COA Assignment: Input/Output, Memory, and Assembly Language

This document provides solutions to a Computer Organization and Architecture assignment, covering a wide array of topics within the field. The solutions address fundamental concepts such as input, output, memory, data path, and control, along with assembly language and MIPS architecture. It explores questions related to number representation, including two's complement and floating-point formats. The assignment delves into operating system concepts such as scheduling, interrupt handling, and virtual memory. Furthermore, it touches upon compiler design, including scanner, parser, and code generation. The document also covers storage solutions like RAID, security measures like disk encryption, and programming languages such as Prolog. Additionally, it offers solutions to questions involving assembly language code, logical operations, and data structures, providing a comprehensive overview of the subject matter.

Running head: COMPUTER ORGANIZATION AND ARCHITECTURE
Computer organization and architecture
Name of the Student:
Name of the University:
Author Note

1
COMPUTER ORGANIZATION AND ARCHITECTURE
Answer 1
Input, Output, Memory, Data path and Control
Answer 2
Inputting, Outputting, Processing, Storing
Answer 3
(D) 32 bits
Answer 4
(C) 2’s complement
Answer 5
True
Answer 6
(B) Terabyte
Answer 7
(A) Less than Zero
Answer 8
Alan Turing – (C) computing theory

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COMPUTER ORGANIZATION AND ARCHITECTURE
L0pht – (D) internet security
J.Presper Eckert – (A) ENIAC
Vinton G. Cerf – (B) TCP/IP
Answer 9
.globl – declares that main is a global symbol that should be visible to code stored in other files
.text – tells the assembler to store the instructions in its text segment
.word – Store the n 32-bit quantities in successive memory words.
.asciiz – stores a null-terminated string in memory
Answer 10
0.01101111010111000011
Answer 11
Prolog is general purpose programming language that is used for implementing logical
programming and artificial intelligence and computational linguistics. They are used for the rule
based logical queries such as voice control systems, searching database and also filling
templates.
Answer 12
HI has a signed value -1, it means that it has all1s (reading in the binary format).

3
COMPUTER ORGANIZATION AND ARCHITECTURE
Answer 13
(A) Provide relocate able object code
(B) Process a program in two passes.
(C) Assign machine memory addresses to symbol labels.
Answer 14
(A) False
(B) True
(C) False
Answer 15
B) Symbols need to be stored in the symbol table.
Answer 16
If the number that is the proper result of such operations cannot be represented by the rightmost
hardware bits, overflow is said to have occurred.
Answer 17
0000 4020 - add $t0, $zero, $zero
0104 4820 - add $t1, $t0, $a0
0024 1009 - lb $t2, 0($t1)
0105 5820 - add $t3, $t0, $a1

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COMPUTER ORGANIZATION AND ARCHITECTURE
0028 1011 - sb $t2, 0($t3)
0054 1000 - beq $t2, $zero, finish
2108 0001 - addi $t0, $t0,1
03E0 0008 - jr $ra
Answer 18
if(R[s4]>R[t1]) PC = AGAIN
Answer 19
Since MIPS supports negative constants, there is no need for subtract immediate in MIPS.
Answer 20
Base – BEQ
Register – ADD
Immediate – ADDI
Pseudo Direct – J LABEL
Program Counter – COUNT
Answer 21
(A) It simplifies exchange of data that includes floating-point numbers;
(B) it simplifies the floating-point arithmetic algorithms to know that numbers will always be in
this form;
(C) it increases the accuracy of the numbers that can be stored in a word

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COMPUTER ORGANIZATION AND ARCHITECTURE
Answer 22
VisiCalc
Answer 23
0.0
Answer 24
0x4608f700
Answer 25
(1) Pass one - a) Translate mnemonic operation codes to machine codes
(1) Pass one - b) Save addresses assigned to labels
(1) Pass one - c) Scan for label definitions
(1) Pass two - d) Write object code
Answer 26
D) Data Segment - e) Data values.
S) Symbol – b) Relative address of global symbol.
H) Header - d) Program name.
T) Text Segment - a) Object instruction code.
R) Relocation - c) Modification information.
Answer 27
No, because the MIPS assemble language only consists of 32 bite registers.
Answer 28
True

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COMPUTER ORGANIZATION AND ARCHITECTURE
Answer 29
True
Answer 30
Guard bit - The possibility of overflow can be eliminated by appending the appropriate number
of guard bits to a binary word. For a two's complement signed value, the guard bits are filled
with either 0's or 1's depending on the value of the most significant bit.
Round bit - The round bit is the second bit after the o bit of the mantissa.
Sticky bit - A sticky bit is a permission bit that is set on a directory that allows only the owner of
the file within that directory or the root user to delete or rename the file.
NaN - NaN is a numeric data type value representing an undefined or unrepresentable value.
Answer 31
Raid 1 is the traditional scheme for tolerating disk failure, called mirroring or shadowing uses
twice as many disks as does RAID 0. Whenever data is written in disk one, data is also written to
a redundant disk, so that there are always two copies of the information. If a disk fails, the
system just goes to the “mirror” and reads its contents to get the desired information. Mirroring
is the most expensive RAID solution, since it requires the most disks.
Answer 32
(A) Is totally machine independent.

