Three Types of Scheduling Job scheduling –Selects processes from the process pool (usually on disk) and loads them into memory Also known as long-term scheduler Process scheduling –Selects processes from the ready queue. Short term scheduler Thread scheduling –Schedules threads from within a process
Why schedule processes? - Increase Efficiency The reason scheduling works is because of the CPU-I/O burst cycle. When processes run, it has been observed that: –CPU has a burst of activity –Then it waits for I/O The Durations of the CPU burst have been measured extensively
CPU Burst Durations 802416 3240 140 120 110 100 80 60 40 20 160 Frequency Burst Duration (ms) [Silberschatz2005]
Reminder- How Operating Systems work Interrupts Dual Mode. Timers
Preemption Non-preemptive Scheduling: Once the system has assigned a CPU to a process, the system cannot remove that CPU from that process Preemptive Scheduling: The system can remove the running process from the CPU
Scheduling Criteria What do we want from a scheduling algorithm?
Scheduling Criteria It is desirable to maximize CPU utilization and throughput, and to minimize turnaround time, waiting time and response time. CPU Utilization – use of the CPU. Throughput – number of processes completed per time unit.
Scheduling Criteria Turnaround time – interval from submission to completion of a process. Waiting time – time spent ready to run but not running. Response time – the time from submission of a request until the first response is produced. –normally used for measuring an interactive system.
Scheduling Algorithms Question….. –Think of some different ways in which we could chose which process to execute next?
Scheduling Algorithms First come First Served SJF non preemptive / preemptive Priority Round robin
First come First Served ProcessBurst Time (ms) P130 P24 P32 Processes arrive at time 0 In the order P1, P2, P3 Uses a FIFO queue Non preemptive An Example FCFS
ProcessBurst Time (ms) P130 P24 P32 P1 0 P3P2 363430 Average wait time = (0+30+34)/3 = 21.33 Turnaround time = (30+34+36)/3 = 33.33 Gant chart Example FCFS
Exercise (5 minutes) ProcessBurst Time (ms) P32 P24 P130 P1P3P2 03626 Average wait time = (0+2+6)/3 = 2.67 Turnaround time = (2+6+36)/3 = 14.67
Analysis of FCFS Average wait time is very variable –If the burst times vary greatly Convoy effect Low device utilisation –If a CPU bound process is active for long periods, other devices are not being used
Shortest job first Assign CPU to the process that has the smallest next CPU-burst time Non Preemptive or Preemptive ProcessArrival Time (ms) Burst Time (ms) P108 P205 P327 P422
ProcessArrival Time (ms) Burst Time (ms) P108 P205 P327 P422 Example: SJF Non Preemptive P2P4 5 0 72214 P1P3 Average Wait Time = (14+0+5+3)/4 = 5.5
Exercise ProcessArrival Time (ms) Burst Time (ms) P108 P205 P327 P422 Using SJF Preemptive What is the average wait time? P2P4 4 0 72214 P1P3 2 P2 Average wait time = (14+2+5+0)/3 = 5.25
Analysis of SJF Optimal with respect to average waiting time How does scheduler know the length of the next CPU-burst time?