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IB Computer Science HL · Paper 1

Computer Science, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 16

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short answer

Questions 1–11 · 25 marks

Question 1

2 marks

Software-as-a-Service benefits and operational tradeoffs

System fundamentals

(a)

1 mark

Benefit: Rapid deployment on demand via predictable subscription pricing, eliminating upfront capital expenditure for dedicated server hardware and in-house maintenance.

Mentor tip

Name subscription elasticity, reduced local infrastructure, or ubiquitous web access.

(b)

1 mark

Drawback: Absolute dependency on continuous network connectivity; an internet outage halts productivity, and sensitive proprietary data is stored on third-party infrastructure.

Mentor tip

Highlight downtime vulnerability during connectivity loss or loss of direct data custody.

Question 2

2 marks

Role of User Sign-off Testing (UAT) in software deployment

System fundamentals

2 marks

User Sign-off Testing (UAT) (UAT) is the final testing phase conducted by intended end-users in real-world operating scenarios.
Its primary objective is verifying that the finished software satisfies business requirements, conforms to agreed specifications, and is viable for production sign-off.

Mentor tip

Emphasize validation by actual end-users against business requirements prior to final sign-off.

Question 3

1 mark

Hardware definition of peripheral devices

Computer organisation

1 mark

A peripheral is any auxiliary hardware device (such as a keyboard, mouse, monitor, or external drive) connected externally or via expansion slots to communicate with the CPU.

Mentor tip

Define peripherals as auxiliary input, output, or secondary storage devices attached to the core computer.

Question 4

2 marks

Mobile usability constraints and interface friction

System fundamentals

2 marks

Issue 1: Limited screen real estate leading to small tap targets, cramped typography, and touch navigation errors.
Issue 2: Sub-optimal contrast and glare under direct sunlight, or touch-screen keyboard typing inaccuracies due to lack of tactile feedback.

Mentor tip

State two concrete physical or software usability constraints (screen size, touch accuracy, glare).

Question 5

2 marks

Accessibility affordances for diverse user cohorts

System fundamentals

2 marks

Method 1: Operating system screen readers and text-to-speech narration paired with high-contrast display palettes for visually impaired users.
Method 2: Speech recognition software or adaptive hardware inputs (e.g. Braille displays and single-switch devices) for users with motor impairments.

Mentor tip

Identify two distinct accommodations: speech synthesis/screen readers and alternative input peripherals.

Question 6

2 marks

Function of the Memory Data Register in instruction cycles

Computer organisation

2 marks

The Memory Data Register (MDR) holds data words or machine instructions during transfers between the CPU and primary memory (RAM).
It temporarily buffers the contents fetched from the memory address specified by the MAR, or holds data staged to be written into RAM via the data bus.

Mentor tip

Pair the MDR directly with the data bus, acting as the two-way holding buffer for data to/from RAM.

Question 7

2 marks

Number representation: decimal to binary and hexadecimal conversions

Computer organisation

(a)

1 mark

Decompose $3 7_{10}$ into powers of 2:
$37 = 32 + 4 + 1 = 2^{5} + 2^{2} + 2^{0}$ .
In 8-bit binary representation: $0010010 1_{2}$ .

3 7_{10} = 0010010 1_{2}

Mentor tip

Sum binary place values: 32 + 4 + 1 = 37, giving 00100101.

(b)

1 mark

Partition the binary byte $0010010 1_{2}$ into 4-bit nibbles:
$001 0_{2} = 2_{16}$
$010 1_{2} = 5_{16}$
Concatenating gives $2 5_{16}$ .

0010010 1_{2} = 2 5_{16}

Mentor tip

Nibbles are 0010 (2) and 0101 (5), yielding 25 in hexadecimal.

Question 8

2 marks

Practical software applications of stack data structures

Abstract data structures

2 marks

Application 1: Call stack management in runtime environments, recording return addresses, parameters, and local activation frames during recursive function calls.
Application 2: Undo/redo command histories in productivity software, and syntax parsing / delimiter matching in language compilers.

Mentor tip

Name two classic LIFO applications: runtime call stack/recursion, undo history, or bracket balancing.

Question 9

4 marks

Operating system resource management mechanisms

Resource management

4 marks

Technique 1: CPU Process Scheduling. The OS scheduler governs process execution states (Ready, Running, Blocked), allocating CPU execution quanta using algorithms like Round Robin or Multi-Level Feedback Queues to ensure equitable processor utilisation.
Technique 2: Virtual Memory Management via Paging. Divides process memory into uniform pages and physical memory into frames, swapping inactive pages to secondary storage swap space to run programs that exceed physical RAM.

Mentor tip

State two concrete OS resource management duties: CPU process scheduling and virtual memory paging.

Question 10

2 marks

Sensor telemetry in domestic washing machine automation

Control systems

2 marks

Sensor type: Water level pressure sensor or float switch.
Purpose: Measures hydraulic pressure inside the tub to determine water fill levels, signaling the microprocessor to close water intake solenoid valves when full and prevent motor activation when empty.

Mentor tip

Pair a specific sensor (water level/pressure sensor) with its operational control role (closing inlet valves).

