Using SJN: How To Calculate Start And Finish Times

Using Sjn How Do I Calculate The Start And Finish Time For Each Of Th

Using SJN, how do I calculate the start and finish time for each of the seven jobs below: Job Arrival Time CPU Cycle A 0 2 B 1 11 C 2 4 D 4 1 E 5 9 F 7 4 G 8 2 Thank you and please explain answer.

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Using Sjn How Do I Calculate The Start And Finish Time For Each Of Th

The Shortest Job Next (SJN), also known as Shortest Job First (SJF), is a non-preemptive scheduling algorithm that selects the process with the shortest CPU burst time from the ready queue for execution. To determine the start and finish times for each job in such a scheduling scenario, it is essential to analyze how the jobs arrive over time and how the scheduler chooses among the jobs based on their CPU cycles.

Given the provided data:

• Job A: Arrival Time = 0, CPU Cycle = 2
• Job B: Arrival Time = 1, CPU Cycle = 11
• Job C: Arrival Time = 2, CPU Cycle = 4
• Job D: Arrival Time = 4, CPU Cycle = 1
• Job E: Arrival Time = 5, CPU Cycle = 9
• Job F: Arrival Time = 7, CPU Cycle = 4
• Job G: Arrival Time = 8, CPU Cycle = 2

To calculate the start and finish times, follow these steps:

1. Identify the initial job to run: At time 0, Job A arrives and is the only job available, so it starts immediately.
2. Determine subsequent jobs based on shortest CPU cycle: After Job A completes at time 2, consider all jobs that have arrived up to that time—Jobs B and C arrive before or at 2, while D, E, F, G arrive later.
3. Apply SJN selection: Among the ready jobs, select the one with the shortest CPU cycle.
4. Iterate this process: Repeat the process at each completion time, considering newly arrived jobs and selecting the shortest job available.

Step-by-Step Calculation

Time 0-2: Only Job A is available; it starts at 0 and finishes at 2. Therefore:

• Start Time of A: 0
• Finish Time of A: 2

Time 2-4: At time 2, Jobs B (arrived at 1) and C (arrived at 2) are available. Job D arrived at 4, so it’s not yet available.

• Between B (CPU 11) and C (CPU 4), select C, the shorter job.
• Start Time of C: 2
• Finish Time of C: 6 (2 + 4)

Time 6-7: Now, at time 6, Jobs B (11), D (1), and newly arrived E (arrived at 5). Job D has the shortest CPU cycle (1), so it starts at 6 and finishes at 7.

• Start Time of D: 6
• Finish Time of D: 7

Time 7-9: Remaining jobs now are B (11), E (9), F (4), G (2). The shortest is G (2).

• Start Time of G: 7
• Finish Time of G: 9 (7 + 2)

Time 9-13: Remaining are B (11), E (9), F (4). Next shortest is F (4).

• Start Time of F: 9
• Finish Time of F: 13 (9 + 4)

Time 13-22: Remaining jobs: B (11), E (9). The shortest is E (9).

• Start Time of E: 13
• Finish Time of E: 22 (13 + 9)

Time 22-33: Remaining job is B (11).

• Start Time of B: 22
• Finish Time of B: 33 (22 + 11)

Summary of Start and Finish Times

Conclusion

Through careful application of the SJN scheduling algorithm, considering job arrivals and CPU cycles, the start and finish times for each job are determined in a sequence that prioritizes shorter jobs and respects arrival times. This method ensures efficient CPU utilization and minimizes waiting time for shorter jobs in a non-preemptive environment.

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