Os module 2 ba

Scheduling
 In multiprogramming systems, the OS
decides which one to activate when there is
more than one runnable process (in the ready
queue).

 The decision is made by the part of the OS
called scheduler using a scheduling
algorithm.

Scheduling
 Set of policies and mechanisms to control the
order of work to be performed by a computer
system.

Scheduling
 OS must keep the CPU busy as much as
possible by executing a (user) process until it
must wait for an event, and then switch to
another process.

 Maximum CPU Utilization obtained with
multiprogramming.

 Processes alternate between consuming
CPU cycles (CPU-burst) and performing I/O
(I/O-Burst)

Alternating Sequence of CPU And I/O Bursts

Types of Scheduler
1. Long-term (job) scheduler
– admits more jobs when the resource utilization is
low and blocks the incoming jobs from entering
ready queue when utilization is high.

2. Medium-term scheduler (swapper)
– releases suspended processes from memory by
swapping it out when the memory becomes over
committed.

Schedulers (cont.)
3. Short-term (CPU) scheduler (dispatcher) –
controls the CPU sharing among the “ready”
processes.
Usually, the next process to execute is selected
under the following circumstances:
 When a process must wait for an event
 When an event occurs (I/O completion, quantum
expired)

Scheduling Criteria
 The goal is to optimize the system performance,
and yet-provide responsive service.
 Criteria used:
 CPU Utilization – the percentage of time that the CPU
is busy to keep the CPU as busy as possible

CPU Utilization = total process burst time x 100%
total CPU execution time
of all processes

Scheduling Criteria
 System throughput – number of processes completed
per time unit.

Throughtput = number of processes
total CPU execution time of all processes

 Turnaround Time – elapsed time from the time the
process is submitted for execution to the time it is
completed.

TAT = end time – arrival time (original)

Scheduling Criteria
4. Waiting time – amount of time a process has been waiting
in the ready queue.

WT = TAT – BT

5. Response Time – amount of time it takes from when a
process was submitted until the first response is produced,
not output (for time sharing environment). This is the time
taken by the system to respond to a user input.

Scheduling Criteria
6. Balanced Utilization – percentage of time
all resources are utilized
7. Predictability – consistency
8. Fairness – the degree to which all processes
are given equal opportunity to execute.
Avoidance of starvation
9. Priorities – give preferential treatment to
processes with higher priorities.

Note:
 Different algorithms may have different
properties and may favor one type of criterion
over the other. So, the design of a scheduler
usually involves a careful balance of all
requirements and constraints.

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

Scheduling Policies
 Non-preemptive scheduling
 Once the CPU is given to a process it cannot be
preempted or interrupted until its CPU Burst is
completed (no suspension)

 Preemptive scheduling
 Force the current executing processes to release
the CPU on certain events such as clock interrupt,
I/O interrupts or a system call. (with suspension).

Common Scheduling Algorithms

Non-Preemptive Preemptive

FCFS RR

Both Preemptive and Non-preemptive:
SJF and Priority

First-Come, First-Ser ved (FCFS)
Scheduling algorithm
 Also known as First in First out (FIFO)
 Simplest scheduling policy
 Arriving jobs are inserted into the tail of the
ready queue and the process to execute next
is removed from the head of the queue.
 Jobs are scheduled in the order in which they
are received.

FCFS Scheduling algorithm
Process Burst Time
P1 24
P2 3
P3 3

 Suppose that the processes arrive in the order: P1 , P2 ,
P3 . The gantt chart for the schedule is:
P1 P2 P3

0 24 27 30

 Waiting time for: P1 = 0; P2 = 24; P3 = 27
 Average waiting time: (0 + 24 + 27)/3 = 17

FCFS Scheduling (Cont.)
Suppose that the processes arrive in the order
P2 , P3 , P1
 The Gantt chart for the schedule is:

P2 P3 P1

0 3 6 30

 Waiting time for P1 = 6; P2 = 0; P3 = 3
 Average waiting time: (6 + 0 + 3)/3 = 3
 Much better than previous case

Note
 A long CPU-bound job may hog the CPU and
may force shorter jobs to wait prolonged
periods. This may lead to a lengthy queue of
ready jobs (convoy effect)

Examples:
1. Suppose that the following processes arrive for
execution, what is the CPU Utilization,
throughput and average TAT and WT for these
processes with the FCFS scheduling algorithm

a) Process BT b) Process BT
A 24 P1 4
B 10 P2 1
C 3 P3 8

Examples:
2. Calculate the TAT and WT of each process.
Consider the process arrival and burst times of
these processes as shown below.

