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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: May 2018
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 15

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2018 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–10 · 25 marks

Question 1

4 marks

Stakeholder requirements elicitation and system prototyping

System fundamentals

(a)

2 marks

Select a stakeholder requirements elicitation method suitable for a school library system:
Direct user interviews (or observation): Conduct structured or semi-structured interviews with the school librarian, administrative staff, and student representatives.
Explain purpose: Discuss current cataloging and borrowing workflows to discover functional requirements, pain points, and specific feature expectations for the new system.

Mentor tip

Name a valid elicitation method (interviews, questionnaires, direct observation) and explicitly connect it to uncovering user needs in context.

(b)

2 marks

Explain why providing a working prototype is beneficial:
1) It gives stakeholders an interactive model of the user interface and core borrowing workflows early in development.
2) Stakeholders can identify missing features, usability issues, or design flaws, providing actionable feedback prior to full implementation.

Mentor tip

Focus on early feedback, visualizing the interface, and refining requirements before costly production coding.

Question 2

2 marks

Function of the memory data register in the machine execution cycle

Computer organisation

(a)

2 marks

Describe the role of the MDR during machine instruction execution:
1) The Memory Data Register (MDR) temporarily holds data or instructions fetched from RAM, or data waiting to be written to RAM.
2) It acts as an electrical buffer between the CPU internal buses and the system data bus, allowing the processor and primary memory to operate independently without being bottlenecked by memory latency.

Mentor tip

State clearly that it holds the data/instruction fetched from or written to memory via the data bus.

Question 3

2 marks

Colour representation using two hexadecimal characters

Computer organisation

(a)

2 marks

Determine total bit capacity and distinct combinations:
Each hexadecimal character represents 4 binary bits (a nibble):
\text{Total bits} = 2 \times 4 = 8\text{ bits}
Calculate the total number of distinct color combinations:
2^8 = 256\text{ distinct colours (or }16^2 = 256\text{)}

Total bits = 2 \times 4 = 8 bits 1 6^{2} = 2^{8} = 256 distinct colours

Mentor tip

Each hex digit corresponds to 4 bits; 2 hex digits yield 8 bits, giving 2^8 = 256 states.

Question 4

3 marks

Importance of operating system memory management

Resource management

(a)

3 marks

Explain the role and significance of primary memory management:
1) Core function: Dynamically allocates and de-allocates distinct address blocks of RAM to active programs and background services.
2) Operational mechanism: Employs paging, segmentation, and virtual memory paging to ensure each running process has adequate contiguous address space to execute.
3) Critical necessity: Prevents concurrent processes from overwriting or corrupting each other's memory space, isolating processes to maintain overall system stability.

Mentor tip

Cover the primary allocation role, prevention of process memory clashes/overwrites, and virtual memory handling.

Question 5

3 marks

Logic diagram construction for Boolean expression

Computer organisation

(a)

3 marks

Break down the compound expression: \text{not } A \text{ or } (A \text{ and } B)
1) Feed input line A through a NOT gate (inverter) to produce intermediate signal \neg A.
2) Feed input lines A and B into a 2-input AND gate to produce intermediate signal (A \land B).
3) Connect both intermediate outputs (\neg A and A \land B) into the inputs of a 2-input OR gate to yield the final output Z.

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

Mentor tip

Show the inverter on A, the 2-input AND on A and B, and both feeding into the final 2-input OR gate.

Question 6

3 marks

Trace algorithm with integer division and modulus conditions

Computational thinking, problem-solving and programming

(a)

3 marks

Trace the execution step by step:
Initial values: K = 47, N = 10.
First output: K is output directly -> 47.
First calculation: K = K div 2 = 47 div 2 = 23.
Condition check: K mod N < 3 -> 23 mod 10 = 3. Is 3 < 3? No (False).
Follow the NO branch: Output K -> 23.
Second calculation: K = K div 2 = 23 div 2 = 11.
Condition check: K mod N < 3 -> 11 mod 10 = 1. Is 1 < 3? Yes (True).
Follow the YES branch: Output K -> 11, then algorithm terminates.
Final sequence of printed outputs: 47, 23, 11.

47 div 2 = 23, 23 mod 10 = 3 \neq < 3 ⟹ NO 23 div 2 = 11, 11 mod 10 = 1 < 3 ⟹ YES

Mentor tip

Notice integer division truncates fractions. 23 mod 10 is 3, which is not strictly less than 3, taking the NO branch.

Question 7

1 mark

Definition of recursion in computer programming

Computational thinking, problem-solving and programming

(a)

1 mark

Define recursive subprogram structure:
A function, procedure, or subroutine that calls itself during its execution until a designated base case termination condition is satisfied.

Mentor tip

State that a method, procedure, or function directly or indirectly invokes itself.

Question 8

2 marks

Fundamental characteristics of a queue data structure

Abstract data structures

(a)

2 marks

Identify key structural properties of a queue:
1) First-In, First-Out (FIFO) discipline: The first element enqueued is guaranteed to be the first element dequeued.
2) Constrained access endpoints: Enqueue insertions occur exclusively at the rear (tail), whereas dequeue deletions occur exclusively at the front (head).

Mentor tip

State FIFO and clarify the distinct operations at opposite ends (enqueue at tail, dequeue at head).

Question 9

3 marks

Linked list versus static array for daily sales recording

Abstract data structures

(a)

3 marks

Compare structural properties in the context of daily business sales transactions:
1) Dynamic size vs fixed capacity: A static array has a fixed upper bound declared at compile time, risking overflow on busy days or wasting memory on slow days. A linked list grows dynamically at runtime node-by-node to accommodate fluctuating transaction volumes.
2) Element access and lookup: A static array provides instant O(1) random index access (e.g., sales[i]) and allows binary search if sorted. A linked list requires sequential O(N) traversal from the head node.
3) Insertion overhead: Inserting records into a linked list requires updating pointer references without memory shifts, whereas inserting into an array requires shifting elements.

