Homework Answers
At time 0, P1 arrives.Round Robin Ready queue is empty then.CPU
starts executing P1
At time 4,P2 arrives. Now ready queue contains P2.
At time 5, a context switch occurs as the time quantum is 5 and P1
is added to ready queue.
Now ready queue contains P1,P2.
Now P2 starts executing. So the ready queue now contains P1.
At time 10,P3 arrives. Now ready queue is P3,P1.
P2 also finishes execution at time 10.
Now a context switch occurs and process P1 starts executing,since
P1 is the first process in ready queue.Now ready queue is P3.
At time 12, the process P1 will go for an I/O operation.Now ready
queue is P3.
So P3 starts executing at 12.Now ready queue is empty.
At time 14, P4 arrives.Now ready queue is P4.
At time 17,P1 finishes I/O operation and P1 is added to ready
queue. Now ready queue is P1,P4.
At that time P3 finishes executing.Since P3 needs 4 time units
more, P3 is added to ready queue.
Now a context switch occurs and process P4 starts executing,since
P4 is the first process in ready queue.Now ready queue is
P3,P1.
P5 arrives at time 21.Now ready queue is P4,P3,P1.
Now balance two units of P1 is executed. Now ready queue is
P4,P3.
Then balance 4 units of P3 is executed.
Now the remaining process is P5.It will be executed
The Gantt chart is
Response time is the time spent when the process is in the ready state and gets the CPU for the first time.
Waiting time is the total time spent by the process in the ready state waiting for CPU.
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question is based on operating system concept. round roubin
algorith
4 5 6 Use Round Robin...
Fill out both charts ROUND ROBIN TQ = 4 process number arrival time service time response...
Fill out both charts
ROUND ROBIN TQ = 4 process number arrival time service time response waiting time total time context switches time P1 02 P24 O 15 gant
Solve by Round Robin Scheduling, Quantum = 3 (preemptive). Calculate the average waiting time: Process Arrival...
Solve by Round Robin Scheduling, Quantum = 3 (preemptive). Calculate the average waiting time: Process Arrival Time Execution Time РО 5 P1 1 3 P2 N 8 P3 3 6
V. Round Robin and SJF comparison 6 Process Arrival Time CPU Burst Time P1 0 9...
V. Round Robin and SJF comparison 6 Process Arrival Time CPU Burst Time P1 0 9 P2 4 4 P3 6 P4 For each of the following algorithms, draw a Gantt chart and calculate the average wait time and turn around time. 1. Round robin with quantum=1 time unit 2. Round-robin with quantum=3 time-units 3. SJF WI preemption (Shortest Remaining Time First), assume quantum=3 4. SJE Wlo preemption 5. ECES
Assume a dynamic queue which is serviced by a Priority Based Round Robin algorithm such that ther...
Assume a dynamic queue which is serviced by a Priority Based Round Robin algorithm such that there exists three priorities (1,2,3) which are used as multipliers of the basic time quantum value with the resulting number being the maximum service time the corresponding job will receive each time it gets the CPU. For a maximum amount of time equal to 1 basic time quantum , a priority 2 job gets the CPU for the maximum amount of time equal to...
16. Consider a single-processor system that uses the round-robin (RR) process scheduling algorithm. Assume that the...
16. Consider a single-processor system that uses the round-robin (RR) process scheduling algorithm. Assume that the time quantum is 10 milliseconds. Consider the five processes with the arrival time and the burst time, in milliseconds, shown in the table below Arrival Time Burst Time (milliseconds) (milliseconds) Process Pl P2 P3 P4 P5 0 24 20 23 24 Which of the five processes is the last to finish its execution? (A) P (B) P2 (C) P3 (D) P4 (E) P5
Could you please add Round Robin CPU scheduling algorithm with quantum 10 in this code? Scheduling.java...
Could you please add Round Robin CPU scheduling algorithm with quantum 10 in this code? Scheduling.java import java.util.ArrayList; import java.util.Random; import java.util.Scanner; public class Scheduling { public static void main(String[] args) { ArrayList<Proc> pr = new ArrayList<Proc>(); ArrayList<Proc> pr1 = new ArrayList<Proc>(); ArrayList<Proc> pr2 = new ArrayList<Proc>(); ArrayList<Proc> pr3 = new ArrayList<Proc>(); int count=0; Random ran = new Random(); boolean unique=false; int num=0; for(int i=0;i<=5;i++){ unique=false; Proc p1=new Proc(); while(!unique){ num=ran.nextInt(10) + 1; if(i==0) break; for(int j=0;j<i;j++){ if(num==pr.get(j).id){ unique=false;...
