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Worked solutions

IB Computer Science HL · Paper 1

Computer Science, Higher Level

Session: May 2015
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 13

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

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

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Hide our working until you've had a go.

Questions 1–8 · 25 marks

Question 1

2 marks

Software update notification methods

System fundamentals

(a)

2 marks

Identify an automatic communication mechanism:
The installed client application can routinely contact the vendor's update server upon startup (or at scheduled intervals in the background).
Explain how notifications and installation instructions reach the user:
When an update is detected, an alert notification dialog or system tray message appears, providing a direct link or automated prompt to download and execute the patch.

Mentor tip

Cover both the discovery mechanism (automated server query/push notification) and the delivery method (in-app notification or email alert containing a download link).

Question 2

4 marks

Truth table for compound Boolean expression

Computer organisation

(a)

4 marks

Break down the compound expression:
Expression: $(A \lor B) \land (\neg C \lor B)$ .
Construct intermediate truth values for each combination of inputs $(A, B, C)$ :
1. For $A = 0, B = 0, C = 0$ : $A \lor B = 0$ , $\neg C \lor B = 1 \lor 0 = 1$ . Result: $0 \land 1 = 0$ .
2. For $A = 0, B = 0, C = 1$ : $A \lor B = 0$ , $\neg C \lor B = 0 \lor 0 = 0$ . Result: $0 \land 0 = 0$ .
3. For $A = 0, B = 1, C = 0$ : $A \lor B = 1$ , $\neg C \lor B = 1 \lor 1 = 1$ . Result: $1 \land 1 = 1$ .
4. For $A = 0, B = 1, C = 1$ : $A \lor B = 1$ , $\neg C \lor B = 0 \lor 1 = 1$ . Result: $1 \land 1 = 1$ .
5. For $A = 1, B = 0, C = 0$ : $A \lor B = 1$ , $\neg C \lor B = 1 \lor 0 = 1$ . Result: $1 \land 1 = 1$ .
6. For $A = 1, B = 0, C = 1$ : $A \lor B = 1$ , $\neg C \lor B = 0 \lor 0 = 0$ . Result: $1 \land 0 = 0$ .
7. For $A = 1, B = 1, C = 0$ : $A \lor B = 1$ , $\neg C \lor B = 1 \lor 1 = 1$ . Result: $1 \land 1 = 1$ .
8. For $A = 1, B = 1, C = 1$ : $A \lor B = 1$ , $\neg C \lor B = 0 \lor 1 = 1$ . Result: $1 \land 1 = 1$ .

A 00001111 B 00110011 C 01010101 A \lor B 00111111 \neg C \lor B 10111011 (A \lor B) \land (\neg C \lor B) 00111011

Mentor tip

Notice that whenever $B = 1$ , both sub-expressions evaluate to 1, immediately making the full output 1.

Question 3

3 marks

Use and function of a virtual private network

Networks

(a)

3 marks

Outline a practical scenario for a Virtual Private Network (VPN):
A remote or travelling employee needs access to internal enterprise servers and database resources from outside the corporate office.
Explain the technical operation:
1. The user establishes an encrypted tunnel over the public internet to the corporate VPN gateway using secure credentials.
2. The gateway assigns a private local corporate IP address, enabling secure communication as if directly wired into the internal local area network while shielding traffic from interception.

Mentor tip

Name the context (telecommuting/remote office connection) and explain both the tunneling/encryption aspect and access to private internal services.

Question 4

4 marks

Algorithm tracing for perfect number verification

Computational thinking, problem-solving and programming

(a)

4 marks

Analyze loop bounds and initial states:
Initial value: $N = 6, SUM = 0$ . The loop iterates $COUNT$ from $1$ to $(N div 2) = 6 div 2 = 3$ .
Step through each iteration:
Iteration 1 (COUNT = 1): $6 (mod 1) = 0$ (True). $SUM = 0 + 1 = 1$ .
Iteration 2 (COUNT = 2): $6 (mod 2) = 0$ (True). $SUM = 1 + 2 = 3$ .
Iteration 3 (COUNT = 3): $6 (mod 3) = 0$ (True). $SUM = 3 + 3 = 6$ .
Evaluate loop termination and output condition:
Loop ends as $COUNT$ reaches upper bound 3.
Condition check: $SUM = N ⟹ 6 = 6$ (True).
Output: "perfect".

COUNT 123 N (mod COUNT) = 0 True True True SUM 01366 SUM = N True Output "perfect"

Mentor tip

A number is perfect if the sum of its proper divisors equals the number itself; for 6, $1 + 2 + 3 = 6$ .

Question 5

3 marks

Descending selection sort pass-by-pass tracing

Computational thinking, problem-solving and programming

(a)

3 marks

Understand selection sort in descending order (highest to lowest):
Initial array: $[12, 52, 16, 42, 88, 86]$ .
At each pass $i$ , search the unsorted slice from index $i$ to the end for the maximum element, then swap it with index $i$ :
Pass 1: Maximum in indices $0..5$ is $88$ (at index 4). Swap with index 0 ( $12$ ). Result: $[88, 52, 16, 42, 12, 86]$ .
Pass 2: Maximum in indices $1..5$ is $86$ (at index 5). Swap with index 1 ( $52$ ). Result: $[88, 86, 16, 42, 12, 52]$ .
Pass 3: Maximum in indices $2..5$ is $52$ (at index 5). Swap with index 2 ( $16$ ). Result: $[88, 86, 52, 42, 12, 16]$ .
Pass 4: Maximum in indices $3..5$ is $42$ (at index 3). Element is already in place; no swap. Result: $[88, 86, 52, 42, 12, 16]$ .
Pass 5: Maximum in indices $4..5$ is $16$ (at index 5). Swap with index 4 ( $12$ ). Result: $[88, 86, 52, 42, 16, 12]$ .

