Computer Systems Assignment: Binary, Systems, and Networking Concepts

This document provides a detailed solution to a computer systems assignment. The assignment covers a range of topics including computer hardware specifications, software components and their roles, the comparison of mainframe and laptop computers, system architecture concepts, and binary number systems. The solution includes conversions between decimal, binary, octal, and hexadecimal number systems, binary arithmetic, and ASCII representation. Furthermore, the document explores topics such as data compression (lossy and lossless), computer architecture, interrupts, DMA, networking concepts, and network protocols. It also touches upon concurrent processing, operating systems, and file systems. The assignment delves into both theoretical concepts and practical applications within computer systems.

Running head:
Computer Systems
Name of the Student
Name of the University
Author Note

1
Task 1:
For buying a personal computer the basic and important features are given below. At first
Random Access Memory (RAM) is the main memory of a computer. It improves the speed of
the computer. It should be of 8Gigabyte for better performance. The hard disk drive should be of
minimum 1Terabyte (TB). If it is of 2Tb, it would be great but having 1Tb storage is okay.
Processor is the main function that controls the computer. It would be great if the processor is i5
with 8th generation. i5 is far better than i3 and this provides much speed than the 7th generation.
Task 2:
The components of a software system are Network services, hardware level operating system,
logical level of operating system, Graphics engine and so on.
Role of network services is it provides a data storage, manipulation and so other things on the
internet.
Role of hardware level operating system is that hides the complexities of the processor. This
includes managing hardware resources like memory, data storage and so on.
Role of logical level operating system is that it provides higher functions like, internet access,
file management.
Role of Graphics engine permits the software framework to build games. They provides features
from animation to artificial intelligence.
Task 3:
The statement implies that each of them has the same facility or speed. But the main difference is
they are built or developed in a same manner but their appearance, flexibility.

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Task 4:
A conceptual model that states the behavior, structure of an architecture is known as system
architecture. As well as it is defines how a process will be executed.
Task 5:
Task 6:
(5278)10= 5*103+2*102+7*101+8*100
Task 7:
The range will be 0-17.
Task 8:
The binary addition is 10101 + 1110= 100011
The binary multiplication is 10101 * 1110= 100100110
Task 9:
For converting decimal to hexadecimal it will be the best way to convert the decimal into binary
and then converting the binary to hexadecimal.
(12345)10 = (11000000111001)2 = (3039)16
From the above solution the binary of 12345 is also provided.
(12345)10 = (11000000111001)2
Task 10:
(212)10 = (324)8

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Task 11:
(117)10= (75)16
Task 12:
(12345)10 = (11000000111001)2
Task 13:
(27.625)10= (11011)2
(4192.37761)10= (1000001100000)2
Task 14:
(0.1001001)10= 0
(1100101.1)10= (101)2
Task 15:
Motion vector captures the differences between video frames. After that DCT (Discrete Consine
Transform) is used to convert information. After that quantization is put on for DCT coefficients.
Then Huffman encoding uses small codes to demonstrate common values.
Task 16:
1010100= 84, 1101000=104, 1101001=105, 1110011=115, 0100000=32
1101001=105, 1110011= 115, 0100000=32, 1000101=69, 1000001=65,
1010011=83, 1011001=89, 0100001=33

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Now we have the ASCII values of each binary value. If the character values of them are
calculated, we can get the string. The string is “This is EASY!”
Task 17:
A number cannot be negative. So there would be no decimal, octal, hexadecimals and ASCII.
Task 18:
Lossy compression Lossless compression
Lossy compression is used in audios, videos
and images.
Lossless compression is used in text or
programme.
Lossy compression is part of data encoding
method.
Lossless compression is part of data
compression algorithm.
Lossy compression’s applications are JPEG,
MP3, and MP4.
Lossless compression’s applications are PNG,
BMP and FLAC.
Lossy compression’s data-holding capacity is
better than lossless.
It has less data-holding capacity.
Lossy compression provides smaller file.
Lost in lossy audio data compression happens when the range of audio decreased and the quality
of audio has reduced.
Lossy data compression is used to reduce the bit size of an image or picture. For that reason the
original file can be restructured. When there should no data loss, in such cases lossy data
compression cannot be used.

