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- Processes P1, P2, and P2 are queued to run. The arrival time of P1 is t = 0, the arrival time of P2 is t = 1, and the arrival time of P3 is t = 2. According to the first come first serve (FCFS) algorithm, processes are run by the operating system. P1 operates 7 time units, P2 4 time units, P3 operates 1 time unit. So what is the average waiting time (in time units) for this system? 0 5 3 7Consider a job mix in which there is one CPU-intensive job, A, that has a single infinitely long CPU burst, and 3 I/O-intensive jobs, B, C, and D. The scheduler uses Round Robin scheduling with a time quantum of 3 ms. Here is an observed execution, where initially A is running and B, C, and D are in the ready queue in this order: AAAoBBoCoDoAAAoBBoAAAoDoAAAoBBoCoDoAAAoBBo Answer the following questions based on the above. Explain and give a specify a part of the execution. a) Could any of B’s I/O-burst time be 6 ms? Explain. b) Could any of B’s I/O-burst time be 4 ms? Explain.An interactive system using round-robin scheduling and swapping tries to giveguaranteed response to trivial requests as follows: After completing a round robin cycleamong all ready processes, the system determines the time slice to allocate to eachready process for the next cycle by dividing a maximum response time by the number ofprocesses requiring service. Is this a reasonable policy?
- a single process, with execution time of BT time units, is detected by the system that uses Round Robin algorithm with time quantum of QT time units. Every time the time quantum elapses; a contect switch is always performed with CS time units. Assume that the process come at time equal to zero(0) If BT is 25, QT is 10 and CS is 2, what is the Cpu utilization of the system? If BT is 20, QT is 6 and CS is 1 what is the CPU utilization of the system? What is the CPU utilization if BT is equal QT and QT is equal to CS? What is the system's throughput if BT is less than QT? What is the system's turnaround time (TAT) if BT is less than QT?An operating system uses Shortest Remaining Time first (SRT) process scheduling algorithm. Consider the arrival times and execution times for the following processes: Process Execution time Arrival time P1 20 P2 25 15 P3 10 30 P4 15 45 What is the total waiting time for process P2?Consider a Multilevel Queue scheduler with two queues, numbered 1 and 2. Queue 1 has higher priority over queue 2. Both queues use RR scheduling. Time Quantum: TQ1 = 3, TQ2 = 4 Process Burst Time Arrival Priority Queue PI 12 1 4 P2 P3 8 5 1 P4 5 12 Ps 10 18 2 Find average turnaround time, average waiting time.
- Let the processes P1, P2, P3, and P4 be given. They arrive in the system at the same time in this order. The processes have the following service times (in time units): Process Service time (CPU burst) P1 53 P2 17 P3 68 P4 24 2) For the scheduling method "Round Robin" (RR), specify the average execution time (average turnaround time) and the number of context switches. The time quantum q is set to 20 time units. You must show your calculations. You will not have to draw Gantt charts. Note: For RR, processes execute in order of their ID (i.e., P1-P2-P3- P4).true/false Consider a group of CPU-time sharing processes P1 , P2, ... Pn with CPU burst times of 1,2,...,n units respectively arrive at the same time. The average waiting time of these processes is always less than n2/6 (regardless of scheduling algorithm)In a timesharing OS we have the following cpu timeline for two tasks X and Y. The timeslice is 1s.Both tasks are available in the system at the same time t=9:00:00.000 and order of arrival is the obviousX followed by Y. (The decimals reflect milliseconds if they showup in an indicated time reference.) There are no other processes (tasks) in the system other than X,Y. 1234567890 XYXYX--YXY The time line 1 indicates that at t=9:00:00s task X starts its execution and when t=9:00:01s is reached task Y takes over. The time line 1 indicates the 'first second' and time line 0 indicates the 'tenth second' above.Task Y completes its execution at t=9:00:10s, the completion of the tenth second since X started its execution. Task X has completed its execution earlier. (a) What is the total number of context switches starting from prior to t=9:00:00s (e.g. t=8:59:59.999) through thecompletion of $Y$? answer is 16s (b) What is the turnaround time for Y? answer is 10s (c) What is…
- Consider three processes, all arriving at time zero, with total execution time of 10, 20 and 30 units, respectively. Each process spends the first 20% of execution time doing I/O, the next 70% of time doing computation, and the last 10% of time doing I/O again. The operating system uses a shortest remaining compute time first scheduling algorithm and schedules a new process either when the running process get blocked on I/O or when the running process finishes its compute burst. Assume that all I/O operations can be overlapped as much as possible. For what percentage of time does the CPU remain idle?Given the three processes p1, p2 and p3. Their execution time are respectively estimated at X, Y and Z such that X < Y < Z. To run these processes we have the choice between the two following schedules: (1) p1, p2, p3, (2) p2, p1, p3. What is the best scheduling compared to the average response time?Given the list of processes, their CPU burst times, arrival times and priorities implement SJF,Priority and Round Robin scheduling algorithms on the processes with preemption. For each ofthe scheduling policies, compute and print the completion Time(CT), Turnaround Time(TAT),and Waiting Time(WT) for each process using C Programming.Waiting time: Processes need to wait in the process queue before execution starts and inexecution while they get preempted. Turnaround time: Time elapsed by each process to get completely served. (Difference betweensubmission time and completion time).