Mentor tip

Address size flexibility, sequential vs direct access speed, and memory usage relative to unknown daily sales volumes.

Question 10

2 marks

Function and handling of CPU hardware and software interrupts

Resource management

(a)

2 marks

Describe the operational role of interrupts in computer systems:
1) An interrupt is an asynchronous signal sent by hardware (e.g., I/O peripheral, timer) or software requesting immediate CPU attention.
2) Upon receiving the signal, the processor suspends execution of the current program, saves its register state onto the call stack, and transfers program control to the dedicated Interrupt Service Routine (ISR) before resuming normal execution.

Mentor tip

Explain that it alerts the CPU to suspend the current task, save context, and execute the interrupt handler.

Questions 11–15 · 75 marks

Question 11

15 marks

Sequential batch processing, backup policies, and medical data governance

Computational thinking, problem-solving and programming
System fundamentals

(a)

5 marks

Outline the algorithmic steps to send appointment SMS reminders using sequential files:
1) Open both the appointments file and the patients file for sequential read access.
2) Loop through each record in the appointments file from beginning to end.
3) Check if the appointment date matches tomorrow's date.
4) If matched, extract the associated PATIENT_ID.
5) Sequentially traverse the patients file from the beginning until a record matching PATIENT_ID is located.
6) Extract the patient's mobile phone number and dispatch the automated reminder SMS message.
7) Repeat until all appointment records for tomorrow have been processed, then close both files.

Mentor tip

Remember both files are unordered and sequential. For each tomorrow appointment, you must search through the patients file for the corresponding ID.

(b)

4 marks

Describe two distinct data recovery and restoration methods:
Method 1: Regular Scheduled Backups (Local external media / offsite secure cloud): Automated daily snapshots of both patient and appointment files written to secondary storage, allowing restoration of the full database state as of the last backup window.
Method 2: Transaction Logging / Journaling: Every booking, modification, or cancellation is appended in real time to an append-only transaction log file. In the event of system failure, the latest baseline backup is restored and subsequent log entries are replayed to recover up-to-the-minute transactions.

Mentor tip

Name two distinct methods (e.g., scheduled offsite backups and transaction log journals) and describe how each restores data.

(c)

6 marks

Discuss considerations regarding access control and local vs cloud storage:
Issue 1: Internal Access Control and Confidentiality:
- Role-Based Access Control (RBAC): Doctors, nurses, and secretaries have distinct functional needs. Doctors and nurses require clinical history and appointment notes, whereas receptionists only require contact and scheduling details.
- Authentication and audit trails: Strong multi-factor authentication (MFA) must be enforced, accompanied by comprehensive logging of every access and modification to comply with healthcare privacy regulations.
Issue 2: Local On-Premise Storage vs Cloud Hosting:
- Cloud benefits: Professional cloud providers deliver enterprise-grade physical security, automated encrypted replication, robust DDoS protection, and high uptime guarantees surpassing a local clinic server.
- Cloud risks: Data in transit over the public internet requires robust TLS encryption. Furthermore, the clinic must ensure cloud data centers comply with local legal jurisdictions and medical data residency regulations (e.g., GDPR, HIPAA), guarding against vendor lock-in or unauthorized third-party inspection.

Mentor tip

Divide your response evenly: 3 marks on access rights / RBAC / authentication, and 3 marks on local vs cloud security / jurisdiction / compliance.

Question 12

15 marks

Gantt scheduling concurrency, WLAN deployment, and VPN tunneling

Resource management
Networks

(a)

1 mark

Define concurrent processing:
The simultaneous or interleaved execution of multiple distinct tasks or processes across overlapping time intervals.

Mentor tip

Emphasize executing multiple tasks at the same time or across overlapping time frames.

(b)

1 mark

Identify concurrent tasks from the Gantt schedule:
'Install connectors on wall of server room' AND 'Install connectors on wall of new office space' (or 'Test the cabling' AND 'Connect the new computers with the cabling').

Mentor tip

Look for two task bars on the Gantt chart that share identical time columns.

(c)

1 mark

Identify two sequential tasks:
'Lay ducts for new cables' followed by 'Lay cables inside the ducts' (or duct laying followed by connector installation).

Mentor tip

Choose any dependency pair where one task strictly commences after the preceding task finishes.

(d)

4 marks

Outline two distinct operational advantages of adopting WLAN:
Advantage 1: Staff Mobility and Dynamic Collaboration: Employees can move between desks, meeting spaces, and adjoining rooms with laptops or tablets while maintaining seamless network connectivity.
Advantage 2: Reduced Physical Cabling and Infrastructure Cost: Eliminates extensive ducting, patch panels, and cable runs to every desk, allowing rapid onboarding of new workstations or office reorganization without structural cabling expenses.

Mentor tip

Explain mobility/collaboration and cost/physical flexibility with clear development.

(e)

4 marks

Discuss two wireless security threats and their mitigation:
Issue 1: Wireless Eavesdropping / Packet Sniffing: Radio frequency signals broadcast beyond office walls and can be intercepted by unauthorized parties within physical range.
Mitigation: Implement enterprise-grade WPA3/WPA2-Enterprise encryption protocols with robust rotating encryption keys and disable legacy unencrypted protocols.
Issue 2: Rogue Device Connection and BYOD Vulnerabilities: Unauthorized personal devices connecting to Wi-Fi can introduce malware or compromise internal resources.
Mitigation: Enforce MAC address filtering, establish a segmented guest VLAN isolated from internal corporate servers, and mandate Mobile Device Management (MDM) authentication profiles.

