Understand primary storage

Data Storage – Primary & Secondary Storage

1. What is Primary Storage?

Primary storage (also called main memory) is the part of a computer that holds data and instructions which the CPU must access rapidly while a program is running. It is directly addressable by the processor, is much faster than secondary storage, and is usually volatile.

2. Key Characteristics of Primary Storage

• Volatility: Most primary memory (e.g., registers, cache, RAM) loses its contents when power is removed.
• Speed: Access times are measured in nanoseconds (ns) – orders of magnitude faster than secondary storage.
• Direct addressability: Every byte has a unique memory address that the CPU can use without any intermediate steps.
• Limited capacity: From a few bytes in micro‑controllers to several gigabytes (GB) in modern PCs.

3. Types of Primary Storage

1. Registers – tiny, ultra‑fast storage inside the CPU for operands and intermediate results.
2. Cache memory – small, fast memory (L1, L2, L3) that holds copies of frequently used RAM data.
3. Random‑Access Memory (RAM) – the main workspace for programs and data while they run.
4. Read‑Only Memory (ROM) and its variants
   • ROM – permanently programmed during manufacture; non‑volatile.
   • PROM (Programmable ROM) – can be programmed once after manufacture.
   • EPROM (Erasable‑Programmable ROM) – can be erased with UV light and re‑programmed.
   • EEPROM (Electrically‑Erasable‑Programmable ROM) – can be erased and re‑programmed electrically.
   • Flash memory – a type of EEPROM used for BIOS/UEFI and solid‑state drives.

4. Memory Hierarchy (Fastest → Slowest)

5. Addressing in Primary Memory

Primary memory is byte‑addressable: each individual byte has its own address. If a system has n address lines, the total number of addressable locations is:

Number of locations = 2ⁿ

Example 1 – 32‑bit address bus: 2³² = 4 294 967 296 bytes ≈ 4 GB of RAM.

Example 2 – 24‑bit address bus (exam style): 2²⁴ = 16 777 216 bytes = 16 MiB of RAM.

6. Volatile vs Non‑volatile Primary Storage

7. Units of Storage (Binary Prefixes)

The IGCSE syllabus uses the binary (base‑2) definitions. All conversions are performed by repeatedly dividing or multiplying by 1 024.

• 1 byte = 8 bits
• 1 KiB (kibibyte) = 2¹⁰ = 1 024 bytes
• 1 MiB (mebibyte) = 2²⁰ = 1 024 KiB = 1 048 576 bytes
• 1 GiB (gibibyte) = 2³⁰ = 1 024 MiB = 1 073 741 824 bytes

8. Primary vs Secondary Storage – Quick Comparison

9. Virtual Memory – Bridging Primary & Secondary Storage

• Virtual memory uses a reserved area of secondary storage (called swap space or page file) as an extension of RAM.
• Memory is divided into fixed‑size blocks called pages (commonly 4 KB). When RAM becomes full, pages that are not currently needed are written to the swap space; needed pages are read back into RAM.
• This paging process allows a computer to run programs that require more memory than the physical RAM provides, at the cost of slower access (disk speed vs RAM speed).
• Key points for the exam:
  • Virtual memory = RAM + part of secondary storage.
  • Pages are moved between RAM and swap space (paging).
  • Purpose: prevent “out‑of‑memory” crashes and enable larger programs.

10. File‑Size Calculations (Exam‑style Examples)

Example A – Colour Image

Calculate the size of a 640 × 480 pixel colour image that uses 24‑bit colour (8 bits for each of red, green, blue).

1. Number of pixels = 640 × 480 = 307 200.
2. Bits per pixel = 24 bits.
3. Total bits = 307 200 × 24 = 7 372 800 bits.
4. Convert to bytes: 7 372 800 ÷ 8 = 921 600 bytes.
5. Convert to KiB: 921 600 ÷ 1 024 ≈ 900 KiB.
6. Convert to MiB: 900 ÷ 1 024 ≈ 0.88 MiB.

Uncompressed size ≈ 0.88 MiB (≈ 900 KiB).

Example B – Mono Audio Clip

Calculate the size of a 30‑second mono audio recording sampled at 44.1 kHz with 16‑bit resolution.

1. Sample rate = 44 100 samples / second.
2. Bits per sample = 16 bits.
3. Total bits = 44 100 × 30 × 16 = 21 196 800 bits.
4. Convert to bytes: 21 196 800 ÷ 8 = 2 649 600 bytes.
5. Convert to KiB: 2 649 600 ÷ 1 024 ≈ 2 588 KiB.
6. Convert to MiB: 2 588 ÷ 1 024 ≈ 2.53 MiB.

Uncompressed size ≈ 2.53 MiB (≈ 2 588 KiB).

11. Data Compression

Compression reduces the amount of storage required for a file.

• Lossless compression – original data can be perfectly reconstructed (e.g., PNG, ZIP). Used where every bit matters, such as text, spreadsheets, or executable files.
• Lossy compression – some data is permanently discarded to achieve higher compression ratios (e.g., JPEG for photos, MP3 for audio). Acceptable when a small loss of quality is not noticeable.

Exam‑style Question – Compression

Question: Identify whether each of the following file types uses lossless or lossy compression.

1. JPEG image
2. PNG image
3. MP3 audio
4. ZIP archive

Answer:

• JPEG – lossy
• PNG – lossless
• MP3 – lossy
• ZIP – lossless

12. Linking to IGCSE Assessment Objectives

13. Key Take‑aways

• Primary storage is fast, directly addressable memory used while programs run.
• The memory hierarchy (registers → cache → RAM → ROM) balances speed, cost, and capacity.
• Address‑bus width determines the maximum addressable memory (2ⁿ locations).
• Binary prefixes (KiB, MiB, GiB) are essential for capacity and file‑size calculations.
• Virtual memory links primary and secondary storage through paging and swap space.
• Compression (lossless vs. lossy) lets large files be stored more efficiently; know the typical examples.
• All sub‑topics map directly to IGCSE assessment objectives (AO1, AO2, AO3), aiding both teaching and exam preparation.