Question

Problem Given the arrival time, and service time for the processes A, B, C, D, E as below: Process A B C D E Arrival Time 0 2 4 6 8 Service Time (T.) 3 6 4 5 2 Work out the Finish Time and Turn Around Time (T.) for each of the following scheduling policies: FCFS, RR with q = 1, SJF, and SRT time Around = finish time -

Answer 1

O First come First Seavice Process Arrival Time Service Time Tu CCT-AT). Time Savice T. Around Time Completion Time (CT) 1 3 2 6 B 1 C1 | d Ilo 4 1 a 13 12 El 8 12 20 FC FS is a non-preemptive scheduling algorithm where process which arcives Host is scheduled first onto CPU. Here process are scheduled in sequence of their avival time. Thus the sequence of execution : A=BC=D E . TABicole 0 3 9 13 18 20 Turn Around Time = Completion Time - Service Time. Kindly refer the above table.

2 Round Robin with TQ=1. Round Robin is a preemptive scheduling where each process is given a time equal to time Quantum to run on CPU. Ready Queue is maintained which keeps track of all process which are ready to be executed in a particular order. C (Finish Time) Twn Around Process | Arrival Time Service Time Completion Time Time (TAT) BT2 T4 18 L 17 20 15 LOH 2 L

RQ: 4,8,6,6,8,8,826,6.$,$.6.,88B.D A BABICBD CB le 101c|B|E 10|C|BD 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20 ☺ SJF Shortest Job First is a non-preemptive scheduling algorithm where job or process with smallest service time is scheduled first and is runned to its completion. - Process Anival Time Service Time completion Tun Around A LOT 3 3 3 B 2 6 a 7 c | 4 | 4 | 15 | ĞI 5 2014 3 mm -- 3 a ii is 20

@ SRT is a preemptive scheduling e first Shortest remaining time first is a preemptive schen agorithm where process are scheduled' whose service time is least and while its execution if another process with a lesses service time arrives, then the running process is preempted and new arrived process is scheduled on 'CPU. Process | Anival Service Time completion | Tun Around mo Dl6 5 E82 T 10 12 IABICE Blo o 3 4 8 10 15 20 Here if there is a tie , then we need to have a tie breaker Here I have used priority to lower process id as tie breaker
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