Practice Question
Duration: 8 min
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The lecture introduces Priority Scheduling, a CPU scheduling algorithm where processes are assigned a priority level. The CPU is allocated to the process with the highest priority, with ties broken using First-Come-First-Served (FCFS) order. The instructor distinguishes between non-preemptive and preemptive versions, explaining that in non-preemptive scheduling, a running process cannot be interrupted by a newly arrived higher-priority process. The session transitions into a numerical problem involving four processes ($P_0, P_1, P_2, P_3$) with specific Arrival Times (AT) and Burst Times (BT). The instructor solves this using non-preemptive logic, constructing a Gantt chart to visualize the execution sequence and subsequently calculating Completion Time (CT), Turnaround Time (TAT), and Waiting Time (WT).
Chapters
0:00 – 2:00 00:00-02:00
The instructor begins by defining Priority Scheduling using the slide text. He explains that at any instance, the CPU is allocated to the available process with the highest priority, noting that priority can be represented by either higher or lower numbers. He clarifies that tie-breaking is handled via FCFS order, independent of seniority or burst time. He explicitly reads the definition of non-preemptive priority scheduling: once a decision is made and a process is scheduled, it cannot be preempted even if a new process with higher priority enters the system.
2:00 – 5:00 02:00-05:00
The focus shifts to the provided table containing processes $P_0$ through $P_3$. The instructor analyzes the priority values (4, 1, 3, 2) and identifies '1' as the highest priority. He begins constructing a Gantt chart at the bottom of the slide. He determines that $P_0$ starts at time 0 with a burst time of 50. Although $P_1$ arrives at time 20 with a higher priority (1), the non-preemptive rule implies $P_0$ continues until completion at time 50. He then draws the subsequent bars for $P_1$ (50-70), $P_3$ (70-110), and $P_2$ (110-210), establishing the execution order based on available priorities at each completion point.
5:00 – 8:20 05:00-08:20
The instructor completes the table by filling in the Completion Time (CT) column based on the Gantt chart end times: 50, 70, 110, and 210. He then calculates Turnaround Time (TAT) using the formula $TAT = CT - AT$ and Waiting Time (WT) using $WT = TAT - BT$. He writes these values into the respective columns for each process. For example, $P_0$ has a TAT of 50 and WT of 0. $P_1$ has a TAT of 50 and WT of 30. $P_3$ has a TAT of 50 and WT of 10. $P_2$ has a TAT of 170 and WT of 70. Finally, he computes the average TAT (80) and average WT (27.5) by summing the individual values and dividing by the number of processes (4), providing the final statistical summary of the scheduling algorithm's performance.
The lesson effectively bridges theoretical definitions with practical application. It starts by establishing the rules of Priority Scheduling, specifically highlighting the non-preemptive constraint where a running process is immune to interruption. The instructor then applies these rules to a concrete dataset, demonstrating how to determine the execution sequence ($P_0 ightarrow P_1 ightarrow P_3 ightarrow P_2$) despite arrivals of higher-priority processes. The final phase focuses on the arithmetic of performance metrics, reinforcing the formulas for TAT and WT to quantify the efficiency of the schedule.