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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Timezone

TZ1TZ2

Session: May 2023
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 May 2023 TZ1 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

Role and core functionality of a web browser

Networks

2 marks

Protocol requests: Issues HTTP/HTTPS requests to remote web servers to retrieve HTML, CSS, JavaScript, and multimedia assets.
Document parsing and rendering: Parses the received HTML Document Object Model (DOM) and applies style rules to display interactive web pages.

Client Browser HTTP GET Web Server HTML/CSS/JS DOM Render

Mentor tip

Explain both fetching web documents via HTTP/HTTPS and rendering HTML/CSS to the display.

Question 2

2 marks

Practical applications of queue data structures in computing

Abstract data structures

2 marks

Print spooling: Queues print jobs sent from multiple workstations so the printer executes them in First-In, First-Out (FIFO) arrival order.
CPU process scheduling: Maintains ready queues for operating system schedulers (such as Round Robin) to sequence process execution.

Mentor tip

Queues model any first-come, first-served buffering scenario: print queues, packet buffers, or CPU ready queues.

Question 3

2 marks

Universal data encoding via the Unicode standard

Computer organisation

2 marks

Global character support: Unlike 7-bit or 8-bit ASCII which is limited to 128 or 256 Latin glyphs, Unicode (UTF-8, UTF-16) supports over 140,000 characters spanning virtually all global written languages, mathematical symbols, and emojis.
Cross-platform text consistency: Prevents encoding errors (mojibake) when exchanging text files across international platforms.

ASCII: 2^{7} = 128 glyphs vs Unicode: > 1.1 \times 1 0^{6} code points

Mentor tip

State that Unicode encodes characters from international alphabets and symbols beyond basic Latin.

Question 4

4 marks

Sensor signal processing: interrupts versus polling mechanisms

Resource management

(a)

1 mark

An interrupt is a hardware or software signal sent to the CPU that temporarily suspends the current execution sequence to invoke a dedicated Interrupt Service Routine (ISR).

Mentor tip

An interrupt alerts the CPU to an event needing immediate processing.

(b)

3 marks

Sequential interrogation: The central computer periodically queries each river pollutant sensor sequentially in round-robin order.
Data status inspection: For each sensor node, the central host checks whether new pollutant readings are available.
Reading retrieval: If data is ready, the computer reads the pollutant values, stores them, and moves to interrogate the next sequential sensor.

Host \to Sensor_{1} \to Sensor_{2} \to \dots \to Sensor_{n} \to Sensor_{1}

Mentor tip

Describe continuous cyclical interrogation of each sensor in turn.

Question 5

4 marks

Truth table construction for compound logic expression

Computer organisation

4 marks

Evaluate expression: $Z = A nand (B nor C)$ .
Sub-expression 1: $B nor C$ is true (1) if and only if $B = 0$ and $C = 0$ ; otherwise 0.
Sub-expression 2: NAND operator produces 0 if and only if both $A = 1$ and $(B nor C) = 1$ ; otherwise 1.
Row-by-row evaluation across all 8 input states:
$A = 0, B = 0, C = 0 : B nor C = 1 ⟹ 0 nand 1 = 1$ .
$A = 0, B = 0, C = 1 : B nor C = 0 ⟹ 0 nand 0 = 1$ .
$A = 0, B = 1, C = 0 : B nor C = 0 ⟹ 0 nand 0 = 1$ .
$A = 0, B = 1, C = 1 : B nor C = 0 ⟹ 0 nand 0 = 1$ .
$A = 1, B = 0, C = 0 : B nor C = 1 ⟹ 1 nand 1 = 0$ .
$A = 1, B = 0, C = 1 : B nor C = 0 ⟹ 1 nand 0 = 1$ .
$A = 1, B = 1, C = 0 : B nor C = 0 ⟹ 1 nand 0 = 1$ .
$A = 1, B = 1, C = 1 : B nor C = 0 ⟹ 1 nand 0 = 1$ .

A 00001111 B 00110011 C 01010101 B nor C 10001000 A nand (B nor C) 11110111

Mentor tip

Notice that only one single row yields 0: when A=1, B=0, and C=0.

Question 6

2 marks

Abstract data structure properties of collections

Abstract data structures

2 marks

Dynamic container: A collection is a container data structure that holds elements of unknown or variable quantity, growing and shrinking dynamically at runtime.
Iterator-driven access: Provides standardized methods (`getNext()`, `hasNext()`, `resetNext()`) to iterate sequentially through items without direct array indexing.

Mentor tip

Highlight dynamic resizing and sequential iterator access.

Question 7

2 marks

Architectural distinction between RAM and ROM

Computer organisation

2 marks

Volatility: RAM is volatile primary storage that loses its stored contents when power is removed, whereas ROM is non-volatile and permanently retains data without electrical power.
Read-write permissions: RAM permits dynamic high-speed read and write operations during runtime, whereas ROM is read-only (used for storing BIOS/boot firmware).

Mentor tip

Compare volatility (volatile vs non-volatile) and write capability (read/write vs read-only).

