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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: November 2015
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2015 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–6 · 25 marks

Question 1

3 marks

System usability and accessibility methods

System fundamentals

(a)

1 mark

Define usability in human-computer interaction:
Usability refers to the extent to which a computer system can be used by specified users to achieve specified goals effectively, efficiently, and satisfactorily in a specified context of use.

Mentor tip

Define usability in terms of ease of use, learnability, efficiency, and meeting user requirements.

(b)

2 marks

Identify two accessibility enhancement features:
1. Voice synthesis and screen readers (text-to-speech) to accommodate visually impaired users.
2. Speech recognition software (voice input control) to assist users with motor impairments or physical disabilities in operating the system hands-free.

Mentor tip

Name two specific assistive technologies such as screen readers, voice recognition, high-contrast display modes, or braille interfaces.

Question 2

4 marks

VPN technologies for secure remote access

Networks

(a)

4 marks

Explain tunneling protocols:
1. Tunneling (such as L2TP or OpenVPN) encapsulates the original private IP packets inside standard public IP packets, establishing a virtual conduit across the public internet.
Explain encryption protocols:
2. Strong cryptographic algorithms and security protocols (such as IPsec or SSL/TLS) encrypt packet contents. If intercepted en route, the data remains undecipherable ciphertext.
Explain VPN gateway authentication:
3. A secure enterprise VPN gateway authenticates the traveling salesperson's digital certificates and login credentials, routing traffic seamlessly into the corporate intranet.

Mentor tip

Mention tunneling protocols (packet encapsulation), encryption protocols (such as IPsec/SSL to prevent eavesdropping), and secure gateway authentication.

Question 3

3 marks

Truth table for compound Boolean expression

Computer organisation

(a)

3 marks

Deconstruct the Boolean expression:
Expression: $(A \land B) NOR C$ .
Compute intermediate term $X = A \land B$ :
$X$ is 1 only when both $A = 1$ and $B = 1$ ; otherwise $X = 0$ .
Apply the NOR operation with $C$ :
Output is $\neg (X \lor C)$ . The output is 1 only when BOTH $X = 0$ AND $C = 0$ :
1. For $A = 0, B = 0, C = 0$ : $X = 0, C = 0 ⟹ 0 NOR 0 = 1$ .
2. For $A = 0, B = 0, C = 1$ : $X = 0, C = 1 ⟹ 0 NOR 1 = 0$ .
3. For $A = 0, B = 1, C = 0$ : $X = 0, C = 0 ⟹ 0 NOR 0 = 1$ .
4. For $A = 0, B = 1, C = 1$ : $X = 0, C = 1 ⟹ 0 NOR 1 = 0$ .
5. For $A = 1, B = 0, C = 0$ : $X = 0, C = 0 ⟹ 0 NOR 0 = 1$ .
6. For $A = 1, B = 0, C = 1$ : $X = 0, C = 1 ⟹ 0 NOR 1 = 0$ .
7. For $A = 1, B = 1, C = 0$ : $X = 1, C = 0 ⟹ 1 NOR 0 = 0$ .
8. For $A = 1, B = 1, C = 1$ : $X = 1, C = 1 ⟹ 1 NOR 1 = 0$ .

A 00001111 B 00110011 C 01010101 A \land B 00000011 (A \land B) NOR C 10101000

Mentor tip

Notice that NOR produces 1 only when both operands are 0. Whenever C = 1, the result is immediately 0.

Question 4

4 marks

Client-server network architecture and security threats

Networks

(a)

1 mark

Define a server in a network context:
A server is a central dedicated computer or software program that manages access to centralized resources, services, files, and applications shared among network clients.

Mentor tip

Highlight that a server provides shared central resources, files, or applications to client devices.

(b)

1 mark

Identify network clients in the school setup:
The client devices comprise the individual workstation desktop computers, laptops, and network printer accessing server resources.

Mentor tip

Name the user workstation computers or printers requesting services from the server.

(c)

2 marks

Identify an external security threat:
External cyber-attacks such as malicious malware/ransomware infection or unauthorized remote intrusion by hackers.
State a specific defensive countermeasure:
Deploy an active network firewall with continuous packet inspection and install regularly updated anti-malware software across all connected endpoints.

Mentor tip

Pair a specific digital threat (malware/viruses or unauthorized hacker intrusion) with its direct defense (firewall or antivirus software).

Question 5

5 marks

Algorithm debugging for digit extraction

Computational thinking, problem-solving and programming

(a)

2 marks

Trace algorithm execution:
In the provided algorithm, the loop checks the condition `(N > 0) and (EVEN = true)`.
Identify the logic defect:
The value of $N$ is never modified or reduced inside the loop body. As a result, the loop repeatedly inspects the same least significant digit `(N mod 10)` indefinitely, resulting in an infinite loop whenever the first examined digit is even.

Mentor tip

Explain that N is never updated inside the loop, causing an infinite loop because the condition remains true.

(b)

3 marks

Formulate algorithm modification:
To process successive digits from right to left, the integer value of $N$ must be truncated after each digit examination.
Identify exact assignment and placement:
Add the integer division statement `N = N div 10` inside the loop body, positioned after the `if` statement block so that each iteration shifts to the next higher-order digit until $N$ reaches 0.

N = N div 10

Mentor tip

Specify adding `N = N div 10` inside the loop body immediately following the if condition.

Question 6

6 marks

Stack representation using arrays and reversal using a queue

Abstract data structures

(a)

2 marks

Describe array-based stack representation:
A static one-dimensional array of fixed size `LIMIT` stores the stack elements from index 0 up to `LIMIT - 1`.
Explain pointer tracking variable:
An integer variable `TOP` tracks the subscript of the most recently inserted element. When empty, `TOP = -1` (or 0); upon each push operation, `TOP` increments and stores the item; upon pop, the item at `TOP` is returned and `TOP` decrements.