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Extra Tasks:
Task 19:
(38)10 = (100110)2
(24)10= (11000)2
After calculating the 1’s complement the values will be 11001, 111
Addition is 100000
Task 20:
(1987)10 = (11111000011)2
For 1’s complement all 1 of binary number will be converted into 0 and 0 will be converted into
1. After 1’s complement the number will be (00000111100). After having the 1’s complement if
1 is added with the binary number, 2’s complement will be calculated. Therefore the 2’s
complement of the number is (00000111101)2.
Task 21:
10’s complement of 24379 is 75621 and
10’s complement of 5098 is 4902.
Addition is (75621+4902) = 80523.
Task 22:
(171.625)= (01010111010000000000000)24
The value should be 24 bit. Therefore rest of the bytes are represented in 0.

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Task 23:
In Little man’s computer, it first goes to the calculator and then checks the number. After that the
number is put into mailbox. But the number in calculation do not have any changes . At last the
specified number replaced with the new one in the calculator.
Task 24:
INP
STA A
INP
STA B
LDA A
SUB B
OUT
HLT
A DAT
B DAT
Task 25:
Task 26:
INP
STA M0
INP

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STA M1
INP
STA M2
SUB M1
BRP J1
LDA M1
STA M2
J1 LDA M2
SUB M0
BRP J2
LDA M0
STA M2
LDA M0
STA M2
JA LDA M2
OUT
HLT
M0 DAT

8
M1 DAT
M2 DAT
Task 27:
IN
STO 99
IN
ADD 99
STO 99
IN
SUN 99
OUT
COB
Task 29:
INP
STA A
INP
STA B
LDA A
SUB B
OUT

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HLT
A DAT
B DAT
Task 30:
IN
STO 99
IN
ADD 99
STO 99
IN
SUB 99
OUT
COB
Task 31:
Interrupt is such a signal that is sent from a particular programme that wants the central
processing unit to be stopped and after that decide what to do next. Interrupts are generally used
for service hardware timers, data transfer, handle mouse and keyboard events.
Most of the microprocessors have multiple hardware timers. This timers also support for
interrupts.

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Sometime in order to transfer data interrupts are used.
Whenever any interrupt is received from keyboard and mouse, they are handled. Interrupt is such
used for creating or managing such interrupts.
Task 32:
The disk system is interrupting the programme code.
Interrupt is basically used to signify and signal that data blocks are either ready or written and is
used to indicate about the completion of the operation. In absence of interrupt, processor has to
add disk transfer of hardware systems.
After an interrupt, stack store the processor state, and after that the stack pointer is updated to
add the saved data. Instruction pointer changed its location so that it can point the interrupt
handler’s location.
Task 33:
Error handling event is such an example in which interrupt is first occurred.
Status event is such an example in which interrupt is occurred at the completion of an event.
In polling interrupt CPU all time checks the device status. It checks whether the device needs the
CPU attention or not. Polled interrupt sends a message to a part of the computer that has the I/O
interface.
In polling interrupt command-ready bit is such a bit that indicates that the processor needs
attention. In polled interrupt a message is sent about the status of the device to repair.

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Task 34:
DMA stands for Direct Memory Access. DMA is such a technique that permits any input output
device to directly access the main memory of a computer.
Capabilities added by DMA-
DMA increase the speed of a computer, without getting help of processor.
It decrease the workload of processor.
DMA required less clock cycles for completing a process.
Extra task:
For the very first time when an interrupt is occurred, CPU gives it the higher priority and looks
upon it. While checking or processing the interrupting if another interrupt is occurred then the
second one will remain pending. After successful completion of the first interrupt CPU checks
whether any more interrupt is pending or not. If any pending interrupt is present then CPU will
execute that interrupt. In this manner, CPU manages many interrupts in a sequential matter.
Purpose of a completion interrupt-
DMA controllers uses interrupts to let the CPU know about the successful data transfer.
Therefore it is important to complete the interrupt after DMA transfer.
Task 35:
In our daily life internet is very important and transferring data from one to another is such an
example. In the social networking sites data communication is must and has great importance.