Software base of a computer system: the operating system 1

SOM02

Task SOM02-Software base of a computer system: the operating system

Part 1. Concept of operating system

1. Complete the following table regarding the evolution of operating systems:

Evolution of operating systems

Technical component
Generation Advances in relation to operating systems
what characterizes it
Generation Lacking a system It works in machine language and is coded to
0 operational hand
Generation Vacuum tubes and Work continues in machine language and
1 plug-in boards hand coding
Generation Transistor and The operating system is used as a transition
2 batch processing between jobs and to show results
Single users, with characteristics of
Integrated circuits generation and multiprogramming, spooling, programming languages
3 multiprogramming
job control, real-time systems
Generation
Microprocessor Multi-user and multitasking
4
Operating systems allow two or
Generation Parallel processing allows multiple processes to share resources at the same time.
5 and artificial intelligence an efficient way. Attempt to resemble the
systems to human capabilities.

2. Make a diagram with the classification of Operating Systems.

Due to its internal structure

a. Monolithic: all functions are implemented in the kernel.

b. Layered: hierarchical structure that is divided into different levels.
c. Virtual machine: presents an interface to each process, showing a machine
which looks identical to the underlying real machine.

According to the number of users

a. Single-user: When all resources are available to a single user.

user
b. Multiuser: Resources can be used simultaneously by several
users.

According to the number of processes

a. Monotasking: The system can only run one process at a time.

b. Multitasking: The system can run more than one process at a time.

For the modes of exploitation

a. In batches: Programs are grouped to be processed sequentially.

by the system.

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b. Multiprogramming: The computational load is distributed among the processors

existing.
b.1. Time sharing: Each process uses CPU cycles until
I finished.
b.2. Real time: The response is immediate or almost immediate after launching the process.

By the way of offering the services

a. Centralized systems: a computer is responsible for processing and the

users operate terminals without memory or processor.
b. Network systems: The operating system connects two or more computers together.
through some means of communication.
c. Distributed systems: Allows for distributing tasks among a set of
processors.

Part 2. Functions of the operating system.

Operating System: what it is, what functions it has, the purpose, what types there are
According to their use, their tasks,..., mention operating systems that you know and for what users they are intended.
indicated....

The operating system is a set of programs, services, and functions that manage and
They coordinate the operation of hardware and software. Acts as an intermediary.
between the user and the computer hardware. On one hand, it tries to make the use of
resources to be efficient and on the other hand try to ensure communication between the machine and the
user be natural and intelligible.
To achieve these objectives, you need to perform a series of functions: management of
processor, memory management, peripheral management, management
of the file system, error handling and processing, security and protection of the
network system and control.
We can find single-tasking or multitasking operating systems depending on the number of
processes, single-user or multi-user operating systems according to the number of users
that interact simultaneously with resources, batch systems, or multiprogramming according to
the modes of exploitation of computational load, centralized systems, network or
distributed according to the way of offering services and monolithic systems, by layers or
virtual machine according to its internal structure.
Some current operating systems are Windows; the most widespread at present and
for the user of any level and for companies; OS X, Apple's operating system
that occupies a small proportion of the market compared to Windows and is aimed at
also to users of any type but has a very simple graphical environment and
easy to handle; Linux is a widely used free system around the world where they have
thousands of programmers have collaborated and it is aimed at all types of users.

4.- Calculate the memory fragmentation if we have seven memory partitions of

1MB and the task queue contains tasks with requirements of 300KB, 1800KB,
1000KB, 900KB, 600Kb, 700KB y 200KB.

Task 1=1024kb-300Kb=724Kb remaining in the first partition

Task 2=1024Kb-1024Kb=0Kb left over from the second partition
Task 2 = 1024Kb - 776Kb (remaining from task 2) = 248Kb left from the third partition

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Task 3 = 1024Kb - 1000Kb = 24Kb remaining from the fourth partition

Task 4=1024Kb-900Kb=124Kb remaining from the fifth partition
Task 5=1024Kb-600Kb=424Kb remaining from the sixth partition
Task 6=1024Kb-700Kb=324Kb left over from the seventh partition
Task 7 cannot be loaded into memory because there are no free partitions left, so
not until there is no free partition will it load, even though the
The fragmentation resulting from loading the first 6 tasks has been 1868Kb.

5.- Explain the difference between internal fragmentation and external fragmentation.