Question 8

5 marks

Binary search tree: leaf identification, postorder traversal, and root deletion

Abstract data structures

(a)

1 mark

Leaf nodes are nodes with no left or right child branches.
Inspecting the binary tree structure, the leaf nodes are: A, C, E, and G.

Leaf nodes = {A, C, E, G}

Mentor tip

Leaf nodes have zero children.

(b)

1 mark

Postorder traversal visits: Left Subtree $\to$ Right Subtree $\to$ Root.
Traverse left subtree of D: visit leaf A $\to$ leaf C $\to$ node B.
Traverse right subtree of D: visit leaf E $\to$ leaf G $\to$ node F.
Finally visit root node D.
Resulting sequence: A C B E G F D.

Postorder: A, C, B, E, G, F, D

Mentor tip

Root D is visited last in postorder traversal.

(c)

3 marks

Deleting root node D with two children requires replacing D with its in-order predecessor (C) or in-order successor (E).
Option 1 - In-order predecessor replacement (C replaces D):
C becomes the new root node.
Left subtree: B remains with left child A (C is detached).
Right subtree: F remains with left child E and right child G.
The resulting binary search tree preserves the BST ordering invariant across all nodes.

C ↙ ↘ B F ↙ ↙ ↘ A E G

Mentor tip

When deleting a node with two children, replace it with the maximum element of the left subtree or the minimum element of the right subtree.

Question 9

2 marks

Architectural benefits and mechanics of virtual memory

Resource management

2 marks

Execution of oversized programs: Virtual memory maps secondary storage (paging file / swap space) into the memory address space, allowing programs larger than physical RAM to execute without 'out of memory' crashes.
Memory isolation: Provides each process with a contiguous virtual address space, preventing unauthorized cross-process memory access.

Virtual Address Space Paging / MMU Physical RAM + Swap Disk

Mentor tip

Focus on allowing oversized processes to run by swapping inactive pages to disk.

Questions 10–14 · 75 marks

Question 10

15 marks

Legacy system migration, requirements engineering, changeover, and training

System fundamentals

(a)

2 marks

User resistance to change: Staff accustomed to legacy interfaces may experience anxiety or reluctance when adapting to modern workflows.
Changing job responsibilities: Automation or revised business logic in the new system may eliminate repetitive roles or necessitate new administrative skills.

Mentor tip

Focus on human factors: resistance to change, retraining demands, or evolving roles.

(b)

2 marks

A legacy system is an outdated computer system, programming language, or software technology that remains in active operational use because it supports critical core business functions and is difficult or expensive to replace.

Mentor tip

Mention both aspects: outdated/obsolete technology and ongoing critical business reliance.

(c)

1 mark

Interviews / Questionnaires / Direct observation / Document analysis of current forms and transaction logs.

Mentor tip

Name one standard stakeholder requirement elicitation method.

(d)

3 marks

Data format and schema incompatibility: Legacy relational or flat-file schemas may use incompatible data types, character encodings, or field lengths compared to the modern target database.
Resolution mechanism: Requires data transformation and cleansing scripts to prevent database corruption, truncated fields, or lost historical records.

Mentor tip

Explain data format incompatibility and the risk of data loss/truncation during migration.

(e)

3 marks

Zero operational downtime risk: In parallel running, both the legacy and new systems operate simultaneously with identical input data.
Instant failback: If the new system encounters critical bugs or crashes, business operations continue uninterrupted on the legacy system without data loss.

Mentor tip

Highlight safety and immediate fallback if the new system fails.

(f)(i)

1 mark

Self-paced online training modules (computer-based training) / Instructor-led hands-on workshops.

Mentor tip

Choose online self-paced training or in-person workshop.

(f)(ii)

3 marks

Advantages of self-paced online training: Employees can complete modules at their own pace, revisit difficult sections anytime, and balance learning around flexible daily schedules.
Disadvantages: Lacks immediate face-to-face interaction with an expert instructor to ask tailored questions, and requires high self-discipline to prevent procrastination.

Mentor tip

Balance flexibility and cost against lack of direct personal guidance.

Question 11

15 marks

VPN protocols, transmission speeds, compression, and remote telecommuting

Networks
System fundamentals

(a)

2 marks

Tunneling protocols (e.g. IPsec, OpenVPN, WireGuard) to encapsulate private network packets within public IP packets.
Cryptographic encryption algorithms (e.g. AES-256) and public key authentication to ensure data confidentiality and integrity.

Mentor tip

Name encryption and tunneling protocols as the two pillars of a VPN.

(b)

2 marks

Bandwidth of the transmission channel: The maximum data-carrying capacity of the physical or wireless medium (e.g. fibre optic vs copper vs Wi-Fi).
Network congestion: The volume of concurrent traffic sharing intermediate routers, switches, and transmission links causing packet queuing delays.

Mentor tip

State physical bandwidth and network traffic congestion.

(c)

3 marks

Reduced payload size: Compression algorithms remove redundancy from data files, significantly reducing the total number of bytes requiring transmission.
Bandwidth conservation and transfer speed: Smaller files transmit across the network in less time, conserving bandwidth and lowering transit latency.

