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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: November 2022
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 14

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2022 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–9 · 25 marks

Question 1

2 marks

Need for translation from high-level to low-level machine code

Computer organisation

2 marks

CPUs execute operations solely via binary machine code instructions (opcodes and operands) implemented in digital circuitry.
High-level programming languages use abstract syntax, structured loops, and human-readable identifiers that hardware cannot directly execute without translation by a compiler or interpreter.

Mentor tip

State that processors exclusively execute binary instructions, requiring high-level code translation.

Question 2

4 marks

Prototyping software development methodology evaluation

System fundamentals

(a)

2 marks

Early stakeholder feedback: A functional prototype gives clients an interactive representation of the proposed user interface, uncovering ambiguous or missing requirements early in the lifecycle.

Mentor tip

Highlight user feedback and early requirement clarification.

(b)

2 marks

Safety-critical or deeply algorithmic systems: In software where formal mathematical correctness, absolute real-time timing, or faultless reliability is paramount (e.g. avionics fly-by-wire or medical pacemaker firmware), quick surface-level prototyping can encourage superficial interface design while neglecting rigorous formal verification.

Mentor tip

Mention safety-critical systems or scenarios where requirements are already strictly fixed.

Question 3

2 marks

Formal definition of the XOR Boolean logic operator

Computer organisation

2 marks

The XOR (exclusive OR) Boolean operator evaluates to true (1) if and only if exactly one of its two input operands is true (1).
If both input operands have identical truth values (both 0 or both 1), XOR evaluates to false (0).

A \oplus B = (A \land \neg B) \lor (\neg A \land B)

Mentor tip

Emphasize that XOR yields true when inputs differ, and false when inputs are identical.

Question 4

4 marks

Ethical and technical implications of unauthorized online media publication

System fundamentals
Networks

(a)

2 marks

Invasion of personal privacy and lack of consent: Publishing photographs and videos of individuals without their explicit permission violates their personal autonomy and can cause social embarrassment or reputational harm.

Mentor tip

Focus on lack of informed consent and potential personal or professional harm.

(b)

2 marks

Incompatible media codecs or excessive file sizes: The viewer's client browser may lack the necessary video decompression codec, or high video bitrates may fail to load over low-bandwidth or unstable network connections.

Mentor tip

Mention codec incompatibility, network bandwidth constraints, or browser format restrictions.

Question 5

2 marks

24-bit Truecolor RGB pixel representation and colour gamut

Computer organisation

(a)

1 mark

Pure blue has hex representation 0000FF:
- Red byte 00 = 00000000
- Green byte 00 = 00000000
- Blue byte FF = 11111111
Binary value: 000000000000000011111111 (or 00000000 00000000 11111111).

0000 FF_{16} = 00000000000000001111111 1_{2}

Mentor tip

Expand each of the three color channels into 8 binary bits.

(b)

1 mark

With 24 bits allocated per pixel, the total number of distinct addressable colors is 2^24 = 16,777,216.

2^{24} = 16777216 colors

Mentor tip

Calculate 2 to the power of 24 (or 256 * 256 * 256).

Question 6

1 mark

Formal definition of a variable in computer programming

Computational thinking, problem-solving and programming

1 mark

A named storage location in primary memory (RAM) designated by an identifier, holding a data value that can be read or modified during program execution.

Mentor tip

Highlight both the named memory location and the capacity for its contents to change during execution.

Question 7

3 marks

Performance benefits of expanding CPU cache capacity

Computer organisation

3 marks

Higher cache hit ratio: Expanding cache capacity allows larger instruction blocks and working data sets to reside on-chip simultaneously.
Reduced bus traffic to main RAM: More requests are resolved directly within the high-speed cache, minimizing memory bus contention.
Elimination of CPU wait states: Accessing cache takes fewer clock cycles than querying slower dynamic RAM, enabling the execution unit to run continuously without pipeline stalls.

Mentor tip

Explain how increased capacity raises the hit rate and cuts memory access latency.

Question 8

3 marks

Queue data structure definition and multitasking scenario queues

Abstract data structures

(a)

1 mark

A linear, dynamic data structure governed by the First-In, First-Out (FIFO) principle, where insertions occur at the rear/tail and deletions occur at the front/head.

Mentor tip

State the FIFO property clearly.

(b)

2 marks

1) Printer spooler queue: Stores print jobs from user applications in arrival order waiting to be transferred to the physical printer.
2) Operating system CPU ready queue: Manages active processes and execution threads waiting for their scheduled time slice on the processor.