Question 11

4 marks

Role of closed-loop feedback in automated microprocessor control

Control systems

4 marks

Sensing and conversion: Sensors continuously sample the physical process variable (e.g. ambient temperature), converting analog readings via an ADC into digital data fed to the microprocessor.
Comparison and error detection: The microprocessor evaluates incoming sensor values against predetermined target setpoints stored in memory, calculating the error deviation.
Actuator adjustment: If a discrepancy exists, the processor outputs corrective control signals to actuators (e.g. energising heating coils or opening cooling vents).
Closed-loop stabilization: The physical change caused by the actuator alters the sensor's next reading, creating a continuous closed feedback loop that stabilizes the system at the desired setpoint.

Mentor tip

Explain the full closed loop: sensor input $\to$ ADC $\to$ setpoint comparison $\to$ error calculation $\to$ actuator adjustment.

Section B · Structured and algorithmic problems

Questions 12–16 · 75 marks

Question 12

15 marks

Transmission media, packet switching dynamics, and VPN social impacts

Networks

(a)(i)

2 marks

High bandwidth and low signal loss: Pulses of light travel through optical glass cores, supporting gigabit-to-terabit data transmission with minimal attenuation over long distances.
Electromagnetic immunity: Optical photons do not emit or conduct electrical current, making fibre lines immune to electromagnetic interference (EMI) and lightning surges.

Mentor tip

State optical photon transmission, high bandwidth over distance, and total EMI immunity.

(a)(ii)

2 marks

Physical mobility: Endpoints transmit data via radio frequency waveforms, enabling roaming without physical tethering.
Susceptibility to interference and attenuation: Signal quality degrades through solid physical obstacles (walls) and experiences RF collisions from ambient wireless devices.

Mentor tip

Contrast freedom of movement against signal degradation from walls and radio interference.

(b)

3 marks

Data encryption translates readable plaintext into unintelligible ciphertext using mathematical cipher algorithms and cryptographic keys.
During transit across unsecure channels, intercepted packets cannot be decoded or altered by unauthorised third parties without possessing the private decryption key.

Mentor tip

Explain plaintext-to-ciphertext translation via mathematical keys and the resulting confidentiality during transit.

(c)

5 marks

Packet decomposition: Messages are broken down into small, standardized units called packets, each formatted with a payload and a header containing sender IP, recipient IP, packet sequence number, and error-checking checksums.
Independent dynamic routing: Intermediate routers inspect the destination IP on each packet and forward it along the optimal path available at that instant; packets in the same session may travel along different physical routes.
Reassembly at destination: The destination host uses header sequence numbers to reassemble packets into their original message sequence, requesting retransmission for any lost or corrupted packets.

Mentor tip

Detail packet headers (IPs, sequence numbers), independent router path selection, and destination reassembly.

(d)

3 marks

Impact: Growth of remote telecommuting and flexible working practices.
Mechanism: VPNs enable secure access to enterprise internal intranets from any residential or remote location.
Social consequence: Enhances work-life balance and eliminates commuting travel times, but can blur the boundary between domestic life and work hours, increasing risks of professional isolation and burnout.

Mentor tip

Connect technical VPN remote access to societal shifts: flexibility/reduced commuting vs boundary erosion/isolation.

Question 13

15 marks

Boolean logic design, sensor thresholds, and vehicle alert algorithms

Computer organisation
Computational thinking, problem-solving and programming

(a)

1 mark

The NOR operator is a Boolean logic function that evaluates to true (1) if and only if all input operands are false (0); it is the logical negation of the OR function.

A 0011 B 0101 NOR (A, B) 1000

Mentor tip

Define NOR as returning 1 exclusively when all inputs are 0.

(b)

4 marks

Formulate conditions from specification:
Condition 1: Car travelling forward ( $B = 1$ ) and distance $< 20 m$ ( $A = 0$ , so $\neg A = 1$ ). Combined with AND: $\neg A \land B$ .
Condition 2: Speed $> 130 km/h$ ( $C = 0$ , so $\neg C = 1$ ).
Combine conditions with OR: $Z = (\neg A \land B) \lor \neg C$ .
Gate layout: Invert $A$ with a NOT gate; route $\neg A$ and $B$ into a 2-input AND gate; invert $C$ with a NOT gate; feed both into a final 2-input OR gate to produce $Z$ .

Z = (\neg A \land B) \lor \neg C

Mentor tip

Notice that $A = 0$ represents distance $< 20 m$ and $C = 0$ represents speed $> 130 km/h$ , requiring NOT gates on A and C.

(c)

2 marks

Assign binary states for input $D$ :
$D = 0$ : Obstruction is less than 3 metres from the rear of the car.
$D = 1$ : No obstruction detected within 3 metres from the rear of the car.

Mentor tip

Define both binary states clearly: 0 for obstacle within 3m, 1 for clear path (or vice versa).

(d)

2 marks

Substitute given values: $F = 5$ , $S = 30$ , $T = true$ .
Evaluate component terms:
$F \geq 25 ⟹ 5 \geq 25$ is false.
$S \geq 10 ⟹ 30 \geq 10$ is true.
$S \leq 130 ⟹ 30 \leq 130$ is true.
$T = true ⟹ true$ .
Combine with AND: $false \land true \land true \land true = false$ .

5 \geq 25 \land 30 \geq 10 \land 30 \leq 130 \land true = false \land true \land true \land true = false

Mentor tip

Because all terms are combined via AND, the first false term ( $5 \geq 25$ ) short-circuits the entire expression to false.

