Burst time Process Priority (msec) 9 2 P1 P2 P3 P4 Ps 1 1 3 4 5 3 te that a smaller priority number implies a higher priority. Draw a timeline (Gant Chart) illustrating the sequence of process executions for the above processes using:

Burst time Process Priority (msec) 9 2 P1 P2 P3 P4 Ps 1 1 3 4 5 3 te that a smaller priority number implies a higher priority. Draw a timeline (Gant Chart) illustrating the sequence of process executions for the above processes using:

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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Shared Variable Problem. Assume there is a shared variable x, initialized to 0, that is operated on by four concurrently executing processes (A, B, C, and D) as shown below: Process A Process B Process C Process D semWait(s);read(x);x = x - 2;write(x);semSignal(s); semWait(s);read(x);x = x - 3;write(x);semSignal(s); semWait(s);read(x);x = x + 2;write(x);semSignal(s); semWait(s);read(x);x = x + 3;write(x);semSignal(s); Note that each process, before reading x, invokes the semWait operation on a counting semaphore s and invokes the semSignal operation on the semaphore s after writing x. The counting semaphore s is initialized to a value of 2. The processes can arrive in any order and can be preempted at any point in their execution by another process. Answer the following questions: a) What is the maximum possible value of xafter all 4 processes complete execution? Show how this value would be achieved and the order of execution necessary for this to occur. b)…
For deadlock to occur, A There must be mutual exclusion in resource allocation to processes There must be hold and wait of resources by processes C There must be circular waiting D All of the above.
ProcessBurst Time Priority Arrival TimeP1 10 3 0P2 1 1 0P3 2 5 0P4 1 4 0P5 5 2 0For the set of processes directly above, draw a Gantt chart for the FCFS Scheduling Algorithm.
1. Two processes, A and B, each need three resources, 1, 2, and 3. Suppose 4 requests them in the order 1, 2, 3. If B also requests them in the same order, deadlock is not possible. However, if B requests them in the order 3, 2, 1, then deadlock is possible. With three resources, there are 3! or six possible orders in which B can request them. For each possible order in which B can request these resources, state whether or not a deadlock is possible for that order. (Assume that once a process acquires a resource, it does not release that resource until the end of that process.) a) 1, 2, 3 b) 1,3,2 c) 2, 1, 3 d) 2, 3, 1 e) 3, 1, 2 f) 3, 2, 1
Given in Table 1 is a set of five (5) processes with the priority, length of the CPU-burst time, and arrival time. Table 1 Process Arrival Time (ms) Burst Time (ms) Priority* P1 3 P2 1 2 1 P3 3 P4 7 P5 8 8 4. *Note: • A smaller priority number implies a higher priority. a) Compute the Waiting Time (WT) of each process in Table 1, and also compute their Average Waiting Time, for each of the following scheduling algorithms. i. Shortest Job First (SJF). ii. Pre-emptive Priority (PP) iii. Round Robin (RR) with quantum = 3 (ms).
Task 1: SJF Scheduling with preemption You can use the following input as sample: 22222 Proc P1 P5 Process P1 P2 P3 P4 Solution in a Gantt chart: P1 P2 P3 AT 3 4 P4 5 18 20 P5 Sample Output Structure: P2 3 P2 BT CT 8 22 5 18 6 20 3 27 3 28 Arrival Time 4 Grant Chart: Pl P5 P2 P1 P5 P2 0 WT 11 9 9 6 5 2 3 4 5 P1 TAT 19 14 15 9 8 7 P1 2 Average waiting time: 8.0 Average turnaround time: 13.0 **Output values are random and do not match the given input. P4 11 Burst Time P5 5 2 7 4 5 16 P3 23 P5 P2 P1 P3 P4 P3 P4
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.
Process PI P2 P3 P4 P5 Buast time 2 1 8 4 5 Priority (2*xx) %5 (2xxx) 5 (2*XX) 0/0 5 (2*XX) do 5 (2*xx) do 5 ✓ 11 Value of XX is 5 All processes Arrives at time + = 0 ms. Draw the Gantt chart for priority scheduling (A larger priority number implies higher priority)
Q-1: Draw the Shortest Remaining Time First (with Preemption) and Round Robin (no Preemption) Gannt chart for the following processes. Calculate average wait time for each algorithm using the processes data below. Round Robin uses a time quanta of 3. Process Number PO P1 P2 P3 P4 P5 Arrival Time Burst Time to Complete 10 2 0 3 4 13 25 26 6 10 1 3
Table 1. 4 processes arriving shown in Operating System Scheduling PDF for Shortest Job Next, non- preemptive Process Arrival Times Execution Times Service Times sequence Wait Times calculations PO 0 12 P1 21 9 P2 41 6 P3 6 3 Total wait time = ? Here the 4 processes of the PDF are repeated in Table 1, with arrival times changed to 0, 2, 4, and 6 respectively. Also, the execution times are changed to 12, 9, 6, and 3. You will be asked to fill table 1 with the numbers like the service times (when is each process serviced), the sequence of executions of these 4 processes in part (a) and part (b) below. Also, you will compute the total wait time. Note for FCFS, it is 1, 2, 3, 4 for P0, P1, P2, and P3; • for the SJN or Shortest Job Next with non-preemptive, PO is #1, P1 may not be #2; for SRT, or shortest remaining time, the preemptive version of SJN, PO is #1, but after 2 time units when P1 comes, P1 kicks out PO since its execution time is 4 < 9- 2 = 7; and P1 may be preempted by P2 later…
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=18
A single-CPU system has four processes, P1, P2, P3 and P4 in the ready queue. The execution times and I/O needs for these processes are given below. All times are in ms. (hint: when a process starts an IO operation, it is removed from the ready queue and put back at the end of the queue only when it completes its IO). Process P1: Arrives at 0ms, needs 23ms of CPU time. Performs I/O for 5ms after 10ms of its execution time, then another 5ms after 20ms of its execution time. Process P2: Arrives at 3ms, needs 12ms of CPU time. Performs I/O for 10ms after 5ms of its execution time. Process P3: Arrives at 1ms, needs 15ms of CPU time. Performs I/O for 1ms after 7ms of its execution time. Process P4: Arrives at 2ms, needs 8ms of CPU time. Performs I/O for 5ms after 4ms of its execution time. We assume that the CPU is idle if no one of these four processes is using it. Using Round Robin scheduling algorithm with a Time Quantum of 4ms and ignoring the context switch time, determine: a) The…