Transmission Time = \frac{Compressed File Size}{Channel Bandwidth}

Mentor tip

Connect reduced byte size to reduced transmission time and lower bandwidth usage.

(d)

2 marks

Ciphertext transformation: Uses mathematical encryption algorithms (ciphers) and secret cryptographic keys to scramble plaintext into unintelligible ciphertext.
Confidentiality during interception: Even if unauthorized third parties intercept packets across public networks, they cannot decipher the contents without the corresponding decryption key.

Plaintext Encrypt (K) Ciphertext Decrypt (K) Plaintext

Mentor tip

Detail mathematical scrambling with keys and protection against eavesdropping.

(e)

2 marks

Traffic inspection and access filtering: Monitored packet headers (source IP, destination port) are compared against authorization rules to block unauthorized connections while permitting authenticated enterprise traffic.

Mentor tip

State rule-based filtering and blocking of unauthorized network traffic.

(f)

4 marks

Positive social impacts: Greater schedule flexibility, improved work-life balance, and elimination of daily commuting times and travel expenses.
Negative social impacts: Blurring of boundaries between domestic and professional life leading to extended working hours, professional isolation, and reduced spontaneous collaborative interaction with colleagues.

Mentor tip

Discuss both benefits (flexibility, reduced commuting) and drawbacks (isolation, blurred work-life boundaries).

Question 12

15 marks

Smart home temperature control: closed-loop feedback, OS, and system topologies

Control systems
Resource management

(a)

5 marks

Data sensing: Temperature sensors continuously sample ambient environmental heat, converting physical thermal conditions into electronic analogue voltages.
Analogue-to-Digital Conversion (ADC): The electronic signal is digitized via an ADC so the microprocessor can interpret numeric readings.
Microprocessor comparison: The microprocessor compares current digitized temperature against a user-defined target setpoint.
Actuator command signal: If ambient temperature deviates from the setpoint threshold, the processor issues a digital control command.
Transducer actuation: An output transducer (relay or electronic switch) activates or deactivates heating elements or air conditioning units.

Sensor \to ADC \to Microprocessor \to Actuator / Heater

Mentor tip

Structure the sequence logically: sensor input, ADC conversion, microprocessor threshold comparison, and output transducer action.

(b)

2 marks

Closed-loop feedback: The output state (temperature change generated by the heater/cooler) directly alters the input reading detected by the temperature sensor.
Dynamic stability: This continuous loop prevents overshoot, automatically turning off heating once target temperature is attained.

Error = T_{setpoint} - T_{measured} ⟹ Drive Output to Zero Error

Mentor tip

Feedback means the system's output directly influences its subsequent inputs.

(c)(i)

2 marks

Hardware resource abstraction: Provides device driver interfaces to manage communications between microcontroller peripherals, memory registers, and network radios.

Mentor tip

Mention device management, memory allocation, or task scheduling.

(c)(ii)

2 marks

Minimal footprint and real-time execution: A dedicated operating system contains only the essential modules required for temperature regulation, reducing RAM/ROM requirements and providing deterministic real-time responsiveness without unnecessary desktop overhead.

Mentor tip

Dedicated embedded OS has minimal footprint and predictable real-time execution.

(d)

4 marks

Centralized control system: A single primary computer processes all room sensor data and controls all heating valves. It is easier to configure and maintain globally, but presents a single point of failure (if the central unit fails, the entire house loses heating).
Distributed control system: Independent microcontrollers are installed in each room, managing localized temperature autonomously. Failure of a single room controller does not affect other zones, but installation and network coordination across multiple nodes is more complex.

Mentor tip

Compare single point of failure and maintenance complexity between centralized and distributed architectures.

Question 13

15 marks

Recursive Fibonacci evaluation, call stack overhead, and stack reversal using queues

Computational thinking, problem-solving and programming
Abstract data structures

(a)

4 marks

Base cases: For $A < 2$ , $rec (0) = 1$ and $rec (1) = 1$ .
Evaluate bottom-up values:
$rec (2) = rec (0) + rec (1) = 1 + 1 = 2$ .
$rec (3) = rec (1) + rec (2) = 1 + 2 = 3$ .
$rec (4) = rec (2) + rec (3) = 2 + 3 = 5$ .
$rec (5) = rec (3) + rec (4) = 3 + 5 = 8$ .
Final returned value is 8.

rec (5) = rec (3) + rec (4) = 3 + 5 = 8

Mentor tip

Build from small values: rec(0)=1, rec(1)=1, rec(2)=2, rec(3)=3, rec(4)=5, rec(5)=8.

(b)

4 marks

Call stack memory consumption: Each recursive invocation allocates a new stack frame storing local variables and return addresses, risking stack overflow on deep recursion.
Execution time overhead: Repeated function call linkage and redundant subproblem re-computation (such as calculating $rec (2)$ multiple times) results in exponential $O (2^{n})$ time complexity compared to iterative loops.

