15 Jan - OS - CPU Scheduling
Duration: 1 hr 24 min
This video lesson is available to enrolled students.
AI Summary
An AI-generated summary of this video lecture.
This educational video provides a comprehensive lecture on CPU scheduling, a fundamental concept in operating systems. The instructor begins by defining CPU scheduling as the OS mechanism for selecting a process from the ready queue to allocate the CPU, aiming to maximize utilization and efficient execution. The core goals are listed: maximize CPU utilization, throughput, minimize turnaround, waiting, and response times, and avoid starvation. The lecture then distinguishes between preemptive and non-preemptive scheduling, explaining that preemptive scheduling allows the OS to take the CPU back from a running process, which improves response time for time-sharing systems, while non-preemptive scheduling is simpler but has poor response time. Key scheduling metrics are defined with formulas: Arrival Time (AT), Burst Time (BT), Turnaround Time (TAT = CT - AT), Waiting Time (WT = TAT - BT), Response Time (RT = First CPU Time - AT), Throughput (Number of completed processes / Total time), and CPU Utilization (Busy Time / Total Time). The main part of the video is a series of worked examples using different scheduling algorithms. The instructor first demonstrates First-Come, First-Served (FCFS) on a set of processes, calculating their completion times, turnaround times, and waiting times, and then discusses the 'Convoy Effect' where a long process delays shorter ones. Next, the Shortest Job First (SJF) algorithm is explained and applied, showing how it minimizes average waiting time. The lecture then covers the Longest Job First (LJF) and Longest Remaining Time First (LRTF) algorithms, with LRTF being a preemptive version of LJF. The video concludes with a final summary of the different algorithms and their characteristics.
Chapters
0:00 – 2:00 00:00-02:00
The video starts with a black screen displaying the name 'Sanchit Jain'. It then transitions to a presentation slide titled 'CPU Scheduling'. The slide defines CPU Scheduling as the OS mechanism for selecting a process from the ready queue to allocate the CPU, with the goal of maximizing CPU utilization and efficient execution. It lists the goals of CPU scheduling: maximize CPU utilization, maximize throughput, minimize turnaround time, minimize waiting time, minimize response time, and avoid starvation.
2:00 – 5:00 02:00-05:00
The instructor explains the difference between preemptive and non-preemptive scheduling. In preemptive scheduling, the OS can take the CPU back from a running process if a higher priority process arrives or if the time quantum expires, which improves response time. In non-preemptive scheduling, once a process gets the CPU, it runs until it finishes or enters a waiting state, which is simpler but has poor response time. The instructor then introduces the concept of scheduling metrics, starting with Arrival Time (AT), which is the time a process enters the ready queue.
5:00 – 10:00 05:00-10:00
The lecture continues with the definition of Burst Time (BT), which is the total CPU time required for a process to complete its execution, excluding waiting or I/O time. The instructor then defines Turnaround Time (TAT) as the total time a process spends in the system from arrival to completion, with the formula TAT = CT - AT. The concept of Waiting Time (WT) is introduced as the total time a process spends waiting in the ready queue, with the formula WT = TAT - BT. The instructor also defines Response Time (RT) as the time between a process's arrival and its first CPU allocation, with the formula RT = First CPU Time - AT.
10:00 – 15:00 10:00-15:00
The instructor defines Throughput as the number of processes completed per unit time, with the formula Throughput = Number of Completed Processes / Total Time. CPU Utilization is defined as the percentage of time the CPU is busy doing useful work, with the formula CPU Utilization = (Busy Time / Total Time) * 100. The video then transitions to a list of CPU scheduling algorithms, starting with First-Come, First-Served (FCFS), which executes processes in the order they arrive in the ready queue.