Index LIMIT - 1 \dots 210 Array Slot Empty \dots Item C \leftarrow TOP Item B Item A

Mentor tip

Show a 1D array indexed 0 to LIMIT-1 with a pointer variable TOP referencing the topmost valid element.

(b)

4 marks

Step 1 - Initialize intermediate FIFO queue:
Create an empty queue `Q`.
Step 2 - Transfer elements from Stack to Queue:
While `STACK.isEmpty()` is false: pop an element from `STACK` and enqueue it into `Q`. Because the stack pops in reverse order (LIFO), elements enter the queue in reverse order.
Step 3 - Transfer elements from Queue back to Stack:
While `Q.isEmpty()` is false: dequeue an element from `Q` and push it onto `STACK`. Because the queue preserves insertion order (FIFO), elements are pushed back in that reversed sequence.
Result:
The elements in the stack are now completely reversed.

Mentor tip

Pop all elements from stack into a queue, then dequeue all elements from queue and push them back onto the stack.

Questions 7–11 · 75 marks

Question 7

15 marks

System implementation, prototyping, data migration, and concurrency

System fundamentals
Resource management

(a)

5 marks

Identify user classes (Part i - 2 marks):
1. Shop sales assistants and administrative office personnel (operational staff).
2. The shop manager/business owner (supervisory management), or external purchasing customers.
Explain role of users in system development (Part ii - 3 marks):
1. Requirement specification: Users provide detailed operational knowledge during initial interviews to define required functional workflows.
2. Iterative feedback: Users evaluate early prototypes to identify usability issues and omissions prior to system finalization.
3. Operational sign-off testing: Users conduct comprehensive user operational sign-off testing (UAT) with real-world scenarios to confirm the software satisfies business needs.

Mentor tip

For part (i) distinguish two user roles (clerks, office staff, manager). For part (ii) explain user involvement in requirements elicitation, prototype reviews, and operational sign-off testing.

(b)

2 marks

Explain benefits of multiple prototypes:
Presenting multiple alternative prototypes allows stakeholders to evaluate different user interface designs and interaction patterns before choosing the most intuitive option, reducing costly post-deployment redesigns.

Mentor tip

Developing multiple prototypes allows comparison of design alternatives and yields superior ergonomics.

(c)

4 marks

Problem 1 - Data schema and format incompatibilities:
The legacy database may store records in incompatible field types, delimiter schemes, or character encodings, requiring complex transformation scripts that risk truncating or corrupting customer records.
Problem 2 - Duplicate or conflicting primary keys:
Legacy records may contain inconsistent, missing, or non-unique identifier keys, creating data integrity errors and primary key collisions during import into the new relational system.

Mentor tip

Explain two concrete data migration issues such as schema/data type mismatches, non-unique/colliding primary keys, or data encoding discrepancies.

(d)

4 marks

Simultaneous multi-user access mechanism (Part i - 2 marks):
The centralized server runs a multi-threaded, multi-user operating system and database management system (DBMS) that allocates concurrent worker threads and uses processor time-slicing to process requests simultaneously.
Data security and isolation between concurrent users (Part ii - 2 marks):
The operating system enforces strict memory isolation between process address spaces in RAM, while the DBMS uses transaction concurrency controls (row-level locking and ACID transaction boundaries) to prevent concurrent writes from overwriting or corrupting another user's session data.

Mentor tip

Explain multi-threading/time-sharing for concurrent access, and process memory protection / database row-level locking for data isolation.

Question 8

15 marks

Computer architecture, registers, operating systems, and CPU structure

Computer organisation

(a)

1 mark

Calculate bit capacity per memory location:
Memory contents are given in hexadecimal, with 8 hex digits per word (e.g., `00EF1079`).
Each hexadecimal character represents 4 bits:
$8 hex digits \times 4 bits/digit = 32 bits$ .

8 \times 4 = 32 bits

Mentor tip

Each hex character is 4 bits; an 8-character hex word equals 32 bits.

(b)

1 mark

Calculate byte size per address:
Addresses are given as 4 hex digits (e.g., `1000`, `1001`).
$4 hex digits \times 4 bits = 16 bits = 2 bytes$ .

\frac{4 \times 4 bits}{8 bits/byte} = 2 bytes

Mentor tip

4 hex digits = 16 bits = 2 bytes.

(c)

4 marks

Function of the Memory Address Register (MAR) (Part i - 2 marks):
The MAR is a hardware register in the CPU that holds the physical address in RAM currently being accessed for a memory read or write operation.
Function of the Memory Data Register (MDR) (Part ii - 2 marks):
The MDR (or Memory Buffer Register) is a CPU register that temporarily holds the actual data or machine instruction read from RAM, or data waiting to be written to RAM.

Mentor tip

Clearly contrast the MAR (holds memory addresses) with the MDR (holds the data/instructions read from or written to memory).

(d)

3 marks

Two core functions of the Operating System (Part i - 2 marks):
1. Memory management: Allocating and tracking physical RAM partitions across active applications and virtual memory swapping.
2. Process and processor scheduling: Managing CPU resource sharing and handling hardware/software interrupts.
Storage location of the OS when powered off (Part ii - 1 mark):
Secondary storage: non-volatile hard disk drive (HDD), solid-state drive (SSD), or read-only memory (ROM).

Mentor tip

List two OS duties (memory management, scheduling, file system) and state that it resides on secondary storage (hard drive/SSD/ROM) when powered down.

