Lecture 2 Part 3 CPU Scheduling Operating System

Lecture 2 Part 3 CPU Scheduling Operating System Concepts Essentials – 2 nd Edition Silberschatz, Galvin and Gagne © 2013

Basic Concepts n Maximum CPU utilization obtained with multiprogramming n CPU–I/O Burst Cycle – Process execution consists of a cycle of CPU execution and I/O wait n CPU burst followed by I/O burst n CPU burst distribution is of main concern Operating System Concepts Essentials – 2 nd Edition 3. 2 Silberschatz, Galvin and Gagne © 2013

Histogram of CPU-burst Times Operating System Concepts Essentials – 2 nd Edition 3. 3 Silberschatz, Galvin and Gagne © 2013

CPU Scheduler n Short-term scheduler selects from among the processes in ready queue, and allocates the CPU to one of them l Queue may be ordered in various ways n 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 n Scheduling under 1 and 4 is nonpreemptive n All other scheduling is preemptive l Consider access to shared data l Consider preemption while in kernel mode l Consider interrupts occurring during crucial OS activities Operating System Concepts Essentials – 2 nd Edition 3. 4 Silberschatz, Galvin and Gagne © 2013

Dispatcher n Dispatcher module gives control of the CPU to the process selected by the short-term scheduler; this involves: l switching context l switching to user mode l jumping to the proper location in the user program to restart that program n Dispatch latency – time it takes for the dispatcher to stop one process and start another running Operating System Concepts Essentials – 2 nd Edition 3. 5 Silberschatz, Galvin and Gagne © 2013

Scheduling Criteria n CPU utilization – keep the CPU as busy as possible n Throughput – # of processes that complete their execution per time unit n Turnaround time – amount of time to execute a particular process n Waiting time – amount of time a process has been waiting in the ready queue n 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) Operating System Concepts Essentials – 2 nd Edition 3. 6 Silberschatz, Galvin and Gagne © 2013

Scheduling Algorithm Optimization Criteria n Max CPU utilization n Max throughput n Min turnaround time n Min waiting time n Min response time Operating System Concepts Essentials – 2 nd Edition 3. 7 Silberschatz, Galvin and Gagne © 2013

First- Come, First-Served (FCFS) Scheduling Process Burst Time P 1 24 P 2 3 P 3 3 n Suppose that the processes arrive in the order: P 1 , P 2 , P 3 The Gantt Chart for the schedule is: n Waiting time for P 1 = 0; P 2 = 24; P 3 = 27 n Average waiting time: (0 + 24 + 27)/3 = 17 Operating System Concepts Essentials – 2 nd Edition 3. 8 Silberschatz, Galvin and Gagne © 2013

FCFS Scheduling (Cont. ) Suppose that the processes arrive in the order: P 2 , P 3 , P 1 n The Gantt chart for the schedule is: n Waiting time for P 1 = 6; P 2 = 0; P 3 = 3 n Average waiting time: (6 + 0 + 3)/3 = 3 n Much better than previous case Operating System Concepts Essentials – 2 nd Edition 3. 9 Silberschatz, Galvin and Gagne © 2013

Example Operating System Concepts Essentials – 2 nd Edition 3. 10 Silberschatz, Galvin and Gagne © 2013

Shortest-Job-First (SJF) Scheduling n Associate with each process the length of its next CPU burst l Use these lengths to schedule the process with the shortest time n SJF is optimal – gives minimum average waiting time for a given set of processes l The difficulty is knowing the length of the next CPU request Could ask the user Two schemes: – Non-preemptive – once CPU assigned, process not preempted until its CPU burst completes – Can be preemptive – if a new process with CPU burst less than remaining time of current, preempt l • Operating System Concepts Essentials – 2 nd Edition 3. 11 Silberschatz, Galvin and Gagne © 2013

Example of SJF (Non-preemptive) Process. Arrival Time Burst Time P 1 0. 0 6 P 2 2. 0 8 P 3 4. 0 7 P 4 5. 0 3 n SJF scheduling chart n Average waiting time = (3 + 16 + 9 + 0) / 4 = 7 Operating System Concepts Essentials – 2 nd Edition 3. 12 Silberschatz, Galvin and Gagne © 2013

Example of Nonpreemptive SJF Process Arrival Time Burst Time P 1 0. 0 7 P 2 2. 0 4 P 3 4. 0 1 P 4 5. 0 4 P 1 P 3 n SJF 0 3 7 P 2 8 P 4 12 16 n Average waiting time = (0 + 6 + 3 + 7)/4 = 4 Operating System Concepts Essentials – 2 nd Edition 3. 13 Silberschatz, Galvin and Gagne © 2013

Example of Preemptive SJF Process Arrival Time Burst Time P 1 0. 0 7 P 2 2. 0 4 P 3 4. 0 1 P 4 5. 0 4 n SJF (preemptive) P 1 0 P 2 2 P 3 4 P 2 5 P 4 7 P 1 11 16 n Average waiting time = (9 + 1 + 0 +2)/4 = 3 Operating System Concepts Essentials – 2 nd Edition 3. 14 Silberschatz, Galvin and Gagne © 2013