(e)

6 marks

Initialize loop control variable or check speed condition.
While the vehicle is moving (e.g. $SPEED > 0$ ), read distance $D$ and current speed $S$ .
If $D < 20$ or $S > 130$ , output warning notification to the driver.
Loop terminates when speed drops to 0.

MOVING = true loop while MOVING input D input S if D < 20 or S > 130 then output ("WARNING: Proximity hazard or overspeed condition!") end if if S = 0 then MOVING = false end if end loop

Mentor tip

Use a clear loop construct governed by vehicle motion, sample both sensor inputs inside the loop, and verify the exit condition.

Question 14

15 marks

Data protection, migration integrity, and mobile resource allocation

System fundamentals
Computer organisation
Networks

(a)

2 marks

Cause 1: Human operational error, such as accidental record deletion or overwriting databases.
Cause 2: Mechanical storage hardware crash, or ransomware infection encrypting files.

Mentor tip

Provide two distinct causes: human error and hardware/malware failure.

(b)

2 marks

Operational paralysis and revenue loss: The hotel cannot identify arriving guests, resulting in double bookings, lost room revenue, damaged commercial reputation, and legal liabilities.

Mentor tip

Link data loss directly to hotel operations: check-in paralysis, double bookings, and financial loss.

(c)

2 marks

Automated off-site cloud backups: Implement scheduled incremental backups mirrored to redundant cloud storage or remote servers to enable rapid disaster recovery.

Mentor tip

Describe automated offsite or cloud backup with a disaster recovery procedure.

(d)

2 marks

Data format incompatibilities and schema mismatch: Legacy database field formats may conflict with the target schema, leading to truncated records, encoding errors, or lost relationships.

Mentor tip

Highlight schema incompatibility, character encoding mismatches, or data loss during ETL conversion.

(e)

3 marks

Primary storage: Desktops feature high-capacity expandable RAM modules (e.g. 16–64 GB), whereas tablets utilize smaller non-expandable soldered RAM (e.g. 4–8 GB) to conserve power.
Secondary storage: Desktops house high-capacity multi-terabyte SSDs or HDDs with modular expansion; tablets rely on compact soldered flash storage (e.g. 64–256 GB) with limited or cloud-reliant capacity.

Mentor tip

Compare both primary (RAM capacity/expandability) and secondary storage (terabyte SSDs vs compact soldered flash) across both form factors.

(f)(i)

1 mark

Central Processing Unit (CPU) architecture (desktop high-power x86 processor vs tablet thermal-efficient ARM processor).

Mentor tip

Identify processor (CPU/GPU) or display/battery hardware.

(f)(ii)

3 marks

Stationary vs mobile design: Desktops remain in fixed locations connected via Gigabit Ethernet cabling, securing high bandwidth and low latency.
Mobility: Tablets are carried by staff across guest areas, requiring wireless Wi-Fi and cellular transceivers to sustain continuous network handoffs while moving.

Mentor tip

Contrast wired Ethernet on stationary desktop terminals with wireless Wi-Fi/cellular connectivity for roaming tablets.

Question 15

15 marks

Dynamic data structures, linked list deletion, and binary tree nodes

Abstract data structures

(a)(i)

2 marks

Dynamic memory sizing: Expands and contracts at runtime based on actual data volume, eliminating memory waste and pre-allocation limits associated with static arrays.

Mentor tip

State dynamic memory allocation that adapts to data volume without fixed compile-time limits.

(a)(ii)

2 marks

Memory overhead and sequential access latency: Each node requires additional pointer memory, and elements cannot be accessed via direct indexing, requiring linear traversal.

Mentor tip

Mention extra pointer storage overhead and the lack of direct $O (1)$ random access indexing.

(b)

5 marks

Step 1: Set pointer `CURRENT` to `HEAD` of the linked list and maintain a `PREVIOUS` pointer.
Step 2: Traverse list by inspecting `CURRENT.data`. If not equal to 19, advance `PREVIOUS = CURRENT` and `CURRENT = CURRENT.next`.
Step 3: When the node containing 19 is reached, update the pointer of the preceding node: `PREVIOUS.next = CURRENT.next` (linking the node holding 11 directly to the node holding 31).
Step 4: Release and deallocate memory assigned to the node containing 19.

Node (11) . next = Node (19) . next = Node (31)

Mentor tip

Detail pointer tracking: maintain a previous pointer, re-route previous.next to current.next, and reclaim current node memory.

(c)

2 marks

A binary tree node consists of a data field holding the stored value, and two pointer references: a left pointer referencing the left subtree child, and a right pointer referencing the right subtree child.

Mentor tip

Define node structure: data payload plus two child pointers (left child, right child).

(d)(i)

3 marks

Insert sequence: 17, 11, 23, 5, 3, 19, 37, 31.
Root: 17
11: $11 < 17 ⟹$ left child of 17.
23: $23 > 17 ⟹$ right child of 17.
5: $5 < 17$ , $5 < 11 ⟹$ left child of 11.
3: $3 < 17, 3 < 11, 3 < 5 ⟹$ left child of 5.
19: $19 > 17, 19 < 23 ⟹$ left child of 23.
37: $37 > 17, 37 > 23 ⟹$ right child of 23.
31: $31 > 17, 31 > 23, 31 < 37 ⟹$ left child of 37.

Inorder: 3 \to 5 \to 11 \to 17 \to 19 \to 23 \to 31 \to 37

Mentor tip

Verify by tracing inorder traversal: numbers must emerge in strictly sorted ascending order.

