Given the following processes with their times and priority (lower number indicates higher priority) Which of the following is a value of average waiting time using priority scheduling algorithm?Process Burst Time Priority P1 10 3 P2 1 1 P3 2 4 P4 1 5 P5 5 2
8.2
In priority scheduling the CPU is given to the ready process with the highest priority (here the lowest number). With no arrival times given, all processes are taken to arrive at time 0, so the schedule is non-preemptive and waiting time = start time.
Step 1 — Order by priority: P2 (1), P5 (2), P1 (3), P3 (4), P4 (5).
| Process | Start | Finish | Waiting |
|---|---|---|---|
| P2 | 0 | 1 | 0 |
| P5 | 1 | 6 | 1 |
| P1 | 6 | 16 | 6 |
| P3 | 16 | 18 | 16 |
| P4 | 18 | 19 | 18 |
Step 2 — Total waiting time = 0 + 1 + 6 + 16 + 18 = 41.
Step 3 — Average waiting time = \(\frac{41}{5} = 8.2\).
The values 9.2, 6.8 and 7.6 do not arise from this priority order (for example, FCFS order P1–P5 gives 9.6 and shortest-job-first gives 3.2).
8.2 option is correct.
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Important Questions from CPU Scheduling
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- $M_2$ uses priority of execution for the processes as, $P_2 > P_3 > P_4 > P_1$, i.e., $P_2$ and $P_1$ have highest and lowest priorities, respectively.
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Process Burst Time Arrival Time P1 09 0 P2 30 0 P3 04 0 P4 08 2 P5 11 6
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