Initial: Pass 1: Pass 2: Pass 3: Pass 4: Pass 5: [12, 52, 16, 42, 88, 86] [88, 52, 16, 42, 12, 86] [88, 86, 16, 42, 12, 52] [88, 86, 52, 42, 12, 16] [88, 86, 52, 42, 12, 16] [88, 86, 52, 42, 16, 12]

Mentor tip

In pass 4, 42 is already the largest in the remaining sub-array; the list remains identical across pass 4.

Question 6

3 marks

GPS positioning principles via trilateration

Networks

(a)

3 marks

Describe satellite signal transmission:
A GPS receiver picks up radio signals transmitted by multiple orbiting Global Positioning System satellites, which include precise orbital parameters and synchronized timestamps.
Explain time-of-flight distance calculation:
The receiver measures the tiny time difference between when each signal was broadcast and when it arrived, multiplying this duration by the speed of light to calculate the distance (range) to each satellite.
Explain positional trilateration:
By combining distance spheres from at least 3 to 4 satellites, the receiver computes the geometric intersection point to determine the person's precise 3D coordinates (latitude, longitude, altitude).

Mentor tip

Explain satellite timestamp broadcast, time delay multiplied by light speed for range determination, and geometric trilateration.

Question 7

3 marks

Operating system network abstraction and virtualization

Computer organisation

(a)

3 marks

Identify an abstraction mechanism provided to the end-user:
The operating system maps remote shared folders, network volumes, or network printers as local drive letters (e.g., drive Z:) or desktop icons.
Explain underlying OS handling:
Device drivers and network file system protocols handle packet packaging, socket connections, error handling, and hardware routing transparently in the background, allowing the user to read and write remote files using standard local GUI file management commands.

Mentor tip

Describe virtual drive mapping or transparent printing where network protocol complexity is handled by background OS drivers.

Question 8

3 marks

Binary tree postorder traversal and node deletion

Abstract data structures

(a)

1 mark

Recall postorder traversal order:
Postorder visits: Left subtree $\to$ Right subtree $\to$ Root node.
Follow the tree structure:
Given postorder traversal result: C, D, L, H, T, P, M.

Postorder: C \to D \to L \to H \to T \to P \to M

Mentor tip

Postorder traverses subtrees recursively (Left, Right, Root), ending at the overall tree root M.

(b)

2 marks

Examine node relationships around H:
Node H has subtrees producing postorder sequence C, D, L then H.
Preserve relative postorder sequence after deleting H:
To maintain the exact postorder traversal of remaining nodes (C, D, L, T, P, M), the child or children of H must be linked directly to H's parent (root M) on its left branch while preserving their postorder relationship.
L becomes the left child of M (or subtree containing C, D, L attached to M's left pointer), leaving the right subtree (T, P) unchanged under M.

Mentor tip

Deleting H requires relinking its subtree directly to the parent node M so the relative postorder visiting order of C, D, L is retained.

Questions 9–13 · 75 marks

Question 9

15 marks

Database processing, mail merge, and data protection

Computational thinking, problem-solving and programming
System fundamentals

(a)

6 marks

Initialize lists and date reference:
Set current date variable CURR_DATE. Initialize LIST1, LIST2, LIST3 as empty lists.
Iterate through records in the customer database:
Loop through each customer record CUST in DATABASE.
Compute payment difference:
Calculate DIFF = CUST.PAYMENT_DATE - CURR_DATE (or DUE_PERIOD = CURR_DATE - CUST.PAYMENT_DATE).
Apply conditional logic to assign categories:
1. If payment is due within the next 30 days (DUE_PERIOD >= -30 and DUE_PERIOD <= 0), add CUST to LIST1.
2. If payment is overdue by more than 14 days and up to 30 days (DUE_PERIOD > 14 and DUE_PERIOD <= 30), add CUST to LIST2.
3. If payment is overdue by more than 30 days (DUE_PERIOD > 30), add CUST to LIST3 and set CUST.FLAGGED_FOR_DELETION = true.
End loop.

Mentor tip

Be precise with boundary conditions (14 < days <= 30 vs > 30) and flag records in list 3 rather than deleting immediately.

(b)

4 marks

Explain mail merge integration:
1. Document templates are designed in the word processor with standardized text and placeholder fields for customer name, address, and overdue balance.
2. The lists generated in part (a) provide the target customer IDs, which query the primary database for full contact information.
3. The word processor merges the corresponding database fields into each placeholder slot, generating personalized documents.
4. The system routes the merged documents to a print spooler for physical postal dispatch or to an SMTP mail engine for automated email transmission.

Mentor tip

Detail template placeholders, database query linkage via customer ID, field substitution, and final dispatch mechanism.

(c)

2 marks

Identify impact on customers:
Customers miss payment reminders, potentially causing their insurance coverage to lapse without their knowledge and leaving them unprotected against claims.
Identify impact on the company:
The insurance firm suffers loss of premium revenue, legal liabilities for wrongful policy terminations, and severe reputational damage.

Mentor tip

Give one direct consequence for the policyholder (loss of insurance coverage) and one for the enterprise (loss of revenue/reputation).

(d)

3 marks

Select a reliable data protection strategy:
Automated offsite backup / mirroring.
Describe implementation and operational process:
1. Daily incremental backups and weekly full system snapshots are encrypted and transmitted to a secure offsite cloud data center.
2. In the event of primary storage corruption or hardware failure, the system restores the latest valid timestamped snapshot, minimizing data recovery point objectives.

Mentor tip

Mention offsite/cloud snapshots with automated scheduling and rapid restore capability.

Question 10

16 marks

Network architectures, protocols, and appointment scheduling

Networks
Computational thinking, problem-solving and programming
Abstract data structures

(a)

3 marks

Define the Open Systems Interconnection (OSI) framework:
The OSI model is a standardized 7-layer architectural framework developed by ISO to standardize telecommunications and computer network communications.
Explain modular layer responsibilities:
Each layer performs a dedicated set of network functions and provides standardized interfaces to the layers immediately above and below it.
Provide an illustrative layer example:
For example, the Physical layer defines hardware electrical signals and cabling, while the Network layer handles logical addressing and packet routing across routers.