Process BT AT Process BT AT

A 10 0 P1 8 0.0
B 5 12 P2 4 0.4
C 2 13 P3 3 8.2
D 1 18 P4 2 14.6
E 5 20 P5 1 16.0

Seatwork
 Calculate the CPU Process AT BT
Utilization, average J1 13 4
TAT and WT for the J2 5 6
execution of the
J3 3 6
process below.
Consider their arrival J4 24 10
times and their burst J5 15 7
times J6 18 7
J7 14 8

Process’ CPU and I/O requests
 CPU and I/O requests
use any of the CPU
scheduling algorithm.

 After CPU burst, the I/O
burst follows.

 CPU burst will always be
the LAST to perform for
the completion of the
execution.

Sample problem using CPU and I/O
 With jobs in the ready queue and I/O queue all
scheduled using FCFS, show the Gantt chart
showing the execution of these processes.

Job ID AT CPU I/O CPU

P1 0 10 12 5
P2 3 5 7 4
P3 10 7 10 2

Shor test-Job-First
Scheduling
 Selects job with the shortest (expected) burst
time first. Shorter jobs are always executed
before long jobs

 One major difficulty with SJF is the need to
know or estimate the burst time of each job

Shortest-Job-First Scheduling
 Another difficulty is long running times may
starve because the CPU has a steady supply
of short jobs

 But SJF is optimal – gives minimum WT for a
given set of processes.

 Preemptive SJF is also known as Shortest
Remaining Time First (SRTF)

Two schemes:
 Non-Preemptive – once the CPU has given to
the process it cannot be preempted until it
completes its CPU burst

 Preemptive – if a new process arrives with
CPU burst less than the remaining time of
current executing process, preempt.

 Also known as the Shortest-Remaining-Time-First
(SRTF)

Example of Non-Preemptive SJF
Process Arrival Time Burst Time
P1 0 7
P2 2 4
P3 4 1
P4 5 4
 SJF (non-preemptive)
P1 P3 P2 P4

0 3 7 8 12 16

 Average waiting time = (0 + 6 + 3 + 7)/4 = 4

Example of Preemptive SJF (SRTF)

Process Arrival Time Burst Time
P1 0 7
P2 2 4
P3 4 1
P4 5 4
 SJF (preemptive)
P1 P2 P3 P2 P4 P1

0 2 4 5 7 11 16

 Average waiting time = (9 + 1 + 0 +2)/4 = 3

Your turn
 Obtain the average waiting time and the
turnaround time, including CPU Utilization and
Gantt Chart for the following set of processes
using Non-P SJF, and SRTF

Process BT AT Process BT AT
A 10 0 J1 6 11
B 6 5 J2 8 0
C 7 7 J3 12 4
D 3 6 J4 2 2
E 1 3 J5 5 5

CPU and I/O request
 With jobs in the ready queue scheduled as SRTF and
the jobs in the I/O queue scheduled as FCFS, show
the Gantt chart showing the execution of these
processes. What is the CPU and I/O utilization?

Job ID AT CPU I/O CPU

P1 0 10 12 5
P2 3 5 7 4
P3 10 7 10 2

Priority Scheduling Algorithm (PSA)
 Priority number (integer) is associated with
each process
 CPU is allocated to the process with the
highest priority (smallest integer ≡ highest
priority)
 Preemptive
 Non-Preemptive

PSA
 SJF is a priority scheduling where priority
is the predicted next CPU burst time
 Problem:
 Starvation – low priority processes may never
execute
 Solution:
 Aging – as time progresses increase the
priority of the process

Sample Problem #1:
Process BT PL (HP = 1)
P1 8 4
P2 1 3
P3 3 2
P4 4 1
P4 P3 P2 P1

1
0 2 4 5 7 8 6
Average WT: ( 8+7+4+0 ) / 4 = 4.75
Average TAT: (16+8+7+4 ) / 4 = 8.75

Sample Problem #2:
Process BT PL AT (HPL =1)
P1 8 4 0
P2 1 3 3
P3 3 2 5
P4 4 1 7

Non-Preemptive PSA: P1 P4 P3 P2

0 8 11 12 15 16

Preemptive PSA: P2 P P3 P3
P1 1 P4 P1

0 3 4 5 7 11 12 16

Round Robin (RR) Scheduling
 Process gets a small unit of CPU time (time
quantum), usually 10-100 milliseconds. After
this time has elapsed, the process is preempted
and added to the end of the ready queue.