(d)(ii)

1 mark

Leaf nodes possess zero child pointers:
Node 3 (no children)
Node 19 (no children)
Node 31 (no children)
Leaves: 3, 19, 31.

Leaf nodes = {3, 19, 31}

Mentor tip

Leaves have neither left nor right children: 3, 19, 31.

Question 16

15 marks

String manipulation primitives, trace execution, and password validation

Computational thinking, problem-solving and programming

(a)

2 marks

Error 1: The logic condition uses AND where it must use OR. A string length cannot simultaneously be $< 1$ and $> 100$ .
Error 2: The loop body outputs $X$ instead of prompting the user with `input X` to re-enter a string.
Corrected pseudocode replaces AND with OR and adds `input X` inside the loop.

input X loop while (lenString (X) < 1) OR (lenString (X) > 100) output "The string’s length is out of range, please try again" input X end loop

Mentor tip

Fix the impossible AND condition to OR, and replace the second output statement with input X.

(b)

5 marks

Iteration 1 ( $C = 1$ ): $X = "MARGORIE", Y = 0, Z = 6$ . $R = subString (\dots) = "MARGOR"$ . $S = revString (R) = "ROGRAM"$ . $J = joinString ("ROGRAM", "") = "ROGRAM"$ .
Iteration 2 ( $C = 2$ ): $X = "SPOT", Y = 1, Z = 3$ . $R = "POT"$ . $S = "TOP"$ . $J = joinString ("TOP", "ROGRAM") = "TOPROGRAM"$ .
Iteration 3 ( $C = 3$ ): $X = "EZEKIEL", Y = 2, Z = 3$ . $R = "EKI"$ . $S = "IKE"$ . $J = joinString ("IKE", "TOPROGRAM") = "IKETOPROGRAM"$ .
Iteration 4 ( $C = 4$ ): $X = "LISA", Y = 0, Z = 2$ . $R = "LI"$ . $S = "IL"$ . $J = joinString ("IL", "IKETOPROGRAM") = "ILIKETOPROGRAM"$ .
Loop terminates. Output displays: 'ILIKETOPROGRAM'.

J "" ROGRAM TOPROGRAM IKETOPROGRAM ILIKETOPROGRAM C 1234 - X MARGORIE SPOT EZEKIEL LISA - Y 0120 - Z 6332 - R MARGOR POT EKI LI - S ROGRAM TOP IKE IL -

Mentor tip

Notice that joinString(S, J) prepends S onto the existing J, assembling 'ILIKETOPROGRAM'.

(c)

8 marks

Step 1: Loop continuously until a valid confirmed password is entered.
Step 2: Prompt for password `PW`. Verify length $\geq 8$ using `lenString(PW)`. If $< 8$ , output warning and repeat.
Step 3: Scan adjacent characters using `subString(PW, POS, 1)` and `subString(PW, POS + 1, 1)` across indices $0$ to $lenString (P W) - 2$ . If identical adjacent characters are detected, output warning and repeat.
Step 4: Prompt for confirmation password `CONFIRM`. If `PW = CONFIRM`, output sign-off and exit loop; otherwise output mismatch message and repeat.

VALIDPW = false loop while not VALIDPW output ("Please enter a password that is at least 8 characters long with no consecutive repeated characters") input PW if lenString (PW) < 8 then output ("Password is too short, try again") else REPEATED = false loop POS from 0 to lenString (PW) - 2 if subString (PW, POS, 1) = subString (PW, POS + 1, 1) then REPEATED = true end if end loop if REPEATED then output ("Password has repeated characters, try again") else output ("Please re-enter your password") input PW2 if PW = PW2 then VALIDPW = true output ("Your password is confirmed") else output ("Passwords do not match, try again") end if end if end if end loop

Mentor tip

Traverse POS up to lenString(PW) - 2 to prevent out-of-bounds indexing when inspecting POS + 1.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Computational thinking, problem-solving and programming carries the most marks on this paper.

Computational thinking, problem-solving and programming23 marks · 23%
Q13, Q16
Networks18 marks · 18%
Q12, Q14
Abstract data structures17 marks · 17%
Q8, Q15
System fundamentals16 marks · 16%
Q1, Q2, Q4, Q5, Q14
Computer organisation16 marks · 16%
Q3, Q6, Q7, Q13, Q14
Control systems6 marks · 6%
Q10, Q11
Resource management4 marks · 4%
Q9

Mentor insights

Traps to watch for

01
Hexadecimal nibble alignment · Q7
Convert $3 7_{10} = 32 + 4 + 1 = 0010010 1_{2}$ . In hex, $001 0_{2} = 2_{16}$ and $010 1_{2} = 5_{16}$ , giving $2 5_{16}$ . Do not confuse the denary digits with hex.
02
Inverted vehicle input signals · Q13
According to the table, $A = 0$ represents distance $< 20 m$ and $C = 0$ represents speed $> 130 km/h$ . Forgetting to invert these yields the exact opposite logic.
03
Linked list deletion pointer splicing · Q15
When deleting node 19, you must splice `previous.next` to `current.next` (pointing from 11 directly to 31). Leaving `previous.next` unchanged orphans the rest of the list.
04
Adjacent character index bounds · Q16
When comparing character at `POS` with `POS + 1`, loop `POS` strictly up to `lenString(PW) - 2`. Looping to `lenString(PW) - 1` causes an out-of-bounds error on the last character.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Timezone

TZ1TZ2

Session: May 2024
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 16

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2024 TZ1 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Section A · Short answer

Questions 1–11 · 25 marks

Question 1

2 marks

Software-as-a-Service benefits and operational tradeoffs

System fundamentals

(a)

1 mark

Benefit: Rapid deployment on demand via predictable subscription pricing, eliminating upfront capital expenditure for dedicated server hardware and in-house maintenance.

