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Consider the following set of processes, with the length of the CPU burst given in milliseconds:Process Burst Time PriorityP1 2 2P2 1 1P3 3 4P4 4 2The processes are assumed to have arrived in the order P1, P2, P3, P4, P5,all at time 0. What is the turnaround time of each process for the nonpreemptive Priority scheduling algorithms (a larger priority number implies a higher priority)?A.5, 10, 3, 9B.2, 3, 6, 10C.3, 1, 6, 10D.2, 3, 8, 10

Question

Consider the following set of processes, with the length of the CPU burst given in milliseconds:Process Burst Time PriorityP1 2 2P2 1 1P3 3 4P4 4 2The processes are assumed to have arrived in the order P1, P2, P3, P4, P5,all at time 0. What is the turnaround time of each process for the nonpreemptive Priority scheduling algorithms (a larger priority number implies a higher priority)?A.5, 10, 3, 9B.2, 3, 6, 10C.3, 1, 6, 10D.2, 3, 8, 10

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Solution

In non-preemptive priority scheduling, the process with the highest priority (largest number) is executed first. If two processes have the same priority, they are executed in their arrival order.

Here is the step-by-step calculation:

  1. At time 0, P1, P2, P3, and P4 arrive. P3 has the highest priority, so it is executed first. It finishes at time 3.

  2. Now, P1, P2, and P4 are waiting. P1 and P4 have the same priority, but P1 arrived first, so it is executed next. It finishes at time 5 (3 from P3 + 2 from P1).

  3. Now, P2 and P4 are waiting. P2 has the highest priority, so it is executed next. It finishes at time 6 (5 from P1 + 1 from P2).

  4. Finally, P4 is executed and it finishes at time 10 (6 from P2 + 4 from P4).

So, the turnaround time for each process is:

  • P1: 5 (it finished at time 5)
  • P2: 6 (it finished at time 6)
  • P3: 3 (it finished at time 3)
  • P4: 10 (it finished at time 10)

Therefore, the correct answer is C.3, 1, 6, 10.

This problem has been solved

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