Opearting system 5. [27pts.] Given the following set of processes, with arrival times, priorities, and the...
Opearting
system
5. [27pts.] Given the following set of processes, with arrival times, priorities, and the length of the CPU burst in ms: Process Arrival Time Burst time Priority P1 P2 P3 2 10 P4 (Note: a lower number means higher priority; processes P2- P4 arrive at the same time, in the given order) a. Draw a Gantt chart showing a FCFS scheduling algorithm. b. Draw a Gantt chart showing a non-preemptive Priority scheduling algorithm. Draw a Gantt chart showing...
The following processes are being scheduled using a pre-emptive, priority-based, round-robin scheduling algorithm. Process Burst Time...
The following processes are being scheduled using a pre-emptive, priority-based, round-robin scheduling algorithm. Process Burst Time Priority Arrival 20 20 0 20 25 45 55 5 5 5 15 Each process is assigned a numerical priority, with a higher number indicating a higher relative priority. The scheduler will execute the highest-priority process. For processes with the same priority, a round-robin scheduler will be used with a time quantum of 10 units. If a process is pre-empted by a higher-priority process,...
Given the following set of processes with corresponding execution times (in ms), arrival times and priority...
Given the following set of processes with corresponding execution times (in ms), arrival times and priority (1 – highest). For each scheduling algorithm: Construct a table showing which process is active and for how long until all processes are completely serviced (as done in class). Calculate the average waiting time and turnaround time. Process ID Burst (ms) Arrival time P1 9 0 P2 12 0 P3 3 0 P4 30 0 P5 20 0 P6 10 0 First Come First Serve...
Answer the following using the processes and their corresponding arrival time and burst time. These processes...
Answer the following using the processes and their corresponding arrival time and burst time. These processes are being scheduled using a preemptive, round-robin scheduling algorithm with a time quantum of 2 units. PID Arrival Time Burst Time P1 0 4 P2 1 5 P3 2 3 P4 3 2 P5 4 6 1. Use a Gantt chart to represent the above processes (5 points) 2. Calculate the average waiting time for each process (5 points) 3. Calculate the completion time...
Fill out both charts
ROUND ROBIN TQ = 4 process number arrival time service time response waiting time total time context switches time P1 02 P24 O 15 gant
Solve by Round Robin Scheduling, Quantum = 3 (preemptive). Calculate the average waiting time: Process Arrival Time Execution Time РО 5 P1 1 3 P2 N 8 P3 3 6
V. Round Robin and SJF comparison 6 Process Arrival Time CPU Burst Time P1 0 9 P2 4 4 P3 6 P4 For each of the following algorithms, draw a Gantt chart and calculate the average wait time and turn around time. 1. Round robin with quantum=1 time unit 2. Round-robin with quantum=3 time-units 3. SJF WI preemption (Shortest Remaining Time First), assume quantum=3 4. SJE Wlo preemption 5. ECES
16. Consider a single-processor system that uses the round-robin (RR) process scheduling algorithm. Assume that the time quantum is 10 milliseconds. Consider the five processes with the arrival time and the burst time, in milliseconds, shown in the table below Arrival Time Burst Time (milliseconds) (milliseconds) Process Pl P2 P3 P4 P5 0 24 20 23 24 Which of the five processes is the last to finish its execution? (A) P (B) P2 (C) P3 (D) P4 (E) P5
Opearting
system
5. [27pts.] Given the following set of processes, with arrival times, priorities, and the length of the CPU burst in ms: Process Arrival Time Burst time Priority P1 P2 P3 2 10 P4 (Note: a lower number means higher priority; processes P2- P4 arrive at the same time, in the given order) a. Draw a Gantt chart showing a FCFS scheduling algorithm. b. Draw a Gantt chart showing a non-preemptive Priority scheduling algorithm. Draw a Gantt chart showing...
The following processes are being scheduled using a pre-emptive, priority-based, round-robin scheduling algorithm. Process Burst Time Priority Arrival 20 20 0 20 25 45 55 5 5 5 15 Each process is assigned a numerical priority, with a higher number indicating a higher relative priority. The scheduler will execute the highest-priority process. For processes with the same priority, a round-robin scheduler will be used with a time quantum of 10 units. If a process is pre-empted by a higher-priority process,...
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