(e)

6 marks

CPU internal structure and data paths (Part i - 4 marks):
The central processing unit consists of three primary functional blocks connected by internal buses:
1. Control Unit (CU): Decodes instructions and issues clock synchronization signals.
2. Arithmetic and Logic Unit (ALU): Executes arithmetic operations and Boolean logic comparisons.
3. Registers (Program Counter, MAR, MDR, Accumulator): Provide high-speed temporary storage during execution.
Data flows from RAM into the MDR across the data bus, moves into the ALU/Accumulator for arithmetic execution, and is steered by the CU.
Component performing instruction decoding (Part ii - 1 mark):
Control Unit (CU).
Component executing instructions (Part iii - 1 mark):
Arithmetic and Logic Unit (ALU).

Mentor tip

Diagram/describe the CU, ALU, registers, and data bus paths. Name the CU for decoding and ALU for execution.

Question 9

15 marks

Automated door control systems, interrupts, and industrial automation

Control systems
Computer organisation

(a)

5 marks

Sensor identification (Part i - 1 mark):
Passive infrared (PIR) motion sensor, radar microwave sensor, or optical break-beam sensor.
Non-sensor hardware components (Part ii - 1 mark):
Actuator (motor drive) or Analog-to-Digital Converter (ADC) / Microprocessor.
Control sequence when a person approaches (Part iii - 3 marks):
1. The motion sensor detects changes in infrared radiation and sends an analog signal.
2. The ADC converts the signal to digital form and transmits it to the microprocessor.
3. The microprocessor evaluates the input and activates the motor actuator to slide open the doors.
4. A timer counts down; once no movement is detected for a preset period, the actuator is signaled to close the doors.

Mentor tip

Identify a sensor (motion/infrared), a non-sensor component (actuator/motor), and trace the sense-process-actuate sequence.

(b)

4 marks

Define interrupt (Part i - 2 marks):
An interrupt is an asynchronous signal sent by hardware or software to the CPU indicating an event that requires immediate processing, causing the CPU to suspend its current execution cycle and invoke an Interrupt Service Routine (ISR).
Describe an interrupt scenario in the automatic door system (Part ii - 2 marks):
If a person or obstruction is detected by an emergency optical sensor while the door is currently closing, a high-priority hardware interrupt immediately halts the motor and reverses the door direction to prevent injury.

Mentor tip

Define an interrupt as an urgent signal causing execution suspension, and illustrate with an obstruction detected during closing.

(c)

6 marks

Economic and productivity advantages:
1. Operating costs and endurance: Automated systems operate 24/7 without fatigue, taking breaks, or requiring overtime pay, leading to significantly higher production throughput and long-term cost savings.
2. Precision and safety: Robots execute repetitive tasks with high precision and consistency, and handle hazardous tasks (toxic fumes, extreme temperatures, heavy lifting) without endangering human lives.
Social and economic disadvantages:
1. Capital expenditure: Designing, procuring, and installing industrial control systems requires substantial upfront capital investment and ongoing specialist maintenance.
2. Workforce disruption: Automation causes structural unemployment for manual laborers, necessitating retraining programs and causing social dislocation.

Mentor tip

Discuss 3 balanced pairs: economic advantages (24/7 productivity, accuracy, hazardous safety) versus drawbacks (high initial cost, technological redundancy, maintenance dependencies).

Question 10

15 marks

Binary search trees, linked lists, and recursive function evaluation

Abstract data structures
Computational thinking, problem-solving and programming

(a)

3 marks

Insert student scores in order of arrival:
Input sequence: Ann Taylor (10), Boris Penn (18), Ivan Troth (8), Peter Hu (9), Mary Looty (7).
1. Root node: Ann Taylor (10).
2. Insert Boris (18): $18 > 10 ⟹$ Right child of Ann (10).
3. Insert Ivan (8): $8 < 10 ⟹$ Left child of Ann (10).
4. Insert Peter (9): $9 < 10 ⟹$ Left of Ann; $9 > 8 ⟹$ Right child of Ivan (8).
5. Insert Mary (7): $7 < 10 ⟹$ Left of Ann; $7 < 8 ⟹$ Left child of Ivan (8).

Ann (10) ↙ ↘ Ivan (8) Boris (18) ↙ ↘ Mary (7) Peter (9)

Mentor tip

Root is Ann (10); right child is Boris (18); left child is Ivan (8) with left child Mary (7) and right child Peter (9).

(b)

3 marks

Organize data in descending score order:
Sorted scores: Boris Penn (18) $\to$ Ann Taylor (10) $\to$ Peter Hu (9) $\to$ Ivan Troth (8) $\to$ Mary Looty (7).
Construct singly linked list representation:
Head pointer $\to$ [Boris Penn, 18 | next] $\to$ [Ann Taylor, 10 | next] $\to$ [Peter Hu, 9 | next] $\to$ [Ivan Troth, 8 | next] $\to$ [Mary Looty, 7 | null].

Head \to [Boris, 18] \to [Ann, 10] \to [Peter, 9] \to [Ivan, 8] \to [Mary, 7] \to null

Mentor tip

Ensure descending order: Boris (18), Ann (10), Peter (9), Ivan (8), Mary (7) ending in null.

(c)

4 marks

Search performance comparison (Part i - 2 marks):
A balanced binary search tree provides binary division of the search space, giving an average search time complexity of $O (lo g n)$ . A singly linked list requires linear sequential traversal with $O (n)$ search time complexity.
Storage requirement comparison (Part ii - 2 marks):
Each node in a binary tree requires two memory pointers (left child and right child), consuming more memory overhead per node than a singly linked list, which only requires a single pointer field to the next node.

Mentor tip

Searching: BST is $O (lo g n)$ binary search vs linked list $O (n)$ linear scan. Storage: BST needs two pointers per node vs one pointer in a singly linked list.

