Scheduling Algorithms of deciding which process in ready queue - PDF document

2/15/2011 Scheduling Algorithms  CPU Scheduling algorithms deal with the problem Scheduling Algorithms of deciding which process in ready queue should be allocated to CPU.  Following are the commonly used scheduling algorithms: Gursharan Singh Tatla professorgstatla@gmail.com www.eazynotes.com 15-Feb-2011 www.eazynotes.com 15-Feb-2011 1 2 First-Come-First-Served Scheduling Algorithms Scheduling (FCFS)  First-Come-First-Served (FCFS)  In this scheduling, the process that requests the CPU first, is allocated the CPU first.  Shortest Job First (SJF)  Thus, the name First-Come-First-Served.  Priority Scheduling  The implementation of FCFS is easily managed  Round-Robin Scheduling (RR) with a FIFO queue.  Multi-Level Queue Scheduling (MLQ) Ready Queue  Multi-Level Feedback Queue Scheduling (MFQ) D C B A CPU T ail of Head of Queue Queue www.eazynotes.com www.eazynotes.com 3 15-Feb-2011 4 15-Feb-2011 First-Come-First-Served Example of FCFS Scheduling Scheduling (FCFS)  When a process enters the ready queue, its PCB  Consider the following set of processes that arrive is linked to the tail of the queue. at time 0 with the length of the CPU burst time in milliseconds:  When CPU is free, it is allocated to the process which is at the head of the queue. Process Burst Time (in milliseconds) P 1 24  FCFS scheduling algorithm is non-preemptive . P 2 3  Once the CPU is allocated to a process, that P 3 3 process keeps the CPU until it releases the CPU, either by terminating or by I/O request. www.eazynotes.com www.eazynotes.com 5 15-Feb-2011 6 15-Feb-2011 1

2/15/2011 Example of FCFS Scheduling Example of FCFS Scheduling  Suppose that the processes arrive in the order: P 1 , P 2 , P 3 . P 1 24  Average Waiting Time = (Total Waiting Time) /  The Gantt Chart for the schedule is: P 2 3 No. of Processes P 3 3 P 1 P 2 P 3 = (0 + 24 + 27) / 3 0 24 27 30 = 51 / 3  Waiting Time for P 1 = 0 milliseconds = 17 milliseconds  Waiting Time for P 2 = 24 milliseconds  Waiting Time for P 3 = 27 milliseconds www.eazynotes.com 15-Feb-2011 www.eazynotes.com 15-Feb-2011 7 8 Example of FCFS Scheduling Example of FCFS Scheduling  Suppose that the processes arrive in the order: P 2 , P 3 , P 1 .  Average Waiting Time = (Total Waiting Time) /  The Gantt chart for the schedule is: No. of Processes = (0 + 3 + 6) / 3 P 2 P 3 P 1 = 9 / 3 0 3 6 30  Waiting Time for P 2 = 0 milliseconds = 3 milliseconds  Waiting Time for P 3 = 3 milliseconds  Thus, the average waiting time depends on the order in which the processes arrive.  Waiting Time for P 1 = 6 milliseconds www.eazynotes.com www.eazynotes.com 9 15-Feb-2011 10 15-Feb-2011 Shortest Job First Scheduling Non-Preemptive SJF (SJF)  In non-preemptive scheduling, CPU is assigned  In SJF, the process with the least estimated execution to the process with least CPU burst time. time is selected from the ready queue for execution.  The process keeps the CPU until it terminates.  It associates with each process, the length of its next CPU burst.  Advantage:  When the CPU is available, it is assigned to the  It gives minimum average waiting time for a given process that has the smallest next CPU burst. set of processes.  If two processes have the same length of next CPU  Disadvantage: burst, FCFS scheduling is used.  It requires knowledge of how long a process will run  SJF algorithm can be preemptive or non-preemptive. and this information is usually not available. www.eazynotes.com www.eazynotes.com 11 15-Feb-2011 12 15-Feb-2011 2