Mentor tip

Identify two distinct queues in the scenario: print spooler queue and OS process ready queue.

Question 9

4 marks

Recursive array traversal and stack unwinding trace

Computational thinking, problem-solving and programming

4 marks

Array NAMES: [0]: 'Zixan', [1]: 'Murali', [2]: 'Eli', [3]: 'Kim'.
Algorithm:
mystery(A, N):
if N > 0 then mystery(A, N - 1)
output(A[3 - N])
Execution trace of mystery(NAMES, 3):
- Level 1: mystery(NAMES, 3) calls mystery(NAMES, 2).
- Level 2: mystery(NAMES, 2) calls mystery(NAMES, 1).
- Level 3: mystery(NAMES, 1) calls mystery(NAMES, 0).
- Level 4: mystery(NAMES, 0) tests N > 0 (0 > 0 is FALSE). No further recursive call.
Outputs A[3 - 0] = A[3] = 'Kim'.
- Unwind to Level 3 (N=1):
Outputs A[3 - 1] = A[2] = 'Eli'.
- Unwind to Level 2 (N=2):
Outputs A[3 - 2] = A[1] = 'Murali'.
- Unwind to Level 1 (N=3):
Outputs A[3 - 3] = A[0] = 'Zixan'.
Complete printed output in exact order:
Kim
Eli
Murali
Zixan

Level 4 (N=0) Level 3 (N=1) Level 2 (N=2) Level 1 (N=3) ⟹ A [3] = Kim ⟹ A [2] = Eli ⟹ A [1] = Murali ⟹ A [0] = Zixan

Mentor tip

Because output occurs after the recursive call, the stack unwinds in reverse call order, printing elements from index 3 down to 0.

Questions 10–14 · 75 marks

Question 10

15 marks

Wired vs Wireless Local Area Networks and VPN security

Networks

(a)

1 mark

Wireless Access Point (WAP) / wireless network interface controller (WNIC).

Mentor tip

A Wireless Access Point is the required hardware bridge connecting wireless devices to the wired backbone.

(b)

4 marks

Wired network transmission reliability:
- Data travels through shielded physical cables (twisted pair or fiber optic) isolated from atmospheric noise, delivering consistent bandwidth, predictable latency, and minimal packet loss.
Wireless network transmission reliability:
- Radio frequency signals attenuate rapidly over distance and are vulnerable to physical obstructions (walls, metal furniture) and electromagnetic interference (microwaves, adjacent Wi-Fi channels), causing variable transmission speeds and packet retransmissions.

Mentor tip

Contrast cable shielding and physical consistency against radio attenuation and environmental interference.

(c)

2 marks

Broadcast medium exposure: Radio waves propagate omnidirectionally through physical walls into public spaces outside the office perimeter.
An eavesdropper within physical range can intercept data packets using a packet sniffer without needing physical access to the building's wiring.

Mentor tip

Emphasize that radio signals cross physical walls and can be captured remotely without physical connection.

(d)

4 marks

Data encryption: All payload data passing through the public network is encrypted using robust cryptographic ciphers (e.g. AES-256), rendering intercepted packets unreadable without the secret decryption key.
Tunneling protocol with IP masking: The original IP packets are encapsulated inside new network packet headers, masking the user's real IP address and origin location from intermediate observers.

Mentor tip

Explain two separate features such as cryptographic payload encryption and packet encapsulation/tunneling.

(e)

4 marks

1) Packetization: The sender segments the original data stream into standardized packets, affixing header metadata (source IP, destination IP, packet sequence number, checksum).
2) Dynamic routing: Each packet travels independently across intermediate routers, which select optimal paths dynamically based on current network congestion.
3) Out-of-order arrival handling: Packets may arrive at the destination via different paths and in scrambled chronological order.
4) Reassembly and verification: The destination host uses packet sequence numbers to reassemble the data in correct sequence and checks error codes, requesting retransmission of any corrupted or dropped packets.

Mentor tip

Detail the full progression: division into packets with headers -> independent dynamic routing -> reassembly by sequence numbers.

Question 11

15 marks

International data migration, change management, and legacy systems

System fundamentals

(a)(i)

1 mark

Data migration is the process of transferring digital data from one computer storage system, format, or database architecture to another.

Mentor tip

State the transfer of data between storage systems, schemas, or environments.

