Perform the following scheduling policy analysis for the above set of processes like the fig. 9.5 or Table 9.5 (on the slide/book)
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- To not let any process wait longer than 500 ms, a system developer programs the Round Robin procedure with a dynamic time quantum size. With ‘n’ processes ready, the time quantum Q is set to 500 ms/n. What do you think of this scheduling strategy? (In which situations is this strategy feasible, and when is it not? Are there edge cases?) Explain your reasoning.Question no 02: Consider the following set of processes, with the length of the CPU burst given in seconds: Process Burst Priority P1 32 4 P2 24 1 P3 4 2 P4 36 2 P5 12 3 NOTE: The processes are assumed to have arrived in the order P1, P2, P3, P4, P5, all at time 0. Draw four Gantt charts that illustrate the execution of these processes using the following scheduling algorithms: FCFS, SJF, non-preemptive priority (a larger priority number implies a higher priority), and Round Robin (quantum = 8). What is the turnaround time of each process for each of the scheduling algorithms in part a? What is the waiting time of each process for each of these scheduling algorithms? Which of the algorithms results in the minimum average waiting time (over all processes)? Which of the algorithms results in the minimum average turnaround time (over all processes)?A real-time system executing four different periodic tasks Ti(p,t) is given, with p and t representing the period and the execution time of the task, respectively. Each task's deadline matches its period p. The system is executing the following set of tasks: S = {T1(10,2), T2(4,1), T3(10,3), T4(7,1)}. Is this real-time system schedulable under Rate-Monotonic (RM) scheduling? Is this real-time system schedulable under Earliest Deadline First (EDF) scheduling? Explain your answers using the concept of overall CPU utilization.
- We are dealing with 6 processes p1;p2; : : : ;p6:(a) Processes p1;p2 in p3 arrive at time 2 ms, respectively. Each of those processesneeds 5 ms of CPU time.(b) Process p4 arrives at time 10 ms and needs 3 ms of CPU time.(c) Processes p5 in p6 arrive at time 19 ms. Process p5 needs 7 ms, while process p6needs 4 ms of CPU time.Scheduling is preemptive, where there process switch takes 1 ms. For the Round-Robinalgorithms with quantum 2 ms illustrate the execution of those processes on a timeline.Then determine the average turnaround time.Consider the following set of processes, with the length of the CPU burst given in seconds: Process Burst time Arrival time P1 10 1 P2 04 2 P3 05 3 P4 03 4 Draw four Gantt charts that illustrate the execution of these processes using the following scheduling algorithms: FCFS, SJF preemptive and Round Robin (quantum = 3). What is the turnaround time of each process for each of the scheduling algorithms in part a? What is the waiting time of each process for each of these scheduling algorithms? Which of the algorithms results in the minimum average waiting time (over all processes)? Which of the algorithms results in the minimum average turnaround time (over all processes)?2. Consider the following processes with their associated CPU burst times. The processes have entered the system in the order that they are presented in the table. Process : BurstTime P1 : 7 P2 : 1 P3 : 13 P4 : 4 P5 : 5 For each of the following scheduling algorithms, draw a Gantt chart that illustrates the order of execution for each of these processes. Also, compute the average waiting time for each algorithm. a. First Come First Served b. Shortest Job First c. Round Robin (with time quantum of 3)
- Question no 02: Consider the following set of processes, with the length of the CPU burst given in seconds: Process Burst Priority P1 32 4 P2 24 1 P3 4 2 P4 36 2 P5 12 3 NOTE: The processes are assumed to have arrived in the order P1, P2, P3, P4, P5, all at time 0. a. Draw four Gantt charts that illustrate the execution of these processes using the following scheduling algorithms: FCFS, SJF, non-preemptive priority (a larger priority number implies a higher priority), and Round Robin (quantum = 8). b. What is the turnaround time of each process for each of the scheduling algorithms in part a? c. What is the waiting time of each process for each of these scheduling algorithms? d. Which of the algorithms results in the minimum average waiting time (over all processes)? e. Which of the algorithms results in the minimum average turnaround time (over all processes)?Consider the following data [Figure].Suppose round robin scheduling is applied on the given data with time quantum=Floor (XX % 3) units. It is also given that scheduling of a process or context switch will take 1 unit of time. What is the completion time and turnaround time of process P4?Where XX is the last two digits of your roll number. XX=18Apply the following scheduling algorithms to the list process. 1. Shortest Remaining Time 2. Round Robin:Quantom =3 For each process scheduling algorithm, provide the following: a. Finish time Turnaround time Waiting time b. A detailed gantt chart c. The average turnaround time d. The average waiting time
- Consider the following set of processes, with the length of the CPU burst given in milliseconds: Process Burst Time Priority Arrival Time P1 20 2 0 P2 10 1 1 P3 80 4 2 P4 40 2 3 P5 50 3 4 Draw the GANTT Chart for the schedule of the processes, the waiting time for each process, the average waiting time, the turnaround time for each process and average turnaround time using the following scheduling algorithm. c.…Consider the following set of processes, with the length of the CPU burst given in milliseconds: Process Burst Time Priority Arrival Time P1 20 2 0 P2 10 1 1 P3 80 4 2 P4 40 2 3 P5 50 3 4 Draw the GANTT Chart for the schedule of the processes, the waiting time for each process, the average waiting time, the turnaround time for each process and average turnaround time using the following scheduling algorithm. b.…Draw Gantt charts that illustrate the execution of these processes using the following scheduling algorithms and calculate the average waiting time. Round-Robin (RR) with Time quantum = 4