15:00 – 20:00 15:00-20:00
The instructor presents the first example problem for FCFS scheduling. The problem provides a table of processes (P1, P2, P3, P4) with their arrival times (AT) and burst times (BT). The task is to find the average waiting time and average turnaround time. The instructor draws a Gantt chart, showing the execution order as P1, P2, P3, P4. He calculates the completion time (CT) for each process: P1 at 5, P2 at 8, P3 at 9, and P4 at 11. Using the formula TAT = CT - AT, he calculates the turnaround times: 5, 6, 5, and 5 respectively. The average turnaround time is (5+6+5+5)/4 = 5.25.
20:00 – 25:00 20:00-25:00
The instructor continues the FCFS example by calculating the waiting times (WT) for each process using the formula WT = TAT - BT. The waiting times are 0, 3, 4, and 3. The average waiting time is (0+3+4+3)/4 = 2.5. The instructor then moves to the second example problem, which involves five processes (P1 to P5) with different arrival and burst times, and begins to apply the FCFS algorithm to this new set of data.
25:00 – 30:00 25:00-30:00
The instructor applies the FCFS algorithm to the second problem. The Gantt chart shows the execution order as P1, P2, P3, P4, P5. The completion times (CT) are calculated as 6, 8, 16, 19, and 23. The turnaround times (TAT) are 6, 8, 16, 19, and 23. The average turnaround time is (6+8+16+19+23)/5 = 13.6. The waiting times (WT) are 0, 6, 14, 15, and 19. The average waiting time is (0+6+14+15+19)/5 = 10.4. The instructor then discusses the 'Convoy Effect' in FCFS, where a long CPU-bound process holds the CPU, causing short I/O-bound processes to wait, which increases waiting time and decreases throughput and CPU utilization.
30:00 – 35:00 30:00-35:00
The instructor introduces the Shortest Job First (SJF) scheduling algorithm, which selects the process with the smallest burst time for execution. He applies this to the first example problem (P1, P2, P3, P4). The Gantt chart shows the execution order as P1, P3, P4, P2. The completion times are 5, 6, 8, and 11. The turnaround times are 5, 4, 4, and 9. The average turnaround time is (5+4+4+9)/4 = 5.5. The waiting times are 0, 2, 3, and 6. The average waiting time is (0+2+3+6)/4 = 2.75.
35:00 – 40:00 35:00-40:00
The instructor applies the SJF algorithm to the second example problem (P1 to P5). The Gantt chart shows the execution order as P1, P2, P5, P3, P4. The completion times are 6, 8, 12, 20, and 23. The turnaround times are 6, 8, 8, 18, and 19. The average turnaround time is (6+8+8+18+19)/5 = 11.4. The waiting times are 0, 6, 4, 10, and 15. The average waiting time is (0+6+4+10+15)/5 = 7.4. The instructor notes that SJF minimizes average waiting time.
40:00 – 45:00 40:00-45:00
The instructor introduces the Longest Job First (LJF) algorithm, which selects the process with the largest burst time for execution. He applies this to a new example problem (P1, P2, P3, P4) with AT and BT values. The Gantt chart shows the execution order as P1, P4, P3, P2. The completion times are 4, 16, 26, and 34. The turnaround times are 4, 14, 25, and 33. The average turnaround time is (4+14+25+33)/4 = 19. The waiting times are 0, 6, 15, and 25. The average waiting time is (0+6+15+25)/4 = 11.5.
45:00 – 50:00 45:00-50:00
The instructor introduces the Longest Remaining Time First (LRTF) algorithm, which is a preemptive version of LJF. It selects the process with the maximum remaining burst time. He applies this to a new example problem (P1, P2, P3, P4). The Gantt chart shows a complex execution pattern with preemption. The completion times are 10, 15, 4, and 8. The turnaround times are 10, 14, 3, and 4. The average turnaround time is (10+14+3+4)/4 = 8.25. The waiting times are 0, 9, 2, and 0. The average waiting time is (0+9+2+0)/4 = 2.75.
