Fundamentals of Computer Systems

Hardware Components

Central Processing Unit (CPU)

The CPU is the primary component that executes instructions. It consists of three main sub-components:

Component	Function
Arithmetic Logic Unit (ALU)	Performs arithmetic (add, subtract, multiply, divide) and logical (AND, OR, NOT, XOR) operations
Control Unit (CU)	Coordinates all activities: fetches instructions, decodes them, and signals other components to execute
Registers	Small, extremely fast storage locations inside the CPU used for temporary data during processing

Registers

Register	Purpose
Program Counter (PC)	Holds the memory address of the next instruction to be fetched
Memory Address Register (MAR)	Holds the address in memory to be read from or written to
Memory Data Register (MDR)	Holds data that has been read from or is about to be written to memory
Accumulator (ACC)	Stores the results of ALU operations
Instruction Register (IR)	Holds the current instruction being decoded and executed
Status Register (Flags)	Stores flags such as Zero, Carry, Negative, Overflow from ALU operations

info

Registers are the fastest memory in a computer system — orders of magnitude faster than RAM. A typical CPU has a small number of general-purpose registers (8--32 in most architectures).

Memory Types

Type	Full Name	Volatile?	Read/Write	Speed	Typical Use
RAM	Random Access Memory	Yes	Both	Fast	Main memory, running programs
ROM	Read Only Memory	No	Read only	Slower than RAM	Boot-up instructions (BIOS/UEFI), firmware

RAM types:

• SRAM (Static RAM): Uses flip-flop circuits. Faster, more expensive. Used for CPU cache (L1, L2, L3).
• DRAM (Dynamic RAM): Uses capacitors. Slower, cheaper, needs refreshing. Used as main memory.

ROM types:

• PROM: Programmable once by the user.
• EPROM: Erasable using UV light, reprogrammable.
• EEPROM: Electrically erasable and reprogrammable.

Secondary Storage

Storage Type	Technology	Speed	Capacity	Cost	Volatility
HDD	Magnetic platters spinning at 5400/7200/10000 RPM	80--160 MB/s	500 GB -- 20 TB	Low	Non-volatile
SSD	NAND flash memory via SATA/NVMe	500 MB/s -- 7 GB/s (NVMe)	256 GB -- 4 TB	Medium-High	Non-volatile
Flash Memory	NAND flash (USB drives, SD cards)	10--300 MB/s	1 GB -- 1 TB	Medium	Non-volatile

:::warning[Exam Tip] When comparing storage, consider all five criteria: speed, capacity, cost per GB, volatility, and durability. HDDs are cheaper per GB but slower and more fragile (moving parts). SSDs are faster with no moving parts but more expensive per GB. :::

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Input Devices

Device	Input Type	Common Use Case
Keyboard	Text, commands	Typing documents, entering data
Mouse	Pointing, clicking	GUI navigation, selecting objects
Scanner	Image, document capture	Digitising photos, OCR (Optical Character Recognition)
Microphone	Audio/sound input	Voice recording, voice commands
Camera / Webcam	Image, video capture	Video conferencing, photography
Touchscreen	Touch gestures	Mobile devices, kiosks, POS systems
Barcode Reader	Light reflection pattern	Retail checkout, inventory management
RFID Reader	Radio frequency signal	Access control, toll collection, tracking

Barcode Reader vs RFID:

Feature	Barcode Reader	RFID Reader
Line of sight required	Yes	No
Read range	Short (contact to ~30 cm)	Up to several metres
Data capacity	Limited (typically a number)	Can store more data
Cost	Lower	Higher
Read multiple at once	No	Yes

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Output Devices

Device	Output Type	Common Use Case
Monitor	Visual display	Primary output for desktops/laptops
Printer	Hard copy (paper)	Inkjet (photo quality), Laser (high volume, fast)
Speaker	Audio output	Music, alerts, multimedia
Projector	Large visual display	Presentations, classrooms, cinemas

Inkjet vs Laser Printer:

Feature	Inkjet	Laser
Speed	Slower	Faster
Print quality (text)	Good	Excellent
Print quality (photos)	Better	Good
Cost per page	Higher	Lower
Initial cost	Lower	Higher
Mechanism	Sprays liquid ink	Uses toner powder, heat

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Von Neumann Architecture

The Von Neumann architecture defines a computer system with:

1. Single shared memory for both instructions and data (stored-program concept).
2. CPU consisting of ALU, CU, and registers.
3. System bus connecting CPU, memory, and I/O devices.

