Operating Systems
CPU Scheduling Algorithms
Note: Some slides and/or pictures are adapted from Lecture slides / Books of
• Dr Mansoor Sarwar.
• Dr Kubiatowicz.
• Dr P. Bhat.
• Dr Hank Levy.
• Dr Indranil Gupta.
• Text Book - OS Concepts by Silberschatz, Galvin, and Gagne.
• Ref Books
• Operating Systems Design and Implementation by Tanenbaum.
• Operating Systems by William Stalling.
• Operating Systems by Colin Ritchie.

• Review of previous Lecture
• What are CPU and IO bursts
• CPU Scheduling and Scheduling Criteria
• Scheduling Algorithms
• FCFS
• SJF & SRTF
• Priority Scheduling
• RR
• VRR
• MQ &MFQ
• Evaluation/Comparison of Scheduling Algorithms
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• Deciding which process/thread should occupy a resource (CPU, disk, etc.)
(CPU (horsepower))
I want to ride it
Whose turn is it?
Process 1
Process 2 Process 3
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• Process execution consists of a cycle of CPU execution and I/O wait
• Processes move back & forth between these two states
• A process execution begins with a CPU burst, followed by an I/O burst and then another CPU burst and so on
• An alternating sequence of CPU & I/O bursts is shown on next slide
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Alternating Sequence of CPU and I/O Bursts
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• When CPU becomes idle the OS must select one of the processes in the Ready Queue to be executed
• CPU Scheduler selects a process from the processes in memory that are ready to execute (from Ready Q), and allocates the CPU to one of them
• The OS code that implements the CPU scheduling algorithm is known as CPU scheduler
• CPU scheduling decisions may take place when a process:
1.
Switches from running to waiting state
2.
Terminates
3.
Switches from running to ready state. (e.g. when time slice of a process expires or an interrupt occurs)
4.
Switches from waiting to ready state. (e.g. on completion of I/O)
5.
On arrival of a new process
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6

In Pre-emptive scheduling the currently running process may be interrupted and moved to the ready state by OS (forcefully)
In Non-preemptive Scheduling , the running process can only lose the processor voluntarily by terminating or by requesting an
I/O. OR, Once CPU given to a process it cannot be preempted until the process completes its CPU burst
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• Dispatcher is an important component involved in CPU scheduling
• The OS code that takes the CPU away from the current process and hands it over to the newly scheduled process is known as the dispatcher
• Dispatcher gives control of the CPU to the process selected by the short-term scheduler
• Dispatcher performs following functions:
– 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 executing another
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• CPU utilization – keep the CPU as busy as possible
• Throughput – # of processes that complete their execution per time unit
• Waiting time – amount of time a process has been waiting in the ready queue
– For Non preemptive Algos = S.T – A.T
– For Preemptive Algos = F.T – A.T – B.T
• Turnaround time – amount of time to execute a particular process.
– Finish Time – Arrival Time
• Response time – amount of time it takes from when a request was submitted until the first response is produced, not output (for time-sharing environment)
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• Max CPU utilization
• Max throughput
• Min turnaround time
• Min waiting time
• Min response time
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– First Come First Serve (FCFS)
– Shortest Job First
– Round Robin
– Priority Scheduling
– Multi Queue & Multi-level Feedback
– Shortest process time
– Guaranteed Scheduling
– Lottery Scheduling
– Fair Sharing Scheduling
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• Simplest CPU scheduling algorithm
• Non preemptive
• The process that requests the CPU first is allocated the
CPU first
• Implemented with a FIFO queue. When a process enters the
Ready Queue, its PCB is linked on to the tail of the Queue.
When the CPU is free it is allocated to the process at the head of the Queue
• Limitations
– FCFS favor long processes as compared to short ones.
(Convoy effect)
– Average waiting time is often quite long
– FCFS is non-preemptive, so it is trouble some for time sharing systems
12
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“
A convoy effect happens when a set of processes need to use a resource for a short time, and one process holds the resource for a long time, blocking all of the other processes. Causes poor utilization of the other resources in the system
”
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13

The process that enters the ready queue first is scheduled first, regardless of the size of its next CPU burst
Example: Process
P1
P2
P3
Burst Time
24
3
3
Suppose that processes arrive into the system in the order:
P1, P2 , P3

Processes are served in the order: P1, P2, P3
The Gantt Chart for the schedule is:
P
1
P
2
P
3
0 24
Waiting times P1 = 0; P2 = 24; P3 = 27
Average waiting time: (0+24+27)/3 = 17
27 30

Suppose that processes arrive in the order: P2 , P3 , P1 .
The Gantt chart for the schedule is:
P
2
P
3
0 3 6
Waiting time for P1 = 6; P2 = 0; P3 = 3
Average waiting time: (6 + 0 + 3)/3 = 3
P
1
30

