Lecturer 5: Process Scheduling
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Process Scheduling
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Criteria & Objectives
Types of Scheduling
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Long term
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Medium term
Short term
CPU Scheduling Algorithms
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First-Come, First-Served (FCFS)
Shortest Job First (SJF)
Priority (PRI)
Round-Robin (RR)
Operating System Concepts

CPU Scheduling
2
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How is the OS to decide which of several tasks to take off a queue?
Scheduling: deciding which threads are given access to resources from moment to moment.
Operating Systems Lec 3

Chapter 5:
CPU Scheduling
3
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Always want to have CPU (or CPU’s) working
Usually many processes in ready queue
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Ready to run on CPU
Focus on a single CPU here
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Need strategies for
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Allocating CPU time
Selecting next process to run
Deciding what happens after a process completes a system call, or completes I/O
Short-term scheduling
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Must not take much CPU time to do the scheduling

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CPU & I/O Bursts
Process execution is nothing but a sequence of CPU & I/O bursts
Processes alternate between these two states
 CPU bursts distribution

CPU Burst & I/O Burst Alternation (Fig.
In general, most programs have short CPU bursts
What happens when the I/O burst occurs?
What is the process state at that time?

Types of Scheduling
• Long-term scheduling the decision to add to pool of processes to be executed
• Mid-term scheduling the decision to add to the number of processes that are partially or fully in memory
• Short-term scheduling decision as to which available process will be executed
• I/O scheduling decision as to which process’s pending request shall be handled by an available I/O device

•
•
•
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Selects from among the processes in memory that are ready to execute, and allocates the CPU to one of them
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
Scheduling under 1 and 4 is non-preemptive
All other scheduling is preemptive

Preemptive & Non-preemptive Scheduling
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SO what is Preemptive & Non-preemptive scheduling?
Non-preemptive
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Once CPU is allocated to a process, it holds it till it
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Terminates (exits)
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Blocks itself to wait for I/O
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Requests some OS service

Preemptive & Non-preemptive Scheduling
 Preemptive
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Currently running process may be interrupted and moved to the ready state by the OS
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Windows 95 introduced preemptive scheduling
Used in all subsequent versions of windows
Mac OS X uses it

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Dispatcher
Dispatcher module gives control of the CPU to the process selected by the short-term scheduler; this involves:
 switching context
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 switching to user mode jumping to the proper location in the user program to restart that program (IP)
 Dispatch latency – time it takes for the dispatcher to stop one process and start another running

Criteria & Objectives For scheduling
•
•
•
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CPU utilization – keep the CPU as busy as possible
Throughput – # of processes that complete their execution per time unit
Turnaround time – amount of time to execute a particular process (total time spent on the system)
Waiting time – amount of time a process has been waiting in the ready queue
Response time – amount of time it takes from when a request was submitted until the first response is produced, not output

Optimization Criteria
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Max CPU utilization
Max throughput
Min turnaround time
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Min waiting time
Min response time

CPU Scheduling Algorithms
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First-Come, First-Served (FCFS)
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Complete the jobs in order of arrival
Shortest Job First (SJF)
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Complete the job with shortest next CPU burst
Priority (PRI)
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Processes have a priority
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Allocate CPU to process with highest priority
Round-Robin (RR)
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Each process gets a small unit of time on CPU (time quantum or time slice)

FCFS: First-Come First-Served
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14
Ready queue data structure: a FIFO queue
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Assuming we have (at least) a multiprogrammed system
Example
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Draw Gantt chart
Compute the average waiting time for processes with the following next CPU burst times and ready queue order:
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P1: 20
P2: 12
P3: 8
P4: 16
P5: 4
Waiting time:
CPU burst times
E.g., time units are milliseconds
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Time period spent in the ready queue (assume processes terminate)
Also calculate average waiting time across processes
Assume 0 context switch time

0
P1
20
P2 P3
32 40
P4 P5
56 60
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Waiting times?
P1: 0
P2: 20
P3: 32
P4: 40
P5: 56
Average wait time: 148/5 = 29.6

FCFS: First-Come First-Served
16
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Advantage: Relatively simple algorithm
Disadvantage: long waiting times

Scheduling Algorithms: First-Come,
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Example 2: Three processes arrive in order P1, P2, P3.
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P1 burst time: 24
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P2 burst time: 3
P3 burst time: 3
Waiting Time
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P1: 0
P2: 24
P3: 27
Completion Time:
0
P1 P2 P3
24 27 30
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P1: 24
P2: 27
P3: 30
Average Waiting Time: (0+24+27)/3 = 17
Average Completion Time: (24+27+30)/3 = 27
Operating Systems Lec 3

Scheduling Algorithms: First-Come,
 What if their order had been P2, P3, P1?
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P1 burst time: 24
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P2 burst time: 3
P3 burst time: 3
Operating Systems Lec 3

Scheduling Algorithms: First-Come,
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What if their order had been P2, P3, P1?
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P1 burst time: 24
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P2 burst time: 3
P3 burst time: 3
Waiting Time
0
P2 P3
3 6
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P1: 0
P2: 3
P3: 6
Completion Time:
P1
30
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P1: 3
P2: 6
P3: 30
Average Waiting Time: (0+3+6)/3 = 3 (compared to 17)
Average Completion Time: (3+6+30)/3 = 13 (compared to 27)
Operating Systems Lec 3

SJF
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Provably shortest average wait time
BUT: What do we need to actually implement this?

