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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, 9

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, 9

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Solution

In non-preemptive priority scheduling, the CPU is allocated to the process with the highest priority. If two processes have the same priority, then the CPU is allocated to the process with the earliest arrival time.

Here, the processes have arrived in the order P1, P2, P3, P4.

Let's calculate the turnaround time for each process:

  1. P1: The burst time is 2ms. Since it's the first process to arrive, it gets the CPU first. So, the turnaround time for P1 is 2ms (completion time - arrival time = 2 - 0 = 2ms).

  2. P2: The burst time is 1ms. It has the highest priority among the remaining processes. So, it gets the CPU next. The CPU has been busy for 2ms executing P1, so P2 starts at 2ms and finishes at 3ms. Therefore, the turnaround time for P2 is 3ms (3 - 0 = 3ms).

  3. P4: The burst time is 4ms. It has the next highest priority. So, it gets the CPU next. The CPU has been busy for 3ms executing P1 and P2, so P4 starts at 3ms and finishes at 7ms. Therefore, the turnaround time for P4 is 7ms (7 - 0 = 7ms).

  4. P3: The burst time is 3ms. It has the lowest priority. So, it gets the CPU last. The CPU has been busy for 7ms executing P1,

This problem has been solved

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