Mentor tip

State that it is a 7-layer standardized reference framework where each layer handles specific abstractions, citing an example layer.

(b)

3 marks

Define communication protocol:
A protocol is a formal set of agreed rules and conventions governing the formatting, transmission, timing, and error control of data exchange between networked devices.
Explain purpose and provide a concrete example:
Protocols ensure interoperability across heterogeneous hardware. For example, TCP (Transmission Control Protocol) manages packet sequencing, connection handshakes, and acknowledgements to ensure reliable, loss-free data delivery.

Mentor tip

Highlight rules/standards for communication plus an explicit network protocol example such as TCP, IP, or HTTP.

(c)

2 marks

Identify required input fields:
1. Client identifier: full client name, contact phone number/email, and client status (new vs returning).
2. Requested appointment details: preferred date/time and the name of the assigned lawyer (for returning clients).

Mentor tip

Include client name/ID, returning/new status, preferred date/time, and lawyer name if returning.

(d)

3 marks

Propose a suitable data structure:
A two-dimensional array (matrix) representing the appointment schedule for a single day.
Describe dimension mappings:
Let columns represent the 6 lawyers (indices 0 to 5) and rows represent discrete time slots across the working day (e.g., indices 0 to 15 representing half-hour intervals from 09:00 to 17:00). Each element contains a client record or null/empty if unbooked.

Mentor tip

A 2D array with lawyers on one axis and time intervals on the other cleanly models a daily appointment grid.

(e)

5 marks

Outline the algorithmic procedure:
1. Load the schedule grid starting from the current or requested date.
2. Branch based on client status:
- If returning client: locate the column corresponding to the client's usual lawyer; scan time slot rows sequentially from earliest available slot to find the first empty entry.
- If new client: scan time slot rows chronologically; for each time slot, check across all 6 lawyer columns to find the first open slot with any available lawyer.
3. If a free slot is identified, insert the client's details into the grid element and confirm booking.
4. If no free slot exists on the current day, advance to the next day's schedule grid and repeat the search until an open slot is allocated.

Mentor tip

Distinguish between searching a specific lawyer's column for returning clients versus scanning across all columns for new clients, then advancing to subsequent days if full.

Question 11

15 marks

Flight display list management, time conversions, and queue comparison

Abstract data structures
Computational thinking, problem-solving and programming

(a)

3 marks

Describe node payload:
Each node in the linked list stores the flight information payload: ID, PLACE, DUE, EXPECTED, and ARRIVED.
Describe pointer mechanism and list boundaries:
Each node contains a reference (next pointer) addressing the subsequent plane node. A external Head pointer references the first display plane, and the final node's pointer contains NULL.
Describe size capacity:
The linked list dynamically maintains up to 20 active plane nodes displayed concurrently on the monitor.

Mentor tip

Detail data fields in the node, the pointer linking to the next node, head pointer, and null termination.

(b)

3 marks

Construct arithmetic algorithm:
Input minute value: CTIME (total minutes past midnight).
Compute hours using integer division: HOURS = CTIME div 60.
Compute remaining minutes using modulo: MINUTES = CTIME mod 60.
Format and return 24-hour time representation: output or return HOURS and MINUTES (formatted as HH:MM).

HOURS MINUTES = CTIME div 60 = CTIME mod 60

Mentor tip

Use integer division (div) for hours and modulus (mod) for minutes.

(c)

4 marks

Describe linear search through the list:
Traverse through the linked list starting at HEAD. For each plane node, compare the current airport clock time with node.ARRIVED.
Check elapsed time condition:
If (CURRENT_TIME - node.ARRIVED) > 30 minutes, initiate removal of that node.
Execute pointer adjustments:
1. If the node to delete is at HEAD: update HEAD = HEAD.next.
2. If the node is in the middle or end: set PREVIOUS.next = CURRENT.next.
3. Sever the removed node's pointer: set CURRENT.next = NULL to release it from the active display chain.

Mentor tip

Explain traversal, condition checking (> 30 min), and bypass pointer modification (PREV.next = CURR.next or updating HEAD).

(d)

5 marks

Analyze Queue characteristics:
A queue follows strict First-In, First-Out (FIFO) discipline. Planes enqueue at the rear and can only dequeue from the front.
Identify limitation of a Queue for this scenario:
Planes do not arrive strictly in the order they were scheduled; a flight in the middle of the display list might arrive and exceed 30 minutes before flights ahead of it, but a queue cannot delete arbitrary internal elements without dumping the entire queue.
Analyze Linked List flexibility:
A linked list allows arbitrary node deletion from any position (head, middle, or tail) in $O (1)$ pointer operations once identified.
Evaluate search and insertion:
Both structures require $O (n)$ search to find matching flight IDs, but the linked list's ability to delete interior nodes makes it far superior for this airport display board.

Mentor tip

Emphasize that queues only remove from the front (FIFO), whereas flights finish and leave from arbitrary display positions, making linked lists necessary.

Question 12

15 marks

Embedded traffic control systems and centralized architectures

Control systems
System fundamentals

(a)

4 marks

Role of sensors:
Inductive loop sensors (or infrared/pressure sensors) embedded in the minor road detect the presence of approaching emergency vehicles.
Signal conversion and transmission:
The analog detector signal is converted to digital format via an ADC and transmitted to the microprocessor.
Processing and actuation:
The microprocessor executes control logic to override the default cycle, triggering actuators to transition the main road lights to red (Stop) and the minor road lights to green (Go).

Mentor tip

Trace from sensor detection $\to$ analog-to-digital conversion $\to$ processor evaluation $\to$ actuator signal to signal lights.

