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CS444/CS544Operating Systems
Scheduling
1/31/2007
Prof. Searleman
jets@clarkson.edu
CS444/CS544 Spring 2007
CPU Scheduling
Reading assignment: Chapter 5
HW#3: done in lab 2-2-2007HW#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-boundDevice1
P1
CPU
P2
CPU
Device2
time quantum
I/O-bound & CPU-bound
P1: CPU-bound
Device1 idle
CPU idle CPU idle
Device1 idle Device1 idle
Turnaround time for P1
I/O-bound & CPU-bound
P2: I/O-bound
Device2 idle
CPU idle CPU idle
Device2 idle
Turnaround time for P2
I/O-bound & CPU-bound
Schedule1: non-preemptive, P1 selected first
Turnaround time for P2
Turnaround time for P1
Without P1
I/O-bound & CPU-bound
Schedule2: non-preemptive, P2 selected first
Turnaround time for P2
Turnaround time for P1
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/O-bound to CPU-bound and back as it works through its Input/Process/Output Cycle
Preemptive vs. Non-Preemptive
t0
ready: P1, P2
t1
ready: P2blocked: P1
t2
Preemptive vs. Non-Preemptive
t3
ready: P2running: P1
t2
ready: P1blocked: P2
Non-Preemptive: must continue to run P1 at t3
Preemptive: can choose between P1 & P2 at t3
New
Ready Running
Waiting
Terminatedadmit dispatch
(4)exit,abort
(2)interrupt
(1)block for I/Oor wait for event
(3)I/O completedor event occurs
• 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)
FCFS
average waiting time = (0 + 18 + 20)/3 = 12.6
P1
18
P2
20
P3
24
Process CPU burst
P1 18
P2 2
P3 4Gantt chart
0
P1, P2, P3ready
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)
SJF
average waiting time = (0 + 2 + 6)/3 = 2.6
P1
24
P2
2
P3
6
Process CPU burst
P1 18
P2 2
P3 4Gantt chart
0
P1, P2, P3ready
SJF
average waiting time = (0 + 6 + 3 + 7)/4 = 4
P1
7
P2
12
P3
80
P1
ready
Process Arrival time
CPU burst
P1 0 7
P2 2 4
P3 4 1
P4 5 4
P4
16
nonpreemptive
2
P1 running
P2 ready
4
P1 running
P2 , P3 ready
5
P1 running
P2 , P3 , P4 ready
SRTF
average waiting time = (9 + 1 + 0 + 2)/4 = 3
P1 P3
0
P1
ready
Process Arrival time
CPU burst
P1 0 7
P2 2 4
P3 4 1
P4 5 4
P4
11
preemptive
2
P1 running
P2 ready
4
P2 running
P1 , P3 ready
5
P3 completes
P1 , P2 , P4 ready
P2 P2
16
P1
7
P1 , P4 ready
P1 ready
SRTF
average waiting time = (9 + 1 + 0 + 2)/4 = 3 average turnaround time = (16 + 5 + 1 + 6)/4
P1 P3
0
Process turnaround
Time
waiting
Time
P1
P2
P3
P4
P4
112 4 5
P2 P2
16
P1
7
16 – 0 = 167 – 2 = 5 15 – 4 = 111 – 5 = 6
9
0
2
preemptive
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