=12
    23-14=9
    2-2=0
3. Next highest response ratio: non premature algorithm
Ratio of response = [waiting time + time for serving]/ serve time
    Procedures
    A
    B
    C
    D
    E
    Time of arrival
    0
    3
    4
    7
    12
    Time of serving
    3
    8
    3
    14
    2
    Time for turnaround
    3
    8
    10
    21
    18
    Time for waiting
    0
    0
    7
    7
    16
a. Gantt chart
A=0-3 at t=0 and only the process A is arrived to execute the A
As A= 0-3
B= 3-11 at t=3 and only B executes to arrive B
At t= 11 the process C and D are taken into account to process and to respond the various ratio relating to C and D.
The users have to execute the highest response to the ratio.
Response ratio = [waiting time + time for serving]/ serve time
= [11+3]/3=4.66
Response ratio = [waiting time + time for serving]/ serve time
= [11+14]/14= 1.04
Response ratio = [waiting time + time for serving]/ serve time
= [14+2]/2= 1.1
Final chart of giant
    A
    0-3
    B
    3-11
    C
    11-14
    D
    14-28
    E
    28-30

b. Time taken for turnaround= time for completion-time of arrival
    A
    B
    C
    D
    E
    3-3=0
    11-3=8
    14-4=10
    18-7=21
    30-12=18
Time for waiting = time taken in turnaround- time of burst
    
    A
    B
    C
    D
    E
    3-3=0
    8-8=0
    10-3=7
    21-14=7
    18-2=16
4. Shortest time remaining: premature version of the respective shortest job

    Process
    A
    B
    C
    D
    E
    Arrival time
    3
    18
    3
    21
    2
    Waiting time
    0
    10
    0
    7
    0
a. giant chart
    A
    0-3
    B
    3-4
    C
    4-7
    D
    7-12
    E
    12-14
    B
    14-21
    D
    21-30
a. Time taken for turnaround= time for completion-time of arrival
    A
    B
    C
    D
    E
    3-3=0
    21-3=18
    7-4=3
    30-7=21
    14-12=2
Time for waiting = time taken in turnaround- time of burst
    A
    B
    C
    D
    E
    3-3=0
    18-8=10
    3-3=0
    21-14=7
    2-2=0
5. Shortest processing time: non premature algorithm
Time taken for turnaround= time for completion-time of arrival
    A
    B
    C
    D
    E
    3-3=0
    11-3=8
    14-4=10
    30-7=23
    16-12=4
    Process
    A
    B
    C
    D
    E
    Arrival time
    3
    8
    10
    23
    4
    Waiting time
    0
    0
    7
    9
    2
Part 2
Management of memory
Management of memory is considered to be one of the most significant and crucial function of the various operating systems. Memories present in the various parts of the operating system help the same to function and run properly. Each and every program working on the operating system is...
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