(b)

3 marks

Continuous sensor polling and timing:
The microprocessor continuously monitors sensor readings on the minor road.
Timer mechanism:
A countdown timer is initialized. Each time a subsequent emergency vehicle triggers the sensor, the timer resets to full duration.
State restoration:
When no vehicles are detected for a predetermined duration (timer expires), the microprocessor transitions the signals back to their normal scheduled cycles.

Mentor tip

Describe a reset timer that counts down once vehicle detection ceases, reverting to normal state on timeout.

(c)

5 marks

Analyze advantages of a central system:
1. Coordinated urban flow: The central computer can synchronize traffic light phases across multiple intersections to create green waves and adapt dynamically to city-wide congestion.
2. Centralized monitoring and maintenance: Telemetry from all intersections is visible on one console, making faults detectable and dispatchable immediately.
Analyze disadvantages of a central system:
1. Single point of failure: If the central server crashes or network connectivity is severed, traffic control across the entire city could fail.
2. Latency and high communication overhead: Transmitting sensor streams from hundreds of intersections across wide-area links introduces latency and dependency on robust networking.
Conclusion / Weighing up:
A hybrid distributed architecture (local microprocessors running autonomously with central supervisory coordination) offers the optimal balance of resilience and coordinated control.

Mentor tip

Provide 2 distinct advantages, 2 distinct disadvantages, and a reasoned concluding appraisal.

(d)

3 marks

Social benefits:
Automated camera monitoring deters reckless driving and speeding, substantially lowering traffic collisions and safeguarding pedestrians.
Privacy concerns:
Continuous video surveillance and automatic license plate recognition enable mass location tracking of individual citizens, raising civil liberty and surveillance concerns.
Synthesize the balance:
Society must balance the public safety benefits of reduced accidents against the erosion of personal privacy and the risk of unauthorized tracking.

Mentor tip

Weigh public safety and accident reduction against privacy invasion and continuous citizen tracking.

Question 13

14 marks

Stack backtracking algorithms and coordinate maze tracing

Abstract data structures
Computational thinking, problem-solving and programming

(a)

2 marks

Explain stack suitability:
A stack is a Last-In, First-Out (LIFO) data structure.
Connect to maze backtracking:
To retrace a path from the maze center back to the starting point, moves must be reversed in the exact inverse order of how they were taken. The most recent move recorded is the first one that must be undone, which directly matches the push/pop behavior of a stack.

Mentor tip

Emphasize LIFO behavior: the last move made must be the first move retraced.

(b)

5 marks

Formulate stack reversal algorithm:
1. Loop while the stack is not empty: while NOT STK.isEmpty()
2. Pop the distance step: STEP = STK.pop()
3. Output forward step instruction: output "Walk " + STEP + " steps"
4. Pop the recorded turn direction: TURN = STK.pop()
5. Invert the turn direction for the return journey:
- If recorded turn was right (TURN == 0), output "Turn left"
- If recorded turn was left (TURN == 1), output "Turn right"
6. Repeat until the stack is exhausted.

Mentor tip

Remember to pop STEP first then TURN, and invert the turn direction (0 right becomes left, 1 left becomes right).

(c)

1 mark

Trace coordinate updates for move 3 Move R8:
Position after move 2: $(10, 5)$ , facing North ( $N = 1, E = 0$ ).
Third move is R8 (TURN = 0, STEP = 8):
$X = X + N \times STEP = 10 + 1 \times 8 = 18$ .
$Y = Y - E \times STEP = 5 - 0 = 5$ .
Updated coordinates: $(18, 5)$ .

(18, 5)

Mentor tip

Applying turn 0 with N=1 gives delta X = 8, so coordinates become (18, 5).

(d)

6 marks

Trace remaining moves using algorithm specification:
Row 1: TURN=0, STEP=8 from (10, 5, 1, 0) $\to$ $X = 18, Y = 5$ . Facing: $N = 0, E = 1$ .
Row 2: TURN=0, STEP=2 from (18, 5, 0, 1) $\to$ $X = 18 + 0 = 18$ , $Y = 5 - 1 \times 2 = 3$ . Facing: since $N = 0$ , new $N = - E = - 1, E = 0$ .
Row 3: TURN=1, STEP=3 from (18, 3, -1, 0) $\to$ $X = 18 - (- 1) \times 3 = 21$ , $Y = 3 + 0 = 3$ . Facing: since $N \neq = 0$ , new $E = - N = - (- 1) = 1, N = 0$ .
Row 4: TURN=0, STEP=0 $\to$ STEP is 0, loop terminates.

TURN 010010 STEP 1058230 X 01010181821 Y 005533 N 1010 - 1 0 E 010101

Mentor tip

Carefully follow the sign formulas for N and E during left (TURN=1) and right (TURN=0) turns.

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 programming35 marks · 35%
Q4, Q5, Q9, Q10, Q11, Q13
Abstract data structures20 marks · 20%
Q8, Q10, Q11, Q13
System fundamentals14 marks · 14%
Q1, Q9, Q12
Networks12 marks · 12%
Q3, Q6, Q10
Control systems12 marks · 12%
Q12
Computer organisation7 marks · 7%
Q2, Q7

Mentor insights

Traps to watch for

01
Short-circuit evaluation in Boolean expressions · Q2
In the expression (A or B) and (not C or B), whenever B is 1, both OR sub-clauses immediately evaluate to 1, producing an output of 1 without needing to evaluate A or C.
02
Selection sort pass tracking with existing maximum · Q5
In pass 4 of the descending selection sort, the element 42 is already the largest among the remaining unsorted sub-array. No swap takes place, so the array sequence in pass 4 remains identical to pass 3.
03
Queue vs linked list for flight departure display · Q11
A FIFO queue can only remove nodes from the front of the queue. Because flights arrive and clear out of chronological order in real life, a linked list is required so interior nodes can be unlinked from arbitrary positions.
04
Inverting moves when backtracking with a stack · Q13
When popping moves from the maze stack to return to the start, each turn direction must be inverted (a recorded right turn requires turning left on the return path, and vice versa).