Mentor tip

Name subscription elasticity, reduced local infrastructure, or ubiquitous web access.

(b)

1 mark

Drawback: Absolute dependency on continuous network connectivity; an internet outage halts productivity, and sensitive proprietary data is stored on third-party infrastructure.

Mentor tip

Highlight downtime vulnerability during connectivity loss or loss of direct data custody.

Question 2

2 marks

Role of User Sign-off Testing (UAT) in software deployment

System fundamentals

2 marks

User Sign-off Testing (UAT) (UAT) is the final testing phase conducted by intended end-users in real-world operating scenarios.
Its primary objective is verifying that the finished software satisfies business requirements, conforms to agreed specifications, and is viable for production sign-off.

Mentor tip

Emphasize validation by actual end-users against business requirements prior to final sign-off.

Question 3

1 mark

Hardware definition of peripheral devices

Computer organisation

1 mark

A peripheral is any auxiliary hardware device (such as a keyboard, mouse, monitor, or external drive) connected externally or via expansion slots to communicate with the CPU.

Mentor tip

Define peripherals as auxiliary input, output, or secondary storage devices attached to the core computer.

Question 4

2 marks

Mobile usability constraints and interface friction

System fundamentals

2 marks

Issue 1: Limited screen real estate leading to small tap targets, cramped typography, and touch navigation errors.
Issue 2: Sub-optimal contrast and glare under direct sunlight, or touch-screen keyboard typing inaccuracies due to lack of tactile feedback.

Mentor tip

State two concrete physical or software usability constraints (screen size, touch accuracy, glare).

Question 5

2 marks

Accessibility affordances for diverse user cohorts

System fundamentals

2 marks

Method 1: Operating system screen readers and text-to-speech narration paired with high-contrast display palettes for visually impaired users.
Method 2: Speech recognition software or adaptive hardware inputs (e.g. Braille displays and single-switch devices) for users with motor impairments.

Mentor tip

Identify two distinct accommodations: speech synthesis/screen readers and alternative input peripherals.

Question 6

2 marks

Function of the Memory Data Register in instruction cycles

Computer organisation

2 marks

The Memory Data Register (MDR) holds data words or machine instructions during transfers between the CPU and primary memory (RAM).
It temporarily buffers the contents fetched from the memory address specified by the MAR, or holds data staged to be written into RAM via the data bus.

Mentor tip

Pair the MDR directly with the data bus, acting as the two-way holding buffer for data to/from RAM.

Question 7

2 marks

Number representation: decimal to binary and hexadecimal conversions

Computer organisation

(a)

1 mark

Decompose $3 7_{10}$ into powers of 2:
$37 = 32 + 4 + 1 = 2^{5} + 2^{2} + 2^{0}$ .
In 8-bit binary representation: $0010010 1_{2}$ .

3 7_{10} = 0010010 1_{2}

Mentor tip

Sum binary place values: 32 + 4 + 1 = 37, giving 00100101.

(b)

1 mark

Partition the binary byte $0010010 1_{2}$ into 4-bit nibbles:
$001 0_{2} = 2_{16}$
$010 1_{2} = 5_{16}$
Concatenating gives $2 5_{16}$ .

0010010 1_{2} = 2 5_{16}

Mentor tip

Nibbles are 0010 (2) and 0101 (5), yielding 25 in hexadecimal.

Question 8

2 marks

Practical software applications of stack data structures

Abstract data structures

2 marks

Application 1: Call stack management in runtime environments, recording return addresses, parameters, and local activation frames during recursive function calls.
Application 2: Undo/redo command histories in productivity software, and syntax parsing / delimiter matching in language compilers.

Mentor tip

Name two classic LIFO applications: runtime call stack/recursion, undo history, or bracket balancing.

Question 9

4 marks

Operating system resource management mechanisms

Resource management

4 marks

Technique 1: CPU Process Scheduling. The OS scheduler governs process execution states (Ready, Running, Blocked), allocating CPU execution quanta using algorithms like Round Robin or Multi-Level Feedback Queues to ensure equitable processor utilisation.
Technique 2: Virtual Memory Management via Paging. Divides process memory into uniform pages and physical memory into frames, swapping inactive pages to secondary storage swap space to run programs that exceed physical RAM.

Mentor tip

State two concrete OS resource management duties: CPU process scheduling and virtual memory paging.

Question 10

2 marks

Sensor telemetry in domestic washing machine automation

Control systems

2 marks

Sensor type: Water level pressure sensor or float switch.
Purpose: Measures hydraulic pressure inside the tub to determine water fill levels, signaling the microprocessor to close water intake solenoid valves when full and prevent motor activation when empty.

Mentor tip

Pair a specific sensor (water level/pressure sensor) with its operational control role (closing inlet valves).

