Lecture 2 -Multitasking In The Operating Systems

Question 1

What is a program?

Answer 1

A program is the code a programmer writes.

Question 2

What is a process?

Answer 2

A process is a program in execution, i.e., a particular instance of a program together with its data

Question 3

Where does the OS store information about each process?

Answer 3

The OS stores information about each process in a process control block (PCB) or process descriptor

Question 4

What is Multitasking?

Answer 4

Simultaneous running of two or more processes.

Question 5

What is created by a user via a command?

Answer 5

• When a user initiates a program, OS creates a PCB data structureto represent the execution of this program.
• OS also allocates resources for the process.
• Process consists of machine code in memory and PCB.

Question 6

What is created by a user process, called spawning a process?

Answer 6

• The process that creates a new process is called the parent while the created process child
• The child can also spawn new processes, forming a tree of processes

Question 7

Why to run two or more processes at the same time?

Answer 7

• For convenience of the user (it is fun to work with a text editor, play chess, and watch a video at the same time)
• For a better CPU usage (when a program sends information to a printer, the CPU stays idle; it is good to load it with another program

Question 8

What is the denition and role of interrupts?

Answer 8

• Interrupts: events that cause the CPU to stop the current execution and to start execute the program handling this interrupt.
• The program (let us call it Dispatcher) performs the operation required by the interrupt (say, sending data to a printer) and decides which process is the next to run.

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Question 9

What is the Dispatcher?

Answer 9

After the I/O system call or interrupt handling, control is passed to the dispatcher or Low Level Scheduler. The dispatcher works as below:

1) Is the current process still the most suitable to run? If so, return control to it; Otherwise . . .
2) Save the state of the current process in the PCB
3) Retrieve the state of the most suitable process
4) Transfer control to the newly selected process, at the point indicated by the restored program counter.
* This action of storing the state of current process and (re-)starting another process is called a context change.

Question 10

What are the two main types of interrupts, states of processes?

Answer 10

Clock interrupt:
Initiated by a clock within CPU.
The means to stop run of the current process in order to give another one the right to run.
The process stopped by the Clock interrupt gets to the READY state

I/O interrupt:
Initiated by CPU that sees an I/O instruction in the program or by the I/O device to signal completion of the I/O operation.
When I/O operation starts the corresponding process gets to the BLOCKED.
When the I/O operation completes, the process becomes READY.

The dispatcher can select a process to run only if it is READY.

The running process is said to be in the RUNNING state

Question 11

What is the story of interrupts?

Answer 11

A process can be in states RUNNING (currently running), READY
(ready to run) and BLOCKED (awaiting of completion of I/O).
Interrupts are used to stop the currently running process.
The clock interrupt causes the process to be READY.
The I/O interrupt causes the process to be BLOCKED, upon
completion it becomes READY
The Dispatcher chooses the next process to run from the queue of
READY processes.
The question of SCHEDULING: which process to choose?

Question 13

What is the Scheduling policy denition and type?

Answer 13

Scheduling policy determines:
The next READY process to run.
The amount of time the process is given to occupy the CPU.
Non-preemptive scheduling:
A process releases CPU only if an I/O occurs or it nishes.
In other words, the process cannot be stopped during a regular
computation
Preemptive scheduling:
A process can be stopped at any moment.
To stop the process without I/O request a clock interrupt is used.

Question 14

What are the 3 effciency measures of scheduling policies?

Answer 14

Turnaround time.
CPU usage.
Response time

Question 15

What is Turnaround Time?

Answer 15

Turnaround time of a particular process is its duration, i.e. the difference between the end and the start times.
Turnaround time of a group of processes is the average of their individual turnaround times.
Example:
We have 3 processes with respective start and end times
(100;200), (150;300) and (200;400).
The individual turnaround times are 100, 150, 200.
The average turnaround time is 150

A good scheduling policy keeps average turnaround time as low as possible.

Question 16

What is CPU Usage?

Answer 16

If all the processes are blocked waiting for completion of their I/O operations, the CPU stays idle.

A good scheduling policy minimises the idleness time.

CPU usage: given some some amount of time, the percentage of time during which the CPU is busy.

Clearly, it is good if the scheduling policy maximises the CPU usage.

Question 17

What is Response time?

Answer 17

The OS can recognise some processes as user commands (requests).
In Linux, user requests are commands like cp, cd, ls, and so on.
The response time is average turnaround time of a user
request.
The turnaround time of the rest of programs is not taken into account.
This measure must be minimised in interactive systems.

Question 18

What are the 2 types of operating systems and what do they do?

Answer 18

Batch systems:
Run a series of programs without human intervention.
Examples: scientic computing, salary calculation in big
companies.
Turnaround time and CPU usage are more important measures than Response time

Interactive systems:
Designed to extensively communicate with the user.
Examples: Linux, Windows.
Response time is more important measure than the other two.

Question 19

What is the First come first served (FCFS) policy?

Answer 19

The policy is non-preemptive.
The rst READY process in the queue is given the right to
run.
The process releases the CPU when it nishes or an I/O
interrupt occurs.
If process gets blocked by the I/O interrupt, it returns to the end of the READY queue when it is unblocked.
Suitable for batch systems only.

Question 20

What are the advantages and disadvantages of the FCFS policy?

Answer 20

Advantages:
Easy to implement.
No indenite waiting (when a process can potentially wait
forever for being run)
Disadvantage: turnaround time of short processes may be much longer than their actual completion time.

Question 21

What is the shortest job first (SJF) policy

Answer 21

The policy is non-preemptive.
Requires information about the runtime of programs (the time
the program runs non-interrupted).
From the queue of READY processes chooses the one whose runtime is smallest. The rest is properties is like for FCFS.

Question 22

What are the advantages and disadvantages of the SJF policy?

Answer 22

Advantage: favourable to program with short runtimes.
Disadvantages:
Requires knowledge about runtimes.

Indenite waiting is possible: a long-time process can wait very long if new short-time processes arrive all the time.

Does not take into account the situation when a long time
process is short before completion (i.e. eectively becomes short one)

Question 23

What is the shortest remaining time rst (SRTF) policy?

Answer 23

This is a preemptive policy.
For every process, requires knowledge of the remaining
runtime if runs non interrupted.
Very similar to SJF.
The differences are:
Chooses process with the shortest remaining time instead shortest overall time.
If a process with shorter remaining time than the currently running one arrives at the queue, the currently running process is preempted.

Question 24

What are the advantages and disadvantages of the SRTF policy.

Answer 24

Advantage: even more sensitive to the short processes that the SJF.
Disadvantages: requires non-trivial knowledge and indenite waiting is still possible.

Question 25

What is the Round-Robin scheduling policy?

Answer 25

This is a preemptive policy used in interactive systems.
Each process in turn gets the CPU for a xed time quantum (typically 10 20 ms).

When the quantum nishes: the timeout event occurs and the process is returned to the back of the queue.

If the process blocks waiting for I/O, it is interrupted as well.

Question 26

Explain what the quantum duration for round-robin is?

Answer 26

Context switch: the procedure of replacement the currently running process by another one.

If the quantum is too short, comparable in time with the
context switch, a lot of time is wasted on context switches.

If the quantum is too long then the response time may
become too long.