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 2015
Marks: 100
Duration: 2 hours 10 min
Calculator: Not allowed
Questions: 13

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

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

Practice mode

Hide our working until you've had a go.

Questions 1–8 · 25 marks

Question 1

2 marks

Software update notification methods

System fundamentals

(a)

2 marks

Identify an automatic communication mechanism:
The installed client application can routinely contact the vendor's update server upon startup (or at scheduled intervals in the background).
Explain how notifications and installation instructions reach the user:
When an update is detected, an alert notification dialog or system tray message appears, providing a direct link or automated prompt to download and execute the patch.

Mentor tip

Cover both the discovery mechanism (automated server query/push notification) and the delivery method (in-app notification or email alert containing a download link).

Question 2

4 marks

Truth table for compound Boolean expression

Computer organisation

(a)

4 marks

Break down the compound expression:
Expression: $(A \lor B) \land (\neg C \lor B)$ .
Construct intermediate truth values for each combination of inputs $(A, B, C)$ :
1. For $A = 0, B = 0, C = 0$ : $A \lor B = 0$ , $\neg C \lor B = 1 \lor 0 = 1$ . Result: $0 \land 1 = 0$ .
2. For $A = 0, B = 0, C = 1$ : $A \lor B = 0$ , $\neg C \lor B = 0 \lor 0 = 0$ . Result: $0 \land 0 = 0$ .
3. For $A = 0, B = 1, C = 0$ : $A \lor B = 1$ , $\neg C \lor B = 1 \lor 1 = 1$ . Result: $1 \land 1 = 1$ .
4. For $A = 0, B = 1, C = 1$ : $A \lor B = 1$ , $\neg C \lor B = 0 \lor 1 = 1$ . Result: $1 \land 1 = 1$ .
5. For $A = 1, B = 0, C = 0$ : $A \lor B = 1$ , $\neg C \lor B = 1 \lor 0 = 1$ . Result: $1 \land 1 = 1$ .
6. For $A = 1, B = 0, C = 1$ : $A \lor B = 1$ , $\neg C \lor B = 0 \lor 0 = 0$ . Result: $1 \land 0 = 0$ .
7. For $A = 1, B = 1, C = 0$ : $A \lor B = 1$ , $\neg C \lor B = 1 \lor 1 = 1$ . Result: $1 \land 1 = 1$ .
8. For $A = 1, B = 1, C = 1$ : $A \lor B = 1$ , $\neg C \lor B = 0 \lor 1 = 1$ . Result: $1 \land 1 = 1$ .

A 00001111 B 00110011 C 01010101 A \lor B 00111111 \neg C \lor B 10111011 (A \lor B) \land (\neg C \lor B) 00111011

Mentor tip

Notice that whenever $B = 1$ , both sub-expressions evaluate to 1, immediately making the full output 1.

Question 3

3 marks

Use and function of a virtual private network

Networks

(a)

3 marks

Outline a practical scenario for a Virtual Private Network (VPN):
A remote or travelling employee needs access to internal enterprise servers and database resources from outside the corporate office.
Explain the technical operation:
1. The user establishes an encrypted tunnel over the public internet to the corporate VPN gateway using secure credentials.
2. The gateway assigns a private local corporate IP address, enabling secure communication as if directly wired into the internal local area network while shielding traffic from interception.

Mentor tip

Name the context (telecommuting/remote office connection) and explain both the tunneling/encryption aspect and access to private internal services.

Question 4

4 marks

Algorithm tracing for perfect number verification

Computational thinking, problem-solving and programming

(a)

4 marks

Analyze loop bounds and initial states:
Initial value: $N = 6, SUM = 0$ . The loop iterates $COUNT$ from $1$ to $(N div 2) = 6 div 2 = 3$ .
Step through each iteration:
Iteration 1 (COUNT = 1): $6 (mod 1) = 0$ (True). $SUM = 0 + 1 = 1$ .
Iteration 2 (COUNT = 2): $6 (mod 2) = 0$ (True). $SUM = 1 + 2 = 3$ .
Iteration 3 (COUNT = 3): $6 (mod 3) = 0$ (True). $SUM = 3 + 3 = 6$ .
Evaluate loop termination and output condition:
Loop ends as $COUNT$ reaches upper bound 3.
Condition check: $SUM = N ⟹ 6 = 6$ (True).
Output: "perfect".

COUNT 123 N (mod COUNT) = 0 True True True SUM 01366 SUM = N True Output "perfect"

Mentor tip

A number is perfect if the sum of its proper divisors equals the number itself; for 6, $1 + 2 + 3 = 6$ .

Question 5

3 marks

Descending selection sort pass-by-pass tracing

Computational thinking, problem-solving and programming

(a)

3 marks

Understand selection sort in descending order (highest to lowest):
Initial array: $[12, 52, 16, 42, 88, 86]$ .
At each pass $i$ , search the unsorted slice from index $i$ to the end for the maximum element, then swap it with index $i$ :
Pass 1: Maximum in indices $0..5$ is $88$ (at index 4). Swap with index 0 ( $12$ ). Result: $[88, 52, 16, 42, 12, 86]$ .
Pass 2: Maximum in indices $1..5$ is $86$ (at index 5). Swap with index 1 ( $52$ ). Result: $[88, 86, 16, 42, 12, 52]$ .
Pass 3: Maximum in indices $2..5$ is $52$ (at index 5). Swap with index 2 ( $16$ ). Result: $[88, 86, 52, 42, 12, 16]$ .
Pass 4: Maximum in indices $3..5$ is $42$ (at index 3). Element is already in place; no swap. Result: $[88, 86, 52, 42, 12, 16]$ .
Pass 5: Maximum in indices $4..5$ is $16$ (at index 5). Swap with index 4 ( $12$ ). Result: $[88, 86, 52, 42, 16, 12]$ .