(d)

5 marks

Trace recursive function calls for $F (5, 11)$ :
Definition:
If $X < Y$ : return $F (X + 1, Y - 2)$ .
Else if $X = Y$ : return $2 \times F (X + 2, Y - 2) - 2$ .
Else: return $2 X + 4 Y$ .
Call 1: $F (5, 11)$ . Since $5 < 11$ , evaluate $F (5 + 1, 11 - 2) = F (6, 9)$ .
Call 2: $F (6, 9)$ . Since $6 < 9$ , evaluate $F (6 + 1, 9 - 2) = F (7, 7)$ .
Call 3: $F (7, 7)$ . Since $X = Y = 7$ , evaluate $2 \times F (7 + 2, 7 - 2) - 2 = 2 \times F (9, 5) - 2$ .
Call 4: $F (9, 5)$ . Since $9 > 5$ , base case executes: $2 (9) + 4 (5) = 18 + 20 = 38$ .
Backtrack and resolve:
$F (7, 7) = 2 \times (38) - 2 = 76 - 2 = 74$ .
Therefore, $F (6, 9) = 74$ and $F (5, 11) = 74$ .

F (5, 11) = F (6, 9) = F (7, 7) = 2 \times F (9, 5) - 2 = 2 \times [2 (9) + 4 (5)] - 2 = 2 \times [18 + 20] - 2 = 2 \times 38 - 2 = 76 - 2 = 74

Mentor tip

Carefully trace each recursive branch: F(5,11) -> F(6,9) -> F(7,7) -> 2*F(9,5) - 2. With F(9,5) = 38, result is 74.

Question 11

15 marks

Population modeling, migration matrix queries, and long-term simulation

Computational thinking, problem-solving and programming

(a)

3 marks

Formulate summation algorithm:
Initialize accumulator variable: TOTAL_POP = 0.
Iterate through all seven region indices:
Loop K from 0 to 6:
TOTAL_POP = TOTAL_POP + Curr_Pop[K]
End loop.
Output total population: output TOTAL_POP.

TOTAL_POP = K = 0 \sum 6 Curr_Pop [K]

Mentor tip

Accumulate elements of Curr_Pop from index 0 through 6 into a running total and output the result.

(b)

5 marks

Percentage of Region G moving to Region A (Part i - 1 mark):
Consult migration table at row G, column A: 0.05%.
Number of people moving from Region B to Region E (Part ii - 1 mark):
Consult migration table at row B, column E: 0.20%.
Population of region B is 2.1 million.
Calculate count: $2.1 million \times \frac{0.20}{100} = 0.0042 million = 4, 200 people$ .
Annual population change formula for Region F (Part iii - 3 marks):
1. Internal natural growth: $Curr_Pop [F] \times \frac{Table [ F ] [ F ]}{100}$ .
2. Total immigration: Sum of inward migration from all other regions $J \neq = F$ : $\sum_{J \neq = F} (Curr_Pop [J] \times \frac{Table [ J ] [ F ]}{100})$ .
3. Total emigration: Sum of outward migration from region $F$ to all other regions $J \neq = F$ : $\sum_{J \neq = F} (Curr_Pop [F] \times \frac{Table [ F ] [ J ]}{100})$ .
Net change in Region F = (Internal growth) + (Total immigration) - (Total emigration).

Inward Outward Δ Pop [F] = J \neq = F \sum (Curr_Pop [J] \times \frac{Table [ J ] [ F ]}{100}) = J \neq = F \sum (Curr_Pop [F] \times \frac{Table [ F ] [ J ]}{100}) = (Curr_Pop [F] \times \frac{Table [ F ] [ F ]}{100}) + Inward - Outward

Mentor tip

Row is 'From' and column is 'To'. Distinguish between internal growth, immigration into F, and emigration out of F.

(c)

7 marks

Construct 10-year simulation algorithm:
1. Outer loop over 10 annual simulation steps: loop YEAR from 1 to 10.
2. Initialize next-year population array: for each region K from 0 to 6, set Future_Pop[K] = 0.
3. For each region K from 0 to 6:
- Add internal growth to current region: Future_Pop[K] = Curr_Pop[K] * (Table[K][K] / 100).
- Loop across all other regions J from 0 to 6:
If J != K then:
// Inflow from J to K
Future_Pop[K] = Future_Pop[K] + (Curr_Pop[J] * (Table[J][K] / 100))
// Outflow from K to J
Future_Pop[K] = Future_Pop[K] - (Curr_Pop[K] * (Table[K][J] / 100))
End if.
4. Update current population array for next year's cycle: for each region K from 0 to 6, set Curr_Pop[K] = Curr_Pop[K] + Future_Pop[K].
5. End 10-year outer loop.
6. Output final values in Curr_Pop.

Mentor tip

Use an outer 10-year loop, zero an intermediate Future_Pop array each cycle, apply internal growth plus inflows minus outflows, and update Curr_Pop before the next annual iteration.

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 programming25 marks · 25%
Q5, Q10, Q11
Computer organisation22 marks · 22%
Q3, Q8, Q9
Abstract data structures16 marks · 16%
Q6, Q10
System fundamentals14 marks · 14%
Q1, Q7
Control systems11 marks · 11%
Q9
Networks8 marks · 8%
Q2, Q4
Resource management4 marks · 4%
Q7

Mentor insights

Traps to watch for

01
NOR logic operation condition · Q3
A NOR operation evaluates to 1 only when BOTH operands are 0. If C = 1, the output is guaranteed to be 0 regardless of A and B.
02
Infinite loop caused by missing digit reduction · Q5
Without N = N div 10 inside the loop body, N remains constant, causing the while loop condition to remain true and never terminate.
03
Recursive evaluation branch tracking in F(X, Y) · Q10
In the recursion for F(5, 11), step 3 reaches F(7, 7) where X = Y, triggering the branch 2 * F(9, 5) - 2. Forgetting to apply the multiplier 2 and subtract 2 leads to an incorrect evaluation.
04
Migration matrix orientation and net population change · Q11
The migration matrix specifies 'From' on rows and 'To' on columns. Inward migration sums down columns, while outward migration sums across rows.