While internal fragmentation refers to the wasted space

Within each partition, external fragmentation refers to the memory that is not
external use to the partition.

6.- A system has a physical memory of 64 Kb divided into page frames of

size 4 Kb. A program has code size of 32768 bytes, data of 16386 bytes and
the stack of 15870 bytes. Can this program be loaded into memory?
What if the page size were 512 bytes?

64 Kb divided by 4 Kb equals 16 page frames.

4kb=4096Kb
32768:4096=8 page frames
16386:4096=4,0004=5 page frames
15870:4096=3,874=4 page frames
A total of 17 page frames are needed, therefore the program cannot be loaded.
in memory.

If the page frames are 512 Bytes

64x1024=65536 Bytes
65536:512=128 page frames
32768:512=64 page frames
16386:512=32,003=33 page frames
15870:512=30,996=31 page frames

64+33+31=128 page frames are needed to load the program into memory.
So much so that it is possible to load the program into memory.

Part 3. Use of the operating system

7.- Compare the interface of an operating system from 10 years ago (for example, a
Windows 3.1) with a current one (for example, Windows 7). What are the
most important differences at a glance?

In Windows 3.1 we do not have a taskbar compared to Windows 7 and

both the icons, dialog boxes, and menus are of a simpler design, fewer colors
and more primitive structuring.

Part 4. Operating system processes.

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8.- Make a drawing/schema of the states that a process goes through. Explaining that
it means each state and explaining the evolution of these processes according to the states
for what it goes through.

Termination

Processor assignment Active process Dead process

Blocking
Appropriation
End of blockade
Prepared process Blocked process
End

The system decides it

Unborn process

Active process: the process is being used in the CPU.

Prepared process: the process is not running but is a candidate to pass to
active state.
Blocked process: the process is waiting for an external event that has caused it
block.
Deadlock: the process has finished its execution or the OS has detected a fatal error and
it has transferred to this state.
Unborn process: the program exists but is not yet known to the OS.

9.- The most commonly used planning criteria are the return time, time of
waiting, performance and efficiency. Explain these concepts in a simple way.

Turnaround time: the time that elapses from the creation of a process until its
completion. It is the sum of the waiting time and the execution time.
Waiting time: it is the time that the process waits until it is granted the
processor.
Performance: indicates the number of processes per unit of time.
Effectiveness: percentage of time the processor is busy.

10.- There are 3 processes P1, P2, and P3 with execution times of 85, 45, and 118. If it acts
the SJF algorithm, determine the order in which they appear in the ready list.
Calculate your return times, wait times, performance, and system efficiency.

The order according to the SJF algorithm would be first the process P2, followed by the process
P1 and finally the process P3.

P2- Tiempo de retorno=45; Tiempo de espera=0; Eficiencia=100%; Rendimiento=0,02

processes per unit of time.
P1- 130
Performance=0.011 processes per unit of time.
P3- Tiempo de retorno=248; Tiempo de espera=130; Eficiencia=47,58%;
Performance = 0.008 processes per unit of time

Average return=82.66

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Average wait = 58.33

70.98
Rendimiento medio=0,013

11.- If the return time of a process is 30 and the actual execution time is 10, what is the
waiting time, the efficiency and the performance of the system?

Tiempo de espera=20
Efficiency=(10/30)x100=33.33%
Performance = 1/30 = 0.033 processes per unit of time

12.- There are 2 processes P1 and P2 with execution times of 25 and 30. If it acts on the
RR algorithm (q = 10) determine the order in which it is in the list of ready processes.
Calculate your turnaround, response, and wait times.

1 2 3 4 5 6 7 8 9 10
P1 P1 P1 P1 P1 P1 P1 P1 P1 P1

11 12 13 14 15 16 17 18 19 20
P2 P2 P2 P2 P2 P2 P2 P2 P2 P2

21 22 23 24 25 26 27 28 29 30
P1 P1 P1 P1 P1 P1 P1 P1 P1 P1

31 32 33 34 35 36 37 38 39 40
P2 P2 P2 P2 P2 P2 P2 P2 P2 P2

41 42 43 44 45 46 47 48 49 50 51
P1 P1 P1 P1 P1 P2 P2 P2 P2 P2 P2

52 53 54 55
P2 P2 P2 P2

P1- Tiempo de retorno=45; Tiempo de espera=0; Tiempo de respuesta=20

P2- Tiempo de retorno=55; Tiempo de espera=10; Tiempo de respuesta=25

13.- There are two processes, P1 with a runtime of 20 and P2 with a runtime
from 15. According to the wheel algorithm with quantum 10. Mark the return time, of
response and waiting.