Initial: Pass 1: Pass 2: Pass 3: Pass 4: Pass 5: [12, 52, 16, 42, 88, 86] [88, 52, 16, 42, 12, 86] [88, 86, 16, 42, 12, 52] [88, 86, 52, 42, 12, 16] [88, 86, 52, 42, 12, 16] [88, 86, 52, 42, 16, 12]

Mentor tip

In pass 4, 42 is already the largest in the remaining sub-array; the list remains identical across pass 4.

Question 6

3 marks

GPS positioning principles via trilateration

Networks

(a)

3 marks

Describe satellite signal transmission:
A GPS receiver picks up radio signals transmitted by multiple orbiting Global Positioning System satellites, which include precise orbital parameters and synchronized timestamps.
Explain time-of-flight distance calculation:
The receiver measures the tiny time difference between when each signal was broadcast and when it arrived, multiplying this duration by the speed of light to calculate the distance (range) to each satellite.
Explain positional trilateration:
By combining distance spheres from at least 3 to 4 satellites, the receiver computes the geometric intersection point to determine the person's precise 3D coordinates (latitude, longitude, altitude).

Mentor tip

Explain satellite timestamp broadcast, time delay multiplied by light speed for range determination, and geometric trilateration.

Question 7

3 marks

Operating system network abstraction and virtualization

Computer organisation

(a)

3 marks

Identify an abstraction mechanism provided to the end-user:
The operating system maps remote shared folders, network volumes, or network printers as local drive letters (e.g., drive Z:) or desktop icons.
Explain underlying OS handling:
Device drivers and network file system protocols handle packet packaging, socket connections, error handling, and hardware routing transparently in the background, allowing the user to read and write remote files using standard local GUI file management commands.

Mentor tip

Describe virtual drive mapping or transparent printing where network protocol complexity is handled by background OS drivers.

Question 8

3 marks

Binary tree postorder traversal and node deletion

Abstract data structures

(a)

1 mark

Recall postorder traversal order:
Postorder visits: Left subtree $\to$ Right subtree $\to$ Root node.
Follow the tree structure:
Given postorder traversal result: C, D, L, H, T, P, M.

Postorder: C \to D \to L \to H \to T \to P \to M

Mentor tip

Postorder traverses subtrees recursively (Left, Right, Root), ending at the overall tree root M.

(b)

2 marks

Examine node relationships around H:
Node H has subtrees producing postorder sequence C, D, L then H.
Preserve relative postorder sequence after deleting H:
To maintain the exact postorder traversal of remaining nodes (C, D, L, T, P, M), the child or children of H must be linked directly to H's parent (root M) on its left branch while preserving their postorder relationship.
L becomes the left child of M (or subtree containing C, D, L attached to M's left pointer), leaving the right subtree (T, P) unchanged under M.

Mentor tip

Deleting H requires relinking its subtree directly to the parent node M so the relative postorder visiting order of C, D, L is retained.

Questions 9–13 · 75 marks

Question 9

15 marks

Database processing, mail merge, and data protection

Computational thinking, problem-solving and programming
System fundamentals

(a)

6 marks

Initialize lists and date reference:
Set current date variable CURR_DATE. Initialize LIST1, LIST2, LIST3 as empty lists.
Iterate through records in the customer database:
Loop through each customer record CUST in DATABASE.
Compute payment difference:
Calculate DIFF = CUST.PAYMENT_DATE - CURR_DATE (or DUE_PERIOD = CURR_DATE - CUST.PAYMENT_DATE).
Apply conditional logic to assign categories:
1. If payment is due within the next 30 days (DUE_PERIOD >= -30 and DUE_PERIOD <= 0), add CUST to LIST1.
2. If payment is overdue by more than 14 days and up to 30 days (DUE_PERIOD > 14 and DUE_PERIOD <= 30), add CUST to LIST2.
3. If payment is overdue by more than 30 days (DUE_PERIOD > 30), add CUST to LIST3 and set CUST.FLAGGED_FOR_DELETION = true.
End loop.

Mentor tip

Be precise with boundary conditions (14 < days <= 30 vs > 30) and flag records in list 3 rather than deleting immediately.

(b)

4 marks

Explain mail merge integration:
1. Document templates are designed in the word processor with standardized text and placeholder fields for customer name, address, and overdue balance.
2. The lists generated in part (a) provide the target customer IDs, which query the primary database for full contact information.
3. The word processor merges the corresponding database fields into each placeholder slot, generating personalized documents.
4. The system routes the merged documents to a print spooler for physical postal dispatch or to an SMTP mail engine for automated email transmission.

Mentor tip

Detail template placeholders, database query linkage via customer ID, field substitution, and final dispatch mechanism.

(c)

2 marks

Identify impact on customers:
Customers miss payment reminders, potentially causing their insurance coverage to lapse without their knowledge and leaving them unprotected against claims.
Identify impact on the company:
The insurance firm suffers loss of premium revenue, legal liabilities for wrongful policy terminations, and severe reputational damage.

Mentor tip

Give one direct consequence for the policyholder (loss of insurance coverage) and one for the enterprise (loss of revenue/reputation).

(d)

3 marks

Select a reliable data protection strategy:
Automated offsite backup / mirroring.
Describe implementation and operational process:
1. Daily incremental backups and weekly full system snapshots are encrypted and transmitted to a secure offsite cloud data center.
2. In the event of primary storage corruption or hardware failure, the system restores the latest valid timestamped snapshot, minimizing data recovery point objectives.

Mentor tip

Mention offsite/cloud snapshots with automated scheduling and rapid restore capability.

Question 10

16 marks

Network architectures, protocols, and appointment scheduling

Networks
Computational thinking, problem-solving and programming
Abstract data structures

(a)

3 marks

Define the Open Systems Interconnection (OSI) framework:
The OSI model is a standardized 7-layer architectural framework developed by ISO to standardize telecommunications and computer network communications.
Explain modular layer responsibilities:
Each layer performs a dedicated set of network functions and provides standardized interfaces to the layers immediately above and below it.
Provide an illustrative layer example:
For example, the Physical layer defines hardware electrical signals and cabling, while the Network layer handles logical addressing and packet routing across routers.

