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CS444/CS544 Operating Systems Scheduling 1/31/2007 Prof. Searleman [email protected] CS444/CS544 Spring 2007 CPU Scheduling Reading assignment: Chapter 5 HW#3: done in lab 2-2-2007 HW#4 posted, due: 2-7-2007 Help for Lab1, 6-7 pm tonight in ITL/COSI Benefits of Concurrency Hide latency of blocking I/O without additional complexity Without concurrency Block whole process Manage complexity of asynchronous I/O (periodically checking to see if it is done so can finish processing) Ability to use multiple processors to accomplish the task Servers often use concurrency to work on multiple requests in parallel User Interfaces often designed to allow interface to be responsive to user input while servicing long operations Scheduling CPU or “short term” scheduler selects process from ready queue (every 10 msec or so) “dispatcher” does the process switching “long-term” scheduler controls “degree of multiprogramming” (number of processes in memory); selects a good “job mix” “job mix” – I/O-bound, CPU-bound, interactive, batch, high priority, background vs. foreground, real-time “non-preemptive” (cooperative) vs. “preemptive” Performance Measures Throughput: #processes/time unit Turnaround time: time completed – time submitted Waiting time: sum of times spent in ready queue Response time: time from submission of a request until the first response is produced Variation of response time (predictability) CPU utilization Disk (or other I/O device) utilization I/O-bound & CPU-bound Device1 P1 CPU time quantum Device2 P2 CPU I/O-bound & CPU-bound P1: CPU-bound Device1 idle Device1 idle CPU idle Device1 idle CPU idle Turnaround time for P1 I/O-bound & CPU-bound P2: I/O-bound Device2 idle Device2 idle CPU idle CPU idle Turnaround time for P2 I/O-bound & CPU-bound Schedule1: non-preemptive, P1 selected first Turnaround time for P1 Turnaround time for P2 Without P1 I/O-bound & CPU-bound Schedule2: non-preemptive, P2 selected first Turnaround time for P1 Turnaround time for P2 I/O-bound & CPU-bound How does the OS know whether a process is I/O-bound or CPU-bound? - can monitor the behavior of a process & save the info in the PCB - example: how much CPU time did the process use in its recent time quanta? (a small fraction => I/O intensive; all of the quantum => CPU intensive) The nature of a typical process changes from I/Obound to CPU-bound and back as it works through its Input/Process/Output Cycle Preemptive vs. Non-Preemptive t0 ready: P1, P2 t2 t1 ready: P2 blocked: P1 Preemptive vs. Non-Preemptive Non-Preemptive: must continue to run P1 at t3 Preemptive: can choose between P1 & P2 at t3 t2 ready: P1 blocked: P2 t3 ready: P2 running: P1 New admit dispatch Ready (2) interrupt (3) I/O completed or event occurs (4) exit, abort Terminated Running (1) block for I/O or wait for event Waiting • nonpreemptive (cooperative): (1) and (4) only • preemptive: otherwise First Come First Serve (FCFS) Also called First In First Out (FIFO) Jobs scheduled in the order they arrive When used, tends to be non-preemptive If you get there first, you get all the resource until you are done “Done” can mean end of CPU burst or completion of job Sounds fair All jobs treated equally No starvation (except for infinite loops that prevent completion of a job) Process CPU burst P1 18 FCFS Gantt chart P2 2 P3 4 P1 0 P3 P2 18 20 P1, P2, P3 ready average waiting time = (0 + 18 + 20)/3 = 12.6 24 Problems with FCFS/FIFO Can lead to poor overlap of I/O and CPU If let first in line run till they are done or block for I/O then can get convoy effect While job with long CPU burst executes, other jobs complete their I/O and the I/O devices sit idle even though they are the “bottleneck” resource and should be kept as busy as possible Also, small jobs wait behind long running jobs (even grocery stores know that) Results in high average turn-around time Shortest Job First (SJF) So if we don’t want short running jobs waiting behind long running jobs, why don’t we let the job with the shortest CPU burst go next Can prove that this results in the minimum (optimal) average waiting time Can be preemptive or non-preemptive Preemptive version called shortest-remaining-time first (SRTF) Process CPU burst P1 18 SJF Gantt chart 0 2 2 P3 4 P1 P3 P2 P2 6 P1, P2, P3 ready average waiting time = (0 + 2 + 6)/3 = 2.6 24 SJF nonpreemptive Process Arrival time CPU burst P1 P2 P3 P4 7 4 1 4 P1 0 2 P1 ready 0 2 4 5 P3 4 5 7 P1 running P2 , P3 ready P2 8 P4 12 P1 running P2 , P3 , P4 ready P1 running P2 ready average waiting time = (0 + 6 + 3 + 7)/4 = 4 16 SRTF preemptive P2 P1 0 2 P1 ready P1 running P2 ready CPU burst P1 P2 P3 P4 7 4 1 4 0 2 4 5 P2 P3 4 Process Arrival time 5 P2 running P1 , P3 ready P4 7 P1 11 16 P1 , P4 ready P3 completes P1 , P2 , P4 ready average waiting time = (9 + 1 + 0 + 2)/4 = 3 P1 ready Process turnaround Time SRTF preemptive 0 2 16 – 0 = 16 7–2=5 5–4=1 11 – 5 = 6 P1 P2 P3 P4 P2 P1 P2 P3 4 5 waiting Time P4 7 9 1 0 2 P1 11 average waiting time = (9 + 1 + 0 + 2)/4 = 3 average turnaround time = (16 + 5 + 1 + 6)/4 16 Problems with SJF First, how do you know which job will have the shortest CPU burst or shortest running time? Can guess based on history but not guaranteed Bigger problem is that it can lead to starvation for long-running jobs If you never got to the head of the grocery queue because someone with a few items was always cutting in front of you