--- title: "Non-preemptive SJF with P4 arriving" subject: "Embedded System" chapter: 6 key: "semester_6/embedded_system:sched-sjf-nonpreemptive-6-8-2" source: https://www.bctengineer.com/q/semester_6/embedded_system/sched-sjf-nonpreemptive-6-8-2 site: BCT Engineering --- # Non-preemptive SJF with P4 arriving Embedded System · Chapter 6: RTOS - Asked in: 2070 Bhadra, 3 marks; 2075 Baisakh, 5 marks (asked 2 times) - Video: [L-2.4: Shortest Job First(SJF) Scheduling Algorithm with Example | Operating System](https://www.youtube.com/watch?v=VCIVXPoiLpU) — Gate Smashers - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/embedded_system/sched-sjf-nonpreemptive-6-8-2 - 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 times 6, 8, 2 milliseconds respectively enter the ready queue together (in the order P1, P2, P3). Process P4 with estimated execution completion time 4 milliseconds enters the ready queue after 1 millisecond. Calculate the waiting time and turn-around-time (TAT) for each process and the average waiting time and TAT in the non-preemptive shortest-job-first scheduling. ## Short answer Non-preemptive SJF selects shortest burst when CPU free. Schedule: P3(0-2), P4(2-6), P1(6-12), P2(12-20). Waiting times: P1=6ms, P2=12ms, P3=0ms, P4=1ms; average=4.75ms. Turnaround times: P1=12ms, P2=20ms, P3=2ms, P4=5ms; average=9.75ms. ## Answer **Non-preemptive SJF:** when the CPU becomes free, the ready process with the shortest burst time is selected, and it runs until it finishes. Waiting time (WT) = start time - arrival time; turnaround time (TAT) = completion time - arrival time = WT + burst time. | Process | Arrival (ms) | Burst (ms) | |---|---|---| | P1 | 0 | 6 | | P2 | 0 | 8 | | P3 | 0 | 2 | | P4 | 1 | 4 | #### Schedule - t = 0: P1, P2, P3 are ready, the shortest is P3 (2 ms), so P3 runs from 0 to 2 (it cannot be interrupted when P4 arrives at t = 1). - t = 2: P1 (6), P2 (8), P4 (4) are ready, the shortest is P4: runs 2 to 6. - t = 6: P1 (6) then P2 (8): P1 runs 6 to 12. - t = 12: P2 runs 12 to 20. ```text | P3 | P4 | P1 | P2 | 0 2 6 12 20 ``` | Process | Completion | TAT = CT - AT | WT = TAT - Burst | |---|---|---|---| | P1 | 12 | 12 - 0 = 12 | 12 - 6 = 6 | | P2 | 20 | 20 - 0 = 20 | 20 - 8 = 12 | | P3 | 2 | 2 - 0 = 2 | 2 - 2 = 0 | | P4 | 6 | 6 - 1 = 5 | 5 - 4 = 1 | Average WT = (6 + 12 + 0 + 1) / 4 = 19 / 4 = **4.75 ms** Average TAT = (12 + 20 + 2 + 5) / 4 = 39 / 4 = **9.75 ms** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At t = 1, P4 (4 ms) arrives while P3 (2 ms) is running. What happens? A. P1 starts B. P2 starts C. P3 continues to finish, because non-preemptive D. P4 pre-empts P3 Answer: C. Non-preemptive SJF cannot interrupt a running process. (Key point: Non-preemptive SJF rule) 2. What is the execution order? A. P3, P4, P1, P2 B. P3, P1, P4, P2 C. P1, P2, P3, P4 D. P3, P4, P2, P1 Answer: A. At t=2 shortest is P4 (4), then P1 (6), then P2 (8). (Key point: Order of execution) 3. What is P4's waiting time? A. 5 ms B. 1 ms C. 4 ms D. 2 ms Answer: B. P4 starts at 2, arrived at 1, so WT = 1 (TAT = 5). (Key point: Waiting time of P4) 4. What is the average WT? A. 5 ms B. 4 ms C. 9.75 ms D. 4.75 ms Answer: D. WT: P1 6, P2 12, P3 0, P4 1; 19/4 = 4.75 (average TAT 9.75). (Key point: Average waiting time) ## 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 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 - [Non-preemptive priority scheduling (10, 5, 7 ms)](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Asched-priority-nonpreemptive-10-5-7) — Embedded System, chapter 6, asked 2× - [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× - [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× - [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 ## Related in other subjects - [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 - [Preemptive priority and SRTF Gantt](https://www.bctengineer.com/question.md?key=semester_6%2Foperating_system%3Apreemptive-priority-srtf-gantt) — Operating System, chapter 2 - [FCFS, SJF, Priority, RR Gantt; max AWT](https://www.bctengineer.com/question.md?key=semester_6%2Foperating_system%3Ascheduling-gantt-fcfs-sjf-priority-rr-q2) — Operating System, chapter 2 - [Real-time scheduling; two-processor numerical](https://www.bctengineer.com/question.md?key=semester_6%2Foperating_system%3Arealtime-two-processor-fcfs-sjf-priority-rr) — Operating System, chapter 2