How Computers Work

Q1. (6 marks)

(a) $45$ to binary: $45 = 32 + 8 + 4 + 1 = 101101_2$. Check: $32+8+4+1 = 45$. ✓ Answer: $101101_2$. (2) — (1 for method, 1 for correct result)

(b) $10110110_2$ to decimal: $128 + 0 + 32 + 16 + 0 + 4 + 2 + 0 = 182$. Answer: $182$. (2)

(c) $10110110_2$ to hex: group into nibbles $1011\,0110 = \text{B}\,6 = \text{B6}_{16}$. Answer: $\text{B6}_{16}$ (= $182$). (2)

Why: positional weights are powers of 2; each group of 4 bits maps to one hex digit.

Q2. (3 marks) Hexadecimal is compact: each hex digit represents exactly 4 bits (one nibble), so one byte = 2 hex digits (1). This makes long binary strings much shorter and easier to read/write without error (1). The conversion between binary and hex is trivial and lossless (grouping bits), unlike decimal which does not align with power-of-2 boundaries (1).

Q3. (4 marks) (a) $1 \oplus 1 = 0$ (1); $1 \oplus 0 = 1$ (1). (XOR outputs 1 only when inputs differ.) (b) Universal gate means any Boolean function / any other gate (AND, OR, NOT, etc.) can be built using only NAND gates (2). (Because NAND is functionally complete.)

Q4. (4 marks) — 0.5 per correct cell (8 cells)

Why: XOR = 1 when inputs differ; NOR = NOT(OR), so 1 only when both are 0.

Q5. (5 marks) (a) Half adder: $\text{Sum} = A \oplus B$ (1); $\text{Carry} = A \cdot B$ (AND) (1). (b) Full adder with $A=1, B=1, C_{in}=1$: total $= 1+1+1 = 3 = 11_2$. So $\text{Sum} = 1$ (1) and $\text{Carry-out} = 1$ (1); correct reasoning $1+1+1=3$ (1).

Q6. (4 marks) (a) A flip-flop stores a single bit of information (a stable 0 or 1); it is a basic memory element that holds state until changed (2). (b) In an SR flip-flop the forbidden input is $S=1, R=1$ (Set and Reset both high) (1), because it drives both outputs to the same value / produces an undefined (indeterminate) state and violates $Q = \overline{Q'}$; the resulting state on release is unpredictable (1).

Q7. (5 marks) Order fastest/smallest → slowest/largest (2 marks, all correct): CPU registers → L1 cache → RAM → SSD → HDD. (Partial: 1 mark if only one adjacent pair swapped.)

Trade-off (3 marks): Faster memory is more expensive per byte and therefore built in smaller quantities; slower memory is cheaper per byte and used in larger capacities (3). Hence the hierarchy balances speed against cost/size.

Q8. (4 marks) — 1 mark per stage named + 1 for full correct ordering

1. Fetch — retrieve the next instruction from memory (address in the program counter). (1)
2. Decode — the control unit interprets the instruction and determines the operation/operands. (1)
3. Execute — the ALU/relevant unit carries out the operation and stores the result. (1) Correct order maintained (1).

Q9. (5 marks) (a) ALU (Arithmetic Logic Unit) (1). (b) Control Unit (1). (c) Machine code is the binary instructions the CPU directly executes (1). Assembly language is a human-readable text form using mnemonics (e.g. MOV, ADD) that maps essentially one-to-one to machine instructions (1) and must be translated (assembled) into machine code before running (1).

[
  {"claim": "45 decimal equals binary 101101", "code": "result = (int('101101',2) == 45)"},
  {"claim": "10110110 binary equals 182 decimal", "code": "result = (int('10110110',2) == 182)"},
  {"claim": "10110110 binary equals hex B6", "code": "result = (format(int('10110110',2),'X') == 'B6')"},
  {"claim": "Full adder 1+1+1 gives sum 1 carry 1", "code": "s=(1+1+1)%2; c=(1+1+1)//2; result = (s==1 and c==1)"},
  {"claim": "XOR column is 0,1,1,0 and NOR column is 1,0,0,0", "code": "xor=[a^b for a in (0,1) for b in (0,1)]; nor=[1-(a|b) for a in (0,1) for b in (0,1)]; result = (xor==[0,1,1,0] and nor==[1,0,0,0])"}
]