Question

luuent Name: 1. (100 points) Three processes P1, P2 and P3 with related information are given in the following table: Process Burst Time (ms) Arrival Time (ms) P1 T 0 P2 30 20 P3 20 T is a positive integer (T>0). Please use a non-preemptive shortest job first scheduling algorithm to make Gantt charts, and calculate different waiting times of three processes for different cases. Please write a professional perfect solution with detailed steps. 30

Answer 1

Process | Burst Time T Burst Time ) Arrival Time 30 20 P3 20 l 30 As it is non-preemptive shortest Job jirst. a) Gantt Charts a case 1 (when I is less than 30) T T+30 T750 as Pi will run as it is non-preemptive shortest job jirst as her Ps will be run as it's amivat hme is less than P3 and it is mon- preemptive in nature complehon time of P - T Pa = T+ 30 Py = T+50 Turn Around time - Complehon hme - Anival time Pi = T-O = 1 Py = T+30-20 - T+10 P3 - 1750 - 30 - T + 20 lalaining time - Turn around time - Burst time P = T-T=0 Pz = (1+10)-30 = 1-20 Pz = (+20) -20 - T waining time - T-20 4T+0 = 21 - 20

b> Gantt chart Yantt charts: 2 Case 2 (when I is greater than 30) T T+20 T+50 as Pi will run as it is non-preemphve in nahres P3 will run before Pz due to shortest Buist time when Pinas buigt time grealer man 30 it will run füst than P2 completion time of PI = T P = T+50 P3 = + 20 Tom around time - complelion ime - Amival time P = T-O=T PL - 1750-20 - T+30 P2 = T +20-30 - T-10 waiting time = Tom around time - Buist hime Pi = T-T=0 P2 = +30-30 = T P2 = T-10 + 20 = T+10 D+T+TUID = 2T+I Avg wailing time - 30