Time Complexity: O (2^{n}) vs Iterative: O (n)

Mentor tip

Identify high memory consumption (call stack frames / risk of overflow) and slow execution time.

(c)

2 marks

Underflow prevention: The `isEmpty()` access method checks whether the stack contains any elements before executing a `pop()` or `peek()` operation, preventing runtime stack underflow errors.

Mentor tip

isEmpty() guards against popping an empty data structure.

(d)

5 marks

Phase 1: Pop each item from TOWNS stack and enqueue into queue TEMP until TOWNS is empty.
Because stack is LIFO and queue is FIFO: the order entering the queue is Cardiff $\to$ Washington DC $\to$ The Hague $\to$ Singapore $\to$ Geneva.
Phase 2: Dequeue each item from queue TEMP and push back onto TOWNS stack until TEMP is empty.
Cardiff is dequeued first and pushed to bottom of TOWNS, followed sequentially by Washington DC, The Hague, Singapore, and finally Geneva is pushed on top.
TOWNS stack now has Geneva at the top and Cardiff at the bottom, achieving full reversal.

loop while not TOWNS.isEmpty () ITEM = TOWNS.pop () TEMP.enqueue (ITEM) end loop loop while not TEMP.isEmpty () ITEM = TEMP.dequeue () TOWNS.push (ITEM) end loop

Mentor tip

LIFO followed by FIFO reverses the sequence completely.

Question 14

15 marks

2D dice matrix calculations: cumulative product, duplicate detection, and minimum row totals

Computational thinking, problem-solving and programming

(a)

3 marks

Initialize product accumulator `PROD = 1`.
Nested loops: outer loop row $R$ from 0 to 6; inner loop column $C$ from 0 to 2.
Multiply each element: $PROD = PROD \times DICEDIAL [R] [C]$ .
Output the final computed product.

PROD = 1 loop R from 0 to 6 loop C from 0 to 2 PROD = PROD * DICEDIAL [R] [C] end loop end loop output (PROD)

Mentor tip

Initialize multiplication accumulators to 1, never 0.

(b)

4 marks

A row contains 3 values: $DICEDIAL [R] [0]$ , $DICEDIAL [R] [1]$ , $DICEDIAL [R] [2]$ .
Check all pairwise combinations:
If column 0 equals column 1, return that repeated value.
Else if column 0 equals column 2, return that repeated value.
Else if column 1 equals column 2, return that repeated value.
If no pair matches, return 0.

sub-program DuplicateNum (DICEDIAL, R) if DICEDIAL [R] [0] == DICEDIAL [R] [1] then return DICEDIAL [R] [0] else if DICEDIAL [R] [0] == DICEDIAL [R] [2] then return DICEDIAL [R] [0] else if DICEDIAL [R] [1] == DICEDIAL [R] [2] then return DICEDIAL [R] [1] else return 0 end if end sub-program

Mentor tip

With only 3 columns, compare the three pairs directly: (0,1), (0,2), and (1,2).

(c)

8 marks

Step 1 - Compute row totals: Store the sum of each row $R$ in a 1D array `ROWTOTALS[0..6]`.
Step 2 - Find minimum sum: Initialize `MIN_SUM = ROWTOTALS[0]` and scan $R$ from 1 to 6 to find the lowest total.
Step 3 - Output results: Print `MIN_SUM`, then scan `ROWTOTALS` again from 0 to 6, outputting the index $R$ of every row where `ROWTOTALS[R] == MIN_SUM`.

sub-program lowestRT (DICEDIAL) loop R from 0 to 6 ROWTOTALS [R] = DICEDIAL [R] [0] + DICEDIAL [R] [1] + DICEDIAL [R] [2] end loop MIN_SUM = ROWTOTALS [0] loop R from 1 to 6 if ROWTOTALS [R] < MIN_SUM then MIN_SUM = ROWTOTALS [R] end if end loop output ("Lowest row total: ", MIN_SUM) output ("Occurs at row index(es):") loop R from 0 to 6 if ROWTOTALS [R] == MIN_SUM then output (R) end if end loop end sub-program

Mentor tip

Save row totals in an array first so you can scan for the minimum and then report all matching row indices.

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, Q14
System fundamentals19 marks · 19%
Q10, Q11
Abstract data structures16 marks · 16%
Q2, Q6, Q8, Q13
Networks13 marks · 13%
Q1, Q11
Control systems11 marks · 11%
Q12
Resource management10 marks · 10%
Q4, Q9, Q12
Computer organisation8 marks · 8%
Q3, Q5, Q7

Mentor insights

Traps to watch for

01
NAND truth evaluation with NOR sub-term · Q5
Remember that NOR is 1 only when both inputs are 0. The only way A NAND (B NOR C) can output 0 is when A=1 and (B NOR C)=1, which occurs uniquely at A=1, B=0, C=0.
02
Binary search tree two-child root deletion · Q8
When deleting a root with two children, replace it with its in-order predecessor or in-order successor. Simply deleting the node breaks the tree into disconnected subtrees.
03
Stack and queue order reversal dynamics · Q13
Popping from a LIFO stack into a FIFO queue and then dequeuing back into the stack completely reverses the element order. Swapping these operations does not achieve reversal.
04
Multiple rows sharing minimum total · Q14
Do not stop after printing the first row with the minimum total. Scan the row totals a second time to output all row indices that share the lowest total.