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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: November 2015
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 11

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2015 global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions 1–6 · 25 marks

Question 1

3 marks

System usability and accessibility methods

System fundamentals

(a)

1 mark

Define usability in human-computer interaction:
Usability refers to the extent to which a computer system can be used by specified users to achieve specified goals effectively, efficiently, and satisfactorily in a specified context of use.

Mentor tip

Define usability in terms of ease of use, learnability, efficiency, and meeting user requirements.

(b)

2 marks

Identify two accessibility enhancement features:
1. Voice synthesis and screen readers (text-to-speech) to accommodate visually impaired users.
2. Speech recognition software (voice input control) to assist users with motor impairments or physical disabilities in operating the system hands-free.

Mentor tip

Name two specific assistive technologies such as screen readers, voice recognition, high-contrast display modes, or braille interfaces.

Question 2

4 marks

VPN technologies for secure remote access

Networks

(a)

4 marks

Explain tunneling protocols:
1. Tunneling (such as L2TP or OpenVPN) encapsulates the original private IP packets inside standard public IP packets, establishing a virtual conduit across the public internet.
Explain encryption protocols:
2. Strong cryptographic algorithms and security protocols (such as IPsec or SSL/TLS) encrypt packet contents. If intercepted en route, the data remains undecipherable ciphertext.
Explain VPN gateway authentication:
3. A secure enterprise VPN gateway authenticates the traveling salesperson's digital certificates and login credentials, routing traffic seamlessly into the corporate intranet.

Mentor tip

Mention tunneling protocols (packet encapsulation), encryption protocols (such as IPsec/SSL to prevent eavesdropping), and secure gateway authentication.

Question 3

3 marks

Truth table for compound Boolean expression

Computer organisation

(a)

3 marks

Deconstruct the Boolean expression:
Expression: $(A \land B) NOR C$ .
Compute intermediate term $X = A \land B$ :
$X$ is 1 only when both $A = 1$ and $B = 1$ ; otherwise $X = 0$ .
Apply the NOR operation with $C$ :
Output is $\neg (X \lor C)$ . The output is 1 only when BOTH $X = 0$ AND $C = 0$ :
1. For $A = 0, B = 0, C = 0$ : $X = 0, C = 0 ⟹ 0 NOR 0 = 1$ .
2. For $A = 0, B = 0, C = 1$ : $X = 0, C = 1 ⟹ 0 NOR 1 = 0$ .
3. For $A = 0, B = 1, C = 0$ : $X = 0, C = 0 ⟹ 0 NOR 0 = 1$ .
4. For $A = 0, B = 1, C = 1$ : $X = 0, C = 1 ⟹ 0 NOR 1 = 0$ .
5. For $A = 1, B = 0, C = 0$ : $X = 0, C = 0 ⟹ 0 NOR 0 = 1$ .
6. For $A = 1, B = 0, C = 1$ : $X = 0, C = 1 ⟹ 0 NOR 1 = 0$ .
7. For $A = 1, B = 1, C = 0$ : $X = 1, C = 0 ⟹ 1 NOR 0 = 0$ .
8. For $A = 1, B = 1, C = 1$ : $X = 1, C = 1 ⟹ 1 NOR 1 = 0$ .

A 00001111 B 00110011 C 01010101 A \land B 00000011 (A \land B) NOR C 10101000

Mentor tip

Notice that NOR produces 1 only when both operands are 0. Whenever C = 1, the result is immediately 0.

Question 4

4 marks

Client-server network architecture and security threats

Networks

(a)

1 mark

Define a server in a network context:
A server is a central dedicated computer or software program that manages access to centralized resources, services, files, and applications shared among network clients.

Mentor tip

Highlight that a server provides shared central resources, files, or applications to client devices.

(b)

1 mark

Identify network clients in the school setup:
The client devices comprise the individual workstation desktop computers, laptops, and network printer accessing server resources.

Mentor tip

Name the user workstation computers or printers requesting services from the server.

(c)

2 marks

Identify an external security threat:
External cyber-attacks such as malicious malware/ransomware infection or unauthorized remote intrusion by hackers.
State a specific defensive countermeasure:
Deploy an active network firewall with continuous packet inspection and install regularly updated anti-malware software across all connected endpoints.

Mentor tip

Pair a specific digital threat (malware/viruses or unauthorized hacker intrusion) with its direct defense (firewall or antivirus software).

Question 5

5 marks

Algorithm debugging for digit extraction

Computational thinking, problem-solving and programming

(a)

2 marks

Trace algorithm execution:
In the provided algorithm, the loop checks the condition `(N > 0) and (EVEN = true)`.
Identify the logic defect:
The value of $N$ is never modified or reduced inside the loop body. As a result, the loop repeatedly inspects the same least significant digit `(N mod 10)` indefinitely, resulting in an infinite loop whenever the first examined digit is even.

Mentor tip

Explain that N is never updated inside the loop, causing an infinite loop because the condition remains true.

(b)

3 marks

Formulate algorithm modification:
To process successive digits from right to left, the integer value of $N$ must be truncated after each digit examination.
Identify exact assignment and placement:
Add the integer division statement `N = N div 10` inside the loop body, positioned after the `if` statement block so that each iteration shifts to the next higher-order digit until $N$ reaches 0.

N = N div 10

Mentor tip

Specify adding `N = N div 10` inside the loop body immediately following the if condition.

Question 6

6 marks

Stack representation using arrays and reversal using a queue

Abstract data structures

(a)

2 marks

Describe array-based stack representation:
A static one-dimensional array of fixed size `LIMIT` stores the stack elements from index 0 up to `LIMIT - 1`.
Explain pointer tracking variable:
An integer variable `TOP` tracks the subscript of the most recently inserted element. When empty, `TOP = -1` (or 0); upon each push operation, `TOP` increments and stores the item; upon pop, the item at `TOP` is returned and `TOP` decrements.

