First Come First Serve Scheduling Algorithm n CPU

 

 What is First Come First Served Scheduling Algorithm ?

  This is the simplest scheduling algorithm which is non preemptive.In this

  algorithm a queue of the process is scheduled and selecting the process with 

  earliest time.

 Example:

                        Process       Arrival Time        Processing Time 

                           P1                  0                             3

                           P2                  2                             3

                           P3                  3                             1

                           P4                  5                             4

                           P5                  8                             2

       If the  process  arrives  as  per  the  arrival  time the Gantt Chart will be:

                    P1       P2         P3         P4             P5

               0         3          6           7             11             13    

 

        Time        Process Completed        Turn Around Time                 Waiting Time    

           0                    -                                       -                                               -

           3                   P1                                    3                                              0

           6                   P2                                    4                                              1

           7                   P3                                    4                                              3

          11                  P4                                    6                                              2

          13                  P5                                    5                                              3

    

     Point To Note

    Turn  around Time : Process Completed  - Process submitted

    Waiting Time- :  Turn around Time - Processing Time

    Average Turn around Time = (3+4+4+6+5)/5  = 4.4

    Average Waiting Time = (0+1+2+3+4)/5 = 1.8

    Processor Utilization = (13/13)*100 = 100%

    Throughput  = 5/13 = 0.38