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 2023
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 May 2023 TZ1 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

Role and core functionality of a web browser

Networks

2 marks

Protocol requests: Issues HTTP/HTTPS requests to remote web servers to retrieve HTML, CSS, JavaScript, and multimedia assets.
Document parsing and rendering: Parses the received HTML Document Object Model (DOM) and applies style rules to display interactive web pages.

Client Browser HTTP GET Web Server HTML/CSS/JS DOM Render

Mentor tip

Explain both fetching web documents via HTTP/HTTPS and rendering HTML/CSS to the display.

Question 2

2 marks

Practical applications of queue data structures in computing

Abstract data structures

2 marks

Print spooling: Queues print jobs sent from multiple workstations so the printer executes them in First-In, First-Out (FIFO) arrival order.
CPU process scheduling: Maintains ready queues for operating system schedulers (such as Round Robin) to sequence process execution.

Mentor tip

Queues model any first-come, first-served buffering scenario: print queues, packet buffers, or CPU ready queues.

Question 3

2 marks

Universal data encoding via the Unicode standard

Computer organisation

2 marks

Global character support: Unlike 7-bit or 8-bit ASCII which is limited to 128 or 256 Latin glyphs, Unicode (UTF-8, UTF-16) supports over 140,000 characters spanning virtually all global written languages, mathematical symbols, and emojis.
Cross-platform text consistency: Prevents encoding errors (mojibake) when exchanging text files across international platforms.

ASCII: 2^{7} = 128 glyphs vs Unicode: > 1.1 \times 1 0^{6} code points

Mentor tip

State that Unicode encodes characters from international alphabets and symbols beyond basic Latin.

Question 4

4 marks

Sensor signal processing: interrupts versus polling mechanisms

Resource management

(a)

1 mark

An interrupt is a hardware or software signal sent to the CPU that temporarily suspends the current execution sequence to invoke a dedicated Interrupt Service Routine (ISR).

Mentor tip

An interrupt alerts the CPU to an event needing immediate processing.

(b)

3 marks

Sequential interrogation: The central computer periodically queries each river pollutant sensor sequentially in round-robin order.
Data status inspection: For each sensor node, the central host checks whether new pollutant readings are available.
Reading retrieval: If data is ready, the computer reads the pollutant values, stores them, and moves to interrogate the next sequential sensor.

Host \to Sensor_{1} \to Sensor_{2} \to \dots \to Sensor_{n} \to Sensor_{1}

Mentor tip

Describe continuous cyclical interrogation of each sensor in turn.

Question 5

4 marks

Truth table construction for compound logic expression

Computer organisation

4 marks

Evaluate expression: $Z = A nand (B nor C)$ .
Sub-expression 1: $B nor C$ is true (1) if and only if $B = 0$ and $C = 0$ ; otherwise 0.
Sub-expression 2: NAND operator produces 0 if and only if both $A = 1$ and $(B nor C) = 1$ ; otherwise 1.
Row-by-row evaluation across all 8 input states:
$A = 0, B = 0, C = 0 : B nor C = 1 ⟹ 0 nand 1 = 1$ .
$A = 0, B = 0, C = 1 : B nor C = 0 ⟹ 0 nand 0 = 1$ .
$A = 0, B = 1, C = 0 : B nor C = 0 ⟹ 0 nand 0 = 1$ .
$A = 0, B = 1, C = 1 : B nor C = 0 ⟹ 0 nand 0 = 1$ .
$A = 1, B = 0, C = 0 : B nor C = 1 ⟹ 1 nand 1 = 0$ .
$A = 1, B = 0, C = 1 : B nor C = 0 ⟹ 1 nand 0 = 1$ .
$A = 1, B = 1, C = 0 : B nor C = 0 ⟹ 1 nand 0 = 1$ .
$A = 1, B = 1, C = 1 : B nor C = 0 ⟹ 1 nand 0 = 1$ .

A 00001111 B 00110011 C 01010101 B nor C 10001000 A nand (B nor C) 11110111

Mentor tip

Notice that only one single row yields 0: when A=1, B=0, and C=0.

Question 6

2 marks

Abstract data structure properties of collections

Abstract data structures

2 marks

Dynamic container: A collection is a container data structure that holds elements of unknown or variable quantity, growing and shrinking dynamically at runtime.
Iterator-driven access: Provides standardized methods (`getNext()`, `hasNext()`, `resetNext()`) to iterate sequentially through items without direct array indexing.

Mentor tip

Highlight dynamic resizing and sequential iterator access.

Question 7

2 marks

Architectural distinction between RAM and ROM

Computer organisation

2 marks

Volatility: RAM is volatile primary storage that loses its stored contents when power is removed, whereas ROM is non-volatile and permanently retains data without electrical power.
Read-write permissions: RAM permits dynamic high-speed read and write operations during runtime, whereas ROM is read-only (used for storing BIOS/boot firmware).