          +---------+
          |   CPU   |
          | +-----+ |
          | | CU  | |
          | +-----+ |
          | | ALU | |
          | +-----+ |
          | |Regs | |
          | +-----+ |
          +----+----+
               |
         System Bus
               |
          +----+----+
          | Memory  |
          | (RAM +  |
          |  ROM)   |
          +---------+
               |
         System Bus
               |
          +----+----+
          |   I/O   |
          | Devices |
          +---------+

Key principles:

• Instructions and data are stored in the same memory.
• Memory is addressed linearly.
• Instructions are executed sequentially unless a branch/jump instruction changes the flow.

Harvard Architecture (Comparison)

Feature	Von Neumann	Harvard
Memory bus	Single shared bus	Separate instruction and data buses
Memory	One memory for both	Separate memories for instructions and data
Speed bottleneck	Yes (bus contention)	No (parallel fetch)
Complexity	Simpler	More complex
Modern usage	Most general-purpose CPUs	DSPs, microcontrollers, CPU caches

The Von Neumann bottleneck arises because the CPU and memory share a single bus. The CPU is often much faster than memory, so it spends time waiting for instructions and data to be fetched. This is why modern CPUs use cache memory (L1, L2, L3) to reduce the impact of the bottleneck.

Cache memory is a small, fast memory between the CPU registers and main memory (RAM):

Cache Level	Location	Size	Speed
L1	Inside CPU core	32--128 KB	Fastest (~1 cycle)
L2	Inside CPU (per core or shared)	256 KB -- 1 MB	Fast (~10 cycles)
L3	Inside CPU (shared among all cores)	2--64 MB	Moderate (~40 cycles)
RAM	Outside CPU on motherboard	4--128 GB	Slowest (~100+ cycles)

When the CPU needs data, it checks L1 first, then L2, then L3, then RAM. If the data is found in cache, it is a cache hit; otherwise it is a cache miss and the CPU must wait for the slower memory.

:::warning[Exam Tip] The DSE syllabus focuses on Von Neumann. Know why it has a bottleneck (CPU waits for memory) and how Harvard architecture addresses this. Most modern CPUs use a modified Harvard architecture internally (separate L1 caches for instructions and data) while presenting a Von Neumann model externally. :::

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The Fetch-Decode-Execute Cycle

This is the fundamental cycle by which the CPU processes every instruction.

Step-by-step

1. Fetch:

   • PC holds the address of the next instruction.
   • Address is copied from PC to MAR.
   • Instruction is fetched from memory address in MAR into MDR.
   • PC is incremented to point to the next instruction.
   • Instruction in MDR is copied to IR.

2. Decode:

   • CU decodes the instruction in IR.
   • The CU determines which operation to perform and which operands are needed.

3. Execute:

   • The instruction is executed (ALU performs calculations, data is moved, etc.).
   • Results are stored in the accumulator or written back to memory.
   • Status flags are updated as needed.
   • Cycle repeats from step 1.