P5 P3
0 4 12
P2
SJF Solution
P4 P1
24 40 60
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Waiting times (how long did process wait before being scheduled on the CPU?):
P1: 40
P2: 12
P3: 4
P4: 24
P5: 0
Average wait time: 16
(Recall: FCFS scheduling had average wait time of 29.6)

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Priority Scheduling
22
Have to decide on a numbering scheme
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0 can be highest or lowest

 Priority scheduling algorithms can suffer from
starvation (indefinite waiting for CPU access)
 In a heavily loaded system, a steady stream of higher-priority processes can result in a low priority process never receiving CPU time
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I.e., it can starve for CPU time
 One solution: aging
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Gradually increasing the priority of a process that waits for a long time

Which CPU Scheduling Algorithms Can be
Preemptive?
24
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FCFS (First-come, First-Served)
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Non-preemptive
SJF (Shortest Job First)
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Can be either
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Choice when a new (shorter) job arrives
Can preempt current job or not
Priority
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Can be either
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Choice when a processes priority changes or when a higher priority process arrives

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RR (Round Robin) Scheduling
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Used in time-sharing or multi-tasking systems
 typical kind of scheduling algorithm in a contemporary general purpose operating system
Method
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Give each process a unit of time (time slice, quantum) of execution on CPU
Then move to next process in ready queue
Continue until all processes completed
Necessarily preemptive
Requires use of timer interrupt
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Time quantum typically between 10 and 100 milliseconds
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Linux default appears to be 100ms

Round Robin (RR) Scheduling
26
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Round Robin Scheme
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Each process gets a small unit of CPU time (time quantum)
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Usually 10-100 ms
After quantum expires, the process is preempted and added to the end of the ready queue
Suppose N processes in ready queue and time quantum is Q ms:
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Each process gets 1/N of the CPU time
In chunks of at most Q ms
What is the maximum wait time for each process?
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No process waits more than (n-1)q time units
Performance Depends on Size of Q
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Small Q => interleaved
Large Q is like FCFS
Q must be large with respect to context switch time, otherwise overhead is too high (spending most of your time context switching!)
Operating Systems Lec 3

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CPU job burst times & order in queue
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P1: 20
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P2: 12
P3: 8
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P4: 16
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P5: 4
Draw Gantt chart, and compute average wait time
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Time quantum of 4
Like our previous examples, assume 0 context switch time

complete s
Solution
28 complete s complete s complete s s
0
P1 P2 P3 P4 P5 P1 P2 P3 P4 P1 P2 P4 P1 P4 P1
4 8 12 16 2
0
2
4
28 3
2
3
6
4
0
44 4
8
52 56 6
0
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Waiting times:
P1: 60 - 20 = 40
P2: 44 - 12 = 32
P3: 32 - 8 = 24
P4: 56 - 16 = 40
P5: 20 - 4 = 16
Average wait time: 30.4

Example 2of RR with Time Quantum = 4
29
Process Burst Time
P
1
P
2
P
3
24
3
3
 The Gantt chart is:
P
1
0 4
P
2
P
3
P
1
P
1
P
1
P
1
P
1
7 10 14 18 22 26 30
Operating Systems Lec 3

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Example of RR with Time Quantum = 4
30
Process Burst Time
P
P
P
1
2
3
24
3
3
P
1
P
2
P
3
P
1
P
1
P
1
P
1
P
1
0 4 7 10 14 18 22 26 30
Waiting Time:
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P1: (10-4) = 6
P2: (4-0) = 4
P3: (7-0) = 7
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Completion Time:
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P1: 30
P2: 7
P3: 10
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Average Waiting Time: (6 + 4 + 7)/3= 5.67
Average Completion Time: (30+7+10)/3=15.67
Operating Systems Lec 3

Performance Characteristics of Scheduling
Algorithms
33
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Different scheduling algorithms will have different performance characteristics
RR (Round Robin)
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Average waiting time often high
Good average response time
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Important for interactive or timesharing systems
SJF
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Best average waiting time
Some overhead with respect to estimates of CPU burst length & ordering ready ‘queue’

Example 3 of RR with Time Quantum = 20
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Waiting Time:
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P1: (68-20)+(112-88) = 72
P2: (20-0) = 20
P3: (28-0)+(88-48)+(125-108) = 85
P4: (48-0)+(108-68) = 88
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Completion Time:
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P1: 125
P2: 28
P3: 153
P4: 112
A process can finish before the time quantum expires, and release the CPU.
Average Waiting Time: (72+20+85+88)/4 = 66.25
Average Completion Time: (125+28+153+112)/4 = 104.5
Operating Systems Lec 3

Comparing FCFS and RR
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35
Assuming zero-cost context switching time, is RR always better than FCFS?
Assume 10 jobs, all start at the same time, and each require
100 seconds of CPU time
Job #
RR scheduler quantum of 1 second
Completion Times (CT)
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Both FCFS and RR finish at the same time
But average response time is much worse under RR!
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Bad when all jobs are same length
1
2
…
9
10
FCFS CT
100
200
…
900
1000
RR CT
1000
Also: cache state must be shared between all jobs with RR but can be devoted to each job with FIFO
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Total time for RR longer even for zero-cost context switch!
991
992
…
999
Operating Systems Lec 3