Mentor tip

Compare volatility (volatile vs non-volatile) and write capability (read/write vs read-only).

Question 8

5 marks

Binary search tree: leaf identification, postorder traversal, and root deletion

Abstract data structures

(a)

1 mark

Leaf nodes are nodes with no left or right child branches.
Inspecting the binary tree structure, the leaf nodes are: A, C, E, and G.

Leaf nodes = {A, C, E, G}

Mentor tip

Leaf nodes have zero children.

(b)

1 mark

Postorder traversal visits: Left Subtree $\to$ Right Subtree $\to$ Root.
Traverse left subtree of D: visit leaf A $\to$ leaf C $\to$ node B.
Traverse right subtree of D: visit leaf E $\to$ leaf G $\to$ node F.
Finally visit root node D.
Resulting sequence: A C B E G F D.

Postorder: A, C, B, E, G, F, D

Mentor tip

Root D is visited last in postorder traversal.

(c)

3 marks

Deleting root node D with two children requires replacing D with its in-order predecessor (C) or in-order successor (E).
Option 1 - In-order predecessor replacement (C replaces D):
C becomes the new root node.
Left subtree: B remains with left child A (C is detached).
Right subtree: F remains with left child E and right child G.
The resulting binary search tree preserves the BST ordering invariant across all nodes.

C ↙ ↘ B F ↙ ↙ ↘ A E G

Mentor tip

When deleting a node with two children, replace it with the maximum element of the left subtree or the minimum element of the right subtree.

Question 9

2 marks

Architectural benefits and mechanics of virtual memory

Resource management

2 marks

Execution of oversized programs: Virtual memory maps secondary storage (paging file / swap space) into the memory address space, allowing programs larger than physical RAM to execute without 'out of memory' crashes.
Memory isolation: Provides each process with a contiguous virtual address space, preventing unauthorized cross-process memory access.

Virtual Address Space Paging / MMU Physical RAM + Swap Disk

Mentor tip

Focus on allowing oversized processes to run by swapping inactive pages to disk.

Questions 10–14 · 75 marks

Question 10

15 marks

Legacy system migration, requirements engineering, changeover, and training

System fundamentals

(a)

2 marks

User resistance to change: Staff accustomed to legacy interfaces may experience anxiety or reluctance when adapting to modern workflows.
Changing job responsibilities: Automation or revised business logic in the new system may eliminate repetitive roles or necessitate new administrative skills.

Mentor tip

Focus on human factors: resistance to change, retraining demands, or evolving roles.

(b)

2 marks

Mentor tip

Mention both aspects: outdated/obsolete technology and ongoing critical business reliance.

(c)

1 mark

Interviews / Questionnaires / Direct observation / Document analysis of current forms and transaction logs.

Mentor tip

Name one standard stakeholder requirement elicitation method.

(d)

3 marks

Data format and schema incompatibility: Legacy relational or flat-file schemas may use incompatible data types, character encodings, or field lengths compared to the modern target database.
Resolution mechanism: Requires data transformation and cleansing scripts to prevent database corruption, truncated fields, or lost historical records.

Mentor tip

Explain data format incompatibility and the risk of data loss/truncation during migration.

(e)

3 marks

Zero operational downtime risk: In parallel running, both the legacy and new systems operate simultaneously with identical input data.
Instant failback: If the new system encounters critical bugs or crashes, business operations continue uninterrupted on the legacy system without data loss.

Mentor tip

Highlight safety and immediate fallback if the new system fails.

(f)(i)

1 mark

Self-paced online training modules (computer-based training) / Instructor-led hands-on workshops.

Mentor tip

Choose online self-paced training or in-person workshop.

(f)(ii)

3 marks

Advantages of self-paced online training: Employees can complete modules at their own pace, revisit difficult sections anytime, and balance learning around flexible daily schedules.
Disadvantages: Lacks immediate face-to-face interaction with an expert instructor to ask tailored questions, and requires high self-discipline to prevent procrastination.

Mentor tip

Balance flexibility and cost against lack of direct personal guidance.

Question 11

15 marks

VPN protocols, transmission speeds, compression, and remote telecommuting

Networks
System fundamentals

(a)

2 marks

Tunneling protocols (e.g. IPsec, OpenVPN, WireGuard) to encapsulate private network packets within public IP packets.
Cryptographic encryption algorithms (e.g. AES-256) and public key authentication to ensure data confidentiality and integrity.

Mentor tip

Name encryption and tunneling protocols as the two pillars of a VPN.

(b)

2 marks

Bandwidth of the transmission channel: The maximum data-carrying capacity of the physical or wireless medium (e.g. fibre optic vs copper vs Wi-Fi).
Network congestion: The volume of concurrent traffic sharing intermediate routers, switches, and transmission links causing packet queuing delays.

Mentor tip

State physical bandwidth and network traffic congestion.

