1. EECE.4810/EECE.5730 Operating Systems
Instructor:
Dr. Michael Geiger
Spring 2019
Lecture 18:
Scheduling

2. Operating Systems: Lecture 18
Lecture outline
Announcements/reminders
Program 2 due 3/20
Exam 2: Monday, 4/1, 3-5 PM, Ball 214
Use same poll as before to request alt. exam (I’ll repost link)
Today’s lecture
Review
Deadlock
Scheduling
Scheduling metrics
Scheduling algorithms
7/9/2019
Operating Systems: Lecture 18

3. Operating Systems: Lecture 18
Review: Deadlock
Cyclical wait for resources, which prevents involved threads from making progress
Necessary conditions
Limited resource: not enough to serve all threads simultaneously
Hold and wait: threads hold resources while waiting to acquire other resources
No preemption: thread system can’t force one thread to give up resources
Cyclical chain of requests
7/9/2019
Operating Systems: Lecture 18

4. Review: Deadlock prevention
Eliminate one of four necessary conditions
Increase resources to decrease waiting
Eliminate hold and wait: move resource acquisition to beginning
If can’t (atomically) get all resources, release everything
Phase 1a: acquire all resources
Phase 1b: while (!done) {work}
Phase 2: release all resources
7/9/2019
Operating Systems: Lecture 18

5. Review: Banker’s Algorithm
Similar to reserving all resources at beginning, but with more concurrency
State maximum resource needs in advance (without acquiring)
May block when thread attempts to acquire resource
General structure
Phase 1a: state maximum resource need
Phase 1b: while (!done) {
acquire some resource (block
if not safe)
work }
Phase 2: release all resources
7/9/2019
Operating Systems: Lecture 18

6. Operating Systems: Lecture 18
Scheduling basics
Multiprogramming maximizes CPU utilization
Process execution is a cycle of CPU execution and I/O wait
CPU burst followed by I/O burst
CPU burst distribution is of main concern
7/9/2019
Operating Systems: Lecture 18

7. Operating Systems: Lecture 18
CPU Scheduler
Short-term scheduler selects from among the processes in ready queue, and allocates the CPU to one of them
Queue may be ordered in various ways
CPU scheduling decisions may take place when a process:
Switches from running to waiting state
Switches from running to ready state
Switches from waiting to ready
Terminates
Scheduling under 1 and 4 is nonpreemptive
Currently running process “chooses” to give up CPU
All other scheduling is preemptive
Scheduler forces current process to give up CPU
7/9/2019
Operating Systems: Lecture 18

8. Operating Systems: Lecture 18
Dispatcher
Dispatcher module gives control of the CPU to the process selected by the short-term scheduler; this involves:
switching context
switching to user mode
jumping to the proper location in the user program to restart that program
Dispatch latency – time it takes for the dispatcher to stop one process and start another running
7/9/2019
Operating Systems: Lecture 18

9. Operating Systems: Lecture 18
Scheduling Criteria
Several possible, often conflicting goals
Want to maximize
CPU utilization: keep CPU as busy as possible
Throughput: rate at which processes complete per time unit
Fairness: ensure CPU shared (relatively) equally
Want to minimize
Turnaround time: amount of time to execute a particular process, from arrival to completion (includes waiting time)
Sometimes called latency or response time …
… although our text defines response time as time to first “response” (output) from program, not completion
Waiting time: amount of time a process has been waiting in the ready queue
Starvation: Thread/process does not get access to resources
Want to avoid, not just minimize!
7/9/2019
Operating Systems: Lecture 18

10. First-Come, First-Served (FCFS) Scheduling
Sometimes called first-in, first-out (FIFO)
Schedule tasks in the order they arrive
Continue running them until they complete or give up the processor—no preemptions
Threads can call yield() or block
Major benefit: simplest scheduling algorithm
On what workloads is FCFS particularly bad?
Always happens to me -- you go to the store to buy a carton of milk, you get stuck behind someone with a huge basket? And insists on paying in pennies. Feature -- gives you time to read the National Enquirer. Computer science professor has space alien's baby.
7/9/2019
Operating Systems: Lecture 18

11. FCFS Scheduling (continued)
Process Burst Time
P1 24
P2 3
P3 3
Suppose that the processes arrive at time 0 in the order: P1 , P2 , P3
What is the waiting time for each process?
What is the turnaround time for each process?
What are the average waiting and turnaround times?
7/9/2019
Operating Systems: Lecture 18

12. FCFS Scheduling (continued)
Process Burst Time
P1 24
P2 3
P3 3
Suppose that the processes arrive at time 0 in the order: P1 , P2 , P3
The Gantt Chart for the schedule is:
Waiting time for P1 = 0; P2 = 24; P3 = 27
Average waiting time: ( )/3 = 17
Average turnaround time: ( )/3 = 27
7/9/2019
Operating Systems: Lecture 18

