1%£0 ll. I.Y:-Aامنة على عمار هاشم - ....exThree processes P1, P2, and P3 arrive at time zero. The total time spent by theprocess in the system is 10ms, 20ms, and 30ms respectively. They spent the first20% of their execution time doing I/O and the rest 80% in CPU processing. What isthe AVERAGE WAITING TIME of the CPU using the FCFS, SJF, and RR schedulingalgorithms?ProcessExecution time1/0 timeCPU timeP1108P220416P330624

Given that, There are three processes P1, P2 and P3. Each process contains the values of execution…

Three processes P1, P2, and P3 arrive at time zero. The total time spent by the process in the system is 10ms, 20ms, and 30ms respectively. They spent the first 20% of their execution time doing 1/0 and the rest 80% in CPU processing. What is the AVERAGE WAITING TIME of the CPU using the FCFS, SJF, and RR scheduling algorithms? Process Execution time 1/0 time CPU time P1 10 8 P2 20 4. 16 P3 30 6. 24 2.

Three processes P1, P2, and P3 arrive at time zero. The total time spent by the process in the system is 10ms, 20ms, and 30ms respectively. They spent the first 20% of their execution time doing 1/0 and the rest 80% in CPU processing. What is the AVERAGE WAITING TIME of the CPU using the FCFS, SJF, and RR scheduling algorithms? Process Execution time 1/0 time CPU time P1 10 8 P2 20 4. 16 P3 30 6. 24 2.

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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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…
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