(c)

3 marks

Reduced payload size: Compression algorithms remove redundancy from data files, significantly reducing the total number of bytes requiring transmission.
Bandwidth conservation and transfer speed: Smaller files transmit across the network in less time, conserving bandwidth and lowering transit latency.

Transmission Time = \frac{Compressed File Size}{Channel Bandwidth}

Mentor tip

Connect reduced byte size to reduced transmission time and lower bandwidth usage.

(d)

2 marks

Ciphertext transformation: Uses mathematical encryption algorithms (ciphers) and secret cryptographic keys to scramble plaintext into unintelligible ciphertext.
Confidentiality during interception: Even if unauthorized third parties intercept packets across public networks, they cannot decipher the contents without the corresponding decryption key.

Plaintext Encrypt (K) Ciphertext Decrypt (K) Plaintext

Mentor tip

Detail mathematical scrambling with keys and protection against eavesdropping.

(e)

2 marks

Mentor tip

State rule-based filtering and blocking of unauthorized network traffic.

(f)

4 marks

Positive social impacts: Greater schedule flexibility, improved work-life balance, and elimination of daily commuting times and travel expenses.
Negative social impacts: Blurring of boundaries between domestic and professional life leading to extended working hours, professional isolation, and reduced spontaneous collaborative interaction with colleagues.

Mentor tip

Discuss both benefits (flexibility, reduced commuting) and drawbacks (isolation, blurred work-life boundaries).

Question 12

15 marks

Smart home temperature control: closed-loop feedback, OS, and system topologies

Control systems
Resource management

(a)

5 marks

Data sensing: Temperature sensors continuously sample ambient environmental heat, converting physical thermal conditions into electronic analogue voltages.
Analogue-to-Digital Conversion (ADC): The electronic signal is digitized via an ADC so the microprocessor can interpret numeric readings.
Microprocessor comparison: The microprocessor compares current digitized temperature against a user-defined target setpoint.
Actuator command signal: If ambient temperature deviates from the setpoint threshold, the processor issues a digital control command.
Transducer actuation: An output transducer (relay or electronic switch) activates or deactivates heating elements or air conditioning units.

Sensor \to ADC \to Microprocessor \to Actuator / Heater

Mentor tip

Structure the sequence logically: sensor input, ADC conversion, microprocessor threshold comparison, and output transducer action.

(b)

2 marks

Closed-loop feedback: The output state (temperature change generated by the heater/cooler) directly alters the input reading detected by the temperature sensor.
Dynamic stability: This continuous loop prevents overshoot, automatically turning off heating once target temperature is attained.

Error = T_{setpoint} - T_{measured} ⟹ Drive Output to Zero Error

Mentor tip

Feedback means the system's output directly influences its subsequent inputs.

(c)(i)

2 marks

Hardware resource abstraction: Provides device driver interfaces to manage communications between microcontroller peripherals, memory registers, and network radios.

Mentor tip

Mention device management, memory allocation, or task scheduling.

(c)(ii)

2 marks

Mentor tip

Dedicated embedded OS has minimal footprint and predictable real-time execution.

(d)

4 marks

Centralized control system: A single primary computer processes all room sensor data and controls all heating valves. It is easier to configure and maintain globally, but presents a single point of failure (if the central unit fails, the entire house loses heating).
Distributed control system: Independent microcontrollers are installed in each room, managing localized temperature autonomously. Failure of a single room controller does not affect other zones, but installation and network coordination across multiple nodes is more complex.

Mentor tip

Compare single point of failure and maintenance complexity between centralized and distributed architectures.

Question 13

15 marks

Recursive Fibonacci evaluation, call stack overhead, and stack reversal using queues

Computational thinking, problem-solving and programming
Abstract data structures

(a)

4 marks

Base cases: For $A < 2$ , $rec (0) = 1$ and $rec (1) = 1$ .
Evaluate bottom-up values:
$rec (2) = rec (0) + rec (1) = 1 + 1 = 2$ .
$rec (3) = rec (1) + rec (2) = 1 + 2 = 3$ .
$rec (4) = rec (2) + rec (3) = 2 + 3 = 5$ .
$rec (5) = rec (3) + rec (4) = 3 + 5 = 8$ .
Final returned value is 8.

rec (5) = rec (3) + rec (4) = 3 + 5 = 8

Mentor tip

Build from small values: rec(0)=1, rec(1)=1, rec(2)=2, rec(3)=3, rec(4)=5, rec(5)=8.

(b)

4 marks

Call stack memory consumption: Each recursive invocation allocates a new stack frame storing local variables and return addresses, risking stack overflow on deep recursion.
Execution time overhead: Repeated function call linkage and redundant subproblem re-computation (such as calculating $rec (2)$ multiple times) results in exponential $O (2^{n})$ time complexity compared to iterative loops.

Time Complexity: O (2^{n}) vs Iterative: O (n)

Mentor tip

Identify high memory consumption (call stack frames / risk of overflow) and slow execution time.

(c)

2 marks

Underflow prevention: The `isEmpty()` access method checks whether the stack contains any elements before executing a `pop()` or `peek()` operation, preventing runtime stack underflow errors.