2/15/2011 Preemptive SJF Example of Non-Preemptive SJF  In preemptive SJF, the process with the smallest  Consider the following set of processes that arrive estimated run-time is executed first. at time 0 with the length of the CPU burst time in milliseconds:  Any time a new process enters into ready queue, the scheduler compares the expected run-time of Process Burst Time (in milliseconds) this process with the currently running process. P 1 6 P 2 8  If the new process’s time is less, then the P 3 7 currently running process is preempted and the P 4 3 CPU is allocated to the new process. www.eazynotes.com 15-Feb-2011 www.eazynotes.com 15-Feb-2011 13 14 Example of Non-Preemptive SJF Example of Non-Preemptive SJF  The Gantt Chart for the schedule is: P 1 6 P 2 8  Average Waiting Time = (Total Waiting Time) / P 1 P 4 P 3 P 2 P 3 7 No. of Processes P 4 3 0 3 9 16 24 = (0 + 3 + 9 + 16 ) / 4  Waiting Time for P 4 = 0 milliseconds = 28 / 4  Waiting Time for P 1 = 3 milliseconds = 7 milliseconds  Waiting Time for P 3 = 9 milliseconds  Waiting Time for P 2 = 16 milliseconds www.eazynotes.com www.eazynotes.com 15 15-Feb-2011 16 15-Feb-2011 Example of Preemptive SJF Example of Preemptive SJF  Consider the following set of processes. These  The Gantt Chart for the schedule is: processes arrived in the ready queue at the times P 1 P 2 P 4 P 1 P 3 given in the table: 5 0 26 1 10 17 Process Arrival Time Burst Time (in milliseconds)  Waiting Time for P 1 = 10 – 1 – 0 = 9 P 1 0 8 P 2 1 4  Waiting Time for P 2 = 1 – 1 = 0 P 3 2 9 P AT BT P 4 3 5  Waiting Time for P 3 = 17 – 2 = 15 P 1 0 8 P 2 1 4  Waiting Time for P 4 = 5 – 3 = 2 P 3 2 9 P 4 3 5 www.eazynotes.com www.eazynotes.com 17 15-Feb-2011 18 15-Feb-2011 3

2/15/2011 Example of Preemptive SJF Explanation of the Example  Process P 1 is started at time 0, as it is the only process in the queue.  Average Waiting Time = (Total Waiting Time) / No. of Processes  Process P 2 arrives at the time 1 and its burst time is 4 milliseconds. = (9 + 0 + 15 + 2) / 4  This burst time is less than the remaining time of = 26 / 4 process P 1 (7 milliseconds). = 6.5 milliseconds  So, process P 1 is preempted and P 2 is scheduled. www.eazynotes.com 15-Feb-2011 www.eazynotes.com 15-Feb-2011 19 20 Explanation of the Example Explanation of the Example  Process P 3 arrives at time 2. Its burst time is 9  After the termination of P 2 , the process with which is larger than remaining time of P 2 (3 shortest next CPU burst i.e. P 4 is scheduled. milliseconds).  After P 4 , processes P 1 (7 milliseconds) and then  So, P 2 is not preempted. P 3 (9 milliseconds) are scheduled.  Process P 4 arrives at time 3. Its burst time is 5. Again it is larger than the remaining time of P 2 (2 milliseconds).  So, P 2 is not preempted. www.eazynotes.com www.eazynotes.com 21 15-Feb-2011 22 15-Feb-2011 Priority Scheduling Priority Scheduling  Priority scheduling can be preemptive or non-  In priority scheduling, a priority is associated with preemptive. all processes.  Preemptive Priority Scheduling:  Processes are executed in sequence according  In this, scheduler allocates the CPU to the new process to their priority. if the priority of new process is higher tan the priority of the running process.  CPU is allocated to the process with highest  Non-Preemptive Priority Scheduling: priority.  The running process is not interrupted even if the new  If priority of two or more processes are equal than process has a higher priority. FCFS is used to break the tie.  In this case, the new process will be placed at the head of the ready queue. www.eazynotes.com www.eazynotes.com 23 15-Feb-2011 24 15-Feb-2011 4

2/15/2011 Priority Scheduling Priority Scheduling  Problem:  Solution:  In certain situations, a low priority process can be  Aging is a technique which gradually increases the blocked infinitely if high priority processes arrive in priority of processes that are victims of starvation. the ready queue frequently.  For e.g.: Priority of process X is 10.  This situation is known as Starvation .  There are several processes with higher priority in the ready queue.  Processes with higher priority are inserted into ready queue frequently.  In this situation, process X will face starvation. www.eazynotes.com 15-Feb-2011 www.eazynotes.com 15-Feb-2011 25 26 Priority Scheduling Example of Priority Scheduling (Cont.):  Consider the following set of processes that arrive at time 0 with the length of the CPU burst time in  The operating system increases priority of a milliseconds. The priority of these processes is process by 1 in every 5 minutes. also given:  Thus, the process X becomes a high priority Process Burst Time Priority process after some time. P 1 10 3  And it is selected for execution by the scheduler. P 2 1 1 P 3 2 4 P 4 1 5 P 5 5 2 www.eazynotes.com www.eazynotes.com 27 15-Feb-2011 28 15-Feb-2011 Example of Priority Scheduling Example of Priority Scheduling  The Gantt Chart for the schedule is: P 2 P 5 P 1 P 3 P 4  Average Waiting Time = (Total Waiting Time) / 0 1 6 16 18 19 No. of Processes = (0 + 1 + 6 + 16 + 18) / 5  Waiting Time for P 2 = 0 P BT Pr P 1 10 3 = 41 / 5  Waiting Time for P 5 = 1 P 2 1 1 P 3 2 4  Waiting Time for P 1 = 6 = 8.2 milliseconds P 4 1 5  Waiting Time for P 3 = 16 P 5 5 2  Waiting Time for P 4 = 18 www.eazynotes.com www.eazynotes.com 29 15-Feb-2011 30 15-Feb-2011 5