Exam 1 result

Operating System (CS - 308)
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Question 1 Answer	Question 3 Answer	Question 5 Answer	Question 7 Answer
Question 2 Answer	Question 4 Answer	Question 6 Answer

Question 1

Give a short definition of an operating system and list a few function that it provides

                Definition:     ____________________________________________________
                                       ____________________________________________________
                                       ____________________________________________________
                                       ____________________________________________________

                Functions:      ____________________________________________________
                                       ____________________________________________________
                                       ____________________________________________________
                                       ____________________________________________________

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Question 2

List a few criteria that are used in evaluating the performance of the CPU

_________

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Question 3

What is a PCB (Process Control Block)? Elaborate.

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Question 4

Given the following processes, use deterministic analysis and Gantt chart to calculate which CPU scheduling algorithm will be optimal among

a . FCFS method
b . SRTF Algorithm (preempted SJF)
c . Priority Algorithm (non-preempted)
d . Round Robin (RR, time slice = 2)
Also, calculate average waiting time

Process	Arrival Time	Burst Time	Priority	(Only for Priority)
P0	0	5	3
P1	1	1	4
P2	2	6	1
P3	3	4	2

Note that if two processes arrive to the ready queue at the same time, you may have to decide arbitrarily which goes first

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Question 5

What is the completion time of each process for each of the scheduling methods in number 4 (fill in the table)

	P0	P1	P2	P3
FCFS
SRTF
PRIORITY
RR

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Question 6

What do you know about starvation? How to remedy?

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Question 7

What do you know about multilevel queue scheduling?

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Answer question 1

Definition:

-*- A program that acts as an intermediary between a user and the computer hardware (and software)

-*- A program that provides an environment in which a user can execute program conveniently and efficiently

-*- A resource allocator and manager

-*- Similar to a government

Functions:
a . Functions provided for the convenience of the user, to make programming task easier

-*- Program execution: Run, terminate

-*- Input / Output operations

-*- File system manipulation: Read, write, open, create, delete . . .

-*- Error detection: Hardware

b . Functions for the efficient operation of the system itself

-*- Resource allocation

-*- Accounting: Read keeping and usage statictics

-*- Protections

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Answer question 2

There are three criteria that used in evaluating the performance of the CPU

Turn around time

Waiting time

Response time

CPU utilization

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Answer question 3

It contains many pieces of information associated with a specific process, including; process state, program counter, CPU register, CPU scheduling information, memory - management information, accounting information, I/O status information

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Answer question 4

Click on hypertext links in the table below for answers. For more details in solving technique, please see Assignment number one. In this part, there are only results, not explaination.

FCFS

SRTF

Priority

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1. FCFS method

FCFS Gantt Chart

wt	= 0 + (5 - 1) + (6 - 2) + ( 12 - 3)
	= 4 + 4 + 9 = 17

awt	= 17 / 4 = 4.25

==>	The average waiting time of given jobs is 4.25

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2. SRTF algorithm

SRTF Gantt Chart

wt	= (2 - 1) + 0 + (10 - 2) + (6 - 3)
	= 12

awt	= 12 / 4 = 3

==>	The average waiting time of the given jobs is 3

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3. Priority method

Priority Gantt Chart

wt	= 0 + (5 - 1) + (6 - 3) + (10 - 2)
	= 4 + 3 + 8 = 15

awt	= 15 / 4 = 3.75

==>	The average waiting time of the given jobs is 3.75

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4. Round robin

Process	Arrival time	Burst time
P0	0	5
P1	1	1
P2	2	6
P0	2	Remain 3
P3	3	4
P2	5	Remain 4
P0	7	Remain 1
P3	9	Remain 2
P2	11	Remain 2

Round Robin Gantt Chart

wt	= (11 - 7) + (5 - 2) + (2 - 1) + (14 - 11) + (9 - 5) + (3 - 2) + (12 - 9) + ( 7 - 3)
	= 4 + 3 + 1 + 3 + 4 + 1 + 3 + 4
	= 23

awt	= 23 / 4 = 5.75

==>	The average waiting time of given jobs is 5.75

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Answer question 5

	P0	P1	P2	P3	Check results in
FCFS	5	6	12	16	FCFS Gantt Chart
SRTF	6	2	16	10	SRTF Gantt Chart
Priority	5	6	16	10	Priority Gantt Chart
RR	12	3	16	14	RR Gantt Chart

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Answer question 6

Stavation of low priority process: Low priority jobs may waits indefinitely the CPU if there are always jobs of greater priority.

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Answer question 7

Multilevel queue scheduling is used to schedule many queue of jobs in defferent levels. It helps CPU run faster and finish the jobs in useful ways

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Exam I Page - Copy right Spring 99
Created by Doan nguyen
Last Update March 23, 1999