Mentor tip

Pair each vulnerability (signal interception, rogue devices) with its practical security solution (WPA3/WPA2 encryption, network segmentation/MAC filtering).

(f)

4 marks

Explain how a Virtual Private Network (VPN) delivers secure remote access:
1) Tunneling: Encapsulates internal company IP packets inside standard public internet transport packets, allowing remote devices to behave as if directly connected to the local office LAN.
2) End-to-End Encryption: Uses protocols such as IPsec or SSL/TLS to encrypt the entire data payload. Any third party intercepting traffic on public networks sees only unreadable ciphertext.
3) Authentication and Gateway Access: Verifies remote employee identity via user credentials and digital certificates before granting access to internal resources, file shares, and intranet servers behind the corporate firewall.

Mentor tip

Discuss tunneling, cryptographic payload encryption, and logical presence on the internal LAN.

Question 13

15 marks

Stack search pseudocode and recursive binary search tree construction

Abstract data structures

(a)

5 marks

Construct pseudocode to search an incoming stack for a name and output its depth position:
searchStack(STACK, TARGET_NAME)
DEPTH = 0
FOUND_POS = -1
while not STACK.isEmpty()
CURRENT = STACK.pop()
if CURRENT == TARGET_NAME then
FOUND_POS = DEPTH
end if
DEPTH = DEPTH + 1
end while
if FOUND_POS != -1 then
output "Position: " + FOUND_POS
else
output TARGET_NAME + " not found"
end if
end searchStack

Mentor tip

Remember stack elements must be accessed via pop() while checking isEmpty(), incrementing depth counter at each step.

(b)

3 marks

Explain search benefits of a Binary Search Tree (BST) over a stack:
1) Time complexity: A stack is an unordered linear structure requiring sequential O(N) popping to search. A balanced BST orders nodes hierarchically, halving the search space at each branch comparison to achieve O(log N) lookup time.
2) Non-destructive inspection: Searching a stack requires popping elements, permanently emptying or altering the stack unless a temporary copy is maintained. In a BST, tree traversal pointers inspect nodes non-destructively without modifying structural contents.

Mentor tip

Highlight O(log N) vs O(N) lookup efficiency and non-destructive tree traversal vs popping stack items.

(c)

4 marks

Describe the recursive insertion procedure for placing a popped name into a BST:
1) Base case (Empty tree/subtree): If the current node pointer is null, allocate a new tree node with the item and set left and right pointers to null; this becomes the root (or child).
2) Lexicographical comparison: Compare the new name alphabetically against the current node's value.
3) Left recursive descent: If the new name precedes the current node's value alphabetically, recursively call the insertion procedure on the left subtree.
4) Right recursive descent: If the new name succeeds the current node's value alphabetically, recursively call the insertion procedure on the right subtree.

Mentor tip

State base case when empty, alphabetical comparison, and recursive call on left or right child pointers.

(d)

3 marks

Construct BST from stack items popped in order [Troy, Mia, Jane, Rick, Ryan, Abed, Zara, Sophie]:
1) 'Troy' is popped first -> Root = 'Troy'.
2) 'Mia' is popped -> 'Mia' < 'Troy' -> Left child of 'Troy'.
3) 'Jane' is popped -> 'Jane' < 'Troy', 'Jane' < 'Mia' -> Left child of 'Mia'.
4) 'Rick' is popped -> 'Rick' < 'Troy', 'Rick' > 'Mia' -> Right child of 'Mia'.
5) 'Ryan' is popped -> 'Ryan' < 'Troy', 'Ryan' > 'Mia', 'Ryan' > 'Rick' -> Right child of 'Rick'.
6) 'Abed' is popped -> 'Abed' < 'Troy', < 'Mia', < 'Jane' -> Left child of 'Jane'.
7) 'Zara' is popped -> 'Zara' > 'Troy' -> Right child of 'Troy'.
8) 'Sophie' is popped -> 'Sophie' < 'Troy', 'Sophie' > 'Mia', 'Sophie' > 'Rick', 'Sophie' > 'Ryan' -> Right child of 'Ryan'.
Verify complete BST hierarchical structure:
Troy
/ \
Mia Zara
/ \
Jane Rick
/ \
Abed Ryan
\
Sophie

Mentor tip

Carefully trace each item popped from top of stack and compare alphabetically from the root down.

Question 14

12 marks

Garden automated lighting and threshold-governed flowerbed irrigation

Control systems

(a)

4 marks

Identify and describe two essential hardware components for fountain illumination:
1) Ambient light sensor (e.g., photoresistor / LDR): Measures light intensity in the garden environment and provides input signals to the microcontroller.
2) Microprocessor / Embedded controller: Executes control logic comparing incoming light sensor readings against sunset and sunrise thresholds to trigger switching signals.
3) Actuator / Output relay transducer: Receives electrical control signals from the processor to switch high-voltage electrical power to the fountain lamps on or off.

Mentor tip

Name the sensor (LDR/light sensor), processor (embedded microcontroller), and actuator/relay with their respective functions.

(b)

3 marks

Explain feedback in computer control systems:
1) Sensor measurement: Continuous sampling of the current physical state of the environment or controlled variable (actual output).
2) Error calculation: The controller compares the measured sensor value against the target setpoint to determine the error difference (Error = Target - Measured).
3) Corrective actuation: The controller adjusts output actuators based on the error value to drive the system toward the target state, forming a closed control loop.

Error (t) = Setpoint - Measured Output (t)

Mentor tip

State that feedback compares the actual measured state against a target setpoint to generate an error signal that adjusts actuator output.