Mentor tip

State that it is a 7-layer standardized reference framework where each layer handles specific abstractions, citing an example layer.

(b)

3 marks

Define communication protocol:
A protocol is a formal set of agreed rules and conventions governing the formatting, transmission, timing, and error control of data exchange between networked devices.
Explain purpose and provide a concrete example:
Protocols ensure interoperability across heterogeneous hardware. For example, TCP (Transmission Control Protocol) manages packet sequencing, connection handshakes, and acknowledgements to ensure reliable, loss-free data delivery.

Mentor tip

Highlight rules/standards for communication plus an explicit network protocol example such as TCP, IP, or HTTP.

(c)

2 marks

Identify required input fields:
1. Client identifier: full client name, contact phone number/email, and client status (new vs returning).
2. Requested appointment details: preferred date/time and the name of the assigned lawyer (for returning clients).

Mentor tip

Include client name/ID, returning/new status, preferred date/time, and lawyer name if returning.

(d)

3 marks

Propose a suitable data structure:
A two-dimensional array (matrix) representing the appointment schedule for a single day.
Describe dimension mappings:
Let columns represent the 6 lawyers (indices 0 to 5) and rows represent discrete time slots across the working day (e.g., indices 0 to 15 representing half-hour intervals from 09:00 to 17:00). Each element contains a client record or null/empty if unbooked.

Mentor tip

A 2D array with lawyers on one axis and time intervals on the other cleanly models a daily appointment grid.

(e)

5 marks

Outline the algorithmic procedure:
1. Load the schedule grid starting from the current or requested date.
2. Branch based on client status:
- If returning client: locate the column corresponding to the client's usual lawyer; scan time slot rows sequentially from earliest available slot to find the first empty entry.
- If new client: scan time slot rows chronologically; for each time slot, check across all 6 lawyer columns to find the first open slot with any available lawyer.
3. If a free slot is identified, insert the client's details into the grid element and confirm booking.
4. If no free slot exists on the current day, advance to the next day's schedule grid and repeat the search until an open slot is allocated.

Mentor tip

Distinguish between searching a specific lawyer's column for returning clients versus scanning across all columns for new clients, then advancing to subsequent days if full.

Question 11

15 marks

Flight display list management, time conversions, and queue comparison

Abstract data structures
Computational thinking, problem-solving and programming

(a)

3 marks

Describe node payload:
Each node in the linked list stores the flight information payload: ID, PLACE, DUE, EXPECTED, and ARRIVED.
Describe pointer mechanism and list boundaries:
Each node contains a reference (next pointer) addressing the subsequent plane node. A external Head pointer references the first display plane, and the final node's pointer contains NULL.
Describe size capacity:
The linked list dynamically maintains up to 20 active plane nodes displayed concurrently on the monitor.

Mentor tip

Detail data fields in the node, the pointer linking to the next node, head pointer, and null termination.

(b)

3 marks

Construct arithmetic algorithm:
Input minute value: CTIME (total minutes past midnight).
Compute hours using integer division: HOURS = CTIME div 60.
Compute remaining minutes using modulo: MINUTES = CTIME mod 60.
Format and return 24-hour time representation: output or return HOURS and MINUTES (formatted as HH:MM).

HOURS MINUTES = CTIME div 60 = CTIME mod 60

Mentor tip

Use integer division (div) for hours and modulus (mod) for minutes.

(c)

4 marks

Describe linear search through the list:
Traverse through the linked list starting at HEAD. For each plane node, compare the current airport clock time with node.ARRIVED.
Check elapsed time condition:
If (CURRENT_TIME - node.ARRIVED) > 30 minutes, initiate removal of that node.
Execute pointer adjustments:
1. If the node to delete is at HEAD: update HEAD = HEAD.next.
2. If the node is in the middle or end: set PREVIOUS.next = CURRENT.next.
3. Sever the removed node's pointer: set CURRENT.next = NULL to release it from the active display chain.

Mentor tip

Explain traversal, condition checking (> 30 min), and bypass pointer modification (PREV.next = CURR.next or updating HEAD).

(d)

5 marks

Analyze Queue characteristics:
A queue follows strict First-In, First-Out (FIFO) discipline. Planes enqueue at the rear and can only dequeue from the front.
Identify limitation of a Queue for this scenario:
Planes do not arrive strictly in the order they were scheduled; a flight in the middle of the display list might arrive and exceed 30 minutes before flights ahead of it, but a queue cannot delete arbitrary internal elements without dumping the entire queue.
Analyze Linked List flexibility:
A linked list allows arbitrary node deletion from any position (head, middle, or tail) in $O (1)$ pointer operations once identified.
Evaluate search and insertion:
Both structures require $O (n)$ search to find matching flight IDs, but the linked list's ability to delete interior nodes makes it far superior for this airport display board.

Mentor tip

Emphasize that queues only remove from the front (FIFO), whereas flights finish and leave from arbitrary display positions, making linked lists necessary.

Question 12

15 marks

Embedded traffic control systems and centralized architectures

Control systems
System fundamentals

(a)

4 marks

Role of sensors:
Inductive loop sensors (or infrared/pressure sensors) embedded in the minor road detect the presence of approaching emergency vehicles.
Signal conversion and transmission:
The analog detector signal is converted to digital format via an ADC and transmitted to the microprocessor.
Processing and actuation:
The microprocessor executes control logic to override the default cycle, triggering actuators to transition the main road lights to red (Stop) and the minor road lights to green (Go).

Mentor tip

Trace from sensor detection $\to$ analog-to-digital conversion $\to$ processor evaluation $\to$ actuator signal to signal lights.

