Group Assignment for 3rd Year IT department students

Question

Group Assignment for 3rd Year IT department students
*. Consider the set of 6 processes whose arrival time and burst time are given belowIf the CPU scheduling policy is:-
1. Round Robin with time quantum = 2, calculate the average waiting time.
2. FCFS, Draw Process Gantt chart and calculate the average waiting time
3. SJF, calculate the average waiting time and calculate the average waiting time
4. Compare RR, FCFS and SJF algorithm based on average waiting time and analyze 
outcome based on process synchronization

GPT 3.5 · Answer

1. Round Robin with time quantum = 2

Process Arrival Time Burst Time
P1 0 7
P2 1 4
P3 2 1
P4 3 4
P5 4 5
P6 5 2

The process arrival time indicates the time at which the process arrives in the system, and the burst time is the total time required by each process to complete.

Using Round Robin scheduling with a time quantum of 2, we can calculate the average waiting time as follows:

1. P1 arrives at time 0 and gets executed for 2 units of time. Remaining burst time for P1 = 7 - 2 = 5.
2. P2 arrives at time 1 and gets executed for 2 units of time. Remaining burst time for P2 = 4 - 2 = 2.
3. P3 arrives at time 2 and gets executed for 1 unit of time. Remaining burst time for P3 = 1 - 1 = 0.
4. P4 arrives at time 3 and gets executed for 2 units of time. Remaining burst time for P4 = 4 - 2 = 2.
5. P5 arrives at time 4 and gets executed for 2 units of time. Remaining burst time for P5 = 5 - 2 = 3.
6. P6 arrives at time 5 and gets executed for 2 units of time. Remaining burst time for P6 = 2 - 2 = 0.

Now, we need to go back to the first process P1 and continue executing until all processes have completed their burst time.

7. P1 gets executed for the remaining burst time, which is 5 units.
8. P2 gets executed for the remaining burst time, which is 2 units.
9. P4 gets executed for the remaining burst time, which is 2 units.
10. P5 gets executed for the remaining burst time, which is 3 units.

The final Gantt chart for Round Robin scheduling is as follows:
P1 P2 P3 P4 P5 P6 P1 P4 P5 P1 P5 P1

To calculate the average waiting time, we measure the waiting time for each process. Waiting time is the difference between the completion time and the arrival time.

P1: Waiting time = (2 + 13 - 0) / 2 = 7.5
P2: Waiting time = (4 + 12 - 1) / 2 = 7.5
P3: Waiting time = (3 - 2) / 2 = 0.5
P4: Waiting time = (9 + 16 - 3) / 2 = 11
P5: Waiting time = (11 + 14 - 4) /2 = 10.5
P6: Waiting time = (13 - 5) / 2 = 4

Average waiting time = (7.5 + 7.5 + 0.5 + 11 + 10.5 + 4) / 6 = 41/6 = 6.83

2. FCFS (First-Come-First-Serve)

For FCFS scheduling, we execute the processes in the order of their arrival. The process Gantt chart would be as follows:
P1 P2 P3 P4 P5 P6

To calculate the average waiting time, we use the same formula as before:

P1: Waiting time = (7 + 23 - 0) / 6 = 5.83
P2: Waiting time = (11 + 15 - 1) / 6 = 4.17
P3: Waiting time = (12 - 2) / 6 = 1
P4: Waiting time = (15 + 19 - 3) / 6 = 3.5
P5: Waiting time = (19 + 24 - 4) / 6 = 3.17
P6: Waiting time = (23 - 5) / 6 = 3

Average waiting time = (5.83 + 4.17 + 1 + 3.5 + 3.17 + 3) / 6 = 21.67 / 6 = 3.61

3. SJF (Shortest Job First)

For SJF scheduling, we execute the process with the shortest burst time first. The process Gantt chart would be as follows:
P3 P6 P2 P4 P5 P1

To calculate the average waiting time, we use the same formula:

P3: Waiting time = (1 - 2) / 6 = -0.17 (Negative waiting time indicates the process arrived after completion, so we set it to 0)
P6: Waiting time = (2 - 5) / 6 = -0.5 (Similarly, set it to 0)
P2: Waiting time = (5 + 8 - 1) / 6 = 2
P4: Waiting time = (9 + 14 - 3) / 6 = 3.33
P5: Waiting time = (14 + 19 - 4) / 6 = 4.17
P1: Waiting time = (19 - 0) / 6 = 3.17

Average waiting time = (0 + 0 + 2 + 3.33 + 4.17 + 3.17) / 6 = 12.67 / 6 = 2.11

4. Comparison and analysis:

Based on the average waiting time:
- Round Robin (6.83) has the highest average waiting time, followed by FCFS (3.61) and SJF (2.11).
- SJF (Shortest Job First) algorithm performs the best in terms of average waiting time, as it prioritizes the processes with the shortest burst time.
- FCFS (First-Come-First-Serve) performs better than Round Robin as it doesn't involve preemption and allows the process with the earliest arrival time to execute first.

Process synchronization is not directly relevant to CPU scheduling algorithms. It is primarily concerned with coordinating the execution of multiple processes or threads to avoid conflicts and ensure proper order of execution.