Determining Length of Next CPU Burst n Can only estimate the length – should be similar to the previous one l Then pick process with shortest predicted next CPU burst n Can be done by using the length of previous CPU bursts, using exponential averaging n Commonly, α set to ½ n Preemptive version called shortest-remaining-time-first Operating System Concepts Essentials – 2 nd Edition 3. 15 Silberschatz, Galvin and Gagne © 2013

Prediction of the Length of the Next CPU Burst Operating System Concepts Essentials – 2 nd Edition 3. 16 Silberschatz, Galvin and Gagne © 2013

Examples of Exponential Averaging n =0 n+1 = n l Recent history does not count n =1 l n+1 = tn l Only the actual last CPU burst counts n If we expand the formula, we get: n+1 = tn+(1 - ) tn -1 + … +(1 - )j tn -j + … l +(1 - )n +1 0 n Since both and (1 - ) are less than or equal to 1, each successive term has less weight than its predecessor Operating System Concepts Essentials – 2 nd Edition 3. 17 Silberschatz, Galvin and Gagne © 2013

Example of Shortest-remaining-time-first n Now we add the concepts of varying arrival times and preemption to the analysis Process. Aarri Arrival Time. T Burst Time P 1 0 8 P 2 1 4 P 3 2 9 P 4 3 5 n Preemptive SJF Gantt Chart n Average waiting time = [(10 -1)+(17 -2)+(5 -3)]/4 = 26/4 = 6. 5 msec Operating System Concepts Essentials – 2 nd Edition 3. 18 Silberschatz, Galvin and Gagne © 2013

Example n Draw a Gantt chart that shows the completion times for each process using shortest-job first (preemptive) CPU scheduling and compute the average waiting time? Operating System Concepts Essentials – 2 nd Edition 3. 19 Silberschatz, Galvin and Gagne © 2013

Priority Scheduling n A priority number (integer) is associated with each process n The CPU is allocated to the process with the highest priority (smallest integer highest priority) l Preemptive l Nonpreemptive n SJF is priority scheduling where priority is the inverse of predicted next CPU burst time n Problem Starvation – low priority processes may never execute n Solution Aging – as time progresses increase the priority of the process Operating System Concepts Essentials – 2 nd Edition 3. 20 Silberschatz, Galvin and Gagne © 2013

Example of Priority Scheduling Process. A arri Burst Time. T Priority P 1 10 3 P 2 1 1 P 3 2 4 P 4 1 5 P 5 5 2 n Priority scheduling Gantt Chart P 2 0 1 P 5 P 1 6 P 3 16 P 4 18 19 n Average waiting time = 8. 2 msec Operating System Concepts Essentials – 2 nd Edition 3. 21 Silberschatz, Galvin and Gagne © 2013

Operating System Concepts Essentials – 2 nd Edition 3. 22 Silberschatz, Galvin and Gagne © 2013

Round Robin (RR) n Each process gets a small unit of CPU time (time quantum q), usually 10 -100 milliseconds. After this time has elapsed, the process is preempted and added to the end of the ready queue. n If there are n processes in the ready queue and the time quantum is q, then each process gets 1/n of the CPU time in chunks of at most q time units at once. No process waits more than (n-1)q time units. n Timer interrupts every quantum to schedule next process Operating System Concepts Essentials – 2 nd Edition 3. 23 Silberschatz, Galvin and Gagne © 2013

Example of RR Process Burst Time P 1 24 P 2 3 P 3 3 n Round Robin, quantum=4, no priority-based preemption n The Gantt chart is: Average wait = ((10 -4)+ 4 + 7 )/3 = 5. 7 msec n Typically, higher average turnaround than SJF, but better response Operating System Concepts Essentials – 2 nd Edition 3. 24 Silberschatz, Galvin and Gagne © 2013

Time Quantum and Context Switch Time Operating System Concepts Essentials – 2 nd Edition 3. 25 Silberschatz, Galvin and Gagne © 2013

Turnaround Time Varies With The Time Quantum 80% of CPU bursts should be shorter than q Operating System Concepts Essentials – 2 nd Edition 3. 26 Silberschatz, Galvin and Gagne © 2013

Example of RR with Arrival Time Process Arrival Burst Time P 1 0 P 2 1 P 3 3 n Round Robin, quantum=2 n The Gantt chart is: P 1 0 p 2 2 p 3 4 3 5 4 p 1 6 p 2 7 p 3 9 p 2 11 12 n Average wait = (6 -2) + (2 -1) + (7 -4) +(11 -9) +(4 -3) + (9 -6)= 4. 7 msec Operating System Concepts Essentials – 2 nd Edition 3. 27 Silberschatz, Galvin and Gagne © 2013

Example of RR n Consider the following processes with arrival time and burst time. Calculate average turnaround time, average waiting time and average response time using round robin with time quantum 3? n Solution: Operating System Concepts Essentials – 2 nd Edition Process id P 1 P 2 P 3 P 4 P 5 P 6 Arrival time 5 4 3 1 2 6 3. 28 Burst time 5 6 7 9 2 3 Silberschatz, Galvin and Gagne © 2013