Index LIMIT - 1 \dots 210 Array Slot Empty \dots Item C \leftarrow TOP Item B Item A

Mentor tip

Show a 1D array indexed 0 to LIMIT-1 with a pointer variable TOP referencing the topmost valid element.

(b)

4 marks

Step 1 - Initialize intermediate FIFO queue:
Create an empty queue `Q`.
Step 2 - Transfer elements from Stack to Queue:
While `STACK.isEmpty()` is false: pop an element from `STACK` and enqueue it into `Q`. Because the stack pops in reverse order (LIFO), elements enter the queue in reverse order.
Step 3 - Transfer elements from Queue back to Stack:
While `Q.isEmpty()` is false: dequeue an element from `Q` and push it onto `STACK`. Because the queue preserves insertion order (FIFO), elements are pushed back in that reversed sequence.
Result:
The elements in the stack are now completely reversed.

Mentor tip

Pop all elements from stack into a queue, then dequeue all elements from queue and push them back onto the stack.

Questions 7–11 · 75 marks

Question 7

15 marks

System implementation, prototyping, data migration, and concurrency

System fundamentals
Resource management

(a)

5 marks

Identify user classes (Part i - 2 marks):
1. Shop sales assistants and administrative office personnel (operational staff).
2. The shop manager/business owner (supervisory management), or external purchasing customers.
Explain role of users in system development (Part ii - 3 marks):
1. Requirement specification: Users provide detailed operational knowledge during initial interviews to define required functional workflows.
2. Iterative feedback: Users evaluate early prototypes to identify usability issues and omissions prior to system finalization.
3. Operational sign-off testing: Users conduct comprehensive user operational sign-off testing (UAT) with real-world scenarios to confirm the software satisfies business needs.

Mentor tip

For part (i) distinguish two user roles (clerks, office staff, manager). For part (ii) explain user involvement in requirements elicitation, prototype reviews, and operational sign-off testing.

(b)

2 marks

Explain benefits of multiple prototypes:
Presenting multiple alternative prototypes allows stakeholders to evaluate different user interface designs and interaction patterns before choosing the most intuitive option, reducing costly post-deployment redesigns.

Mentor tip

Developing multiple prototypes allows comparison of design alternatives and yields superior ergonomics.

(c)

4 marks

Problem 1 - Data schema and format incompatibilities:
The legacy database may store records in incompatible field types, delimiter schemes, or character encodings, requiring complex transformation scripts that risk truncating or corrupting customer records.
Problem 2 - Duplicate or conflicting primary keys:
Legacy records may contain inconsistent, missing, or non-unique identifier keys, creating data integrity errors and primary key collisions during import into the new relational system.

Mentor tip

Explain two concrete data migration issues such as schema/data type mismatches, non-unique/colliding primary keys, or data encoding discrepancies.

(d)

4 marks

Simultaneous multi-user access mechanism (Part i - 2 marks):
The centralized server runs a multi-threaded, multi-user operating system and database management system (DBMS) that allocates concurrent worker threads and uses processor time-slicing to process requests simultaneously.
Data security and isolation between concurrent users (Part ii - 2 marks):
The operating system enforces strict memory isolation between process address spaces in RAM, while the DBMS uses transaction concurrency controls (row-level locking and ACID transaction boundaries) to prevent concurrent writes from overwriting or corrupting another user's session data.

Mentor tip

Explain multi-threading/time-sharing for concurrent access, and process memory protection / database row-level locking for data isolation.

Question 8

15 marks

Computer architecture, registers, operating systems, and CPU structure

Computer organisation

(a)

1 mark

Calculate bit capacity per memory location:
Memory contents are given in hexadecimal, with 8 hex digits per word (e.g., `00EF1079`).
Each hexadecimal character represents 4 bits:
$8 hex digits \times 4 bits/digit = 32 bits$ .

8 \times 4 = 32 bits

Mentor tip

Each hex character is 4 bits; an 8-character hex word equals 32 bits.

(b)

1 mark

Calculate byte size per address:
Addresses are given as 4 hex digits (e.g., `1000`, `1001`).
$4 hex digits \times 4 bits = 16 bits = 2 bytes$ .

\frac{4 \times 4 bits}{8 bits/byte} = 2 bytes

Mentor tip

4 hex digits = 16 bits = 2 bytes.

(c)

4 marks

Function of the Memory Address Register (MAR) (Part i - 2 marks):
The MAR is a hardware register in the CPU that holds the physical address in RAM currently being accessed for a memory read or write operation.
Function of the Memory Data Register (MDR) (Part ii - 2 marks):
The MDR (or Memory Buffer Register) is a CPU register that temporarily holds the actual data or machine instruction read from RAM, or data waiting to be written to RAM.

Mentor tip

Clearly contrast the MAR (holds memory addresses) with the MDR (holds the data/instructions read from or written to memory).

(d)

3 marks

Two core functions of the Operating System (Part i - 2 marks):
1. Memory management: Allocating and tracking physical RAM partitions across active applications and virtual memory swapping.
2. Process and processor scheduling: Managing CPU resource sharing and handling hardware/software interrupts.
Storage location of the OS when powered off (Part ii - 1 mark):
Secondary storage: non-volatile hard disk drive (HDD), solid-state drive (SSD), or read-only memory (ROM).

Mentor tip

List two OS duties (memory management, scheduling, file system) and state that it resides on secondary storage (hard drive/SSD/ROM) when powered down.

