Three batch jobs, A through E, arrive at a computer center at almost the same time. They have estimated running times of 10, 6, 2, 4 and 8 minutes. Their priorities are 3, 5, 2, 1 and 4 respectively, with 5 being the highest priority.
(a) For each of the following scheduling algorithms, determine the mean turnaround time. Ignore any process switching overhead.
1. Round Robin
2. Priority scheduling
3. First come, First served (run in order 10, 6, 2, 4, 8)
4. Shortest job first
For (a), assume that the system is multiprogrammed, and that each job gets its fair share of the CPU. For (b) through (d) assume that only one job runs at a time, until it finishes. All jobs are completely CPU bound.

1. The average turn around time is 21.2 minute using round robin scheduling algorithm with a quantum time of 1 minute.

2. The average turn around time is 20 minute using Priority scheduling algorithm.

3. The average turn around time is 19.2 minute using First Come First Serve scheduling algorithm.

4. The average turn around time is 14 minute using Shortest Job First scheduling algorithm.

Explanation:

Gantt chart was used to solve the scheduling problem.

Image showing the Gantt chart and working solution is attached.

Round Robin scheduling algorithm is a type of scheduling that uses a time slice for each process/job. Once the time slice is complete, the process leave to join the queue again if it has not finished execution.

Priority scheduling algorithm is a type of scheduling algorithm that allows a process/job to execute/complete based on the order of priority. In this case, the higher the number, the higher the priority. With 5 having the highest priority.

First Come First Serve scheduling algorithm is a type of scheduling algorithm that allows execution of a process/job based on first come first server i.e based on order of arrival time. In this case the order is ABCDE.

Shortest Job First scheduling algorithm is a type of scheduling algorithm that allocate the CPU to the process/job having the shortest completion time first. In this case, Job C has the shortest completion time.

answer; ///400 ibs;