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Module 6:  CPU Scheduling_Module 6: CPU SchedulingB
Updated:2011-11-29 Category:CPU
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Module 6: CPU Scheduling
    Basic Concepts
      Scheduling Criteria
        Scheduling Algorithms
          Multiple-Processor Scheduling
            Real-Time Scheduling
              Algorithm Evaluation Basic Concepts
                Maximum CPU utilization obtained with multiprogramming
                  CPU–I/O Burst Cycle – Process execution consists of a cycle of CPU execution and I/O wait.
                    CPU burst distribution Alternating Sequence of CPU And I/O Bursts Histogram of CPU-burst Times CPU Scheduler
                      Selects from among the processes in memory that are ready to execute, and allocates the CPU to one of them.
                        CPU scheduling decisions may take place when a process: 1.Switches from running to waiting state. 2.Switches from running to ready state. 3.Switches from waiting to ready. 4.Terminates.
                          Scheduling under 1 and 4 is nonpreemptive.
                            All other scheduling is preemptive. Dispatcher
                              Dispatcher module gives control of the CPU to the process selected by the short-term scheduler; this involves:
                                switching context
                                  switching to user mode
                                    jumping to the proper location in the user program to restart that program
                                      Dispatch latency – time it takes for the dispatcher to stop one process and start another running. Scheduling Criteria
                                        CPU utilization – keep the CPU as busy as possible
                                          Throughput – # of processes that complete their execution per time unit
                                            Turnaround time – amount of time to execute a particular process
                                              Waiting time – amount of time a process has been wiating in the ready queue
                                                Response time – amount of time it takes from when a request was submitted until the first response is produced, not output (for time-sharing environment) Optimization Criteria
                                                  Max CPU utilization
                                                    Max throughput
                                                      Min turnaround time
                                                        Min waiting time
                                                          Min response time First-Come, First-Served (FCFS) Scheduling
                                                            Example:Process Burst Time P1 24 P2 3 P3 3

                                                            P1

                                                            P2

                                                            P3

                                                            24

                                                            27

                                                            30

                                                            0

                                                              Suppose that the processes arrive in the order: P1 , P2 , P3 The Gantt Chart for the schedule is:
                                                                Waiting time for P1 = 0; P2 = 24; P3 = 27
                                                                  Average waiting time: (0 + 24 + 27)/3 = 17 FCFS Scheduling (Cont.) Suppose that the processes arrive in the order P2 , P3 , P1 .

                                                                  P1

                                                                  P3

                                                                  P2

                                                                  6

                                                                  3

                                                                  30

                                                                  0

                                                                    The Gantt chart for the schedule is:
                                                                      Waiting time for P1 = 6; P2 = 0; P3 = 3
                                                                        Average waiting time: (6 + 0 + 3)/3 = 3
                                                                          Much better than previous case.
                                                                            Convoy effect short process behind long process Shortest-Job-First (SJR) Scheduling
                                                                              Associate with each process the length of its next CPU burst. Use these lengths to schedule the process with the shortest time.
                                                                                Two schemes:
                                                                                  nonpreemptive – once CPU given to the process it cannot be preempted until completes its CPU burst.
                                                                                    Preemptive – if a new process arrives with CPU burst length less than remaining time of current executing process, preempt. This scheme is know as the Shortest-Remaining-Time-First (SRTF).
                                                                                      SJF is optimal – gives minimum average waiting time for a given set of processes. Example of Non-Preemptive SJF Process Arrival Time Burst Time P1 0.0 7 P2 2.0 4 P3 4.0 1 P4 5.0 4
                                                                                        SJF (non-preemptive)

                                                                                        P1

                                                                                        P3

                                                                                        P2

                                                                                        7

                                                                                        3

                                                                                        16

                                                                                        0

                                                                                        P4

                                                                                        8

                                                                                        12

                                                                                          Average waiting time = (0 + 6 + 3 + 7)/4 - 4 Example of Preemptive SJF Process Arrival Time Burst Time P1 0.0 7 P2 2.0 4 P3 4.0 1 P4 5.0 4
                                                                                            SJF (preemptive)

                                                                                            P1

                                                                                            P3

                                                                                            P2

                                                                                            4

                                                                                            2

                                                                                            11

                                                                                            0

                                                                                            P4

                                                                                            5

                                                                                            7

                                                                                            P2

                                                                                            P1

                                                                                            16

                                                                                              Average waiting time = (9 + 1 + 0 +2)/4 - 3 Determining Length of Next CPU Burst
                                                                                                Can only estimate the length.
                                                                                                  Can be done by using the length of previous CPU bursts, using exponential averaging. Examples of Exponential Averaging
                                                                                                     =0
                                                                                                      n+1 = n
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