Example Trace

Given memory starting at address 100:

Address	Instruction
100	LOAD 5
101	ADD 3
102	STORE 6

Execution trace:

Step	Action	PC	MAR	MDR	IR	ACC
Fetch	PC(100) -> MAR; Mem[MAR] -> MDR; PC = 101; MDR -> IR	101	100	LOAD 5	LOAD 5	?
Decode	CU decodes LOAD 5	101	100	LOAD 5	LOAD 5	?
Execute	Mem[5] -> ACC	101	100	LOAD 5	LOAD 5	M[5]
Fetch	PC(101) -> MAR; Mem[MAR] -> MDR; PC = 102; MDR -> IR	102	101	ADD 3	ADD 3	M[5]
Decode	CU decodes ADD 3	102	101	ADD 3	ADD 3	M[5]
Execute	ACC = ACC + Mem[3]	102	101	ADD 3	ADD 3	M[5]+M[3]
Fetch	PC(102) -> MAR; Mem[MAR] -> MDR; PC = 103; MDR -> IR	103	102	STORE 6	STORE 6	M[5]+M[3]
Decode	CU decodes STORE 6	103	102	STORE 6	STORE 6	M[5]+M[3]
Execute	ACC -> Mem[6]	103	102	STORE 6	STORE 6	M[5]+M[3]

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Software Types

System Software

Software that manages and controls hardware and provides a platform for application software.

Type	Description	Examples
Operating System (OS)	Manages all hardware and software resources	Windows, macOS, Linux, Android, iOS
Utility Programs	Perform specific maintenance tasks	Disk defragmenter, antivirus, file manager, backup tool

Application Software

Software designed for end-users to perform specific tasks.

Type	Description	Examples
General-purpose	Widely used across many domains	Word processors, spreadsheets, web browsers
Special-purpose	Designed for a specific field	Accounting software, CAD, medical imaging
Custom/bespoke	Written for a specific organisation	A company's payroll system

Operating System Functions

Function	Description
Memory Management	Allocates and deallocates memory space, uses virtual memory (swap space on disk to extend RAM), manages paging and segmentation
Process Management	Schedules CPU time among processes, handles multitasking (time-sharing), manages process creation and termination
File Management	Organises files in directories/folders, handles file naming, access control, and storage allocation
User Interface	Provides CLI (Command Line Interface) or GUI (Graphical User Interface) for user interaction
Device Management	Uses device drivers to communicate with hardware, manages I/O operations
Security	User authentication, access control, firewall integration

Types of User Interfaces

Command Line Interface (CLI):

• User types commands using a keyboard.
• Requires memorisation of commands and syntax.
• Efficient for experienced users; supports scripting and automation.
• Examples: Windows Command Prompt, Linux bash, macOS Terminal.

Graphical User Interface (GUI):

• Users interact with visual elements (windows, icons, menus, buttons).
• Intuitive and easy to learn; does not require memorising commands.
• Uses pointing devices (mouse, touchscreen).
• Consumes more system resources than CLI.
• Examples: Windows Explorer, macOS Finder, Android/iOS home screens.

Menu-Driven Interface:

• User selects options from a predefined list of menus.
• Common in ATMs, self-service kiosks, and embedded systems.
• Limited flexibility but very easy to use for specific tasks.

Virtual Memory

Virtual memory is a memory management technique that uses secondary storage (hard disk/SSD) as an extension of RAM. When physical RAM is full, the operating system moves less frequently used pages (data blocks, typically 4 KB) from RAM to a designated area on disk called the swap space or page file. When those pages are needed again, they are swapped back into RAM.

Aspect	Advantage	Disadvantage
Capacity	Allows running more/larger programs than physical RAM alone	Disk access is much slower than RAM (thousands of times slower)
Cost	Effectively increases memory without buying more RAM	Excessive swapping (thrashing) severely degrades performance
Implementation	Transparent to the user and applications	Requires disk space to be reserved

Thrashing occurs when the system spends more time swapping pages in and out of memory than executing actual instructions. This happens when the system is overloaded with too many processes competing for insufficient RAM.

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Number Systems

Binary (Base 2)

Digits: 0, 1. Each digit is a bit. 8 bits = 1 byte.

Decimal	Binary
0	0000 0000
1	0000 0001
2	0000 0010
5	0000 0101
10	0000 1010
255	1111 1111

Hexadecimal (Base 16)

Digits: 0--9, A(10), B(11), C(12), D(13), E(14), F(15). Each hex digit represents exactly 4 bits.