(a)(ii)

4 marks

1) Data corruption and loss: Transmission interruptions, truncated records, or character encoding mismatches during mass migration can compromise data integrity.
2) Schema and format incompatibility: Divergent database structures, missing relational fields, or inconsistent primary keys between legacy and target systems require complex transformation logic, risking record misinterpretation.

Mentor tip

Detail two distinct risks such as data corruption/loss and schema incompatibility.

(b)

4 marks

User training and re-skilling: Delivering comprehensive instruction programs to ensure employees acquire the necessary competence and confidence to navigate the new system effectively.
Stakeholder communication and organizational alignment: Maintaining transparent dialogue regarding project milestones, revised business workflows, and temporary transition schedules to minimize employee disruption and resistance.

Mentor tip

Cover user education and proactive organizational communication.

(c)

3 marks

Scarcity of maintenance expertise: Older legacy programming languages and architectures (e.g. COBOL or proprietary mainframe databases) have dwindling pools of skilled developers, making bug fixes and routine maintenance exceptionally expensive.
Additionally, legacy vendors frequently discontinue security patches, exposing the organization to critical vulnerabilities and severe integration hurdles with modern cloud infrastructure.

Mentor tip

Highlight lack of vendor support, scarce specialized developer skills, and rising maintenance expenses.

(d)

3 marks

Redundant workforce labor: Operating two systems simultaneously requires personnel to duplicate all operational transactions in both platforms, effectively doubling staffing hours.
Infrastructure maintenance overhead: The organization must simultaneously fund operating licenses, dual server hosting environments, electricity, and continuous synchronization maintenance for both systems.

Mentor tip

Explain the dual expenditure of staffing duplication and concurrent infrastructure operational costs.

Question 12

15 marks

Smart building automated control systems and OS role

Control systems
Computer organisation

(a)

4 marks

1) Climate and HVAC regulation: Monitoring indoor temperature, humidity, and CO2 levels to dynamically adjust heating, cooling, and air ventilation dampers.
2) Intelligent lighting control: Using occupancy infrared sensors and daylight harvesting to activate, dim, or extinguish luminaires based on human presence and natural sunlight.

Mentor tip

Describe two separate operations such as HVAC/environmental management and automated lighting or security alarm dispatch.

(b)

4 marks

Sensors: Proximity/RFID sensors or optical cameras detect an approaching person or credential badge and send analog/digital signals to the processor.
Processing: The embedded processor compares input signals against authorization access lists or predefined motion criteria to decide whether the door should cycle open.
Output transducers: When authorized, the processor sends control voltage signals to electromechanical actuators (e.g. electric motor drives or magnetic solenoid locks) to physically retract latches and swing doors open.
Timing and closure: Timer sensors monitor door clearance, after which the processor instructs the actuator to gently shut and re-engage magnetic locks.

Mentor tip

Clearly explain the role of sensors (data collection), processor (decision logic), and transducers (physical actuation).

(c)(i)

2 marks

Hardware I/O device management: Providing drivers and interfaces to communicate with diverse sensor and actuator buses.
Process scheduling and multitasking: Prioritizing real-time emergency events (e.g. fire alarms) over routine telemetry logging.

Mentor tip

State two standard OS functions tailored to control systems, like device management and task prioritization.

(c)(ii)

2 marks

Hardware Interrupt: The peripheral hardware device asserts an electrical interrupt line to the processor when an event occurs, causing the OS to suspend current tasks and execute a dedicated Interrupt Service Routine (ISR) immediately.

Mentor tip

Interrupts or periodic status polling are both valid; explain the mechanism concisely.

(d)

3 marks

Energy conservation and reduced utility expenses: Automated HVAC and lighting cutoffs eliminate energy waste in unoccupied office zones.
Enhanced physical security and audit tracking: Centralized access logs track employee movements and automatically lock sensitive zones after hours, safeguarding corporate assets.
Predictive maintenance: Continuous telemetry allows facility teams to detect failing motors or filters before catastrophic equipment breakdowns.

Mentor tip

Emphasize financial savings through energy efficiency, heightened security, and automated maintenance.

Question 13

15 marks

Parallel array sorting and binary search tree construction

Computational thinking, problem-solving and programming
Abstract data structures

(a)(i)

5 marks

Bubble sort descending pass sequence:
1) Outer loop executes N - 1 passes across the array length.
2) Inner loop compares adjacent score elements SCORES[J] and SCORES[J + 1].
3) For descending order, if SCORES[J] < SCORES[J + 1], swap the two scores using a temporary variable.
4) Crucially, whenever SCORES[J] and SCORES[J + 1] are swapped, also swap NAMES[J] and NAMES[J + 1] to preserve data correspondence.
5) Continue comparisons until the end of the pass; repeat until no swaps occur or all passes complete.