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Process
P1
P2
P3
Duration/B.T
24
3
4
Order
1
2
3
Arrival Time
0
3
4
0
P2 (3) P3 (4)
24 27
P1 waiting time: 0-0
P2 waiting time: 24-3
P3 waiting time: 27-4
The average waiting time:
(0+21+23)/3 = 14.667
17

• Draw the graph (Gantt chart) and compute average waiting time for the following processes using FCFS
Scheduling algorithm.
Process Arrival time Burst Time
P1 1
P2 5
16
3
P3 6 4
P4 9 2
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• When the CPU is available it is assigned to the process that has the smallest next CPU burst
• If two processes have the same length next CPU bursts,
FCFS scheduling is used to break the tie
Comes in two flavors
• Shortest Job First (SJF)
– It’s a non preemptive algorithm.
• Shortest Remaining Time First (SRTF)
– It’s a Preemptive algorithm.
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Process
P1
P2
P3
P4
0
P4 (3) P1 (6)
3
P4 waiting time: 0-0
P1 waiting time: 3-0
P3 waiting time: 9-0
P2 waiting time: 16-0
Duration/B.T
6
8
7
3
Order
1
2
3
4
Arrival Time
0
0
0
0
P3 (7) P2 (8)
9 16 24
The total running time is: 24
The average waiting time (AWT):
(0+3+9+16)/4 = 7 time units
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Process
P1
P2
Duration
10
2
Order
1
2
Arrival Time
0
2
P1 (2)
0
P2 (2)
P1 (8)
2 4
12
P1 waiting time: 4-2 =2
P2 waiting time: 0
The average waiting time (AWT):
(0+2)/2 = 1
Now run this using SJF!
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Draw the graph (Gantt chart) and compute waiting time and turn around time for the following processes using SJF & SRTF
Scheduling algorithm. For SJF consider no arrival time and consider all processes arrive at time 0 in sequence P1, P2, P3, P4.
Process
P1
P2
P3
P4
Arrival Time
0
1
2
3
4
9
5
Burst Time
8
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Draw the graph (Gantt chart) and compute waiting time and turn around time for the following processes using
SJF & SRTF Scheduling algorithm.
Process
P1
P2
P3
P4
Arrival Time
0
1
2
3
2
3
Burst Time
5
1
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Draw the graph (Gantt chart) and compute waiting time and turn around time for the following processes using
SJF & SRTF Scheduling algorithm.
Process
P1
P2
P3
P4
Arrival Time
0
3
6
9
6
2
Burst Time
9
1
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$100 QUESTION
How to compute the next CPU burst?
• This algorithm cannot be implemented at the level of short-term CPU scheduling. There is no way to know the length of the next CPU burst
• One approach is to try to approximate. We may not know the length of the next CPU burst, but we may be able to predict its value. We expect that the next CPU burst will be similar in length to the previous ones. Thus, by computing an approximation of the length of the next CPU burst, we can pick the process with the shortest predicted CPU burst
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Exponential Averaging
Estimation based on historical data t n
 n
= Actual length of
= Estimate for n th n th CPU burst
CPU burst
 n+1
= Estimate for n+1

, 0 ≤  ≤1 st CPU burst
 n+1
=
 t n
+ (1-

)
 n
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 n+1
=
 t n
+ (1-

)
 n
• Plugging in value for  n
, we get
 n+1
=
 t n
+ (1-

) [
 t n-1
+ (1-

)
 n-1
]
 n+1
=
 t n
+ (1-

)
 t n-1
+ (1-

) 2
 n-1
• Again plugging in value for  n-1
, we get

 n+1 n+1
=
=

 t t n n
+ (1-
+ (1-


)
)

 t t n-1 n-1
+ (1-

) 2 [
 t n-2
+ (1-

)
• Continuing like this results in
2
 t n-2
+ (1-

)

(1-

) 3
 n-2 n-2
]
 n+1
=
 t n
+ (1 -

)
 t n-1
+ …+ (1 
) j
 t n-j
+ … + (1 
) n+1

0
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27

 n+1
=
 t n
+ (1-

)
 n
Lets take two extreme values of 
If  = 0
 n+1
=  n
Next CPU burst estimate will be exactly equal to previous CPU burst estimate.
If  = 1
 n+1
= t n
Next CPU burst estimate will be equal to previous actual CPU burst.
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 n+1
=
 t n
+ (1 -

)
 t n-1
+ …+ (1 
) j
 t n-j
+ … + (1 
) n+1

0
Typical value used for  is ½. With this value, our
(n+1) st estimate is
 n+1
= t n
/2 + t n-1
/2 2 + t n-2
/2 3 + t n-3
/2 4 + …
Note that as we move back in time we are giving less weight-age to the previous CPU bursts. Older histories are given exponentially less weight-age
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