(e)

6 marks

CPU internal structure and data paths (Part i - 4 marks):
The central processing unit consists of three primary functional blocks connected by internal buses:
1. Control Unit (CU): Decodes instructions and issues clock synchronization signals.
2. Arithmetic and Logic Unit (ALU): Executes arithmetic operations and Boolean logic comparisons.
3. Registers (Program Counter, MAR, MDR, Accumulator): Provide high-speed temporary storage during execution.
Data flows from RAM into the MDR across the data bus, moves into the ALU/Accumulator for arithmetic execution, and is steered by the CU.
Component performing instruction decoding (Part ii - 1 mark):
Control Unit (CU).
Component executing instructions (Part iii - 1 mark):
Arithmetic and Logic Unit (ALU).

Mentor tip

Diagram/describe the CU, ALU, registers, and data bus paths. Name the CU for decoding and ALU for execution.

Question 9

15 marks

Automated door control systems, interrupts, and industrial automation

Control systems
Computer organisation

(a)

5 marks

Sensor identification (Part i - 1 mark):
Passive infrared (PIR) motion sensor, radar microwave sensor, or optical break-beam sensor.
Non-sensor hardware components (Part ii - 1 mark):
Actuator (motor drive) or Analog-to-Digital Converter (ADC) / Microprocessor.
Control sequence when a person approaches (Part iii - 3 marks):
1. The motion sensor detects changes in infrared radiation and sends an analog signal.
2. The ADC converts the signal to digital form and transmits it to the microprocessor.
3. The microprocessor evaluates the input and activates the motor actuator to slide open the doors.
4. A timer counts down; once no movement is detected for a preset period, the actuator is signaled to close the doors.

Mentor tip

Identify a sensor (motion/infrared), a non-sensor component (actuator/motor), and trace the sense-process-actuate sequence.

(b)

4 marks

Define interrupt (Part i - 2 marks):
An interrupt is an asynchronous signal sent by hardware or software to the CPU indicating an event that requires immediate processing, causing the CPU to suspend its current execution cycle and invoke an Interrupt Service Routine (ISR).
Describe an interrupt scenario in the automatic door system (Part ii - 2 marks):
If a person or obstruction is detected by an emergency optical sensor while the door is currently closing, a high-priority hardware interrupt immediately halts the motor and reverses the door direction to prevent injury.

Mentor tip

Define an interrupt as an urgent signal causing execution suspension, and illustrate with an obstruction detected during closing.

(c)

6 marks

Economic and productivity advantages:
1. Operating costs and endurance: Automated systems operate 24/7 without fatigue, taking breaks, or requiring overtime pay, leading to significantly higher production throughput and long-term cost savings.
2. Precision and safety: Robots execute repetitive tasks with high precision and consistency, and handle hazardous tasks (toxic fumes, extreme temperatures, heavy lifting) without endangering human lives.
Social and economic disadvantages:
1. Capital expenditure: Designing, procuring, and installing industrial control systems requires substantial upfront capital investment and ongoing specialist maintenance.
2. Workforce disruption: Automation causes structural unemployment for manual laborers, necessitating retraining programs and causing social dislocation.

Mentor tip

Discuss 3 balanced pairs: economic advantages (24/7 productivity, accuracy, hazardous safety) versus drawbacks (high initial cost, technological redundancy, maintenance dependencies).

Question 10

15 marks

Binary search trees, linked lists, and recursive function evaluation

Abstract data structures
Computational thinking, problem-solving and programming

(a)

3 marks

Insert student scores in order of arrival:
Input sequence: Ann Taylor (10), Boris Penn (18), Ivan Troth (8), Peter Hu (9), Mary Looty (7).
1. Root node: Ann Taylor (10).
2. Insert Boris (18): $18 > 10 ⟹$ Right child of Ann (10).
3. Insert Ivan (8): $8 < 10 ⟹$ Left child of Ann (10).
4. Insert Peter (9): $9 < 10 ⟹$ Left of Ann; $9 > 8 ⟹$ Right child of Ivan (8).
5. Insert Mary (7): $7 < 10 ⟹$ Left of Ann; $7 < 8 ⟹$ Left child of Ivan (8).

Ann (10) ↙ ↘ Ivan (8) Boris (18) ↙ ↘ Mary (7) Peter (9)

Mentor tip

Root is Ann (10); right child is Boris (18); left child is Ivan (8) with left child Mary (7) and right child Peter (9).

(b)

3 marks

Organize data in descending score order:
Sorted scores: Boris Penn (18) $\to$ Ann Taylor (10) $\to$ Peter Hu (9) $\to$ Ivan Troth (8) $\to$ Mary Looty (7).
Construct singly linked list representation:
Head pointer $\to$ [Boris Penn, 18 | next] $\to$ [Ann Taylor, 10 | next] $\to$ [Peter Hu, 9 | next] $\to$ [Ivan Troth, 8 | next] $\to$ [Mary Looty, 7 | null].

Head \to [Boris, 18] \to [Ann, 10] \to [Peter, 9] \to [Ivan, 8] \to [Mary, 7] \to null

Mentor tip

Ensure descending order: Boris (18), Ann (10), Peter (9), Ivan (8), Mary (7) ending in null.

(c)

4 marks

Search performance comparison (Part i - 2 marks):
A balanced binary search tree provides binary division of the search space, giving an average search time complexity of $O (lo g n)$ . A singly linked list requires linear sequential traversal with $O (n)$ search time complexity.
Storage requirement comparison (Part ii - 2 marks):
Each node in a binary tree requires two memory pointers (left child and right child), consuming more memory overhead per node than a singly linked list, which only requires a single pointer field to the next node.

Mentor tip

Searching: BST is $O (lo g n)$ binary search vs linked list $O (n)$ linear scan. Storage: BST needs two pointers per node vs one pointer in a singly linked list.