Hex	Binary	Decimal
0	0000	0
1	0001	1
9	1001	9
A	1010	10
F	1111	15
10	0001 0000	16
FF	1111 1111	255
100	0001 0000 0000	256

Binary-Coded Decimal (BCD)

Each decimal digit (0--9) is represented by its 4-bit binary equivalent.

Decimal	BCD
0	0000
9	1001
15	0001 0101
127	0001 0010 0111

:::warning[Exam Tip] BCD is different from pure binary. The decimal number 15 in pure binary is 1111, but in BCD it is 0001 0101 (each decimal digit encoded separately). BCD wastes some bit patterns (1010--1111 are invalid) but is useful for displays and financial calculations where each decimal digit must be preserved exactly. :::

Conversions

Decimal to Binary: Repeatedly divide by 2, record remainders from bottom to top.

Decimal to Hexadecimal: Repeatedly divide by 16, record remainders.

Binary to Hexadecimal: Group bits in groups of 4 from the right, convert each group.

Hexadecimal to Binary: Replace each hex digit with its 4-bit binary equivalent.

Decimal to BCD: Replace each decimal digit with its 4-bit binary equivalent.

Worked Example: Decimal 185 to Binary, Hex, and BCD

To binary:

$185 \div 2 = 92$ R $1$ $92 \div 2 = 46$ R $0$ $46 \div 2 = 23$ R $0$ $23 \div 2 = 11$ R $1$ $11 \div 2 = 5$ R $1$ $5 \div 2 = 2$ R $1$ $2 \div 2 = 1$ R $0$ $1 \div 2 = 0$ R $1$

Reading remainders from bottom to top: $185_{10} = 10111001_2$

To hexadecimal:

$185 \div 16 = 11$ R $9$. $11 = B$. So $185_{10} = B9_{16}$

To BCD:

$185 \to 1 \to 8 \to 5 \to 0001\ 1000\ 0101$

$185_{10} = 000110000101_{BCD}$

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Data Representation

Text Representation

ASCII (American Standard Code for Information Interchange)

• 7-bit code, extended to 8-bit (Extended ASCII).
• Represents 128 characters (0--127): uppercase letters (65--90), lowercase letters (97--122), digits (48--57), control characters (0--31), symbols.
• Each character stored as 1 byte (8 bits) with the MSB unused or used for parity.

Character	ASCII (decimal)	ASCII (binary)
'A'	65	0100 0001
'Z'	90	0101 1010
'a'	97	0110 0001
'0'	48	0011 0000
Space	32	0010 0000

Unicode

• Supports characters from all languages, symbols, and emoji.
• UTF-8 encoding: variable-length (1--4 bytes). Backward-compatible with ASCII (first 128 characters identical).
• UTF-16: 2 or 4 bytes per character.
• UTF-32: fixed 4 bytes per character.

info

Key difference: ASCII uses 1 byte per character and covers only English and basic symbols. Unicode covers all writing systems but uses more storage. UTF-8 is the most widely used encoding on the internet.

Image Representation

A bitmap image is a grid of pixels (picture elements). Each pixel is assigned a colour value.

Image resolution: Width $\times$ Height in pixels (e.g., $1920 \times 1080$).

Colour depth: Number of bits used to represent each pixel's colour.

Colour Depth	Number of Colours	Calculation
1 bit	2 (black and white)	$2^1$
8 bits	256	$2^8$
16 bits	65 536 (High Colour)	$2^{16}$
24 bits	16 777 216 (True Colour)	$2^{24}$

Image file size (uncompressed):

$$\mathrm{File Size (bits)} = \mathrm{Width} \times \mathrm{Height} \times \mathrm{Colour Depth}$$

In bytes:

$$\mathrm{File Size (bytes)} = \frac{\mathrm{Width} \times \mathrm{Height} \times \mathrm{Colour Depth}}{8}$$

Worked Example: Image File Size Calculation

A $1024 \times 768$ image with 24-bit colour depth.