Mentor tip

Emphasize that parallel arrays require swapping both SCORES and NAMES simultaneously whenever a score inversion occurs.

(a)(ii)

1 mark

Selection sort (or Insertion sort / Quicksort / Merge sort).

Mentor tip

Name any recognized alternative sorting algorithm.

(b)

3 marks

Data payload / value field: Stores the student name and associated numerical test score.
Left child pointer: A reference holding the memory address of the left sub-tree node.
Right child pointer: A reference holding the memory address of the right sub-tree node.

Mentor tip

List the three essential components: data value/fields, left pointer, and right pointer.

(c)

6 marks

Insertion algorithm into binary search tree for descending inorder output (Left -> Root -> Right):
To ensure an inorder traversal prints scores in descending order (highest to lowest), the tree must be structured such that higher scores branch to the left and lower scores branch to the right (or standard traversal inverted):
1) If the tree is empty, instantiate the new node with the student data and designate it as the root.
2) If the tree is not empty, start comparison at the root node.
3) Compare the incoming student's score with the current node's score.
4) If incoming score > current node score, traverse to the left child pointer.
- If the left pointer is null, attach the new node here.
- Otherwise, update current node to left child and repeat comparison.
5) If incoming score <= current node score, traverse to the right child pointer.
- If the right pointer is null, attach the new node here.
- Otherwise, update current node to right child and repeat comparison.

Mentor tip

Carefully define how higher scores navigate left and lower scores navigate right to yield descending order during standard inorder traversal.

Question 14

15 marks

Number classification algorithms: Proper divisors, abundant, deficient, and perfect

Computational thinking, problem-solving and programming

(a)

2 marks

X is a proper divisor of Y if X divides Y without remainder and X is strictly less than Y.
Logical expression: (Y mod X = 0) and (X < Y).

(Y mod X = 0) \land (X < Y)

Mentor tip

Include both conditions: zero remainder and X strictly less than Y.

(b)

3 marks

Initialize an accumulator variable SUM to 0.
Iterate a loop variable D from 1 up to N - 1 (or N div 2).
If N mod D = 0, add D to SUM.
After the loop finishes, return SUM.

subprogram sumPD (N) SUM = 0 loop D from 1 to N - 1 if N mod D = 0 then SUM = SUM + D end if end loop return SUM end subprogram

Mentor tip

Iterate strictly up to N - 1 so that N itself is excluded from the proper divisors sum.

(c)

10 marks

1) Initialize index tracking pointers: A_IDX = 0, D_IDX = 0, P_IDX = 0.
2) Traverse the 2D array MAT using nested loops: outer row R from 0 to 3, inner column C from 0 to 5.
3) For each element NUM = MAT[R][C]:
- Compute proper divisor sum: S = sumPD(NUM).
- Classify according to definitions:
* If S = NUM: add NUM to PERFECT[P_IDX], increment P_IDX.
* Else if S > NUM: add NUM to ABUNDANT[A_IDX], increment A_IDX.
* Else (S < NUM): add NUM to DEFICIENT[D_IDX], increment D_IDX.
4) Close all conditionals and loops.

A_IDX = 0, D_IDX = 0, P_IDX = 0 loop R from 0 to 3 loop C from 0 to 5 NUM = MAT [R] [C] S = sumPD (NUM) if S = NUM then PERFECT [P_IDX] = NUM P_IDX = P_IDX + 1 else if S > NUM then ABUNDANT [A_IDX] = NUM A_IDX = A_IDX + 1 else DEFICIENT [D_IDX] = NUM D_IDX = D_IDX + 1 end if end loop end loop

Mentor tip

Maintain three separate destination index counters and invoke sumPD() for each element.

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 programming26 marks · 26%
Q6, Q9, Q13, Q14
System fundamentals21 marks · 21%
Q2, Q4, Q11
Networks17 marks · 17%
Q4, Q10
Computer organisation13 marks · 13%
Q1, Q3, Q5, Q7, Q12
Abstract data structures12 marks · 12%
Q8, Q13
Control systems11 marks · 11%
Q12

Mentor insights

Traps to watch for

01
Stack unwinding in post-recursion printing · Q9
Because output occurs after the recursive call mystery(A, N-1), the function calls delve all the way to N=0 before any output occurs. Printing begins on stack unwinding with N=0 (outputting A[3]), printing the array in reverse order.
02
Parallel array synchronization during sorting · Q13
When swapping elements in the SCORES array, you must execute the exact same swap on the NAMES array at identical indices, otherwise names become attached to incorrect student scores.
03
Proper divisor definition exclusion of N · Q14
Proper divisors of N explicitly exclude N itself. The divisor loop must stop at N - 1 (or N div 2); including N would cause every number to be misclassified as abundant.