13. FCFS Scheduling (continued)
Suppose that the same processes arrive at time 0 in the order:
P2 , P3 , P1
What is the waiting time for each process?
What is the turnaround time for each process?
What are the average waiting and turnaround times?
7/9/2019
Operating Systems: Lecture 18

14. FCFS Scheduling (continued)
Suppose that the processes arrive at time 0 in the order:
P2 , P3 , P1
The Gantt chart for the schedule is:
Waiting time for P1 = 6; P2 = 0; P3 = 3
Average waiting time: ( )/3 = 3
Average turnaround time: ( )/3 = 13
Much better than previous case
Convoy effect - short process behind long process
Consider one CPU-bound and many I/O-bound processes
7/9/2019
Operating Systems: Lecture 18

15. Shortest Job First (SJF)
Always do the task that has the shortest remaining amount of work to do
Suppose we have five tasks arrive one right after each other, but the first one is much longer than the others
Which completes first in FCFS? Next?
Which completes first in SJF? Next?
Need to be careful: optimal wrt average response time.
Recall: only preemptive schedulers.
7/9/2019
Operating Systems: Lecture 18

16. SJF Scheduling (continued)
Associate with each process the length of its next CPU burst
Use these lengths to schedule the process with the shortest time
SJF is “optimal” – gives minimum average waiting time for a given set of processes
Is SJF optimal by all metrics?
Difficulty: knowing length of the next CPU burst
7/9/2019
Operating Systems: Lecture 18

17. Operating Systems: Lecture 18
FIFO vs. SJF
Now you can see why SJF improves average response time – it runs short jobs first. Effect on the short jobs is huge; effect on the long job is small.
7/9/2019
Operating Systems: Lecture 18

18. Operating Systems: Lecture 18
SJF Example
ProcessAr Time Burst Time P P P P
What is the waiting time and turnaround time for each process?
What are the average waiting and turnaround times?
7/9/2019
Operating Systems: Lecture 18

19. Operating Systems: Lecture 18
SJF Example
ProcessAr Time Burst Time P P P P
SJF scheduling chart
Average waiting time = ( ) / 4 = 7
Average turnaround time = ( ) / 4 = 13
7/9/2019
Operating Systems: Lecture 18

20. Determining Length of Next CPU Burst
Can only estimate the length – should be similar to the previous one
Then pick process with shortest predicted next CPU burst
Can be done by using the length of previous CPU bursts, using exponential averaging
Commonly, α set to ½
Preemptive version: shortest-remaining-time-first (SRTF) or shortest time to completion first (STCF)
7/9/2019
Operating Systems: Lecture 18

21. Operating Systems: Lecture 18
STCF example
Now we add the concepts of varying arrival times and preemption to the analysis
ProcessA arri Arrival TimeT Burst Time
P1 0 8
P2 1 4
P3 2 9
P4 3 5
What are the waiting and turnaround times for each process?
What are the average waiting and turnaround times?
7/9/2019
Operating Systems: Lecture 18

22. Operating Systems: Lecture 18
STCF example
Now we add the concepts of varying arrival times and preemption to the analysis
ProcessA arri Arrival TimeT Burst Time
P1 0 8
P2 1 4
P3 2 9
P4 3 5
Preemptive SJF (STCF) Gantt Chart
Average waiting time = [(10-1)+(1-1)+(17-2)+(5-3)]/4 = 26/4 = 6.5
Average turnaround time = [17 + (5-1) + (26-2) + (10 – 3)]/4 = 13
7/9/2019
Operating Systems: Lecture 18

23. Operating Systems: Lecture 18
Priority Scheduling
Priority number (integer) associated with process
Can be preemptive or non-preemptive
CPU allocated to the process with the highest priority (smallest integer = highest priority)
SJF is priority scheduling where priority is inverse of predicted next CPU burst time
Potential problems? Solution?
Problem: starvation for low-priority jobs
Solution: aging  priority increases longer job stays in queue
7/9/2019
Operating Systems: Lecture 18

24. Priority Scheduling Example
ProcessA arri Burst TimeT Priority
P1 10 3
P2 1 1
P3 2 4
P4 1 5
P5 5 2
Priority scheduling Gantt Chart
Average waiting time = 8.2
Average turnaround time = 12 P2 P5
7/9/2019
Operating Systems: Lecture 18

25. Operating Systems: Lecture 18
Round Robin (RR)
Each task gets resource for a fixed period of time (time quantum)
On order of ms
If task doesn’t complete, it goes back in line
If n processes in ready queue and time quantum is q, each process gets 1/n of the CPU time in chunks of <= q time units at once.
No process waits more than (n-1)q time units.
Need to pick a time quantum
What if time quantum is too long?
Degenerates to FCFS
What if time quantum is too short?
Context switching overhead causes delays
Want tradeoff—quantum long enough to dwarf switching time
How can we impose priority in round robin?
Add task to ready queue multiple times  multiple quanta
Can we combine best of both worlds? RR approximates SJF by moving long tasks to the end of the line.
7/9/2019
Operating Systems: Lecture 18

