FCFS ( First Come First Serve ) in Operating System

Question 1)

                  Process No.              AT(Arrival Time)               BT (Burst Time)

                         P1                                    0                                         4
                         P2                                    1                                         3
                         P3                                    2                                         1
                         P4                                    3                                         2
                         P5                                    4                                         5


Solution:- First Create a Gantt chart.Here given a Process
     
                               Criteria :- Arrival Time
                     Mode :- Non-Preemptive

 

Here P1 process is starting time 0 and completion time is 4 because 4 is burst time.

Then,P2 starting time is 4 and completion time is 7 because 4+3=7 and 3 is a burst time of P2

Next , P3 is starting time is 7 and completion time is 8, because 7+1=8 and 1 is burst time of P3

Same here apply condition on P4 and P5.

Then,Next find out completion time , Turn around time , Waiting time and Average Waiting time.

     CT = Completion time

     TAT= Turn Around time

     WT= Waiting time

     AWT= Average Waiting time

         CT :-

                  P1= 4

                  P2= 7

                  P3= 8

                  P4= 10

                  P5= 15

       TAT:-

                        Formula :- TAT = CT - AT

                   TAT (P1) = 4 - 0 = 4 

                           (P2) = 7 - 1 = 6

                           (P3) = 8 - 2 = 6

                           (P4) = 10 - 3 = 7

                           (P5) = 15 - 4 = 11

            WT :-

                        Formula :-  WT = TAT - BT

                    WT(P1) = 4 - 4 = 0

                           (P2) = 6 - 3 =3

                           (P3) = 6 - 1 =5

                           (P4) = 7 - 2 =5

                           (P5) = 11 - 5 = 6

         AWT (Average Waiting time):-
                         

                                      ( 0 + 3 + 5 + 5 + 6 ) /5  = 19 /5  =  3.8

Processes      AT              BT                 CT                    TAT                   WT

    P1               0                  4                   4                         4                     0

    P2               1                  3                   7                         6                     3

    P3               2                  1                   8                         6                     5

    P4               3                  2                   10                       7                      5

    P5               4                  5                   15                       11                     6