Mentor tip

isEmpty() guards against popping an empty data structure.

(d)

5 marks

Phase 1: Pop each item from TOWNS stack and enqueue into queue TEMP until TOWNS is empty.
Because stack is LIFO and queue is FIFO: the order entering the queue is Cardiff $\to$ Washington DC $\to$ The Hague $\to$ Singapore $\to$ Geneva.
Phase 2: Dequeue each item from queue TEMP and push back onto TOWNS stack until TEMP is empty.
Cardiff is dequeued first and pushed to bottom of TOWNS, followed sequentially by Washington DC, The Hague, Singapore, and finally Geneva is pushed on top.
TOWNS stack now has Geneva at the top and Cardiff at the bottom, achieving full reversal.

loop while not TOWNS.isEmpty () ITEM = TOWNS.pop () TEMP.enqueue (ITEM) end loop loop while not TEMP.isEmpty () ITEM = TEMP.dequeue () TOWNS.push (ITEM) end loop

Mentor tip

LIFO followed by FIFO reverses the sequence completely.

Question 14

15 marks

2D dice matrix calculations: cumulative product, duplicate detection, and minimum row totals

Computational thinking, problem-solving and programming

(a)

3 marks

Initialize product accumulator `PROD = 1`.
Nested loops: outer loop row $R$ from 0 to 6; inner loop column $C$ from 0 to 2.
Multiply each element: $PROD = PROD \times DICEDIAL [R] [C]$ .
Output the final computed product.

PROD = 1 loop R from 0 to 6 loop C from 0 to 2 PROD = PROD * DICEDIAL [R] [C] end loop end loop output (PROD)

Mentor tip

Initialize multiplication accumulators to 1, never 0.

(b)

4 marks

A row contains 3 values: $DICEDIAL [R] [0]$ , $DICEDIAL [R] [1]$ , $DICEDIAL [R] [2]$ .
Check all pairwise combinations:
If column 0 equals column 1, return that repeated value.
Else if column 0 equals column 2, return that repeated value.
Else if column 1 equals column 2, return that repeated value.
If no pair matches, return 0.

sub-program DuplicateNum (DICEDIAL, R) if DICEDIAL [R] [0] == DICEDIAL [R] [1] then return DICEDIAL [R] [0] else if DICEDIAL [R] [0] == DICEDIAL [R] [2] then return DICEDIAL [R] [0] else if DICEDIAL [R] [1] == DICEDIAL [R] [2] then return DICEDIAL [R] [1] else return 0 end if end sub-program

Mentor tip

With only 3 columns, compare the three pairs directly: (0,1), (0,2), and (1,2).

(c)

8 marks

Step 1 - Compute row totals: Store the sum of each row $R$ in a 1D array `ROWTOTALS[0..6]`.
Step 2 - Find minimum sum: Initialize `MIN_SUM = ROWTOTALS[0]` and scan $R$ from 1 to 6 to find the lowest total.
Step 3 - Output results: Print `MIN_SUM`, then scan `ROWTOTALS` again from 0 to 6, outputting the index $R$ of every row where `ROWTOTALS[R] == MIN_SUM`.

sub-program lowestRT (DICEDIAL) loop R from 0 to 6 ROWTOTALS [R] = DICEDIAL [R] [0] + DICEDIAL [R] [1] + DICEDIAL [R] [2] end loop MIN_SUM = ROWTOTALS [0] loop R from 1 to 6 if ROWTOTALS [R] < MIN_SUM then MIN_SUM = ROWTOTALS [R] end if end loop output ("Lowest row total: ", MIN_SUM) output ("Occurs at row index(es):") loop R from 0 to 6 if ROWTOTALS [R] == MIN_SUM then output (R) end if end loop end sub-program

Mentor tip

Save row totals in an array first so you can scan for the minimum and then report all matching row indices.

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, Q14
System fundamentals19 marks · 19%
Q10, Q11
Abstract data structures16 marks · 16%
Q2, Q6, Q8, Q13
Networks13 marks · 13%
Q1, Q11
Control systems11 marks · 11%
Q12
Resource management10 marks · 10%
Q4, Q9, Q12
Computer organisation8 marks · 8%
Q3, Q5, Q7

Mentor insights

Traps to watch for

01
NAND truth evaluation with NOR sub-term · Q5
Remember that NOR is 1 only when both inputs are 0. The only way A NAND (B NOR C) can output 0 is when A=1 and (B NOR C)=1, which occurs uniquely at A=1, B=0, C=0.
02
Binary search tree two-child root deletion · Q8
When deleting a root with two children, replace it with its in-order predecessor or in-order successor. Simply deleting the node breaks the tree into disconnected subtrees.
03
Stack and queue order reversal dynamics · Q13
Popping from a LIFO stack into a FIFO queue and then dequeuing back into the stack completely reverses the element order. Swapping these operations does not achieve reversal.
04
Multiple rows sharing minimum total · Q14
Do not stop after printing the first row with the minimum total. Scan the row totals a second time to output all row indices that share the lowest total.

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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