An operating system uses Shortest RemainingTime first (SRTF) process scheduling algorithm.Consider the arrival time and burst times for thefollowing processesProcessBurst TimeArrival TimeP₁100P₂84P3152P446What is the waiting time of the processP3(answer up to 1 decimal place)?

It is a preemptive scheduling policy in which the process with the minimum remaining time to…

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hortest Remaining cheduling algorithm dburst times for th- ne Arrival Time 0 4 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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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?
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?
It is a computer science question based on Operating systems Need the answer and explained solution for "PART (b) ONLY"
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?
Consider a system consisting of m resources of the same type being shared by n processes, n > m. Each process has a maximum need of m/2 resources. Initially, each process has no resource requests. A process can request or release only one resource at a time. With n=3k and m=2k, for some integer k, show that the system is deadlock free.
A system uses proportional allocation (according to process size) when allocating frames to processes. There are 3 processes with sizes: P1: 20 frames P2: 15 frames P3: 5 frames. The system posses 32 frames, from which 16 frames are reserved for the system itself and the rest are to be allocated to processes. How many frames process P2 will be allocated Select one: 12 O 6 O 15 O 10
We define an m-section to be a sequence of code that can be run concurrently by maximum m threads. There are n threads in a process. Each thread executes a thread function doWork() that calls doCriticalWork() in an infinite loop. Function doCriticalWork() requires that at most m threads run it concurrently. The enter() and leave() functions are used to limit the number of threads within the m- section to a maximum of m and are the only functions that deal with synchronization. The pseudo-code algorithm for the thread function is this: void doWork(...) { while (true) { enter(...); I/ limit access to m threads Il execute m-section doCriticalWork(.); Il run by max. m threads leave(.); I/ leave m-section Il do more work Function enter() returns immediately only if there are less than m threads in the m-section. Otherwise, the calling thread will be blocked. A thread calls leave() to indicate it has finished the m-section. If there was another thread blocked (in enter() waiting to enter the…
Consider a computer whose RAM has the following memory partition: 4K, 8K, 20K, 2K. At time T, a snapshot of the process queue is taken and is given below. Queue Head Tail Process PO P1 P2 P3 P4 P5 P6 P7 Memory 2K 14K 3K 6K 6K 10K 7K 20K Usage time 4ms 10ms 2ms 8ms 4ms 1ms 8ms 6ms If the processes are allocated using FCFS algorithm with best-fit memory allocation, calculate the time at which P6 finishes its execution. Assume there is no additional time overhead for allocating these processes.
For the RR scheduling algorithm, Burst time of 5 processes (p0,p1,22,23,24) are as follows (2ms, 3ms, 15ms, 1ms, 2ms). what is the most suitable time quantum-- Select one: a. 2ms b. 1ms c. 4ms d. 16mg
Consider 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.
Suppose there are 8 processes in a system which are waiting to be assigned to the ready queue. The processes have the following properties. Process P1 has burst time 15, priority 6 Process P2 has burst time 12, priority 4 Process P3 has burst time 6, priority 12 Process P4 has burst time 5, priority 11 Process P5 has burst time 8, priority 9 Process P6 has burst time 9, priority 2 Process P7 has burst time 2, priority 20 Process P8 has burst time 17, priority 6 You have to use Multilevel Feedback Queue. It has three queues. The first queue takes processes with priority value from 1 to 5, the second queue takes processes with priority value of 6 to 10 and rest of the processes go to the third queue. First queue follows Round Robin Scheduling with time quantum= 4 Second queue follows Round Robin Scheduling with time quantum = 6 Third queue follows FCFC scheduling. Draw the gantt chart and calculate the average waiting 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).