7
COMPUTER ORGANIZATION AND ARCHITECTURE
Answer 33
Pass One is generally used to adding literals to literal table and address resolution of local
symbols.
Answer 34
True
Answer 35
The program that uses larger amount of CPU and the smaller amount of I/O would get the
priority as the CPU does not need to wait for the I/O in the first process, and the process would
start automatically when the CPU receives the required Input.
Answer 36
(A) First come First Server.
(B) Shortest Job First.
Answer 37
(A) SEMANTICS
Answer 38
True

8
COMPUTER ORGANIZATION AND ARCHITECTURE
Answer 39
(B) Enable the operating system to efficiently process several concurrent programs.
Answer 40
lw $t0, 1200, $t1
Answer 41
Manage the computer's resources, such as the central processing unit, memory, disk drives, and
printers.
Answer 42
(1) : = #1 Indx
(2) JGT Indx #25 (14)
(3) - Indx #1 t1
(4) * t1 #10 t2
(5) * #2 MLK t3
(6) * t3 #3 t4
(7) – t4 #1 t5
(8) + t2 t5 t6
(9) * t6 #4 t7
(10) : = ZYX[t13] CBA[t7]
(11) + #1 Indx t14
(12) : = t14 Indx

9
COMPUTER ORGANIZATION AND ARCHITECTURE
(13) JMP ( 2 )
(14)
Answer 43
SCANNER – The scanner function is used for scanning the user defined input for the system.
PARSER – The parser is a compiler / interpreter component that breaks data into smaller elements
for easy translation into another language. It takes input in the form of a sequence of tokens or
program instructions and usually builds a data structure in the form of a parse tree or an abstract
syntax tree.
CODE GENERATOR – code generation is the process by which a compiler's code generator
converts some intermediate representation of source code into a form.
Answer 44
Ubuntu is an operating system. This is a linux based system that supports shell based
programming.
Answer 45
(A) Eliminating Redundancy
(B) Eliminating useless and unreachable code
Answer 46
1) CHANNEL PROGRAM - G) Saving / restoring registers by Interrupt Processor.
2) CONTEXT SWITCHING - E) Select next program to process by CPU

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COMPUTER ORGANIZATION AND ARCHITECTURE
3) DEAD – LOCK - A) Set of programs in a Circular Wait.
4) DISPATCHING - D) Set of instructions for special i/o
5) STORAGE PROTECTION KEY - C) Register containing interrupt mask
6) INHIBIT INTERRUPT - B) Set mask to prevent Interrupt Processing
7) PAGE TABLE MAP - F) Data area used by the Memory Manager
8) PROGRAM STATUS WORD - H) Half byte used for memory access control.
Answer 47
Yes, using virtual memory hardware, each page can reside in any location of the computer's
physical memory, or be flagged as being protected.
Answer 48
If timer tick priority wasn't the lowest interrupt priority in the system, it may occurs situation
when CPU is in ISR with lower priority than system tick interrupt and enables further interrupts
and user process, that we need for potentially scheduling, is stored somewhere on the stack,
where we can't find it in system tick routine.
Answer 49
(A) Aborts are more expensive than waits.
(B) Unnecessary aborts result in wasted system resources.
Answer 50
(A) The system should respond to the interrupt.
(B) The system should provide the required resource for the process.

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COMPUTER ORGANIZATION AND ARCHITECTURE
Answer 51
Yes, to rescue the hardware from interrupts, MIPS programmers agreed to reserve registers $k0
and $k1 for the operating system.
Answer 52
(Instruction/Program X Clock cycles/Instruction X Seconds/Clock Cycle)
Answer 53
< stmt-list > : : = < stmt > { ; < stmt > }
< stmt > : : = < assign > | < read > | < write > | < for >
< assign > : : = id : = < exp >
< exp > : : = < term > { ex<term>|+ < term > | - < term > }
< term > : : = < factor > { * < factor > | DIV < factor > }
< factor > : : = id | int | ( < exp > )
Answer 54
Operand Argument 1 Argument 2 Result
* 3 EX t1
* 5 WHY t2
/ EX WHY t3
+ t1 t3 t5
- t5 t2 t6

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COMPUTER ORGANIZATION AND ARCHITECTURE
= t6 ZEE
Answer 55
True
Answer 56
Disk encryption
Backups
Data masking
Data Erasure