(c)

5 marks

Outline the control algorithm for the automated flowerbed irrigation system:
1) Continuously sample the ambient light intensity sensor (LIGHT_VAL) and soil moisture sensor (SOIL_WATER).
2) Check if ambient light is below the activation threshold (LIGHT_VAL < THRESHOLD_LIGHT):
3) If light condition is met, check if moisture has dropped below minimum (SOIL_WATER < MIN_WATER):
4) If true, activate the irrigation valve/pump actuator to commence watering.
5) Maintain watering while monitoring soil moisture until SOIL_WATER reaches MAX_WATER.
6) Deactivate the water valve/pump immediately once MAX_WATER is reached.
7) If ambient light rises above THRESHOLD_LIGHT or soil water exceeds allowable levels, ensure the watering valve remains closed.

Mentor tip

Ensure both environmental conditions are checked: dark enough to water, and moisture between minimum and maximum bounds.

Question 15

18 marks

Train passenger collection matrix transformations and ticket revenue analytics

Computational thinking, problem-solving and programming

(a)

4 marks

Construct pseudocode to populate 2D_ARRAY[30][7] from PASSENGERS collection:
PASSENGERS.resetNext()
loop WEEK from 0 to 29
loop DAY from 0 to 6
if PASSENGERS.hasNext() then
2D_ARRAY[WEEK][DAY] = PASSENGERS.getNext()
end if
end loop
end loop

Mentor tip

Use nested loops (WEEK 0 to 29, DAY 0 to 6) and fetch items sequentially using PASSENGERS.getNext().

(b)

4 marks

Construct procedure total(N) returning column sum across all 30 weeks:
procedure total(N)
SUM = 0
loop WEEK from 0 to 29
SUM = SUM + 2D_ARRAY[WEEK][N]
end loop
return SUM
end procedure

Mentor tip

Iterate row index WEEK from 0 to 29 while holding column index fixed at N.

(c)

3 marks

Construct pseudocode to find and output the day with highest average passenger count:
MAX_AVG = -1
BEST_DAY_NAME = ""
loop DAY from 0 to 6
CURRENT_AVG = total(DAY) / 30
if CURRENT_AVG > MAX_AVG then
MAX_AVG = CURRENT_AVG
BEST_DAY_NAME = convert(DAY)
end if
end loop
output "Busiest day: " + BEST_DAY_NAME + " with average: " + MAX_AVG

Average (d) = \frac{total ( d )}{30}

Mentor tip

Call total(DAY)/30 for each day 0 to 6, track the maximum, and apply convert(DAY).

(d)

7 marks

Construct procedure salesCalculate(ROW1, COL1, ROW2, COL2) across arbitrary calendar bounds:
procedure salesCalculate(ROW1, COL1, ROW2, COL2)
TOTAL_REVENUE = 0
// Case 1: Start and end in the exact same week
if ROW1 == ROW2 then
loop C from COL1 to COL2
FEE_RATE = FEES[0]
if C >= 5 then FEE_RATE = FEES[1] end if
TOTAL_REVENUE = TOTAL_REVENUE + 2D_ARRAY[ROW1][C] * FEE_RATE
end loop
else
// Case 2a: Remaining days of initial week ROW1
loop C from COL1 to 6
FEE_RATE = FEES[0]
if C >= 5 then FEE_RATE = FEES[1] end if
TOTAL_REVENUE = TOTAL_REVENUE + 2D_ARRAY[ROW1][C] * FEE_RATE
end loop
// Case 2b: All full intermediate weeks between ROW1+1 and ROW2-1
loop W from ROW1 + 1 to ROW2 - 1
loop C from 0 to 6
FEE_RATE = FEES[0]
if C >= 5 then FEE_RATE = FEES[1] end if
TOTAL_REVENUE = TOTAL_REVENUE + 2D_ARRAY[W][C] * FEE_RATE
end loop
end loop
// Case 2c: Concluding week ROW2 up to day COL2
loop C from 0 to COL2
FEE_RATE = FEES[0]
if C >= 5 then FEE_RATE = FEES[1] end if
TOTAL_REVENUE = TOTAL_REVENUE + 2D_ARRAY[ROW2][C] * FEE_RATE
end loop
end if
output TOTAL_REVENUE
end procedure

Revenue = \sum (Passengers \times Fee)

Mentor tip

Handle the three distinct spans: partial first week (COL1 to 6), intermediate full weeks (0 to 6), and partial last week (0 to COL2). Use FEES[0] for weekdays (0-4) and FEES[1] for weekends (5-6).

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 programming27 marks · 27%
Q6, Q7, Q11, Q15
Abstract data structures20 marks · 20%
Q8, Q9, Q13
System fundamentals14 marks · 14%
Q1, Q11
Networks12 marks · 12%
Q12
Control systems12 marks · 12%
Q14
Resource management8 marks · 8%
Q4, Q10, Q12
Computer organisation7 marks · 7%
Q2, Q3, Q5

Mentor insights

Traps to watch for

01
Hexadecimal digit bit length vs distinct values · Q3
Two hexadecimal characters correspond to 2 × 4 = 8 bits, not 2 × 8 = 16 bits. Total representable colours is 2^8 = 256.
02
Strict inequality check in modulus loop · Q6
When K = 23, K mod 10 evaluates to 3. The condition check is K mod 10 < 3. Because 3 < 3 evaluates to False, the algorithm follows the NO branch to print 23 before dividing.
03
Stack popping alters structure during linear search · Q13
Searching an item in a stack via pop() empties and modifies the stack. To preserve the original stack, a working copy or reconstruction queue must be maintained.
04
Revenue summation across non-contiguous calendar weeks · Q15
When calculating ticket sales between arbitrary date endpoints, week 1 requires iterating from COL1 to 6, intermediate weeks span 0 to 6, and the final week spans 0 to COL2. Applying a uniform 0 to 6 loop to all weeks causes boundary calculation errors.