$$\mathrm{Size (bits)} = 1024 \times 768 \times 24 = 18874368 \mathrm{ bits}$$$$\mathrm{Size (bytes)} = \frac{18874368}{8} = 2359296 \mathrm{ bytes}$$$$\mathrm{Size (KB)} = \frac{2359296}{1024} = 2304 \mathrm{ KB}$$$$\mathrm{Size (MB)} = \frac{2304}{1024} = 2.25 \mathrm{ MB}$$

Bitmap vs Vector Graphics:

Feature	Bitmap	Vector
Storage method	Grid of pixels	Mathematical descriptions of shapes
Scalability	Loses quality when enlarged	Scales without quality loss
File size for diagrams	Larger	Smaller
Suitability for photos	Excellent	Not suitable
Suitability for logos/diagrams	Depends on resolution	Ideal
Examples	JPEG, PNG, BMP	SVG, EPS
Editing	Pixel-level editing	Shape and path editing

Image compression:

Type	Method	Quality Loss	Example
Lossless	Reduces redundancy without discarding data	None	PNG, GIF, BMP
Lossy	Discards data that is less perceptible to the human eye	Yes (irreversible)	JPEG, WebP

JPEG compression works by dividing the image into $8 \times 8$ pixel blocks, applying a Discrete Cosine Transform (DCT), and then quantising (rounding) the frequency coefficients. Higher compression ratios discard more high-frequency detail, resulting in smaller files with visible artefacts (blockiness, blurring).

Sound Representation

Sound is an analogue wave that must be converted to digital form via sampling.

Steps in sound digitisation:

1. Sampling: Measure the amplitude of the sound wave at regular intervals (sampling rate).
2. Quantisation: Map each sampled amplitude to the nearest value in a set of discrete levels.
3. Encoding: Convert each quantised value to binary.

Sampling rate: Number of samples per second, measured in Hz. CD quality = 44 100 Hz (44.1 kHz).

Bit depth: Number of bits per sample. CD quality = 16 bits per sample. Higher bit depth = more accurate representation of the analogue signal. Common values: 8-bit (telephone quality), 16-bit (CD), 24-bit (studio recording).

Nyquist theorem: To accurately capture a sound with maximum frequency $f_{\max}$, the sampling rate must be at least $2f_{\max}$. Since human hearing ranges up to approximately $20\mathrm{ kHz}$, the CD sampling rate of $44\,100\mathrm{ Hz}$ is sufficient (Nyquist frequency = $22\,050\mathrm{ Hz}$).

File size calculation:

$$\mathrm{File Size (bits)} = \mathrm{Sampling Rate} \times \mathrm{Bit Depth} \times \mathrm{Duration (s)} \times \mathrm{Channels}$$

• Mono: 1 channel. Stereo: 2 channels.

Worked Example: Sound File Size Calculation

A 3-minute stereo recording at CD quality (44 100 Hz, 16-bit).

$$\mathrm{Duration} = 3 \times 60 = 180 \mathrm{ seconds}$$$$\mathrm{Size (bits)} = 44100 \times 16 \times 180 \times 2 = 254016000 \mathrm{ bits}$$$$\mathrm{Size (bytes)} = \frac{254016000}{8} = 31752000 \mathrm{ bytes}$$$$\mathrm{Size (MB)} = \frac{31752000}{1024 \times 1024} \approx 30.28 \mathrm{ MB}$$

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Common Pitfalls

1. Confusing RAM and ROM: RAM is volatile (loses data on power off) and read-write. ROM is non-volatile and read-only (or mostly read-only).

2. Fetch-decode-execute order: The PC is incremented during the fetch phase, not after execution. The order matters for exam questions.

3. Binary vs BCD: 15 in binary is 1111; 15 in BCD is 0001 0101. They are different representations.

4. Colour depth and number of colours: 24-bit colour gives $2^{24} = 16\ 777\ 216$ colours, not 24 colours.

5. Image file size formula: Do not forget to divide by 8 to convert bits to bytes, and by 1024 for KB/MB.

6. Sound file size: Remember to multiply by the number of channels (2 for stereo). Forgetting this halves the file size.

7. ASCII range: Uppercase 'A' is 65, lowercase 'a' is 97. The difference is 32. Digits '0'--'9' are 48--57. These values are frequently tested.