Question 11

4 marks

Role of closed-loop feedback in automated microprocessor control

Control systems

4 marks

Sensing and conversion: Sensors continuously sample the physical process variable (e.g. ambient temperature), converting analog readings via an ADC into digital data fed to the microprocessor.
Comparison and error detection: The microprocessor evaluates incoming sensor values against predetermined target setpoints stored in memory, calculating the error deviation.
Actuator adjustment: If a discrepancy exists, the processor outputs corrective control signals to actuators (e.g. energising heating coils or opening cooling vents).
Closed-loop stabilization: The physical change caused by the actuator alters the sensor's next reading, creating a continuous closed feedback loop that stabilizes the system at the desired setpoint.

Mentor tip

Explain the full closed loop: sensor input $\to$ ADC $\to$ setpoint comparison $\to$ error calculation $\to$ actuator adjustment.

Section B · Structured and algorithmic problems

Questions 12–16 · 75 marks

Question 12

15 marks

Transmission media, packet switching dynamics, and VPN social impacts

Networks

(a)(i)

2 marks

High bandwidth and low signal loss: Pulses of light travel through optical glass cores, supporting gigabit-to-terabit data transmission with minimal attenuation over long distances.
Electromagnetic immunity: Optical photons do not emit or conduct electrical current, making fibre lines immune to electromagnetic interference (EMI) and lightning surges.

Mentor tip

State optical photon transmission, high bandwidth over distance, and total EMI immunity.

(a)(ii)

2 marks

Physical mobility: Endpoints transmit data via radio frequency waveforms, enabling roaming without physical tethering.
Susceptibility to interference and attenuation: Signal quality degrades through solid physical obstacles (walls) and experiences RF collisions from ambient wireless devices.

Mentor tip

Contrast freedom of movement against signal degradation from walls and radio interference.

(b)

3 marks

Data encryption translates readable plaintext into unintelligible ciphertext using mathematical cipher algorithms and cryptographic keys.
During transit across unsecure channels, intercepted packets cannot be decoded or altered by unauthorised third parties without possessing the private decryption key.

Mentor tip

Explain plaintext-to-ciphertext translation via mathematical keys and the resulting confidentiality during transit.

(c)

5 marks

Packet decomposition: Messages are broken down into small, standardized units called packets, each formatted with a payload and a header containing sender IP, recipient IP, packet sequence number, and error-checking checksums.
Independent dynamic routing: Intermediate routers inspect the destination IP on each packet and forward it along the optimal path available at that instant; packets in the same session may travel along different physical routes.
Reassembly at destination: The destination host uses header sequence numbers to reassemble packets into their original message sequence, requesting retransmission for any lost or corrupted packets.

Mentor tip

Detail packet headers (IPs, sequence numbers), independent router path selection, and destination reassembly.

(d)

3 marks

Impact: Growth of remote telecommuting and flexible working practices.
Mechanism: VPNs enable secure access to enterprise internal intranets from any residential or remote location.
Social consequence: Enhances work-life balance and eliminates commuting travel times, but can blur the boundary between domestic life and work hours, increasing risks of professional isolation and burnout.

Mentor tip

Connect technical VPN remote access to societal shifts: flexibility/reduced commuting vs boundary erosion/isolation.

Question 13

15 marks

Boolean logic design, sensor thresholds, and vehicle alert algorithms

Computer organisation
Computational thinking, problem-solving and programming

(a)

1 mark

The NOR operator is a Boolean logic function that evaluates to true (1) if and only if all input operands are false (0); it is the logical negation of the OR function.

A 0011 B 0101 NOR (A, B) 1000

Mentor tip

Define NOR as returning 1 exclusively when all inputs are 0.

(b)

4 marks

Formulate conditions from specification:
Condition 1: Car travelling forward ( $B = 1$ ) and distance $< 20 m$ ( $A = 0$ , so $\neg A = 1$ ). Combined with AND: $\neg A \land B$ .
Condition 2: Speed $> 130 km/h$ ( $C = 0$ , so $\neg C = 1$ ).
Combine conditions with OR: $Z = (\neg A \land B) \lor \neg C$ .
Gate layout: Invert $A$ with a NOT gate; route $\neg A$ and $B$ into a 2-input AND gate; invert $C$ with a NOT gate; feed both into a final 2-input OR gate to produce $Z$ .

Z = (\neg A \land B) \lor \neg C

Mentor tip

Notice that $A = 0$ represents distance $< 20 m$ and $C = 0$ represents speed $> 130 km/h$ , requiring NOT gates on A and C.

(c)

2 marks

Assign binary states for input $D$ :
$D = 0$ : Obstruction is less than 3 metres from the rear of the car.
$D = 1$ : No obstruction detected within 3 metres from the rear of the car.

Mentor tip

Define both binary states clearly: 0 for obstacle within 3m, 1 for clear path (or vice versa).

(d)

2 marks

Substitute given values: $F = 5$ , $S = 30$ , $T = true$ .
Evaluate component terms:
$F \geq 25 ⟹ 5 \geq 25$ is false.
$S \geq 10 ⟹ 30 \geq 10$ is true.
$S \leq 130 ⟹ 30 \leq 130$ is true.
$T = true ⟹ true$ .
Combine with AND: $false \land true \land true \land true = false$ .

5 \geq 25 \land 30 \geq 10 \land 30 \leq 130 \land true = false \land true \land true \land true = false

Mentor tip

Because all terms are combined via AND, the first false term ( $5 \geq 25$ ) short-circuits the entire expression to false.

