Operating System Course 1. Consider the FIVE (5) process with their CPU burst time and arrival time.ProcessArrival TimeCPU burst timePO3P122P24P33P4Based on the above data, answer the followinga) Draw a time line for the following algorithm:I.Shortest Job First (SJF)I.Shortest Remaining Time (SRT)b) Calculate the turn-around time for SRT algorithm.c) Calculate average turn-around time for SRT algorithm.d) Calculate the response time for SJF.e) Calculate the waiting time for SJF.f) Which algorithm is the best among SJF and SRT? Justify your answer.

Given five processes namely p0,p1,p2,p3,p4 and their Arrival time and burst time are mentioned. a)…

Answered: 1. Consider the FIVE (5) process with…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Consider the following 5 processes which are labelled as A, B, C, D and E. Process Arrival Time CPU Cycle PriorityA 0 5 3B 1 2 2C 4 1 5D 5 4 1E 5 2 4 List of processes You are required to draw a timeline Gantt Chart for each of the following CPU schedulingalgorithms. Calculate the average turnaround time and average waiting time for each of thescheduling algorithm respectively.(Note: Assume that larger priority number implies a higher priority) (i) Shortest Remaining Time First (SRTF), Round Robin (assume that time quantum in 3 time slices) and…
Obtain the average waiting time, and the turnaround time, including CPU Utilization and throughput, and draw the chart for the following set of processes using Non-P SJF, and SRTF. Process Burst Time Arrival Time A 10 0 B 6 5 C 7 7 D 3 5 E 1 3
Consider four processes with the following start and end times: Process Start Time End Time ABCD 3 5 1 7 6 12 4 9 For each pair of proceses, say whether they run concurrently (Y) or not (N). (a) AB (b) AC (c) AD (d) BC (e) BD (f) CD
Consider the following set of processes, the length of the CPU burst time given in milliseconds: Process Burst time: P1 6 P2 8 P3 7 P4 3 Assuming the above process being scheduled with the SJF scheduling algorithm: The waiting time for process P1 is 3ms. The waiting time for process P1 is 0ms. The waiting time for process P1 is 16ms. The waiting time for process P1 is 9ms.
Consider the following set of processes, the length of the CPU burst time given in milliseconds: Process Burst Time P1 6 P2 8 P2 7 P4 3 Assuming the above processes are being scheduled with the Shortest Job First (SJF) scheduling algorithm. The waiting time for process P1 is 3ms The waiting time for process P1 is 0ms The waiting time for process P1 is 16ms The waiting time for process P1 is 9ms From the following list which is the deadlock avoidance algorithm? Banker’s algorithm Round-robin algorithm Elevator algorithm Karn’s algorithm A problem encountered in multitasking when a process is perpetually denied necessary resources is called Deadlock Starvation Inversion Aging To break a deadlock: Abort one or more processes to break the circular wait Abort all the process in the system Preempt all resources from all processes None of the mentioned Termination of the process terminates First thread of the process First two threads…
NOTE: "Exekveringstid(ms)" means execution time or burst time in milliseconds Assume that a system with a CPU at one time has the following processes queued in the CPU's READY queue: Calculate the waiting time for the seven processes: 1) If FCFS is being used 2) If LRU is being used 3) If RR with time quantum of 8ms
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…
please help with parts a and b with steps/explanation!
Direction: Draw a Gantt chart showing the execution of CPU for the following processes implementing Round robin (RR) as indicated below. Also, compute the turnaround times (TAT) and waiting times (WT) of all processes using this scheduling algorithm. The quantum time (QT) is set to 6. Question 1 10 pts Write your answers on the spaces provided. Process Burst Time (BT) Arrival time (AT) RR TAT RR WT P1 15 P2 10 5 P3 P4 3 10 P5 5 20
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Execution time in a multiprocessor system can be split into computing time for the routines plus routing time spent sending data from one processor to another. Consider the execution time and routing time given in the following table. In this case, the routing time is an important component of the total time. a) For each doubling of the number of processors, determine the ratio of new to old computing time and the ratio of new to old routing time. b) Find the computing time and routing time for a system with one processor.
We have 4 periodical processes II₁, II, II and II, that process a video. We know their periods P and their running times C; (in milliseconds). For: (a) the rate monotonic scheduling (RMS) algorithm and (b) the earliest deadline first (EDF) algorithm draw diagrams that displays the schedule for those processes for a period of 120 ms. The times are the following: Process P Ci II1 40 15 II2 35 5 II3 60 15 II4 45 5 In the case of equal priorities the process that is already running has the priority. If this is not the case then take the process with the lowest ID. Does any of the above algorithms fail?

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