8. Volatility: "Volatile" means data is lost when power is removed. RAM is volatile; ROM, HDD, SSD, and flash memory are non-volatile.

9. SSD vs Flash: An SSD is a type of flash memory device. All SSDs use flash memory, but not all flash memory devices are SSDs (e.g., USB drives, SD cards).

10. ALU vs CU: The ALU performs calculations; the CU coordinates and controls operations. Do not mix them up.

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Practice Problems

Question 1: Storage Comparison

A school needs to store 500 GB of student records. They are considering an HDD vs SSD solution.

(a) Calculate how many 1 TB HDDs would be needed to store 500 GB of data.

(b) Give two advantages and two disadvantages of using SSDs instead of HDDs for this purpose.

Answer:

(a) $500 \mathrm{ GB} = 0.5 \mathrm{ TB}$. One 1 TB HDD is sufficient.

(b) Advantages of SSD: Faster read/write speeds, no moving parts (more durable, less prone to mechanical failure), quieter operation, lower power consumption.

Disadvantages of SSD: Higher cost per GB, limited write cycles (though modern SSDs mitigate this with wear levelling).

Question 2: Fetch-Decode-Execute Trace

Given the following program in memory:

Address	Instruction
200	LOAD 10
201	ADD 11
202	SUB 12
203	STORE 13

Memory values: M[10] = 25, M[11] = 17, M[12] = 8, M[13] = 0.

Trace the fetch-decode-execute cycle for the first two instructions, showing the state of PC, MAR, MDR, IR, and ACC after each phase.

Answer:

Instruction 1: LOAD 10

Phase	PC	MAR	MDR	IR	ACC
Fetch	201	200	LOAD 10	LOAD 10	?
Decode	201	200	LOAD 10	LOAD 10	?
Execute	201	200	LOAD 10	LOAD 10	25

Instruction 2: ADD 11

Phase	PC	MAR	MDR	IR	ACC
Fetch	202	201	ADD 11	ADD 11	25
Decode	202	201	ADD 11	ADD 11	25
Execute	202	201	ADD 11	ADD 11	42

ACC = 25 + 17 = 42.

Question 3: Number Conversions

(a) Convert $11010110_2$ to decimal.

(b) Convert $234_{10}$ to binary and hexadecimal.

(c) Convert the decimal number 593 to BCD.

Answer:

(a) $1 \times 128 + 1 \times 64 + 0 \times 32 + 1 \times 16 + 0 \times 8 + 1 \times 4 + 1 \times 2 + 0 \times 1 = 128 + 64 + 16 + 4 + 2 = 214_{10}$

(b) To binary: $234 \div 2 = 117$ R $0$ $117 \div 2 = 58$ R $1$ $58 \div 2 = 29$ R $0$ $29 \div 2 = 14$ R $1$ $14 \div 2 = 7$ R $0$ $7 \div 2 = 3$ R $1$ $3 \div 2 = 1$ R $1$ $1 \div 2 = 0$ R $1$

$234_{10} = 11101010_2$

To hexadecimal: $234 \div 16 = 14$ R $10$. $14 = E$, $10 = A$. So $234_{10} = EA_{16}$

(c) $593 \to 5 \to 9 \to 3 \to 0101\ 1001\ 0011_{BCD}$

Question 4: Image Representation

A digital camera produces images of $4000 \times 3000$ pixels with 32-bit colour depth.

(a) Calculate the uncompressed file size of one image in MB.

(b) If the camera has a 16 GB storage card, how many such images can it store?