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

Session: November 2022
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 14

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2022 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–9 · 25 marks

Question 1

2 marks

Need for translation from high-level to low-level machine code

Computer organisation

2 marks

CPUs execute operations solely via binary machine code instructions (opcodes and operands) implemented in digital circuitry.
High-level programming languages use abstract syntax, structured loops, and human-readable identifiers that hardware cannot directly execute without translation by a compiler or interpreter.

Mentor tip

State that processors exclusively execute binary instructions, requiring high-level code translation.

Question 2

4 marks

Prototyping software development methodology evaluation

System fundamentals

(a)

2 marks

Early stakeholder feedback: A functional prototype gives clients an interactive representation of the proposed user interface, uncovering ambiguous or missing requirements early in the lifecycle.

Mentor tip

Highlight user feedback and early requirement clarification.

(b)

2 marks

Mentor tip

Mention safety-critical systems or scenarios where requirements are already strictly fixed.

Question 3

2 marks

Formal definition of the XOR Boolean logic operator

Computer organisation

2 marks

The XOR (exclusive OR) Boolean operator evaluates to true (1) if and only if exactly one of its two input operands is true (1).
If both input operands have identical truth values (both 0 or both 1), XOR evaluates to false (0).

A \oplus B = (A \land \neg B) \lor (\neg A \land B)

Mentor tip

Emphasize that XOR yields true when inputs differ, and false when inputs are identical.

Question 4

4 marks

Ethical and technical implications of unauthorized online media publication

System fundamentals
Networks

(a)

2 marks

Mentor tip

Focus on lack of informed consent and potential personal or professional harm.

(b)

2 marks

Mentor tip

Mention codec incompatibility, network bandwidth constraints, or browser format restrictions.

Question 5

2 marks

24-bit Truecolor RGB pixel representation and colour gamut

Computer organisation

(a)

1 mark

Pure blue has hex representation 0000FF:
- Red byte 00 = 00000000
- Green byte 00 = 00000000
- Blue byte FF = 11111111
Binary value: 000000000000000011111111 (or 00000000 00000000 11111111).

0000 FF_{16} = 00000000000000001111111 1_{2}

Mentor tip

Expand each of the three color channels into 8 binary bits.

(b)

1 mark

With 24 bits allocated per pixel, the total number of distinct addressable colors is 2^24 = 16,777,216.

2^{24} = 16777216 colors

Mentor tip

Calculate 2 to the power of 24 (or 256 * 256 * 256).

Question 6

1 mark

Formal definition of a variable in computer programming

Computational thinking, problem-solving and programming

1 mark

A named storage location in primary memory (RAM) designated by an identifier, holding a data value that can be read or modified during program execution.

Mentor tip

Highlight both the named memory location and the capacity for its contents to change during execution.

Question 7

3 marks

Performance benefits of expanding CPU cache capacity

Computer organisation

3 marks

Higher cache hit ratio: Expanding cache capacity allows larger instruction blocks and working data sets to reside on-chip simultaneously.
Reduced bus traffic to main RAM: More requests are resolved directly within the high-speed cache, minimizing memory bus contention.
Elimination of CPU wait states: Accessing cache takes fewer clock cycles than querying slower dynamic RAM, enabling the execution unit to run continuously without pipeline stalls.

Mentor tip

Explain how increased capacity raises the hit rate and cuts memory access latency.

Question 8

3 marks

Queue data structure definition and multitasking scenario queues

Abstract data structures

(a)

1 mark

A linear, dynamic data structure governed by the First-In, First-Out (FIFO) principle, where insertions occur at the rear/tail and deletions occur at the front/head.

Mentor tip

State the FIFO property clearly.

(b)

2 marks

1) Printer spooler queue: Stores print jobs from user applications in arrival order waiting to be transferred to the physical printer.
2) Operating system CPU ready queue: Manages active processes and execution threads waiting for their scheduled time slice on the processor.

Mentor tip

Identify two distinct queues in the scenario: print spooler queue and OS process ready queue.