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: May 2018
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 15

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

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2018 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–10 · 25 marks

Question 1

4 marks

Stakeholder requirements elicitation and system prototyping

System fundamentals

(a)

2 marks

Select a stakeholder requirements elicitation method suitable for a school library system:
Direct user interviews (or observation): Conduct structured or semi-structured interviews with the school librarian, administrative staff, and student representatives.
Explain purpose: Discuss current cataloging and borrowing workflows to discover functional requirements, pain points, and specific feature expectations for the new system.

Mentor tip

Name a valid elicitation method (interviews, questionnaires, direct observation) and explicitly connect it to uncovering user needs in context.

(b)

2 marks

Explain why providing a working prototype is beneficial:
1) It gives stakeholders an interactive model of the user interface and core borrowing workflows early in development.
2) Stakeholders can identify missing features, usability issues, or design flaws, providing actionable feedback prior to full implementation.

Mentor tip

Focus on early feedback, visualizing the interface, and refining requirements before costly production coding.

Question 2

2 marks

Function of the memory data register in the machine execution cycle

Computer organisation

(a)

2 marks

Describe the role of the MDR during machine instruction execution:
1) The Memory Data Register (MDR) temporarily holds data or instructions fetched from RAM, or data waiting to be written to RAM.
2) It acts as an electrical buffer between the CPU internal buses and the system data bus, allowing the processor and primary memory to operate independently without being bottlenecked by memory latency.

Mentor tip

State clearly that it holds the data/instruction fetched from or written to memory via the data bus.

Question 3

2 marks

Colour representation using two hexadecimal characters

Computer organisation

(a)

2 marks

Determine total bit capacity and distinct combinations:
Each hexadecimal character represents 4 binary bits (a nibble):
\text{Total bits} = 2 \times 4 = 8\text{ bits}
Calculate the total number of distinct color combinations:
2^8 = 256\text{ distinct colours (or }16^2 = 256\text{)}

Total bits = 2 \times 4 = 8 bits 1 6^{2} = 2^{8} = 256 distinct colours

Mentor tip

Each hex digit corresponds to 4 bits; 2 hex digits yield 8 bits, giving 2^8 = 256 states.

Question 4

3 marks

Importance of operating system memory management

Resource management

(a)

3 marks

Explain the role and significance of primary memory management:
1) Core function: Dynamically allocates and de-allocates distinct address blocks of RAM to active programs and background services.
2) Operational mechanism: Employs paging, segmentation, and virtual memory paging to ensure each running process has adequate contiguous address space to execute.
3) Critical necessity: Prevents concurrent processes from overwriting or corrupting each other's memory space, isolating processes to maintain overall system stability.

Mentor tip

Cover the primary allocation role, prevention of process memory clashes/overwrites, and virtual memory handling.

Question 5

3 marks

Logic diagram construction for Boolean expression

Computer organisation

(a)

3 marks

Break down the compound expression: \text{not } A \text{ or } (A \text{ and } B)
1) Feed input line A through a NOT gate (inverter) to produce intermediate signal \neg A.
2) Feed input lines A and B into a 2-input AND gate to produce intermediate signal (A \land B).
3) Connect both intermediate outputs (\neg A and A \land B) into the inputs of a 2-input OR gate to yield the final output Z.

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

Mentor tip

Show the inverter on A, the 2-input AND on A and B, and both feeding into the final 2-input OR gate.

Question 6

3 marks

Trace algorithm with integer division and modulus conditions

Computational thinking, problem-solving and programming

(a)

3 marks

Trace the execution step by step:
Initial values: K = 47, N = 10.
First output: K is output directly -> 47.
First calculation: K = K div 2 = 47 div 2 = 23.
Condition check: K mod N < 3 -> 23 mod 10 = 3. Is 3 < 3? No (False).
Follow the NO branch: Output K -> 23.
Second calculation: K = K div 2 = 23 div 2 = 11.
Condition check: K mod N < 3 -> 11 mod 10 = 1. Is 1 < 3? Yes (True).
Follow the YES branch: Output K -> 11, then algorithm terminates.
Final sequence of printed outputs: 47, 23, 11.

47 div 2 = 23, 23 mod 10 = 3 \neq < 3 ⟹ NO 23 div 2 = 11, 11 mod 10 = 1 < 3 ⟹ YES

Mentor tip

Notice integer division truncates fractions. 23 mod 10 is 3, which is not strictly less than 3, taking the NO branch.

Question 7

1 mark

Definition of recursion in computer programming

Computational thinking, problem-solving and programming

(a)

1 mark

Define recursive subprogram structure:
A function, procedure, or subroutine that calls itself during its execution until a designated base case termination condition is satisfied.

Mentor tip

State that a method, procedure, or function directly or indirectly invokes itself.

Question 8

2 marks

Fundamental characteristics of a queue data structure

Abstract data structures

(a)

2 marks

Identify key structural properties of a queue:
1) First-In, First-Out (FIFO) discipline: The first element enqueued is guaranteed to be the first element dequeued.
2) Constrained access endpoints: Enqueue insertions occur exclusively at the rear (tail), whereas dequeue deletions occur exclusively at the front (head).

Mentor tip

State FIFO and clarify the distinct operations at opposite ends (enqueue at tail, dequeue at head).

Question 9

3 marks

Linked list versus static array for daily sales recording

Abstract data structures

(a)

3 marks

Compare structural properties in the context of daily business sales transactions:
1) Dynamic size vs fixed capacity: A static array has a fixed upper bound declared at compile time, risking overflow on busy days or wasting memory on slow days. A linked list grows dynamically at runtime node-by-node to accommodate fluctuating transaction volumes.
2) Element access and lookup: A static array provides instant O(1) random index access (e.g., sales[i]) and allows binary search if sorted. A linked list requires sequential O(N) traversal from the head node.
3) Insertion overhead: Inserting records into a linked list requires updating pointer references without memory shifts, whereas inserting into an array requires shifting elements.