Answer:

(a)

$$\mathrm{Size (bits)} = 4000 \times 3000 \times 32 = 384000000 \mathrm{ bits}$$$$\mathrm{Size (bytes)} = \frac{384000000}{8} = 48000000 \mathrm{ bytes}$$$$\mathrm{Size (MB)} = \frac{48000000}{1024 \times 1024} \approx 45.77 \mathrm{ MB}$$

(b)

$$16 \mathrm{ GB} = 16 \times 1024 \mathrm{ MB} = 16384 \mathrm{ MB}$$$$\mathrm{Number of images} = \left\lfloor \frac{16384}{45.77} \right\rfloor = 358 \mathrm{ images}$$

Question 5: Sound Representation

A 2-minute mono audio recording is digitised at a sampling rate of 22 050 Hz with 8-bit resolution.

(a) Calculate the file size in MB.

(b) Explain the effect on file size and sound quality if the sampling rate is doubled to 44 100 Hz.

Answer:

(a)

$$\mathrm{Duration} = 2 \times 60 = 120 \mathrm{ seconds}$$$$\mathrm{Size (bits)} = 22050 \times 8 \times 120 \times 1 = 21168000 \mathrm{ bits}$$$$\mathrm{Size (bytes)} = \frac{21168000}{8} = 2646000 \mathrm{ bytes}$$$$\mathrm{Size (MB)} = \frac{2646000}{1024 \times 1024} \approx 2.52 \mathrm{ MB}$$

(b) Doubling the sampling rate doubles the file size to approximately 5.05 MB. Sound quality improves because more samples per second capture the original waveform more accurately, reducing aliasing and better reproducing higher frequencies (the Nyquist frequency doubles from ~11 kHz to ~22 kHz).

Question 6: ASCII and Character Encoding

(a) The ASCII code for 'H' is 72. What is the ASCII code for 'h'?

(b) A text file contains 1000 characters, all ASCII. What is the file size in KB?

(c) Explain why Unicode is preferred over ASCII for international applications.

Answer:

(a) 'h' = 72 + 32 = 104. (Lowercase letters are 32 positions after their uppercase counterparts.)

(b) Each ASCII character is 1 byte. $1000 \mathrm{ bytes} / 1024 = 0.98 \mathrm{ KB}$.

(c) ASCII only supports 128 characters (English alphabet, digits, basic symbols). Unicode supports over 149 000 characters covering all major writing systems, symbols, and emoji. This makes Unicode essential for applications serving users who speak different languages or need special characters.

Question 7: Operating System Functions

(a) Explain the difference between memory management and process management in an operating system.

(b) Describe what virtual memory is and why it is used.

(c) Give an example of a utility program and explain its purpose.

Answer:

(a) Memory management allocates and deallocates RAM among running programs, ensuring each process has sufficient memory and that processes do not interfere with each other's memory space. Process management schedules CPU time among processes, handles creation/termination of processes, and enables multitasking.

(b) Virtual memory uses a portion of the hard disk as an extension of RAM. When RAM is full, the OS moves less frequently used data (pages) from RAM to disk (swapping/paging). This allows the system to run more programs than would fit in physical RAM, at the cost of slower access to swapped data.

(c) Disk defragmenter: Rearranges fragmented file data on an HDD so that related data blocks are stored contiguously. This improves read/write performance by reducing disk head movement. (Note: defragmentation is not needed for SSDs.)

Question 8: Software Types

Classify each of the following as system software or application software, and explain your reasoning:

(a) Microsoft Word (b) Windows Defender (c) Linux (d) Adobe Photoshop

Answer:

(a) Application software -- it is a general-purpose word processor designed for end-user tasks.

(b) System software (utility program) -- it is an antivirus utility that protects the system by detecting and removing malware. It performs a maintenance/security task for the system.

(c) System software (operating system) -- it manages all hardware resources, provides memory/process/file management, and serves as a platform for other software.

(d) Application software -- it is a special-purpose image editing program designed for end-users.

Question 9: Data Representation -- Extended

(a) Convert $3F_{16}$ to decimal.

(b) Convert $11001100_2$ to hexadecimal.

(c) A bitmap image has dimensions $800 \times 600$ and uses 16-bit colour. Calculate the file size in KB. If the image is compressed using a lossless technique with a 3:2 compression ratio, what is the compressed file size?