Question 9

4 marks

Recursive array traversal and stack unwinding trace

Computational thinking, problem-solving and programming

4 marks

Array NAMES: [0]: 'Zixan', [1]: 'Murali', [2]: 'Eli', [3]: 'Kim'.
Algorithm:
mystery(A, N):
if N > 0 then mystery(A, N - 1)
output(A[3 - N])
Execution trace of mystery(NAMES, 3):
- Level 1: mystery(NAMES, 3) calls mystery(NAMES, 2).
- Level 2: mystery(NAMES, 2) calls mystery(NAMES, 1).
- Level 3: mystery(NAMES, 1) calls mystery(NAMES, 0).
- Level 4: mystery(NAMES, 0) tests N > 0 (0 > 0 is FALSE). No further recursive call.
Outputs A[3 - 0] = A[3] = 'Kim'.
- Unwind to Level 3 (N=1):
Outputs A[3 - 1] = A[2] = 'Eli'.
- Unwind to Level 2 (N=2):
Outputs A[3 - 2] = A[1] = 'Murali'.
- Unwind to Level 1 (N=3):
Outputs A[3 - 3] = A[0] = 'Zixan'.
Complete printed output in exact order:
Kim
Eli
Murali
Zixan

Level 4 (N=0) Level 3 (N=1) Level 2 (N=2) Level 1 (N=3) ⟹ A [3] = Kim ⟹ A [2] = Eli ⟹ A [1] = Murali ⟹ A [0] = Zixan

Mentor tip

Because output occurs after the recursive call, the stack unwinds in reverse call order, printing elements from index 3 down to 0.

Questions 10–14 · 75 marks

Question 10

15 marks

Wired vs Wireless Local Area Networks and VPN security

Networks

(a)

1 mark

Wireless Access Point (WAP) / wireless network interface controller (WNIC).

Mentor tip

A Wireless Access Point is the required hardware bridge connecting wireless devices to the wired backbone.

(b)

4 marks

Wired network transmission reliability:
- Data travels through shielded physical cables (twisted pair or fiber optic) isolated from atmospheric noise, delivering consistent bandwidth, predictable latency, and minimal packet loss.
Wireless network transmission reliability:
- Radio frequency signals attenuate rapidly over distance and are vulnerable to physical obstructions (walls, metal furniture) and electromagnetic interference (microwaves, adjacent Wi-Fi channels), causing variable transmission speeds and packet retransmissions.

Mentor tip

Contrast cable shielding and physical consistency against radio attenuation and environmental interference.

(c)

2 marks

Broadcast medium exposure: Radio waves propagate omnidirectionally through physical walls into public spaces outside the office perimeter.
An eavesdropper within physical range can intercept data packets using a packet sniffer without needing physical access to the building's wiring.

Mentor tip

Emphasize that radio signals cross physical walls and can be captured remotely without physical connection.

(d)

4 marks

Data encryption: All payload data passing through the public network is encrypted using robust cryptographic ciphers (e.g. AES-256), rendering intercepted packets unreadable without the secret decryption key.
Tunneling protocol with IP masking: The original IP packets are encapsulated inside new network packet headers, masking the user's real IP address and origin location from intermediate observers.

Mentor tip

Explain two separate features such as cryptographic payload encryption and packet encapsulation/tunneling.

(e)

4 marks

1) Packetization: The sender segments the original data stream into standardized packets, affixing header metadata (source IP, destination IP, packet sequence number, checksum).
2) Dynamic routing: Each packet travels independently across intermediate routers, which select optimal paths dynamically based on current network congestion.
3) Out-of-order arrival handling: Packets may arrive at the destination via different paths and in scrambled chronological order.
4) Reassembly and verification: The destination host uses packet sequence numbers to reassemble the data in correct sequence and checks error codes, requesting retransmission of any corrupted or dropped packets.

Mentor tip

Detail the full progression: division into packets with headers -> independent dynamic routing -> reassembly by sequence numbers.

Question 11

15 marks

International data migration, change management, and legacy systems

System fundamentals

(a)(i)

1 mark

Data migration is the process of transferring digital data from one computer storage system, format, or database architecture to another.

Mentor tip

State the transfer of data between storage systems, schemas, or environments.

(a)(ii)

4 marks

1) Data corruption and loss: Transmission interruptions, truncated records, or character encoding mismatches during mass migration can compromise data integrity.
2) Schema and format incompatibility: Divergent database structures, missing relational fields, or inconsistent primary keys between legacy and target systems require complex transformation logic, risking record misinterpretation.