(e)

6 marks

Initialize loop control variable or check speed condition.
While the vehicle is moving (e.g. $SPEED > 0$ ), read distance $D$ and current speed $S$ .
If $D < 20$ or $S > 130$ , output warning notification to the driver.
Loop terminates when speed drops to 0.

MOVING = true loop while MOVING input D input S if D < 20 or S > 130 then output ("WARNING: Proximity hazard or overspeed condition!") end if if S = 0 then MOVING = false end if end loop

Mentor tip

Use a clear loop construct governed by vehicle motion, sample both sensor inputs inside the loop, and verify the exit condition.

Question 14

15 marks

Data protection, migration integrity, and mobile resource allocation

System fundamentals
Computer organisation
Networks

(a)

2 marks

Cause 1: Human operational error, such as accidental record deletion or overwriting databases.
Cause 2: Mechanical storage hardware crash, or ransomware infection encrypting files.

Mentor tip

Provide two distinct causes: human error and hardware/malware failure.

(b)

2 marks

Operational paralysis and revenue loss: The hotel cannot identify arriving guests, resulting in double bookings, lost room revenue, damaged commercial reputation, and legal liabilities.

Mentor tip

Link data loss directly to hotel operations: check-in paralysis, double bookings, and financial loss.

(c)

2 marks

Automated off-site cloud backups: Implement scheduled incremental backups mirrored to redundant cloud storage or remote servers to enable rapid disaster recovery.

Mentor tip

Describe automated offsite or cloud backup with a disaster recovery procedure.

(d)

2 marks

Data format incompatibilities and schema mismatch: Legacy database field formats may conflict with the target schema, leading to truncated records, encoding errors, or lost relationships.

Mentor tip

Highlight schema incompatibility, character encoding mismatches, or data loss during ETL conversion.

(e)

3 marks

Primary storage: Desktops feature high-capacity expandable RAM modules (e.g. 16–64 GB), whereas tablets utilize smaller non-expandable soldered RAM (e.g. 4–8 GB) to conserve power.
Secondary storage: Desktops house high-capacity multi-terabyte SSDs or HDDs with modular expansion; tablets rely on compact soldered flash storage (e.g. 64–256 GB) with limited or cloud-reliant capacity.

Mentor tip

Compare both primary (RAM capacity/expandability) and secondary storage (terabyte SSDs vs compact soldered flash) across both form factors.

(f)(i)

1 mark

Central Processing Unit (CPU) architecture (desktop high-power x86 processor vs tablet thermal-efficient ARM processor).

Mentor tip

Identify processor (CPU/GPU) or display/battery hardware.

(f)(ii)

3 marks

Stationary vs mobile design: Desktops remain in fixed locations connected via Gigabit Ethernet cabling, securing high bandwidth and low latency.
Mobility: Tablets are carried by staff across guest areas, requiring wireless Wi-Fi and cellular transceivers to sustain continuous network handoffs while moving.

Mentor tip

Contrast wired Ethernet on stationary desktop terminals with wireless Wi-Fi/cellular connectivity for roaming tablets.

Question 15

15 marks

Dynamic data structures, linked list deletion, and binary tree nodes

Abstract data structures

(a)(i)

2 marks

Dynamic memory sizing: Expands and contracts at runtime based on actual data volume, eliminating memory waste and pre-allocation limits associated with static arrays.

Mentor tip

State dynamic memory allocation that adapts to data volume without fixed compile-time limits.

(a)(ii)

2 marks

Memory overhead and sequential access latency: Each node requires additional pointer memory, and elements cannot be accessed via direct indexing, requiring linear traversal.

Mentor tip

Mention extra pointer storage overhead and the lack of direct $O (1)$ random access indexing.

(b)

5 marks

Step 1: Set pointer `CURRENT` to `HEAD` of the linked list and maintain a `PREVIOUS` pointer.
Step 2: Traverse list by inspecting `CURRENT.data`. If not equal to 19, advance `PREVIOUS = CURRENT` and `CURRENT = CURRENT.next`.
Step 3: When the node containing 19 is reached, update the pointer of the preceding node: `PREVIOUS.next = CURRENT.next` (linking the node holding 11 directly to the node holding 31).
Step 4: Release and deallocate memory assigned to the node containing 19.

Node (11) . next = Node (19) . next = Node (31)

Mentor tip

Detail pointer tracking: maintain a previous pointer, re-route previous.next to current.next, and reclaim current node memory.

(c)

2 marks

Mentor tip

Define node structure: data payload plus two child pointers (left child, right child).

(d)(i)

3 marks

Insert sequence: 17, 11, 23, 5, 3, 19, 37, 31.
Root: 17
11: $11 < 17 ⟹$ left child of 17.
23: $23 > 17 ⟹$ right child of 17.
5: $5 < 17$ , $5 < 11 ⟹$ left child of 11.
3: $3 < 17, 3 < 11, 3 < 5 ⟹$ left child of 5.
19: $19 > 17, 19 < 23 ⟹$ left child of 23.
37: $37 > 17, 37 > 23 ⟹$ right child of 23.
31: $31 > 17, 31 > 23, 31 < 37 ⟹$ left child of 37.

Inorder: 3 \to 5 \to 11 \to 17 \to 19 \to 23 \to 31 \to 37

Mentor tip

Verify by tracing inorder traversal: numbers must emerge in strictly sorted ascending order.

