Question

Suppose that the following processes arrive for execution at the times indicated. Each process will run for the amount of time listed. In answering the questions, use non-preemptive scheduling, and base all decisions on the information you have at the time the decision must be made.

Process Arrival Time Burst Time
P1 0.0 8
P2 0.4 4
P3 1.0 1
a) What is the average turnaround time for these processes with the FCFS scheduling algorithm?

b) What is the average turnaround time for these processes with the SJF scheduling algorithm?

Answer 1

Answer:

a) 10.53

b) 9.53

Explanation:

a) Average Turnaround Time: ( (8-0)+(12-0.4)+(13-1.0) ) / 3 = 10.53

b) Average Turnaround Time: ( (8-0)+(13-0.4)+(9-1.0) ) / 3 = 9.53

Answer 2

Following are the calculation to the average turnaround time:

• The length of time required to process a request is known as "turn-around time".
• It is defined as the total amount of time spent between the end date and the starting time.

• The essential times are associated with processes inside the OS:$\text{"arrival time, waiting time, response time, burst time, completion time, and turn-around time".}$

Given:

Process                         Arrival Time                     Burst Time

$P1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8\\\\P2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\\\\P3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1$

Using formula:

$\to \bold{TAT= CT- AT}$

For point a:

Process  AT            BT           CT         TAT

$P1\ \ \ \ \ \ \ \ \ \ 0.0 \ \ \ \ \ \ \ \ \ \ 8 \ \ \ \ \ \ \ \ \ \ 8 \ \ \ \ \ \ \ \ \ \ 8.0 \\\\P2\ \ \ \ \ \ \ \ \ \ 0.4\ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ 12 \ \ \ \ \ \ \ \ \ \ 11.6 \\\\ P3\ \ \ \ \ \ \ \ \ \ 1.0 \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ 13 \ \ \ \ \ \ \ \ \ \ 12.0$

$\to \bold{TAT=\frac{P1+P2+P3}{3}}$

             $=\frac{8.0+11.6+12.0}{3}\\\\=\frac{31.6}{3}\\\\=10.53$

For point b:

Process  AT            BT           CT         TAT

$P1\ \ \ \ \ \ \ \ \ \ 0.0 \ \ \ \ \ \ \ \ \ \ 8 \ \ \ \ \ \ \ \ \ \ 8 \ \ \ \ \ \ \ \ \ \ 8.0 \\\\P2\ \ \ \ \ \ \ \ \ \ 0.4\ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ 13 \ \ \ \ \ \ \ \ \ \ 12.6 \\\\ P3\ \ \ \ \ \ \ \ \ \ 1.0 \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ 8.0$

$\to \bold{TAT=\frac{P1+P2+P3}{3}}$

             $=\frac{8.0+12.6+8.0}{3}\\\\=\frac{28.6}{3}\\\\=9.53$

Therefore, the final answer is "10.53  and 9.53".

Learn more:

brainly.com/question/16045350