(b)

3 marks

Continuous sensor polling and timing:
The microprocessor continuously monitors sensor readings on the minor road.
Timer mechanism:
A countdown timer is initialized. Each time a subsequent emergency vehicle triggers the sensor, the timer resets to full duration.
State restoration:
When no vehicles are detected for a predetermined duration (timer expires), the microprocessor transitions the signals back to their normal scheduled cycles.

Mentor tip

Describe a reset timer that counts down once vehicle detection ceases, reverting to normal state on timeout.

(c)

5 marks

Analyze advantages of a central system:
1. Coordinated urban flow: The central computer can synchronize traffic light phases across multiple intersections to create green waves and adapt dynamically to city-wide congestion.
2. Centralized monitoring and maintenance: Telemetry from all intersections is visible on one console, making faults detectable and dispatchable immediately.
Analyze disadvantages of a central system:
1. Single point of failure: If the central server crashes or network connectivity is severed, traffic control across the entire city could fail.
2. Latency and high communication overhead: Transmitting sensor streams from hundreds of intersections across wide-area links introduces latency and dependency on robust networking.
Conclusion / Weighing up:
A hybrid distributed architecture (local microprocessors running autonomously with central supervisory coordination) offers the optimal balance of resilience and coordinated control.

Mentor tip

Provide 2 distinct advantages, 2 distinct disadvantages, and a reasoned concluding appraisal.

(d)

3 marks

Social benefits:
Automated camera monitoring deters reckless driving and speeding, substantially lowering traffic collisions and safeguarding pedestrians.
Privacy concerns:
Continuous video surveillance and automatic license plate recognition enable mass location tracking of individual citizens, raising civil liberty and surveillance concerns.
Synthesize the balance:
Society must balance the public safety benefits of reduced accidents against the erosion of personal privacy and the risk of unauthorized tracking.

Mentor tip

Weigh public safety and accident reduction against privacy invasion and continuous citizen tracking.

Question 13

14 marks

Stack backtracking algorithms and coordinate maze tracing

Abstract data structures
Computational thinking, problem-solving and programming

(a)

2 marks

Explain stack suitability:
A stack is a Last-In, First-Out (LIFO) data structure.
Connect to maze backtracking:
To retrace a path from the maze center back to the starting point, moves must be reversed in the exact inverse order of how they were taken. The most recent move recorded is the first one that must be undone, which directly matches the push/pop behavior of a stack.

Mentor tip

Emphasize LIFO behavior: the last move made must be the first move retraced.

(b)

5 marks

Formulate stack reversal algorithm:
1. Loop while the stack is not empty: while NOT STK.isEmpty()
2. Pop the distance step: STEP = STK.pop()
3. Output forward step instruction: output "Walk " + STEP + " steps"
4. Pop the recorded turn direction: TURN = STK.pop()
5. Invert the turn direction for the return journey:
- If recorded turn was right (TURN == 0), output "Turn left"
- If recorded turn was left (TURN == 1), output "Turn right"
6. Repeat until the stack is exhausted.

Mentor tip

Remember to pop STEP first then TURN, and invert the turn direction (0 right becomes left, 1 left becomes right).

(c)

1 mark

Trace coordinate updates for move 3 Move R8:
Position after move 2: $(10, 5)$ , facing North ( $N = 1, E = 0$ ).
Third move is R8 (TURN = 0, STEP = 8):
$X = X + N \times STEP = 10 + 1 \times 8 = 18$ .
$Y = Y - E \times STEP = 5 - 0 = 5$ .
Updated coordinates: $(18, 5)$ .

(18, 5)

Mentor tip

Applying turn 0 with N=1 gives delta X = 8, so coordinates become (18, 5).

(d)

6 marks

Trace remaining moves using algorithm specification:
Row 1: TURN=0, STEP=8 from (10, 5, 1, 0) $\to$ $X = 18, Y = 5$ . Facing: $N = 0, E = 1$ .
Row 2: TURN=0, STEP=2 from (18, 5, 0, 1) $\to$ $X = 18 + 0 = 18$ , $Y = 5 - 1 \times 2 = 3$ . Facing: since $N = 0$ , new $N = - E = - 1, E = 0$ .
Row 3: TURN=1, STEP=3 from (18, 3, -1, 0) $\to$ $X = 18 - (- 1) \times 3 = 21$ , $Y = 3 + 0 = 3$ . Facing: since $N \neq = 0$ , new $E = - N = - (- 1) = 1, N = 0$ .
Row 4: TURN=0, STEP=0 $\to$ STEP is 0, loop terminates.

TURN 010010 STEP 1058230 X 01010181821 Y 005533 N 1010 - 1 0 E 010101

Mentor tip

Carefully follow the sign formulas for N and E during left (TURN=1) and right (TURN=0) turns.

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 programming35 marks · 35%
Q4, Q5, Q9, Q10, Q11, Q13
Abstract data structures20 marks · 20%
Q8, Q10, Q11, Q13
System fundamentals14 marks · 14%
Q1, Q9, Q12
Networks12 marks · 12%
Q3, Q6, Q10
Control systems12 marks · 12%
Q12
Computer organisation7 marks · 7%
Q2, Q7

Mentor insights

Traps to watch for

01
Short-circuit evaluation in Boolean expressions · Q2
In the expression (A or B) and (not C or B), whenever B is 1, both OR sub-clauses immediately evaluate to 1, producing an output of 1 without needing to evaluate A or C.
02
Selection sort pass tracking with existing maximum · Q5
In pass 4 of the descending selection sort, the element 42 is already the largest among the remaining unsorted sub-array. No swap takes place, so the array sequence in pass 4 remains identical to pass 3.
03
Queue vs linked list for flight departure display · Q11
A FIFO queue can only remove nodes from the front of the queue. Because flights arrive and clear out of chronological order in real life, a linked list is required so interior nodes can be unlinked from arbitrary positions.
04
Inverting moves when backtracking with a stack · Q13
When popping moves from the maze stack to return to the start, each turn direction must be inverted (a recorded right turn requires turning left on the return path, and vice versa).

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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