(d)(ii)

1 mark

Leaf nodes possess zero child pointers:
Node 3 (no children)
Node 19 (no children)
Node 31 (no children)
Leaves: 3, 19, 31.

Leaf nodes = {3, 19, 31}

Mentor tip

Leaves have neither left nor right children: 3, 19, 31.

Question 16

15 marks

String manipulation primitives, trace execution, and password validation

Computational thinking, problem-solving and programming

(a)

2 marks

Error 1: The logic condition uses AND where it must use OR. A string length cannot simultaneously be $< 1$ and $> 100$ .
Error 2: The loop body outputs $X$ instead of prompting the user with `input X` to re-enter a string.
Corrected pseudocode replaces AND with OR and adds `input X` inside the loop.

input X loop while (lenString (X) < 1) OR (lenString (X) > 100) output "The string’s length is out of range, please try again" input X end loop

Mentor tip

Fix the impossible AND condition to OR, and replace the second output statement with input X.

(b)

5 marks

Iteration 1 ( $C = 1$ ): $X = "MARGORIE", Y = 0, Z = 6$ . $R = subString (\dots) = "MARGOR"$ . $S = revString (R) = "ROGRAM"$ . $J = joinString ("ROGRAM", "") = "ROGRAM"$ .
Iteration 2 ( $C = 2$ ): $X = "SPOT", Y = 1, Z = 3$ . $R = "POT"$ . $S = "TOP"$ . $J = joinString ("TOP", "ROGRAM") = "TOPROGRAM"$ .
Iteration 3 ( $C = 3$ ): $X = "EZEKIEL", Y = 2, Z = 3$ . $R = "EKI"$ . $S = "IKE"$ . $J = joinString ("IKE", "TOPROGRAM") = "IKETOPROGRAM"$ .
Iteration 4 ( $C = 4$ ): $X = "LISA", Y = 0, Z = 2$ . $R = "LI"$ . $S = "IL"$ . $J = joinString ("IL", "IKETOPROGRAM") = "ILIKETOPROGRAM"$ .
Loop terminates. Output displays: 'ILIKETOPROGRAM'.

J "" ROGRAM TOPROGRAM IKETOPROGRAM ILIKETOPROGRAM C 1234 - X MARGORIE SPOT EZEKIEL LISA - Y 0120 - Z 6332 - R MARGOR POT EKI LI - S ROGRAM TOP IKE IL -

Mentor tip

Notice that joinString(S, J) prepends S onto the existing J, assembling 'ILIKETOPROGRAM'.

(c)

8 marks

Step 1: Loop continuously until a valid confirmed password is entered.
Step 2: Prompt for password `PW`. Verify length $\geq 8$ using `lenString(PW)`. If $< 8$ , output warning and repeat.
Step 3: Scan adjacent characters using `subString(PW, POS, 1)` and `subString(PW, POS + 1, 1)` across indices $0$ to $lenString (P W) - 2$ . If identical adjacent characters are detected, output warning and repeat.
Step 4: Prompt for confirmation password `CONFIRM`. If `PW = CONFIRM`, output sign-off and exit loop; otherwise output mismatch message and repeat.

VALIDPW = false loop while not VALIDPW output ("Please enter a password that is at least 8 characters long with no consecutive repeated characters") input PW if lenString (PW) < 8 then output ("Password is too short, try again") else REPEATED = false loop POS from 0 to lenString (PW) - 2 if subString (PW, POS, 1) = subString (PW, POS + 1, 1) then REPEATED = true end if end loop if REPEATED then output ("Password has repeated characters, try again") else output ("Please re-enter your password") input PW2 if PW = PW2 then VALIDPW = true output ("Your password is confirmed") else output ("Passwords do not match, try again") end if end if end if end loop

Mentor tip

Traverse POS up to lenString(PW) - 2 to prevent out-of-bounds indexing when inspecting POS + 1.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Computational thinking, problem-solving and programming carries the most marks on this paper.

Computational thinking, problem-solving and programming23 marks · 23%
Q13, Q16
Networks18 marks · 18%
Q12, Q14
Abstract data structures17 marks · 17%
Q8, Q15
System fundamentals16 marks · 16%
Q1, Q2, Q4, Q5, Q14
Computer organisation16 marks · 16%
Q3, Q6, Q7, Q13, Q14
Control systems6 marks · 6%
Q10, Q11
Resource management4 marks · 4%
Q9

Mentor insights

Traps to watch for

01
Hexadecimal nibble alignment · Q7
Convert $3 7_{10} = 32 + 4 + 1 = 0010010 1_{2}$ . In hex, $001 0_{2} = 2_{16}$ and $010 1_{2} = 5_{16}$ , giving $2 5_{16}$ . Do not confuse the denary digits with hex.
02
Inverted vehicle input signals · Q13
According to the table, $A = 0$ represents distance $< 20 m$ and $C = 0$ represents speed $> 130 km/h$ . Forgetting to invert these yields the exact opposite logic.
03
Linked list deletion pointer splicing · Q15
When deleting node 19, you must splice `previous.next` to `current.next` (pointing from 11 directly to 31). Leaving `previous.next` unchanged orphans the rest of the list.
04
Adjacent character index bounds · Q16
When comparing character at `POS` with `POS + 1`, loop `POS` strictly up to `lenString(PW) - 2`. Looping to `lenString(PW) - 1` causes an out-of-bounds error on the last character.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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