Mentor tip

Address size flexibility, sequential vs direct access speed, and memory usage relative to unknown daily sales volumes.

Question 10

2 marks

Function and handling of CPU hardware and software interrupts

Resource management

(a)

2 marks

Describe the operational role of interrupts in computer systems:
1) An interrupt is an asynchronous signal sent by hardware (e.g., I/O peripheral, timer) or software requesting immediate CPU attention.
2) Upon receiving the signal, the processor suspends execution of the current program, saves its register state onto the call stack, and transfers program control to the dedicated Interrupt Service Routine (ISR) before resuming normal execution.

Mentor tip

Explain that it alerts the CPU to suspend the current task, save context, and execute the interrupt handler.

Questions 11–15 · 75 marks

Question 11

15 marks

Sequential batch processing, backup policies, and medical data governance

Computational thinking, problem-solving and programming
System fundamentals

(a)

5 marks

Outline the algorithmic steps to send appointment SMS reminders using sequential files:
1) Open both the appointments file and the patients file for sequential read access.
2) Loop through each record in the appointments file from beginning to end.
3) Check if the appointment date matches tomorrow's date.
4) If matched, extract the associated PATIENT_ID.
5) Sequentially traverse the patients file from the beginning until a record matching PATIENT_ID is located.
6) Extract the patient's mobile phone number and dispatch the automated reminder SMS message.
7) Repeat until all appointment records for tomorrow have been processed, then close both files.

Mentor tip

Remember both files are unordered and sequential. For each tomorrow appointment, you must search through the patients file for the corresponding ID.

(b)

4 marks

Describe two distinct data recovery and restoration methods:
Method 1: Regular Scheduled Backups (Local external media / offsite secure cloud): Automated daily snapshots of both patient and appointment files written to secondary storage, allowing restoration of the full database state as of the last backup window.
Method 2: Transaction Logging / Journaling: Every booking, modification, or cancellation is appended in real time to an append-only transaction log file. In the event of system failure, the latest baseline backup is restored and subsequent log entries are replayed to recover up-to-the-minute transactions.

Mentor tip

Name two distinct methods (e.g., scheduled offsite backups and transaction log journals) and describe how each restores data.

(c)

6 marks

Discuss considerations regarding access control and local vs cloud storage:
Issue 1: Internal Access Control and Confidentiality:
- Role-Based Access Control (RBAC): Doctors, nurses, and secretaries have distinct functional needs. Doctors and nurses require clinical history and appointment notes, whereas receptionists only require contact and scheduling details.
- Authentication and audit trails: Strong multi-factor authentication (MFA) must be enforced, accompanied by comprehensive logging of every access and modification to comply with healthcare privacy regulations.
Issue 2: Local On-Premise Storage vs Cloud Hosting:
- Cloud benefits: Professional cloud providers deliver enterprise-grade physical security, automated encrypted replication, robust DDoS protection, and high uptime guarantees surpassing a local clinic server.
- Cloud risks: Data in transit over the public internet requires robust TLS encryption. Furthermore, the clinic must ensure cloud data centers comply with local legal jurisdictions and medical data residency regulations (e.g., GDPR, HIPAA), guarding against vendor lock-in or unauthorized third-party inspection.

Mentor tip

Divide your response evenly: 3 marks on access rights / RBAC / authentication, and 3 marks on local vs cloud security / jurisdiction / compliance.

Question 12

15 marks

Gantt scheduling concurrency, WLAN deployment, and VPN tunneling

Resource management
Networks

(a)

1 mark

Define concurrent processing:
The simultaneous or interleaved execution of multiple distinct tasks or processes across overlapping time intervals.

Mentor tip

Emphasize executing multiple tasks at the same time or across overlapping time frames.

(b)

1 mark

Identify concurrent tasks from the Gantt schedule:
'Install connectors on wall of server room' AND 'Install connectors on wall of new office space' (or 'Test the cabling' AND 'Connect the new computers with the cabling').

Mentor tip

Look for two task bars on the Gantt chart that share identical time columns.

(c)

1 mark

Identify two sequential tasks:
'Lay ducts for new cables' followed by 'Lay cables inside the ducts' (or duct laying followed by connector installation).

Mentor tip

Choose any dependency pair where one task strictly commences after the preceding task finishes.

(d)

4 marks

Outline two distinct operational advantages of adopting WLAN:
Advantage 1: Staff Mobility and Dynamic Collaboration: Employees can move between desks, meeting spaces, and adjoining rooms with laptops or tablets while maintaining seamless network connectivity.
Advantage 2: Reduced Physical Cabling and Infrastructure Cost: Eliminates extensive ducting, patch panels, and cable runs to every desk, allowing rapid onboarding of new workstations or office reorganization without structural cabling expenses.

Mentor tip

Explain mobility/collaboration and cost/physical flexibility with clear development.

(e)

4 marks

Discuss two wireless security threats and their mitigation:
Issue 1: Wireless Eavesdropping / Packet Sniffing: Radio frequency signals broadcast beyond office walls and can be intercepted by unauthorized parties within physical range.
Mitigation: Implement enterprise-grade WPA3/WPA2-Enterprise encryption protocols with robust rotating encryption keys and disable legacy unencrypted protocols.
Issue 2: Rogue Device Connection and BYOD Vulnerabilities: Unauthorized personal devices connecting to Wi-Fi can introduce malware or compromise internal resources.
Mitigation: Enforce MAC address filtering, establish a segmented guest VLAN isolated from internal corporate servers, and mandate Mobile Device Management (MDM) authentication profiles.