 If there are n processes in the ready queue and
the time quantum is q, then each process gets 1/
n of the CPU time in chunks of at most q time
units at once. No process waits more than
(n-1)q time units.

Round Robin (RR)
 Performance
 q large ⇒ FIFO
 q small ⇒ q must be large with respect to
context switch, otherwise overhead is too high

Time quantum and context switches

Sample Problem:
Process Burst Time
P1 53
P2 17
P3 68
P4 24

P1 P2 P3 P4 P1 P3 P4 P1 P3 P3
0 20 37 57 77 97 117 121 134 154 162

Your turn:
 Given the following Job AT BT PL
jobs, schedule using HPL=1

PPSA and RR 1 0 18 3
(QT=4). 2 8 8 2
 Answer the queries 3 10 5 2
on the next slide. 4 17 12 1
5 25 8 3
6 30 6 1

Using PPSA, answer the following:
1. What job is currently allocated to the CPU @
time 18?
2. How many jobs are finished at time 24?
3. What is the last job to finish?
4. When did job 2 start executing?
5. When did job 4 start executing?
6. After job 2 totally finished, which job was
allocated to the CPU?
7. What is the waiting time of job 5?
8. What is the finish time of job 6?

Using RR, answer the following:
1. When was job 2 first assigned to the CPU?
2. How many jobs are finished at time 22?
3. Which job was allocated the CPU at time 19?
4. Which job was allocated the CPU at time 29?
5. After job 3 was totally finished, which job was
allocated to the CPU?
6. What is the finish time of job 3?
7. What is the last job to finish?
8. What is the turnaround time of job 1?

Comparison between PPSA and RR:
 Comparing the 2 algorithms:
1. Which algorithm did job 2 start the earliest?
2. Which algorithm did job 5 start the latest?
3. Which algorithm did job 1 finish the earliest?
4. Which algorithm did job 1 finish the latest?

Multilevel Queue
 Ready queue is partitioned into separate
queues:
foreground (interactive) & background
(batch)
 Each queue has its own scheduling
algorithm
 foreground – RR
 background – FCFS
 Serves foreground first then background
Possibility of starvation.

MLQ Example
Job AT BT Type
 Assume that the CPU
1 0 5 B
scheduler uses MLQ
with 2 levels: foreground 2 7 20 B
queue for type F (jobs 3 8 1 F
uses SJF); and 4 9 12 F
background queue for 5 50 10 B
type B (jobs uses FCFS).
6 75 15 B
Scheduling algorithms
between queues is fixed 7 76 2 F
preemptive priority ( F 8 78 2 F
being the highest
priority)

Activity
Assuming the ready queue is partitioned into
2 queues: Background queue - processes
with priorities from 3-5 (uses FCFS); and J BT AT P
Foreground - processes with priorities 1
and 2. (uses RR with Q=6. Algorithm A 12 17 1
between queues is preemptive priority.
Draw the Gantt chart for the CPU. B 15 18 4
C 7 1 2
D 11 20 2
E 4 5 3
F 24 0 5

Multilevel Feedback Queue
 A process can move between the various
queues; aging can be implemented this way
 Multilevel-feedback-queue scheduler defined by
the following parameters:
 number of queues
 scheduling algorithms for each queue
 method used to determine when to upgrade a process
 method used to determine when to demote a process
 method used to determine which queue a process will
enter when that process needs service

Example of Multilevel Feedback Queue
 Three queues:
 Q0 – RR with time quantum 8 milliseconds
 Q1 – RR time quantum 16 milliseconds
 Q2 – FCFS

Multilevel Feedback Queue: Scheduling

 New job enters queue Q0 which is served FCFS. When it
gains CPU, job receives 8 milliseconds. If it does not
finish in 8 milliseconds, job is moved to queue Q1.

 At Q1 job is again served FCFS and receives 16
additional milliseconds. If it still does not complete, it is
preempted and moved to queue Q2.

MLFQ Example
 Schedule using an
Job BT AT
MLFQ:
A 10 4
Q1: RR Q=3 B 13 3
Q2: RR Q=5
C 5 0
Q3: FCFS
D 6 1

E 18 1

Try it!
Job BT AT
 Schedule using an
MLFQ: A 24 32

B 36 15
Q1: RR with Q=4
Q2: RR with Q=10 C 17 0

Q3: FCFS D 9 1

E 15 17

Os module 2 ba

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Os module 2 ba