26. Round Robin (continued)
Approximation depends on granularity
7/9/2019
Operating Systems: Lecture 18

27. Operating Systems: Lecture 18
RR example (quantum = 4)
Process Burst Time
P1 24
P2 3
P3 3
The Gantt chart is:
Typically, higher average turnaround than SJF, but better response
q should be large compared to context switch time
q usually 10ms to 100ms, context switch < 10 usec
7/9/2019
Operating Systems: Lecture 18

28. Time Quantum and Context Switch Time
7/9/2019
Operating Systems: Lecture 18

29. Operating Systems: Lecture 18
Round Robin vs. FCFS
Assuming zero-cost time slice, is Round Robin always better than FCFS?
When would FCFS be better?
Same-sized jobs
Time-slicing for no reason
Every single job finishes late!
Show of hands!
What’s the worst case for RR? Same sized jobs – then you are time-slicing for no purpose. Worse, this is nearly pessimal for average response time.
7/9/2019
Operating Systems: Lecture 18

30. Operating Systems: Lecture 18
Round Robin vs. FIFO
Everything is fair, but average response time in this case is awful – everyone finishes very late! In fact, this case is exactly when FIFO is optimal, RR is poor.
On the other hand, if we’re running streaming video, RR is great – everything happens in turn. SJF maximizes variance. But RR minimizes it.
7/9/2019
Operating Systems: Lecture 18

31. Operating Systems: Lecture 18
Example
Consider following processes with the length of the CPU burst time given in milliseconds:
Process Burst Priority P P P P P
Assume processes arrive at same time, in order P1  P5
What is turnaround time of each process for:
FCFS
Round Robin (quantum = 1)
SJF
Non-preemptive priority
7/9/2019
Operating Systems: Lecture 18

32. Operating Systems: Lecture 18
Solution
FCFS RR
SJF
Priority P1
1  10
1  19
10  19
7  16 P2
11 11
2  2
1  1 P3
12  13
3  7
3  4
17  18 P4
14  14
4  4
19  19 P5
15  19
5  14
5  9
2  6
Each entry shows start/end time under given scheduling algorithm
Start at beginning of given time step, end at end of given time step
So process with burst time of 1 (i.e., P2, P4) will appear to start and end in same “cycle”
7/9/2019
Operating Systems: Lecture 18

33. Operating Systems: Lecture 18
Example 2
Process
Burst
Priority
Arrival time P1 10 1 P2 3 4 P3 7 2 P4 P5 5 6
Now consider processes with different arrival times
Assume all times in ms; process can start 1 ms after it arrives (e.g., process arriving at time 0 can start at time 1)
What is turnaround time of each process for:
FCFS
SJF
STCF (remember, this scheme is preemptive!)
RR (time quantum: 1 ms; assume ready queue ordered based on arrival order)
Non-preemptive priority
7/9/2019
Operating Systems: Lecture 18

34. Operating Systems: Lecture 18
Example solution
FCFS
SJF
STCF RR
Priority
Proc St
End TT P1 1 10 17 26 P2 11 13 3 2 24 P3 14 20 18 4 8 9 23 21 15 P4 7 5 P5 22 12 16 19
7/9/2019
Operating Systems: Lecture 18

35. Operating Systems: Lecture 18
Example solution (2)
Detailed STCF schedule:
Time 4 5
Process (time)
P2 (1-3)
P3 (1)
P4 (1)
P3 (2-7)
P5 (1-5)
P1 (1-10)
Detailed RR schedule:
Time 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Proc P1 P2 P3 P4 P5 17 18 19 20 21 22 23 24 25 26
7/9/2019
Operating Systems: Lecture 18

36. Operating Systems: Lecture 18
Real-time scheduling
Focus thus far on average wait time/turnaround time
Real-time systems require tasks to meet deadlines
Video or audio output
Control of physical systems
Requires worst-case analysis
How do we schedule for deadlines in life?
7/9/2019
Operating Systems: Lecture 18

37. Earliest Deadline First Scheduling (EDF)
Priorities assigned according to deadlines
Preempt current job if new job arrives with earlier deadline
Optimal: will meet all deadlines if possible
7/9/2019
Operating Systems: Lecture 18

38. Operating Systems: Lecture 18
Final notes
Next time
Memory management (likely starting Lec. 21)
Reminders:
Program 2 due 3/20
Exam 2: Monday, 4/1, 3-5 PM, Ball 214
Use same poll as before to request alt. exam (I’ll repost link)
7/9/2019
Operating Systems: Lecture 18

39. Operating Systems: Lecture 18
Acknowledgements
These slides are adapted from the following sources:
Silberschatz, Galvin, & Gagne, Operating Systems Concepts, 9th edition
Anderson & Dahlin, Operating Systems: Principles and Practice, 2nd edition
Chen & Madhyastha, EECS 482 lecture notes, University of Michigan, Fall 2016
7/9/2019
Operating Systems: Lecture 18