Mentor tip

Pair each vulnerability (signal interception, rogue devices) with its practical security solution (WPA3/WPA2 encryption, network segmentation/MAC filtering).

(f)

4 marks

Explain how a Virtual Private Network (VPN) delivers secure remote access:
1) Tunneling: Encapsulates internal company IP packets inside standard public internet transport packets, allowing remote devices to behave as if directly connected to the local office LAN.
2) End-to-End Encryption: Uses protocols such as IPsec or SSL/TLS to encrypt the entire data payload. Any third party intercepting traffic on public networks sees only unreadable ciphertext.
3) Authentication and Gateway Access: Verifies remote employee identity via user credentials and digital certificates before granting access to internal resources, file shares, and intranet servers behind the corporate firewall.

Mentor tip

Discuss tunneling, cryptographic payload encryption, and logical presence on the internal LAN.

Question 13

15 marks

Stack search pseudocode and recursive binary search tree construction

Abstract data structures

(a)

5 marks

Construct pseudocode to search an incoming stack for a name and output its depth position:
searchStack(STACK, TARGET_NAME)
DEPTH = 0
FOUND_POS = -1
while not STACK.isEmpty()
CURRENT = STACK.pop()
if CURRENT == TARGET_NAME then
FOUND_POS = DEPTH
end if
DEPTH = DEPTH + 1
end while
if FOUND_POS != -1 then
output "Position: " + FOUND_POS
else
output TARGET_NAME + " not found"
end if
end searchStack

Mentor tip

Remember stack elements must be accessed via pop() while checking isEmpty(), incrementing depth counter at each step.

(b)

3 marks

Explain search benefits of a Binary Search Tree (BST) over a stack:
1) Time complexity: A stack is an unordered linear structure requiring sequential O(N) popping to search. A balanced BST orders nodes hierarchically, halving the search space at each branch comparison to achieve O(log N) lookup time.
2) Non-destructive inspection: Searching a stack requires popping elements, permanently emptying or altering the stack unless a temporary copy is maintained. In a BST, tree traversal pointers inspect nodes non-destructively without modifying structural contents.

Mentor tip

Highlight O(log N) vs O(N) lookup efficiency and non-destructive tree traversal vs popping stack items.

(c)

4 marks

Describe the recursive insertion procedure for placing a popped name into a BST:
1) Base case (Empty tree/subtree): If the current node pointer is null, allocate a new tree node with the item and set left and right pointers to null; this becomes the root (or child).
2) Lexicographical comparison: Compare the new name alphabetically against the current node's value.
3) Left recursive descent: If the new name precedes the current node's value alphabetically, recursively call the insertion procedure on the left subtree.
4) Right recursive descent: If the new name succeeds the current node's value alphabetically, recursively call the insertion procedure on the right subtree.

Mentor tip

State base case when empty, alphabetical comparison, and recursive call on left or right child pointers.

(d)

3 marks

Construct BST from stack items popped in order [Troy, Mia, Jane, Rick, Ryan, Abed, Zara, Sophie]:
1) 'Troy' is popped first -> Root = 'Troy'.
2) 'Mia' is popped -> 'Mia' < 'Troy' -> Left child of 'Troy'.
3) 'Jane' is popped -> 'Jane' < 'Troy', 'Jane' < 'Mia' -> Left child of 'Mia'.
4) 'Rick' is popped -> 'Rick' < 'Troy', 'Rick' > 'Mia' -> Right child of 'Mia'.
5) 'Ryan' is popped -> 'Ryan' < 'Troy', 'Ryan' > 'Mia', 'Ryan' > 'Rick' -> Right child of 'Rick'.
6) 'Abed' is popped -> 'Abed' < 'Troy', < 'Mia', < 'Jane' -> Left child of 'Jane'.
7) 'Zara' is popped -> 'Zara' > 'Troy' -> Right child of 'Troy'.
8) 'Sophie' is popped -> 'Sophie' < 'Troy', 'Sophie' > 'Mia', 'Sophie' > 'Rick', 'Sophie' > 'Ryan' -> Right child of 'Ryan'.
Verify complete BST hierarchical structure:
Troy
/ \
Mia Zara
/ \
Jane Rick
/ \
Abed Ryan
\
Sophie

Mentor tip

Carefully trace each item popped from top of stack and compare alphabetically from the root down.

Question 14

12 marks

Garden automated lighting and threshold-governed flowerbed irrigation

Control systems

(a)

4 marks

Identify and describe two essential hardware components for fountain illumination:
1) Ambient light sensor (e.g., photoresistor / LDR): Measures light intensity in the garden environment and provides input signals to the microcontroller.
2) Microprocessor / Embedded controller: Executes control logic comparing incoming light sensor readings against sunset and sunrise thresholds to trigger switching signals.
3) Actuator / Output relay transducer: Receives electrical control signals from the processor to switch high-voltage electrical power to the fountain lamps on or off.

Mentor tip

Name the sensor (LDR/light sensor), processor (embedded microcontroller), and actuator/relay with their respective functions.

(b)

3 marks

Explain feedback in computer control systems:
1) Sensor measurement: Continuous sampling of the current physical state of the environment or controlled variable (actual output).
2) Error calculation: The controller compares the measured sensor value against the target setpoint to determine the error difference (Error = Target - Measured).
3) Corrective actuation: The controller adjusts output actuators based on the error value to drive the system toward the target state, forming a closed control loop.

Error (t) = Setpoint - Measured Output (t)

Mentor tip

State that feedback compares the actual measured state against a target setpoint to generate an error signal that adjusts actuator output.