Answer:

(a) $3 \times 16 + 15 = 48 + 15 = 63_{10}$

(b) Group into nibbles: $1100\ 1100$. $C$ and $C$. So $11001100_2 = CC_{16}$

(c)

$$\mathrm{Uncompressed (bits)} = 800 \times 600 \times 16 = 7680000 \mathrm{ bits}$$$$\mathrm{Uncompressed (bytes)} = \frac{7680000}{8} = 960000 \mathrm{ bytes}$$$$\mathrm{Uncompressed (KB)} = \frac{960000}{1024} = 937.5 \mathrm{ KB}$$$$\mathrm{Compressed (KB)} = \frac{937.5}{3} \times 2 = 625 \mathrm{ KB}$$

Question 10: Input/Output Devices -- Scenario

A library wants to automate its book checkout system. For each scenario, recommend the most appropriate input or output device and justify your choice.

(a) Identifying a book when it is checked out. (b) Identifying a library member. (c) Printing a receipt for the member. (d) Displaying the library catalogue to members on a self-service kiosk.

Answer:

(a) Barcode reader: Each book has a barcode on the cover containing a unique identifier. Scanning is fast and accurate.

(b) RFID reader / barcode reader: Library cards often have barcodes or RFID chips. RFID allows contactless scanning without removing the card from a wallet.

(c) Laser printer: Fast, low cost per page, suitable for printing simple text receipts in high volume.

(d) Touchscreen monitor: Allows members to interact directly with the catalogue. Combines input (touch) and output (display) in one device, ideal for self-service kiosks.

Question 11: Comparing CPU Components

(a) Describe the role of the Control Unit (CU) in the fetch-decode-execute cycle.

(b) Describe the role of the Arithmetic Logic Unit (ALU).

(c) Explain why registers are necessary in a CPU. Why can the CPU not operate using only RAM?

Answer:

(a) The CU coordinates and controls all operations during the fetch-decode-execute cycle. During the fetch phase, it sends control signals to copy the PC value to the MAR and trigger a memory read. During decode, it interprets the instruction in the IR and determines which operation to perform. During execute, it sends the appropriate control signals to the ALU, registers, or memory to carry out the instruction.

(b) The ALU performs all arithmetic calculations (addition, subtraction, multiplication, division) and logical operations (AND, OR, NOT, XOR, comparisons). It receives operands from registers, performs the operation as directed by the CU, and stores the result in the accumulator or a specified register.

(c) Registers are necessary because they provide the fastest possible storage access (1 clock cycle), whereas RAM access takes many clock cycles (100+ cycles for L3 cache, even more for RAM). If the CPU had to fetch every operand from RAM for every instruction, processing would be impractically slow. Registers allow the CPU to keep frequently used data (current instruction, memory addresses, intermediate results) immediately accessible, maximising throughput.

Question 12: System Performance Factors

A student claims that a computer with a faster CPU clock speed will always perform better than one with a slower clock speed. Evaluate this claim by considering at least three other factors that affect system performance.

Answer:

The claim is not always correct. While CPU clock speed is one factor affecting performance, the following factors are also critical:

1. Number of CPU cores: A 2.0 GHz quad-core processor can outperform a 3.0 GHz dual-core processor for multitasking and parallel workloads, since it can execute more instructions simultaneously.

2. Cache size: Larger caches reduce the frequency of cache misses, meaning the CPU spends less time waiting for data from RAM. A CPU with a large L3 cache may outperform a faster CPU with a small cache.

3. RAM capacity and speed: Insufficient RAM causes the system to rely heavily on virtual memory (swapping to disk), which is extremely slow. A system with adequate fast RAM will outperform one with insufficient RAM regardless of CPU speed.

4. Storage type: An SSD is orders of magnitude faster than an HDD for boot times, application loading, and file operations. A system with an SSD but slower CPU may feel more responsive than one with a fast CPU and HDD.

5. Bus speed and width: The speed of the system bus (front-side bus) affects how quickly data can transfer between the CPU, RAM, and other components.