1 2 3 4 5 6 7 8 9 10
P1 P1 P1 P1 P1 P1 P1 P1 P1 P1

11 12 13 14 15 16 17 18 19 20
P2 P2 P2 P2 P2 P2 P2 P2 P2 P2

21 22 23 24 25 26 27 28 29 30
P1 P1 P1 P1 P1 P1 P1 P1 P1 P1

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31 32 33 34 35
P2 P2 P2 P2 P2

P1- Tiempo de retorno=30; Tiempo de espera=0; Tiempo de respuesta=10

P2- Tiempo de retorno=35; Tiempo de espera=10; Tiempo de respuesta=20

14.- The following cargo description; assume the indicated order of arrival

WorkTime CPU=UPriority
1 4 3
2 2 2
3 3 2
4 2 1
Represent the CPU access by using a Gantt chart when applying scheduling.
FCFS (First Come First Serve: First to arrive, first to be served), RR (Robin
Round Robin (RR) (q = 1), SJF (Shortest Job First: the shortest task first)) and by
priority. In each case, calculate the efficiency=U/R*100, the return time and of
Wait for each job. Create a summary table with the different results and indicate
the one that has the best results overall.

FCFS Algorithm
1 2 3 4 5 6 7 8 9 10 11
P1 P1 P1 P1 P2 P2 P3 P3 P3 P4 P4

P1- Tiempo de retorno= 4; Tiempo de espera=0; Eficiencia=100%

P2- Tiempo de retorno=6; Tiempo de espera=4; Eficiencia=33,33%
P3- Tiempo de retorno=9; Tiempo de espera=6; Eficiencia=33,33%
Return time=11

Average return time = 7.5

Average waiting time = 4.75
Average efficiency=46.21%

RR Algorithm
1 2 3 4 5 6 7 8 9 10 11
P1 P2 P3 P4 P1 P2 P3 P4 P1 P3 P1

P1- Tiempo de retorno=11; Tiempo de espera=0; Eficiencia=36,36%

P2- Tiempo de retorno=6; Tiempo de espera=1; Eficiencia=33,33%
P3- Tiempo de retorno=10; Tiempo de espera=2; Eficiencia=30%
P4- Tiempo de retorno=8; Tiempo de espera=3; Eficiencia=25%

Average return time=8.75

1.5
Eficiencia=31,17

SFJ Algorithm

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1 2 3 4 5 6 7 8 9 10 11
P2 P2 P4 P4 P3 P3 P3 P1 P1 P1 P1

P2- Tiempo de retorno=2; Tiempo de espera=0; Eficiencia=100%

P4- Tiempo de retorno=4; Tiempo de espera=2; Eficiencia=50%
P3- Tiempo de retorno=7, Tiempo es espera=4; Eficiencia=42,85%
P1- Tiempo de retorno=11, Tiempo de espera=7; Eficiencia=36,36%

Average return time=6

Average waiting time = 3.25
57.30%

Priority algorithm
1 2 3 4 5 6 7 8 9 10 11
P4 P4 P2 P2 P3 P3 P3 P1 P1 P1 P1

P4- Tiempo de retorno=2; Tiempo de espera=0; Eficiencia=100%

P2- Tiempo de retorno=4; Tiempo de espera=2; Eficiencia=50%
P3- Tiempo de retorno=7; Tiempo de espera=4; Eficiencia=42,85%
P1- Tiempo de retorno=11, Tiempo de espera=7; Eficiencia=36,36%

Mean return time = 6

Tiempo de espera medio=3,25
Efficiency=57.30

Summary of results:

FCFS RR SFJ PRIORITY

T. waits half 7.5 8.75 6 6
average return 4.75 1.5 3.25 3.25
Efficiency 46,21% 31,17% 57,30% 57.30%

The best results are from the SFJ AND PRIORITY algorithm since both in
Overall, they have the lowest average times and highest average efficiency.

15.- Let the following load description be; assume the indicated order of arrival.

WorkTime CPU=Priority
1 6 3
2 2 2
3 3 4
4 1 1
5 4 2

Represent the access to the CPU by using a Gantt chart when applying scheduling.
FCFS, RR (q = 1), SFJ and by priority. In each case, calculate the efficiency, the time of
return and waiting for each job.