(c)

5 marks

Outline the control algorithm for the automated flowerbed irrigation system:
1) Continuously sample the ambient light intensity sensor (LIGHT_VAL) and soil moisture sensor (SOIL_WATER).
2) Check if ambient light is below the activation threshold (LIGHT_VAL < THRESHOLD_LIGHT):
3) If light condition is met, check if moisture has dropped below minimum (SOIL_WATER < MIN_WATER):
4) If true, activate the irrigation valve/pump actuator to commence watering.
5) Maintain watering while monitoring soil moisture until SOIL_WATER reaches MAX_WATER.
6) Deactivate the water valve/pump immediately once MAX_WATER is reached.
7) If ambient light rises above THRESHOLD_LIGHT or soil water exceeds allowable levels, ensure the watering valve remains closed.

Mentor tip

Ensure both environmental conditions are checked: dark enough to water, and moisture between minimum and maximum bounds.

Question 15

18 marks

Train passenger collection matrix transformations and ticket revenue analytics

Computational thinking, problem-solving and programming

(a)

4 marks

Construct pseudocode to populate 2D_ARRAY[30][7] from PASSENGERS collection:
PASSENGERS.resetNext()
loop WEEK from 0 to 29
loop DAY from 0 to 6
if PASSENGERS.hasNext() then
2D_ARRAY[WEEK][DAY] = PASSENGERS.getNext()
end if
end loop
end loop

Mentor tip

Use nested loops (WEEK 0 to 29, DAY 0 to 6) and fetch items sequentially using PASSENGERS.getNext().

(b)

4 marks

Construct procedure total(N) returning column sum across all 30 weeks:
procedure total(N)
SUM = 0
loop WEEK from 0 to 29
SUM = SUM + 2D_ARRAY[WEEK][N]
end loop
return SUM
end procedure

Mentor tip

Iterate row index WEEK from 0 to 29 while holding column index fixed at N.

(c)

3 marks

Construct pseudocode to find and output the day with highest average passenger count:
MAX_AVG = -1
BEST_DAY_NAME = ""
loop DAY from 0 to 6
CURRENT_AVG = total(DAY) / 30
if CURRENT_AVG > MAX_AVG then
MAX_AVG = CURRENT_AVG
BEST_DAY_NAME = convert(DAY)
end if
end loop
output "Busiest day: " + BEST_DAY_NAME + " with average: " + MAX_AVG

Average (d) = \frac{total ( d )}{30}

Mentor tip

Call total(DAY)/30 for each day 0 to 6, track the maximum, and apply convert(DAY).

(d)

7 marks

Construct procedure salesCalculate(ROW1, COL1, ROW2, COL2) across arbitrary calendar bounds:
procedure salesCalculate(ROW1, COL1, ROW2, COL2)
TOTAL_REVENUE = 0
// Case 1: Start and end in the exact same week
if ROW1 == ROW2 then
loop C from COL1 to COL2
FEE_RATE = FEES[0]
if C >= 5 then FEE_RATE = FEES[1] end if
TOTAL_REVENUE = TOTAL_REVENUE + 2D_ARRAY[ROW1][C] * FEE_RATE
end loop
else
// Case 2a: Remaining days of initial week ROW1
loop C from COL1 to 6
FEE_RATE = FEES[0]
if C >= 5 then FEE_RATE = FEES[1] end if
TOTAL_REVENUE = TOTAL_REVENUE + 2D_ARRAY[ROW1][C] * FEE_RATE
end loop
// Case 2b: All full intermediate weeks between ROW1+1 and ROW2-1
loop W from ROW1 + 1 to ROW2 - 1
loop C from 0 to 6
FEE_RATE = FEES[0]
if C >= 5 then FEE_RATE = FEES[1] end if
TOTAL_REVENUE = TOTAL_REVENUE + 2D_ARRAY[W][C] * FEE_RATE
end loop
end loop
// Case 2c: Concluding week ROW2 up to day COL2
loop C from 0 to COL2
FEE_RATE = FEES[0]
if C >= 5 then FEE_RATE = FEES[1] end if
TOTAL_REVENUE = TOTAL_REVENUE + 2D_ARRAY[ROW2][C] * FEE_RATE
end loop
end if
output TOTAL_REVENUE
end procedure

Revenue = \sum (Passengers \times Fee)

Mentor tip

Handle the three distinct spans: partial first week (COL1 to 6), intermediate full weeks (0 to 6), and partial last week (0 to COL2). Use FEES[0] for weekdays (0-4) and FEES[1] for weekends (5-6).

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 programming27 marks · 27%
Q6, Q7, Q11, Q15
Abstract data structures20 marks · 20%
Q8, Q9, Q13
System fundamentals14 marks · 14%
Q1, Q11
Networks12 marks · 12%
Q12
Control systems12 marks · 12%
Q14
Resource management8 marks · 8%
Q4, Q10, Q12
Computer organisation7 marks · 7%
Q2, Q3, Q5

Mentor insights

Traps to watch for

01
Hexadecimal digit bit length vs distinct values · Q3
Two hexadecimal characters correspond to 2 × 4 = 8 bits, not 2 × 8 = 16 bits. Total representable colours is 2^8 = 256.
02
Strict inequality check in modulus loop · Q6
When K = 23, K mod 10 evaluates to 3. The condition check is K mod 10 < 3. Because 3 < 3 evaluates to False, the algorithm follows the NO branch to print 23 before dividing.
03
Stack popping alters structure during linear search · Q13
Searching an item in a stack via pop() empties and modifies the stack. To preserve the original stack, a working copy or reconstruction queue must be maintained.
04
Revenue summation across non-contiguous calendar weeks · Q15
When calculating ticket sales between arbitrary date endpoints, week 1 requires iterating from COL1 to 6, intermediate weeks span 0 to 6, and the final week spans 0 to COL2. Applying a uniform 0 to 6 loop to all weeks causes boundary calculation errors.

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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