(d)

5 marks

Trace recursive function calls for $F (5, 11)$ :
Definition:
If $X < Y$ : return $F (X + 1, Y - 2)$ .
Else if $X = Y$ : return $2 \times F (X + 2, Y - 2) - 2$ .
Else: return $2 X + 4 Y$ .
Call 1: $F (5, 11)$ . Since $5 < 11$ , evaluate $F (5 + 1, 11 - 2) = F (6, 9)$ .
Call 2: $F (6, 9)$ . Since $6 < 9$ , evaluate $F (6 + 1, 9 - 2) = F (7, 7)$ .
Call 3: $F (7, 7)$ . Since $X = Y = 7$ , evaluate $2 \times F (7 + 2, 7 - 2) - 2 = 2 \times F (9, 5) - 2$ .
Call 4: $F (9, 5)$ . Since $9 > 5$ , base case executes: $2 (9) + 4 (5) = 18 + 20 = 38$ .
Backtrack and resolve:
$F (7, 7) = 2 \times (38) - 2 = 76 - 2 = 74$ .
Therefore, $F (6, 9) = 74$ and $F (5, 11) = 74$ .

F (5, 11) = F (6, 9) = F (7, 7) = 2 \times F (9, 5) - 2 = 2 \times [2 (9) + 4 (5)] - 2 = 2 \times [18 + 20] - 2 = 2 \times 38 - 2 = 76 - 2 = 74

Mentor tip

Carefully trace each recursive branch: F(5,11) -> F(6,9) -> F(7,7) -> 2*F(9,5) - 2. With F(9,5) = 38, result is 74.

Question 11

15 marks

Population modeling, migration matrix queries, and long-term simulation

Computational thinking, problem-solving and programming

(a)

3 marks

Formulate summation algorithm:
Initialize accumulator variable: TOTAL_POP = 0.
Iterate through all seven region indices:
Loop K from 0 to 6:
TOTAL_POP = TOTAL_POP + Curr_Pop[K]
End loop.
Output total population: output TOTAL_POP.

TOTAL_POP = K = 0 \sum 6 Curr_Pop [K]

Mentor tip

Accumulate elements of Curr_Pop from index 0 through 6 into a running total and output the result.

(b)

5 marks

Percentage of Region G moving to Region A (Part i - 1 mark):
Consult migration table at row G, column A: 0.05%.
Number of people moving from Region B to Region E (Part ii - 1 mark):
Consult migration table at row B, column E: 0.20%.
Population of region B is 2.1 million.
Calculate count: $2.1 million \times \frac{0.20}{100} = 0.0042 million = 4, 200 people$ .
Annual population change formula for Region F (Part iii - 3 marks):
1. Internal natural growth: $Curr_Pop [F] \times \frac{Table [ F ] [ F ]}{100}$ .
2. Total immigration: Sum of inward migration from all other regions $J \neq = F$ : $\sum_{J \neq = F} (Curr_Pop [J] \times \frac{Table [ J ] [ F ]}{100})$ .
3. Total emigration: Sum of outward migration from region $F$ to all other regions $J \neq = F$ : $\sum_{J \neq = F} (Curr_Pop [F] \times \frac{Table [ F ] [ J ]}{100})$ .
Net change in Region F = (Internal growth) + (Total immigration) - (Total emigration).

Inward Outward Δ Pop [F] = J \neq = F \sum (Curr_Pop [J] \times \frac{Table [ J ] [ F ]}{100}) = J \neq = F \sum (Curr_Pop [F] \times \frac{Table [ F ] [ J ]}{100}) = (Curr_Pop [F] \times \frac{Table [ F ] [ F ]}{100}) + Inward - Outward

Mentor tip

Row is 'From' and column is 'To'. Distinguish between internal growth, immigration into F, and emigration out of F.

(c)

7 marks

Construct 10-year simulation algorithm:
1. Outer loop over 10 annual simulation steps: loop YEAR from 1 to 10.
2. Initialize next-year population array: for each region K from 0 to 6, set Future_Pop[K] = 0.
3. For each region K from 0 to 6:
- Add internal growth to current region: Future_Pop[K] = Curr_Pop[K] * (Table[K][K] / 100).
- Loop across all other regions J from 0 to 6:
If J != K then:
// Inflow from J to K
Future_Pop[K] = Future_Pop[K] + (Curr_Pop[J] * (Table[J][K] / 100))
// Outflow from K to J
Future_Pop[K] = Future_Pop[K] - (Curr_Pop[K] * (Table[K][J] / 100))
End if.
4. Update current population array for next year's cycle: for each region K from 0 to 6, set Curr_Pop[K] = Curr_Pop[K] + Future_Pop[K].
5. End 10-year outer loop.
6. Output final values in Curr_Pop.

Mentor tip

Use an outer 10-year loop, zero an intermediate Future_Pop array each cycle, apply internal growth plus inflows minus outflows, and update Curr_Pop before the next annual iteration.

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 programming25 marks · 25%
Q5, Q10, Q11
Computer organisation22 marks · 22%
Q3, Q8, Q9
Abstract data structures16 marks · 16%
Q6, Q10
System fundamentals14 marks · 14%
Q1, Q7
Control systems11 marks · 11%
Q9
Networks8 marks · 8%
Q2, Q4
Resource management4 marks · 4%
Q7

Mentor insights

Traps to watch for

01
NOR logic operation condition · Q3
A NOR operation evaluates to 1 only when BOTH operands are 0. If C = 1, the output is guaranteed to be 0 regardless of A and B.
02
Infinite loop caused by missing digit reduction · Q5
Without N = N div 10 inside the loop body, N remains constant, causing the while loop condition to remain true and never terminate.
03
Recursive evaluation branch tracking in F(X, Y) · Q10
In the recursion for F(5, 11), step 3 reaches F(7, 7) where X = Y, triggering the branch 2 * F(9, 5) - 2. Forgetting to apply the multiplier 2 and subtract 2 leads to an incorrect evaluation.
04
Migration matrix orientation and net population change · Q11
The migration matrix specifies 'From' on rows and 'To' on columns. Inward migration sums down columns, while outward migration sums across rows.

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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