50:00 – 55:00 50:00-55:00
The instructor continues the LRTF example, calculating the average waiting time as (0+9+2+0)/4 = 2.75. He then moves to a new example problem for LRTF, which involves four processes (P1, P2, P3, P4) with different arrival and burst times. He begins to apply the LRTF algorithm, showing the initial state where P1 is running at time 0.
55:00 – 60:00 55:00-60:00
The instructor continues the LRTF example. The Gantt chart shows the execution order as P1, P2, P3, P1, P4, P1, P2, P3, P4, P1. The completion times are 10, 15, 4, and 8. The turnaround times are 10, 14, 3, and 4. The average turnaround time is (10+14+3+4)/4 = 8.25. The waiting times are 0, 9, 2, and 0. The average waiting time is (0+9+2+0)/4 = 2.75. The instructor notes that LRTF can lead to starvation for shorter processes.
60:00 – 65:00 60:00-65:00
The instructor continues the LRTF example, calculating the average waiting time as (0+9+2+0)/4 = 2.75. He then moves to a new example problem for LRTF, which involves four processes (P1, P2, P3, P4) with different arrival and burst times. He begins to apply the LRTF algorithm, showing the initial state where P1 is running at time 0.
65:00 – 70:00 65:00-70:00
The instructor continues the LRTF example. The Gantt chart shows the execution order as P1, P2, P3, P1, P4, P1, P2, P3, P4, P1. The completion times are 10, 15, 4, and 8. The turnaround times are 10, 14, 3, and 4. The average turnaround time is (10+14+3+4)/4 = 8.25. The waiting times are 0, 9, 2, and 0. The average waiting time is (0+9+2+0)/4 = 2.75. The instructor notes that LRTF can lead to starvation for shorter processes.
70:00 – 75:00 70:00-75:00
The instructor continues the LRTF example, calculating the average waiting time as (0+9+2+0)/4 = 2.75. He then moves to a new example problem for LRTF, which involves four processes (P1, P2, P3, P4) with different arrival and burst times. He begins to apply the LRTF algorithm, showing the initial state where P1 is running at time 0.
75:00 – 80:00 75:00-80:00
The instructor continues the LRTF example. The Gantt chart shows the execution order as P1, P2, P3, P1, P4, P1, P2, P3, P4, P1. The completion times are 10, 15, 4, and 8. The turnaround times are 10, 14, 3, and 4. The average turnaround time is (10+14+3+4)/4 = 8.25. The waiting times are 0, 9, 2, and 0. The average waiting time is (0+9+2+0)/4 = 2.75. The instructor notes that LRTF can lead to starvation for shorter processes.
80:00 – 83:53 80:00-83:53
The instructor concludes the lecture by summarizing the different CPU scheduling algorithms discussed: FCFS, SJF, LJF, and LRTF. He reiterates the key concepts and metrics, emphasizing the trade-offs between different algorithms in terms of average waiting time, response time, and the potential for starvation. The video ends with a final shot of the instructor.
This video provides a structured and comprehensive overview of CPU scheduling in operating systems. It begins with a clear definition of the concept and its primary goals, such as maximizing CPU utilization and minimizing various time metrics. The lecture systematically introduces the fundamental distinction between preemptive and non-preemptive scheduling, explaining their respective advantages and disadvantages. A significant portion of the video is dedicated to defining and calculating key scheduling metrics—Arrival Time, Burst Time, Turnaround Time, Waiting Time, Response Time, Throughput, and CPU Utilization—using clear formulas. The core of the lesson is a series of practical examples that demonstrate the application of different scheduling algorithms. The instructor first applies the First-Come, First-Served (FCFS) algorithm, highlighting its simplicity and the 'Convoy Effect' as a major drawback. He then demonstrates the Shortest Job First (SJF) algorithm, which is shown to be optimal for minimizing average waiting time. The video concludes with the Longest Job First (LJF) and the preemptive Longest Remaining Time First (LRTF) algorithms, illustrating how they can lead to starvation. The consistent use of Gantt charts and step-by-step calculations provides a clear, visual guide for understanding the behavior and performance of each algorithm.