--- title: "Non-preemptive priority scheduling (10, 5, 7 ms)" subject: "Embedded System" chapter: 6 key: "semester_6/embedded_system:sched-priority-nonpreemptive-10-5-7" source: https://www.bctengineer.com/q/semester_6/embedded_system/sched-priority-nonpreemptive-10-5-7 site: BCT Engineering --- # Non-preemptive priority scheduling (10, 5, 7 ms) Embedded System · Chapter 6: RTOS - Asked in: 2078 Poush, 6 marks - Video: [Non Preemptive Priority Scheduling Algorithm| Operating System | OS](https://www.youtube.com/watch?v=WPQP4n3GIkw) — Sudhakar Atchala - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/embedded_system/sched-priority-nonpreemptive-10-5-7 - Whole chapter: https://www.bctengineer.com/md/semester_6/embedded_system/chapter-6 ## Question Three processes with process IDs P1, P2, P3 with estimated completion time 10, 5, 7 ms and priorities 0, 2, 3 (0-highest, 3-lowest) respectively enter the ready queue together in order P1, P2, P3 (assume only P1 is present in the Ready queue when the scheduler picks it up and P2 & P3 enter ready queue after that). Now a process P4 with estimated completion time 6 ms and priority 1 enters the ready queue after 6 ms of execution of P1. Calculate waiting time and TAT for each process and average waiting time and average TAT. Assume there is no I/O waiting for the processes and priority-based non-preemptive scheduling. ## Short answer Non-preemptive priority gives order P1, P4, P2, P3. Waiting times: P1=0, P2=16, P3=21, P4=4 ms. Average waiting time = 10.25 ms, average turnaround time = 17.25 ms. ## Answer In non-preemptive priority scheduling a running process is never interrupted; when it ends, the ready process with the highest priority (lowest number) is chosen. P1 is alone when picked, so it starts at 0 and P2, P3 join the queue just after. P4 arrives at 6 ms (6 ms into P1). #### Gantt chart ```text | P1 | P4 | P2 | P3 | 0 10 16 21 28 ``` #### Trace - 0-10: P1 runs to completion (no preemption, although P4 arrives at 6). - At 10 the ready queue holds P2 (priority 2), P3 (priority 3), P4 (priority 1). P4 is chosen: 10-16. - 16-21: P2 (priority 2). 21-28: P3 (priority 3). #### Calculation TAT = completion time - arrival time; WT = TAT - burst time. | Process | AT | BT | CT | TAT = CT - AT | WT = TAT - BT | |---|---|---|---|---|---| | P1 | 0 | 10 | 10 | 10 | 0 | | P2 | 0 | 5 | 21 | 21 | 16 | | P3 | 0 | 7 | 28 | 28 | 21 | | P4 | 6 | 6 | 16 | 10 | 4 | AWT = (0 + 16 + 21 + 4) / 4 = 41/4 = 10.25 ms ATAT = (10 + 21 + 28 + 10) / 4 = 69/4 = 17.25 ms **Answer: AWT = 10.25 ms, ATAT = 17.25 ms.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does P4 not interrupt P1 at 6 ms? A. P4 has lowest priority B. P4 arrived late C. Non-preemptive scheduling never interrupts a running process D. P1 has equal priority Answer: C. P1 runs to completion at 10. (Key point: Non-preemptive) 2. Who is chosen at t = 10? A. P2 (priority 2) B. P4 (priority 1) C. P3 (priority 3) D. The shortest burst Answer: B. Among P2, P3, P4 the lowest priority number wins. (Key point: Next at 10) 3. What is P4's waiting time? A. 4 ms B. 10 ms C. 6 ms D. 0 ms Answer: A. TAT = 16 - 6 = 10, WT = 10 - 6 = 4. (Key point: P4 waiting) 4. What are AWT and ATAT? A. 17.25 ms and 10.25 ms B. 11.25 ms and 18.25 ms C. 41 ms and 69 ms D. 10.25 ms and 17.25 ms Answer: D. AWT = 41/4; ATAT = 69/4. (Key point: Averages) ## Close variants set in other exams - [Priority scheduling (7, 8, 5 ms) with P4](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Asched-priority-7-8-5-p4) — Embedded System, chapter 6, asked 2× ## Related in Embedded System - [Preemptive priority scheduling with P4 (6 ms)](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Asched-priority-10-5-7-p4) — Embedded System, chapter 6, asked 2× - [Preemptive priority scheduling (4, 10, 5 ms)](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Asched-priority-4-10-5-p4) — Embedded System, chapter 6, asked 2× - [Preemptive priority scheduling (4, 6, 5 ms)](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Asched-priority-4-6-5) — Embedded System, chapter 6 - [Preemptive SJF/SRT scheduling (10, 5, 7 ms)](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Asched-srt-10-5-7) — Embedded System, chapter 6 - [Priority and SJF scheduling with P4 and P5](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Asched-priority-and-sjf-5-process) — Embedded System, chapter 6, asked 2× - [Non-preemptive SJF with P4 arriving](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Asched-sjf-nonpreemptive-6-8-2) — Embedded System, chapter 6, asked 2× ## Related in other subjects - [Preemptive vs non-preemptive; real-time scheduling](https://www.bctengineer.com/question.md?key=semester_6%2Foperating_system%3Apreemptive-nonpreemptive-realtime) — Operating System, chapter 2 - [Preemptive priority and SRTF Gantt](https://www.bctengineer.com/question.md?key=semester_6%2Foperating_system%3Apreemptive-priority-srtf-gantt) — Operating System, chapter 2 - [RR, preemptive priority, preemptive SJF](https://www.bctengineer.com/question.md?key=semester_6%2Foperating_system%3Aprocess-rr-priority-preemptive-sjf) — Operating System, chapter 2