Cont. Process id Arrival time Burst time Completion time Turnaround time Waiting time Response time P 1 5 5 32 27 22 10 P 2 4 6 27 23 17 5 P 3 3 7 33 30 23 3 P 4 1 9 30 29 20 0 P 5 2 2 6 4 2 2 P 6 6 3 21 15 12 12 Average turnaround time= (27+23+30+29+4+15) / 6= 21. 33 Average waiting time= (22+17+23+20+2+12) / 6= 16 Average response time= (10+5+3+0+2+12) / 6= 5. 33 Operating System Concepts Essentials – 2 nd Edition 3. 29 Silberschatz, Galvin and Gagne © 2013

Multilevel Queue n Ready queue is partitioned into separate queues, eg: l foreground (interactive) l background (batch) n Process permanently in a given queue n Each queue has its own scheduling algorithm: l foreground – RR l background – FCFS n Scheduling must be done between the queues: l Fixed priority scheduling; (i. e. , serve all from foreground then from background). Possibility of starvation. l Time slice – each queue gets a certain amount of CPU time which it can schedule amongst its processes; i. e. , 80% to foreground in RR l 20% to background in FCFS Operating System Concepts Essentials – 2 nd Edition 3. 30 Silberschatz, Galvin and Gagne © 2013

Multilevel Queue Scheduling Operating System Concepts Essentials – 2 nd Edition 3. 31 Silberschatz, Galvin and Gagne © 2013

Multilevel Feedback Queue n A process can move between the various queues; aging can be implemented this way n Multilevel-feedback-queue scheduler defined by the following parameters: l number of queues l scheduling algorithms for each queue l method used to determine when to upgrade a process l method used to determine when to demote a process l method used to determine which queue a process will enter when that process needs service Operating System Concepts Essentials – 2 nd Edition 3. 32 Silberschatz, Galvin and Gagne © 2013

Example of Multilevel Feedback Queue n Three queues: l Q 0 – RR with time quantum 8 milliseconds l Q 1 – RR time quantum 16 milliseconds l Q 2 – FCFS n Scheduling l l A new job enters queue Q 0 which is served FCFS 4 When it gains CPU, job receives 8 milliseconds 4 If it does not finish in 8 milliseconds, job is moved to queue Q 1 At Q 1 job is again served FCFS and receives 16 additional milliseconds 4 If it still does not complete, it is preempted and moved to queue Q 2 Operating System Concepts Essentials – 2 nd Edition 3. 33 Silberschatz, Galvin and Gagne © 2013

Algorithm Evaluation n How to select CPU-scheduling algorithm for an OS? n Determine criteria, then evaluate algorithms n Deterministic modeling l Type of analytic evaluation l Takes a particular predetermined workload and defines the performance of each algorithm for that workload n Consider 5 processes arriving at time 0: Operating System Concepts Essentials – 2 nd Edition 3. 34 Silberschatz, Galvin and Gagne © 2013

Deterministic Evaluation n For each algorithm, calculate minimum average waiting time n Simple and fast, but requires exact numbers for input, applies only to those inputs l FCS is 28 ms: l Non-preemptive SFJ is 13 ms: l RR is 23 ms: Operating System Concepts Essentials – 2 nd Edition 3. 35 Silberschatz, Galvin and Gagne © 2013

Queueing Models n Describes the arrival of processes, and CPU and I/O bursts probabilistically l Commonly exponential, and described by mean l Computes average throughput, utilization, waiting time, etc n Computer system described as network of servers, each with queue of waiting processes l Knowing arrival rates and service rates l Computes utilization, average queue length, average wait time, etc Operating System Concepts Essentials – 2 nd Edition 3. 36 Silberschatz, Galvin and Gagne © 2013

Little’s Formula n n = average queue length n W = average waiting time in queue n λ = average arrival rate into queue n Little’s law – in steady state, processes leaving queue must equal processes arriving, thus: n=λx. W l Valid for any scheduling algorithm and arrival distribution n For example, if on average 7 processes arrive per second, and normally 14 processes in queue, then average wait time per process = 2 seconds Operating System Concepts Essentials – 2 nd Edition 3. 37 Silberschatz, Galvin and Gagne © 2013

Simulations n Queueing models limited n Simulations more accurate l Programmed model of computer system l Clock is a variable l Gather statistics indicating algorithm performance l Data to drive simulation gathered via 4 Random number generator according to probabilities 4 Distributions 4 Trace defined mathematically or empirically tapes record sequences of real events in real systems Operating System Concepts Essentials – 2 nd Edition 3. 38 Silberschatz, Galvin and Gagne © 2013

Evaluation of CPU Schedulers by Simulation Operating System Concepts Essentials – 2 nd Edition 3. 39 Silberschatz, Galvin and Gagne © 2013

End of Chapter 6 Operating System Concepts Essentials – 2 nd Edition Silberschatz, Galvin and Gagne © 2013