FCFS Algorithm

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
P1 P1 P1 P1 P1 P1 P2 P2 P3 P3 P3 P4 P5 P5 P5 P5

P1- Tiempo de retorno=6; Tiempo de espera=0; Eficiencia=100%

P2- Tiempo de retorno=8; Tiempo de espera=6; Eficiencia=25%
P3- Tiempo de retorno=11; Tiempo de espera=8; Eficiencia=27,27%
P4- Tiempo de retorno=12; Tiempo de espera=11; Eficiencia=8,3%
P5- Tiempo de retorno=16; Tiempo de espera=12; Eficiencia=25%

Mean return time=10.6

7.4
37.11%

RR Algorithm
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
P1 P2 P3 P4 P5 P1 P2 P3 P5 P1 P3 P5 P1 P5 P1 P1

P1- Tiempo de retorno=16; Tiempo de espera=0; Eficiencia=37,5%

P2- Tiempo de retorno=7; Tiempo de espera=1; Eficiencia=28,57%
P3- Tiempo de retorno=11; Tiempo de espera=2; Eficiencia=27,27%
P4- Tiempo de retorno=4; Tiempo de espera=3; Eficiencia=25%
P5- Tiempo de retorno=14; Tiempo de espera=4; Eficiencia=28,57%

Average return time=10.4

Average wait time=2
29.38%

SFJ Algorithm
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
P4 P2 P2 P3 P3 P3 P5 P5 P5 P5 P1 P1 P1 P1 P1 P1

P4- Tiempo de retorno=1; Tiempo de espera=0, Eficiencia=100%

P2- Tiempo de retorno=3; Tiempo de espera=1; Eficiencia=66,67%
P3- Tiempo de retorno=6; Tiempo de espera=3; Eficiencia=50%
P5- Tiempo de retorno=10; Tiempo de espera=6; Eficiencia=40%
P1- Tiempo de retorno=16; Tiempo de espera=10, Eficiencia=31,25%

Mean return time=7.2

Tiempo de espera medio=4
Average efficiency = 57.58%

Priority Algorithm
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
P4 P2 P2 P5 P5 P5 P5 P1 P1 P1 P1 P1 P1 P3 P3 P3

P4- Tiempo de retorno=1; Tiempo de espera=0; Eficiencia=100%

P2- Tiempo de retorno=3; Tiempo de espera=1; Eficiencia=66,67%
P5- Tiempo de retorno=7; Tiempo de espera=3; Eficiencia=57,14%
P1- Tiempo de retorno=13; Tiempo de espera=7; Eficiencia=46,15%

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P3- Tiempo de retorno=16; Tiempo de espera=13; Eficiencia=18,75%

Tiempo de retorno medio=8

Tiempo de espera medio=4,8
Average efficiency=57.74%

16.- Based on the following schedules; draw the Gantt chart and calculate the
Ui, Ri and Ei: and indicate the algorithm used in each case.

----------
A
--
-----------------------
B
-----
---------------
C
---
-------------------
D
----
-----
E
--

FCFS Algorithm

A- Tiempo de ejecución=2; Tiempo de retorno=2; Tiempo de espera=0

5
C- Tiempo de ejecución=3; Tiempo de retorno=10; Tiempo de espera=7
D- Tiempo de ejecución=4; Tiempo de retorno=14; Tiempo de espera=10
1

Average execution time=3

Average return time=9.6
6.6

A -- -- --
B -- -- -- -- -- --
C -- -- -- --
D -- -- -- -- --
E -- --

RR algorithm with a quantum of 1

A- Tiempo de ejecución=3; Tiempo de retorno=11; Tiempo de espera=0

B- Tiempo de ejecución=6; Tiempo de retorno=20; Tiempo de espera=1
C- Tiempo de ejecución=4; Tiempo de retorno=16, Tiempo de espera=2
D- Tiempo de ejecución=5; Tiempo de retorno=19; Tiempo de espera=3
E- Tiempo de ejecución=2; Tiempo de retorno=10; Tiempo de espera=4

Average runtime=4

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Mean return time=15.2

Average waiting time=2

A ------- --
B ------- -------
C ------- -------
D ------- ------- --
E ------- -------

RR algorithm with a quantum of 3

A- Tiempo de ejecución=4; Tiempo de retorno=16; Tiempo de espera=0

B- Tiempo de ejecución=6; Tiempo de retorno=19; Tiempo de espera=3
C- Tiempo de ejecución=6; Tiempo de retorno=22; Tiempo de espera=6
D- Tiempo de ejecución=7; Tiempo de retorno=29; Tiempo de espera=9
6