Mentor tip

Detail two distinct risks such as data corruption/loss and schema incompatibility.

(b)

4 marks

User training and re-skilling: Delivering comprehensive instruction programs to ensure employees acquire the necessary competence and confidence to navigate the new system effectively.
Stakeholder communication and organizational alignment: Maintaining transparent dialogue regarding project milestones, revised business workflows, and temporary transition schedules to minimize employee disruption and resistance.

Mentor tip

Cover user education and proactive organizational communication.

(c)

3 marks

Scarcity of maintenance expertise: Older legacy programming languages and architectures (e.g. COBOL or proprietary mainframe databases) have dwindling pools of skilled developers, making bug fixes and routine maintenance exceptionally expensive.
Additionally, legacy vendors frequently discontinue security patches, exposing the organization to critical vulnerabilities and severe integration hurdles with modern cloud infrastructure.

Mentor tip

Highlight lack of vendor support, scarce specialized developer skills, and rising maintenance expenses.

(d)

3 marks

Redundant workforce labor: Operating two systems simultaneously requires personnel to duplicate all operational transactions in both platforms, effectively doubling staffing hours.
Infrastructure maintenance overhead: The organization must simultaneously fund operating licenses, dual server hosting environments, electricity, and continuous synchronization maintenance for both systems.

Mentor tip

Explain the dual expenditure of staffing duplication and concurrent infrastructure operational costs.

Question 12

15 marks

Smart building automated control systems and OS role

Control systems
Computer organisation

(a)

4 marks

1) Climate and HVAC regulation: Monitoring indoor temperature, humidity, and CO2 levels to dynamically adjust heating, cooling, and air ventilation dampers.
2) Intelligent lighting control: Using occupancy infrared sensors and daylight harvesting to activate, dim, or extinguish luminaires based on human presence and natural sunlight.

Mentor tip

Describe two separate operations such as HVAC/environmental management and automated lighting or security alarm dispatch.

(b)

4 marks

Sensors: Proximity/RFID sensors or optical cameras detect an approaching person or credential badge and send analog/digital signals to the processor.
Processing: The embedded processor compares input signals against authorization access lists or predefined motion criteria to decide whether the door should cycle open.
Output transducers: When authorized, the processor sends control voltage signals to electromechanical actuators (e.g. electric motor drives or magnetic solenoid locks) to physically retract latches and swing doors open.
Timing and closure: Timer sensors monitor door clearance, after which the processor instructs the actuator to gently shut and re-engage magnetic locks.

Mentor tip

Clearly explain the role of sensors (data collection), processor (decision logic), and transducers (physical actuation).

(c)(i)

2 marks

Hardware I/O device management: Providing drivers and interfaces to communicate with diverse sensor and actuator buses.
Process scheduling and multitasking: Prioritizing real-time emergency events (e.g. fire alarms) over routine telemetry logging.

Mentor tip

State two standard OS functions tailored to control systems, like device management and task prioritization.

(c)(ii)

2 marks

Mentor tip

Interrupts or periodic status polling are both valid; explain the mechanism concisely.

(d)

3 marks

Energy conservation and reduced utility expenses: Automated HVAC and lighting cutoffs eliminate energy waste in unoccupied office zones.
Enhanced physical security and audit tracking: Centralized access logs track employee movements and automatically lock sensitive zones after hours, safeguarding corporate assets.
Predictive maintenance: Continuous telemetry allows facility teams to detect failing motors or filters before catastrophic equipment breakdowns.

Mentor tip

Emphasize financial savings through energy efficiency, heightened security, and automated maintenance.

Question 13

15 marks

Parallel array sorting and binary search tree construction

Computational thinking, problem-solving and programming
Abstract data structures

(a)(i)

5 marks

Bubble sort descending pass sequence:
1) Outer loop executes N - 1 passes across the array length.
2) Inner loop compares adjacent score elements SCORES[J] and SCORES[J + 1].
3) For descending order, if SCORES[J] < SCORES[J + 1], swap the two scores using a temporary variable.
4) Crucially, whenever SCORES[J] and SCORES[J + 1] are swapped, also swap NAMES[J] and NAMES[J + 1] to preserve data correspondence.
5) Continue comparisons until the end of the pass; repeat until no swaps occur or all passes complete.

Mentor tip

Emphasize that parallel arrays require swapping both SCORES and NAMES simultaneously whenever a score inversion occurs.

(a)(ii)

1 mark

Selection sort (or Insertion sort / Quicksort / Merge sort).

