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Three processes arrive at time zero with CPU bursts of 16, 20 and 10 milliseconds. If the scheduler has prior knowledge about the length of the CPU bursts, the minimum achievable average waiting time for these three processes in a non-preemptive scheduler (rounded to nearest integer) is _____________ milliseconds.
- a)12
- b)36
- c)46
- d)10
Correct answer is option 'A'. Can you explain this answer?
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Three processes arrive at time zero with CPU bursts of 16, 20 and 10 m...
Use SRTF, for the minimum achievable average waiting time : Gantt chart is
Since, TAT = CT - AT and WT = TAT - BT, so WT = CT - AT - BT = CT - (AT+BT) Therefore, Avg, WT = {(26-0-16) + (46-0-20) + (10-0-10)} / 3 = {10 + 26 + 0} / 3 = 36 / 3 = 12
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Three processes arrive at time zero with CPU bursts of 16, 20 and 10 m...
Given Information:
- Three processes arrive at time zero.
- The CPU bursts for the three processes are 16, 20, and 10 milliseconds.
Objective:
To find the minimum achievable average waiting time for these three processes in a non-preemptive scheduler.
Approach:
In a non-preemptive scheduler, a process is not interrupted once it starts its execution. The scheduler selects the next process to run based on its arrival time and CPU burst length. To minimize the average waiting time, we need to schedule the processes in such a way that the shorter burst time processes are executed first.
Steps:
1. Sort the processes based on their CPU burst lengths in ascending order.
2. Assign a unique identifier to each process (P1, P2, P3) to track their order.
3. Calculate the waiting time for each process using the formula (waiting time = cumulative waiting time of previous processes).
4. Calculate the average waiting time by summing up the waiting times of all processes and dividing it by the number of processes.
Calculations:
1. Sort the processes based on their CPU burst lengths:
- Process P1: Burst time = 10 ms
- Process P2: Burst time = 16 ms
- Process P3: Burst time = 20 ms
2. Assign a unique identifier to each process:
- P1 (10 ms), P2 (16 ms), P3 (20 ms)
3. Calculate the waiting time for each process:
- Waiting time for P1 = 0 ms (No previous processes)
- Waiting time for P2 = Burst time of P1 = 10 ms
- Waiting time for P3 = Burst time of P1 + Burst time of P2 = 26 ms
4. Calculate the average waiting time:
- Average waiting time = (Waiting time for P1 + Waiting time for P2 + Waiting time for P3) / Number of processes
- Average waiting time = (0 + 10 + 26) / 3
- Average waiting time = 36 / 3
- Average waiting time = 12 ms (rounded to nearest integer)
Conclusion:
The minimum achievable average waiting time for the three processes in a non-preemptive scheduler is 12 milliseconds. Hence, option A is the correct answer.
- Three processes arrive at time zero.
- The CPU bursts for the three processes are 16, 20, and 10 milliseconds.
Objective:
To find the minimum achievable average waiting time for these three processes in a non-preemptive scheduler.
Approach:
In a non-preemptive scheduler, a process is not interrupted once it starts its execution. The scheduler selects the next process to run based on its arrival time and CPU burst length. To minimize the average waiting time, we need to schedule the processes in such a way that the shorter burst time processes are executed first.
Steps:
1. Sort the processes based on their CPU burst lengths in ascending order.
2. Assign a unique identifier to each process (P1, P2, P3) to track their order.
3. Calculate the waiting time for each process using the formula (waiting time = cumulative waiting time of previous processes).
4. Calculate the average waiting time by summing up the waiting times of all processes and dividing it by the number of processes.
Calculations:
1. Sort the processes based on their CPU burst lengths:
- Process P1: Burst time = 10 ms
- Process P2: Burst time = 16 ms
- Process P3: Burst time = 20 ms
2. Assign a unique identifier to each process:
- P1 (10 ms), P2 (16 ms), P3 (20 ms)
3. Calculate the waiting time for each process:
- Waiting time for P1 = 0 ms (No previous processes)
- Waiting time for P2 = Burst time of P1 = 10 ms
- Waiting time for P3 = Burst time of P1 + Burst time of P2 = 26 ms
4. Calculate the average waiting time:
- Average waiting time = (Waiting time for P1 + Waiting time for P2 + Waiting time for P3) / Number of processes
- Average waiting time = (0 + 10 + 26) / 3
- Average waiting time = 36 / 3
- Average waiting time = 12 ms (rounded to nearest integer)
Conclusion:
The minimum achievable average waiting time for the three processes in a non-preemptive scheduler is 12 milliseconds. Hence, option A is the correct answer.
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Three processes arrive at time zero with CPU bursts of 16, 20 and 10 milliseconds. If the scheduler has prior knowledge about the length of the CPU bursts, the minimum achievable average waiting time for these three processes in a non-preemptive scheduler (rounded to nearest integer) is _____________ milliseconds.a)12b)36c)46d)10Correct answer is option 'A'. Can you explain this answer?
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