BATCH SCHEDULING
ALGORITHMS
Module 2.5
COP4600 – Operating Systems
Richard Newman
SCHEDULING IN BATCH SYSTEMS
• Algorithms
• First-Come First-Served
• Shortest Job First
• Shortest Remaining Time Next
• Non-preemptive Priority
• Preemptive Priority
• Metrics
• Mean Turnaround Time
• Processor Utilization
Tanenbaum & Bos, Modern Operating Systems:4th ed., (c) 2013 Prentice-Hall, Inc. All rights reserved.
FIRST-COME FIRST-SERVE (FCFS)
A
A arrives at time 0, take 9 units
B arrives at time 1, takes 4 units
B
C
C arrives at time 2, takes 3 units
What is FCFS Schedule?
B arrives at time 1 C arrives at time 2
C must wait
B must wait
A finishes at time 9 B finishes at time 13
A running
Now B can run
Now C can run
B waiting
B running
C finishes at time 16
C waiting
C runs
All done!
Schedule
A arrives at time 0
A gets CPU
A
B
C
MTT= (9+12+14)/3= 35/3
TT(A) = 9 – 0 = 9
TT(B) = 13 – 1 = 12
TT(C) = 16 – 2 = 14
SHORTEST JOB FIRST (SJF)
A arrives at time 0, take 9 units
A
B arrives at time 1, takes 4 units
B
C arrives at time 2, takes 3 units
C
What is SJF Schedule?
Schedule
A arrives at time 0
A gets CPU
C arrives at time 2
C must wait
B arrives at time 1
B must wait
A
A A
B
B
C
C goes next – why?
C has shorter run time
A finishes at time 9
Who gets CPU next?
A running
B waiting
wait…
C waiting
C runs
TT(A) = 9 – 0 = 9
C finishes at time 12
Now B runs..
And finishes at time 16
B running
TT(C) = 12 – 2 = 10
MTT= (9+10+15)/3= 34/3
TT(B) = 16 – 1 = 15
SJF IS OPTIMAL… IF ALL JOBS READY
Assume all jobs arrive at time 0 (i.e., all jobs are ready)
Let jobs be numbered in order of time required
C1 < C2 < … < CN
SJF schedule is J1, J2, … , JN
Suppose “optimal” schedule S is not SJF
At some point, there must be long job before shorter job
S
J1
J2
…
SJF
J1
J2
…
…
Ji+k
Ji
…
Ji
Ji+k
…
…
Swap job Ji+k with Ji in “optimal” schedule S
All jobs before or after swap finish at same time as before
Ji+k finishes at same time as Ji finished in S
Ji finishes at earlier time in SJF than Ji+k finished in S
So do all jobs between Ji+k and Ji in S!
SJF has smaller MTT – contradiction!
OPTIMAL NON-PREEMPTIVE
A arrives at time 0, take 9 units
A
B arrives at time 1, takes 4 units
B
C arrives at time 2, takes 3 units
C
What is Optimal Schedule? (using MTT)
Schedule
A arrives at time 0
A waits! – Why?
C arrives at time 2
C must wait – Why?
B arrives at time 1
B runs
A
B
C
A A A waits A waits
B finishes at time 5
Who gets CPU next?
A running
B B runs
C finishes at time 8
Now A finally runs…
and finishes at time 17
Sure beats 34/3!
C waits C runs
TT(B) = 5 – 1 = 4
C goes next – why?
C has shorter run time
TT(C) = 8 – 2 = 6
MTT= (4+6+17)/3= 27/3
TT(A) = 17 – 0 = 17
Tanenbaum & Bos, Modern Operating Systems:4th ed., (c) 2013 Prentice-Hall, Inc. All rights reserved.
SHORTEST REMAINING TIME FIRST (SRTF)
A arrives at time 0, take 9 units
A
B arrives at time 1, takes 4 units
B
C arrives at time 2, takes 3 units
C
SRTF is Optimal Preemptive Schedule (using MTT)
Schedule
A arrives at time 0
A runs
C arrives at time 2
C should wait – Why?
B arrives at time 1
B runs – why?
A
B
C
A A A waits A waits
C goes next – why?
C has shorter run time
B finishes at time 5
Who gets CPU next?
A running
C finishes at time 8
Now A finally runs again
and finishes at time 16
B B runs
C waits C runs
TT(B) = 5 – 1 = 4
TT(C) = 8 – 2 = 6
MTT= (4+6+16)/3= 26/3
TT(A) = 16 – 0 = 16
Tanenbaum & Bos, Modern Operating Systems:4th ed., (c) 2013 Prentice-Hall, Inc. All rights reserved.
SRTF IS OPTIMAL
Similar argument to SJF optimality
But no need to constrain arrival times!
Main assumption: context switching time is zero
Not necessarily a good assumption!
Example: Suppose context switching takes 1 time unit
A arrives at time 0, take 10 units
A
B arrives at time 2, takes 4 units
B
A
A A
A waits 6
B
B B runs 4
OS S
S
A runs 9
S
TT(B) = 7 – 2 = 5
TT(A) = 17 – 0 = 17
MTT = 17 + 5 = 22/2 = 11
Exercises:
First – what is FCFS MTT for example?
Second: Come up with job set for which SRFT is NOT optimal!
FCFS, SJF, SRTF SCHEDULING
Process Name:
Arrival Time (AT):
CPU Burst Length (CT):
A
0
8
B
2
5
C
3
1
D
5
2
E
7
2
FCFS Schedule: A A A A A A A A B B B B B C D D E E
Turnaround times: A:8-0, B:13-2, C:14-3, D:16-5, E: 18-7
Average TT = (8+11+11+11+11)/5 = 52/5 = 10.4
SJF Schedule: A A A A A A A A C D D E E B B B B B
Turnaround times: A:8-0, B:18-2, C:9-3, D:11-5, E: 13-7
Average TT = (8+16+6+6+6)/5 = 42/5 = 8.4
SRTF Schedule: A A B C B D D E E B B B A A A A A A
Turnaround times: A:18-0, B:12-2, C:4-3, D:7-5, E: 9-7
Average TT = (18+10+1+2+2)/5 = 33/5 = 6.6
SJF SCHEDULING
Figure 2-41. An example of shortest job first scheduling.
(a) Running four jobs in the original order (FCFS).
(b) Running them in shortest job (SJF) first order.
What is the mean turn-around time for the set of jobs?
FCFS
A:8, B:12, C:16, D:20
Mean = 56/4 = 14
SJF
A:20, B:4, C:8, D:12
Mean = 44/4 = 11
Tanenbaum & Bos, Modern Operating Systems:4th ed., (c) 2013 Prentice-Hall, Inc. All rights reserved.
PRIORITY SCHEDULING
• Processes assigned a priority
• May be given externally
• May be computed internally
• Non-preemptive
• Current process runs until burst done
• Preemptive
• Newly ready process can preempt
current process, which resumes later
• Generality
• Priority function can be used to model
other algorithms
SUMMARY
• Metrics
• Mean Turnaround Time
•
•
•
•
FCFS
SJF
SRTF
Priority