5.8
Mean return time=22.8
Average wait time=6

-------------------
A
---
----------
B
--
------
C
-
--------------
D
---
------------------------
E
----

SJF Algorithm

A- Tiempo de ejecución=4; Tiempo de retorno=10; Tiempo de espera=6

B- Tiempo de ejecución=2; Tiempo de retorno=3; Tiempo de espera=1
C- Tiempo de ejecución=1; Tiempo de retorno=1; Tiempo de espera=0
D- Tiempo de ejecución=3; Tiempo de retorno=6; Tiempo de espera=3
E- Tiempo de ejecución=5; Tiempo de retorno=15; Tiempo de espera=10

Average runtime=3
Mean return time=7
Average wait time=4

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Part 5. Current operating systems

17.- Make a list of the latest operating systems, as well as their versions, of
the systems, Macintosh, Windows, and Linux. Learn about them and explain what improvements
they contribute compared to their predecessors.

Windows 8.1 - Replaces version 8 and has the novelty of the return of the button
start, traditional start mode so that the user can choose whether to begin with
the view of the traditional desktop, greater customization of the desktop, listing of
all applications, faster organization of the Tiles, tasks and multiple views,
universal search, automatic updates in the background without any
notification, complete integration with Skydrive, new Internet Explorer, better control with
mouse and keyboard, new configuration options for active corners, changes in
Windows Explorer, on-screen keyboard, photo slideshow on the screen of
blocking, improvements in system applications, compatibility with 3D printers.

Mac OS X Mavericks replaces the Mountain Lion version and has the following new features: To
save a file, users can add tags that will be displayed on the bar
Finder's sidebar, to assist in the search for files; The tabs will allow users to
users open multiple windows in different tabs and multiple screens; they can
open full-screen applications on multiple monitors and switch between them
easily; The new operating system also allows the use of an HDTV as
second monitor thanks to Apple TV; Apple Maps arrives manifestly on Macs
Improved, there is the possibility of syncing the application with other devices, something
very useful for finding directions at home and then syncing them; Maps is also
integrated into Mail, Contacts, and the redesigned Calendar application; Mavericks
introduce iBooks for the first time for Mac devices. Cloud integration; The
users will be able to access notifications even if their Mac is in sleep mode without
unlock the screen.

Ubuntu 13.10 - Precedes Ubuntu 13.04 and has new search features.
smart to the board, new voice recognition; Contacts will be able to edit the
contact information and will also be able to delete it; The document viewer receives
a new toolbar, Simple Scan will receive brightness and contrast options;
Rythmbox will have textless play buttons and options in the bottom bar for
add files, playlists and check connected devices, new
graphics server.

Part 6. File systems

18.- Answer the following questions:

a. What is understood by the file system?

It is the way each OS permanently stores information in the

external supports.

b. Do all operating systems use the same file system?

No, each OS uses its own file system.

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c. Why do we need to know the appropriate file system for the OS?

To format the hard drive that contains it in that format, as well as other media.
of storage must also be formatted in the file system that
recognize the OS.

d. State different file systems.

NTFS, FAT32 for Windows; HFS, HFS+ for MAC OS; EXT, EXT2, EXT3 for
Linux; CDFS, UDF, LFS for CD, DVD, BLU-RAY DISC.

Part 7. Selection of a file system

19.- If there is a computer with the MS-DOS operating system and with this FAT:

0123456789101112131415161718192021
810721816612141301720150 0 0

a. Explain what FAT is.

It is the file allocation table and contains an index that indicates the sectors.
free and those that contain information, all of which are identified through records of
directions.

b. According to the FAT, how many files does it contain and why?

It contains 4 files because the 0s mark the ends of the files and there are 4 zeros.
therefore 4 files.

c. What blocks does the file occupy whose first block is number 1?

Blocks 1, 6, 8, 18.

d. And the file whose first block is number 4?

Blocks 2, 4, 10, 12, 13.

e. Is there any other file on the disk? What blocks do they occupy?

Yes, two more, one occupies blocks 3, 7, 15, 16, 20 and the other occupies blocks 11, 14,
17.

f. What blocks are available for the OS?

Blocks 0, 5, 9, 19, 21.

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