Mentor tip

Name any recognized alternative sorting algorithm.

(b)

3 marks

Data payload / value field: Stores the student name and associated numerical test score.
Left child pointer: A reference holding the memory address of the left sub-tree node.
Right child pointer: A reference holding the memory address of the right sub-tree node.

Mentor tip

List the three essential components: data value/fields, left pointer, and right pointer.

(c)

6 marks

Insertion algorithm into binary search tree for descending inorder output (Left -> Root -> Right):
To ensure an inorder traversal prints scores in descending order (highest to lowest), the tree must be structured such that higher scores branch to the left and lower scores branch to the right (or standard traversal inverted):
1) If the tree is empty, instantiate the new node with the student data and designate it as the root.
2) If the tree is not empty, start comparison at the root node.
3) Compare the incoming student's score with the current node's score.
4) If incoming score > current node score, traverse to the left child pointer.
- If the left pointer is null, attach the new node here.
- Otherwise, update current node to left child and repeat comparison.
5) If incoming score <= current node score, traverse to the right child pointer.
- If the right pointer is null, attach the new node here.
- Otherwise, update current node to right child and repeat comparison.

Mentor tip

Carefully define how higher scores navigate left and lower scores navigate right to yield descending order during standard inorder traversal.

Question 14

15 marks

Number classification algorithms: Proper divisors, abundant, deficient, and perfect

Computational thinking, problem-solving and programming

(a)

2 marks

X is a proper divisor of Y if X divides Y without remainder and X is strictly less than Y.
Logical expression: (Y mod X = 0) and (X < Y).

(Y mod X = 0) \land (X < Y)

Mentor tip

Include both conditions: zero remainder and X strictly less than Y.

(b)

3 marks

Initialize an accumulator variable SUM to 0.
Iterate a loop variable D from 1 up to N - 1 (or N div 2).
If N mod D = 0, add D to SUM.
After the loop finishes, return SUM.

subprogram sumPD (N) SUM = 0 loop D from 1 to N - 1 if N mod D = 0 then SUM = SUM + D end if end loop return SUM end subprogram

Mentor tip

Iterate strictly up to N - 1 so that N itself is excluded from the proper divisors sum.

(c)

10 marks

1) Initialize index tracking pointers: A_IDX = 0, D_IDX = 0, P_IDX = 0.
2) Traverse the 2D array MAT using nested loops: outer row R from 0 to 3, inner column C from 0 to 5.
3) For each element NUM = MAT[R][C]:
- Compute proper divisor sum: S = sumPD(NUM).
- Classify according to definitions:
* If S = NUM: add NUM to PERFECT[P_IDX], increment P_IDX.
* Else if S > NUM: add NUM to ABUNDANT[A_IDX], increment A_IDX.
* Else (S < NUM): add NUM to DEFICIENT[D_IDX], increment D_IDX.
4) Close all conditionals and loops.

A_IDX = 0, D_IDX = 0, P_IDX = 0 loop R from 0 to 3 loop C from 0 to 5 NUM = MAT [R] [C] S = sumPD (NUM) if S = NUM then PERFECT [P_IDX] = NUM P_IDX = P_IDX + 1 else if S > NUM then ABUNDANT [A_IDX] = NUM A_IDX = A_IDX + 1 else DEFICIENT [D_IDX] = NUM D_IDX = D_IDX + 1 end if end loop end loop

Mentor tip

Maintain three separate destination index counters and invoke sumPD() for each element.

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 programming26 marks · 26%
Q6, Q9, Q13, Q14
System fundamentals21 marks · 21%
Q2, Q4, Q11
Networks17 marks · 17%
Q4, Q10
Computer organisation13 marks · 13%
Q1, Q3, Q5, Q7, Q12
Abstract data structures12 marks · 12%
Q8, Q13
Control systems11 marks · 11%
Q12

Mentor insights

Traps to watch for

01
Stack unwinding in post-recursion printing · Q9
Because output occurs after the recursive call mystery(A, N-1), the function calls delve all the way to N=0 before any output occurs. Printing begins on stack unwinding with N=0 (outputting A[3]), printing the array in reverse order.
02
Parallel array synchronization during sorting · Q13
When swapping elements in the SCORES array, you must execute the exact same swap on the NAMES array at identical indices, otherwise names become attached to incorrect student scores.
03
Proper divisor definition exclusion of N · Q14
Proper divisors of N explicitly exclude N itself. The divisor loop must stop at N - 1 (or N div 2); including N would cause every number to be misclassified as abundant.

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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