IGCSE Computer Science — Theory: Bits, Boxes and Consequences
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Lesson overview
Binary and hexadecimal, the size of sound and image files, hardware and the fetch-execute cycle, networks and their threats, and the effects of being connected.
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Pacing
~1 minute per mark; match answer depth strictly to `[1]`, `[2]`, `[3]`, or `[4]` mark allocations.
How IGCSE marks this
Command words govern mark schemes: 'State' = 1 concise fact; 'Describe' = what happens; 'Explain' = why/how using scientific terminology.
Watch out for
Vague everyday wording ('it gets bigger') fails Cambridge mark schemes where exact terms ('rate of collisions increases') are required.
Lesson content
The theory paper is a set of small calculations
Most of the marks on Paper 1 are earned by candidates who can do four things precisely: convert between bases, compute a file size from a stated resolution and bit depth, trace a loop without losing a pass, and name the trade-off a hardware or network choice makes. None of them needs a memory of a definition; all of them need practice, and all of them are the questions candidates get wrong while believing they knew the topic. So treat each as an arithmetic habit with a check. Binary conversion: build up from 128, 64, 32, 16, 8, 4, 2, 1, and write a 1 where the value is used, then verify by converting back. File size: pixels × bits per pixel ÷ 8 for bytes, and be explicit about whether 'kilobyte' in the question means 1,000 or 1,024 — the syllabus convention governs, so state which you used. Loop trace: two columns, one for the variable and one for the count, and stop when the test fails. Trade-off: name what is gained and what is lost, since 'faster' without a cost is not an explanation.
Data representation, honestly quantified
Everything a computer stores is a number, and the paper asks the candidate to say which. Text uses a character set and a coding scheme: the set defines which symbols exist, the encoding assigns each a bit pattern, and a larger set with a longer code covers more scripts at the price of size. Images are grids of pixels, each holding a colour taken from a palette whose depth is bits per pixel; sound is sampled at a rate, with each sample held at a bit depth, so a higher rate and depth reproduce more of the original waveform and store more bits. The consequence is a single multiplication, and it is examinable in every session: more resolution, more colour depth, a higher sample rate or a greater bit depth each improve fidelity, each increase the file, and each slow down transmission and filling of storage. Compression trades one against the other: lossless removes redundancy so the original can be recovered exactly, and suits text, code and a spreadsheet; lossy discards data the human senses are least likely to notice, gains far smaller files, and cannot be undone — which is why it is used for delivery and never for the master. State the trade-off, and then apply it to the example the question gives.
Two file sizes, computed in front of the marker
(a) An uncompressed image of 800 × 600 pixels with 24-bit colour. (b) A two-minute mono audio recording sampled at 8,000 times per second with 8 bits per sample. (a) 800 × 600 = 480,000 pixels; 24 bits each = 480,000 × 24 = 11,520,000 bits = 1,440,000 bytes. That is just under 1.4 MB in decimal units, or about 1.37 MiB if the answer divides by 1,024 twice — so write the divisor you used. (b) 120 s × 8,000 samples/s × 8 bits = 7,680,000 bits = 960,000 bytes. Then the sentence that earns the follow-up marks: halving the colour depth to 12 bits halves the image, and halving the sample rate halves the audio but puts an audible limit on the highest frequency that can be captured, because sampling at 8,000 times a second can represent frequencies only up to about 4 kHz — which is why telephony sounds like telephony. Note the habit: the units are stated, the arithmetic is shown, and the effect of changing one parameter is named rather than asserted.
Both directions of the conversion, and the check
Convert 45 to binary, and hexadecimal 2C to denary. Then explain one reason an address might be written in hexadecimal rather than binary. 45 = 32 + 8 + 4 + 1, so the bits for 32, 16, 8, 4, 2 and 1 are 1 0 1 1 0 1: 101101, and as a byte 00101101. Check it: 64×0 + 32×1 + 16×0 + 8×1 + 4×1 + 2×0 + 1×1 = 45 ✓. For 2C: 2 × 16 + 12 = 44, since C is the tenth letter of the hex digits above 9. The reason for hexadecimal is compactness and readability for a human: one hex digit is exactly four bits, so a byte is two digits, a mistake is visible, and a memory address of 0xF0A1 is far less error-prone to copy than 1111000010100001. The point a candidate should make in the exam is that four-to-one mapping — it is why hex and binary convert without arithmetic, by grouping bits into fours.
Four habits for Paper 1
Show the conversion and check it in the other direction; state the units and the divisor in any file-size answer; trace a loop in a table instead of in your head; and finish every 'explain' or 'discuss' with the cost of the thing you have recommended — speed against capacity, security against convenience, lossy against lossless. The named components, the protocols and the current technology the syllabus lists are set for your examination year, and the paper's options change with the revision. Check the syllabus for what is examinable this cycle; the arithmetic, the trace and the trade-off do not change.
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